*As this is a challenge, and to give you guys time to think about the answer I won't reply here to all your comments here say if you are right or wrong. But I WILL publish another video in a couple of days or so discussing your answers and revealing the correct answer, and why*
I don't know a lot about electronics, but while trying to fix my own stuff I've ran across resistors that don't have a tolerance band. So that would make this resistor a 4,500 Ohm resistor. I think anyways. The yellow would be 4, then green is 5, then black is 0, then your brown would be another zero added on. I think that's how the numbers would work. I just figured the value with the last band being the multiplier instead of the tolerance. I have no idea how to plug that value into a formula to see if it would actually function with that value. All I know is that I've seen those type of resistors that just don't have a tolerance band at all, don't know why. Hopefully this helps maybe. If not, I'm sorry, I hope you can figure it out soon. 🙂
That must be 475 ohms. This is a common value with metal film 1% resistors. The purple ring (number 7) has changed color due to the high temperature and it is seen as black. So, a resistor of 470 ohm / 0.25 -0.5W does the job. Also, the resistor was damaged first, then the electrolytic capacitor, because the voltage is higher if there is no load.
Yellow Gray Green Black Brown. - 485 Ohms 1%. Still and odd value but the grey could have originally been violet which is 475 ohms ! Look carefully at the burned section you can clearly see the band.
@@TheCrakkle There must be a leakage on the 470nf capacitor, no way to burn that resistor with the expected current of around 33 ma. (or maybe the dark glue...)
You can see that there's a bigger gap between the brown & black, indicating that the brown band is the tolerance (1%). As pointed out by others, there's another band where the burn mark is. It's quite obvious, as it has a straight edge, unlike a burn mark which is uneven. It looks like a black band, but that would give a value that doesn't exist in the E96 series. I can only think that it's a discoloured violet, giving a value of 475 Ohms.
I suspect it's not detecting brownout but is there to ensure the amplifier output doesn't suddenly shift when mains is applied, causing speaker thump - a soft power-on. The optocoupler output probably disables part of the output stages until the rest of the circuit has stabilised after power-up.
I'm going to suggest 475 Ohms. As others have pointed out already, there appears to be a purple band under the charring. Great video for working things out from first principle.
My first go to to get a rough idea was what you did at the end… the test would have been about 2/3’s the actual value, so likely a burnt off band. Give the 1%, the only 4_5 will be 475 so it’s burnt off the violet second band.
The opto-isolator will have a current specification, you can calculate the resistor value based on that. Or, use a potentiometer and simply adjust for maximum smoke.
I agree with 5 band with 475R value. I think function is to prevent switch on thump, possibly switch off thump controlling amp bias or loudspeaker relay. Given capacitor tolerances I would have opted for higher voltage on the electrolytic. Also, a resistor across the electrolytic/opto diode would have provided more consistent turn off by the opto. I also wondered if all four bridge rectifier diodes are ok, failure could result in much higher voltage across the electrolytic but I guess it would have exploded in that case!
That's a 5-band resistor. The burned area hides another band. Scratch it slightly then look through microscope and see if you can get the real color of it. Pretty sure it has 5 bands.
Richard, I think you're on the right track. I did a quick simulation, and a resistor in the 400 ohm range will limit the voltage across the 470uF output capacitor to around 12VDC. And yeah, it looks like it's a 5 band resistor with a value in the 400 ohm range. And somewhat suprisingly if you increase it even to around 1k ohms the output capacitor voltage (assuming a 310 volt peak 50Hz utility voltage) will already be up around 26VDC. So yeah, around 400 ohms makes sense, and limits the optocoupler input amps to around 35mA. So I'm guessing a voltage transient (lightning strike?) got into the circuit, and since the same voltage is across both the output cap and resistor they both exploded. I suppose that's what you get when you save a few $$ by using a cheap capacitor dropper rather than a transformer. But the bigger question I have is what does this circuit do? If my simulation is right, it looks like you need to get the input AC down around 20 volts to turn off the optocoupler. So why didn't they just use a simple AC relay to detect loss of the AC input? Or maybe they're using this circuit to somehow limit audio transients?
observe band spacing closely when dealing with burnt resistors to avoid such mishaps, tolerance band usually has distinct spacing with value bands being equidistant or close to it
@@Manticore1960 Yes there is, definitely. 🙂Who actually has them available in stock in small quantities could be another matter, but that value is available. Vishay, for example. Not a common value, granted, but seemingly dumb values like that can be purchased. There is a black band there.
Yes - you can get a 405 ohms resistor - but it will be meaningless to specify such a special component in this application. The resistor was most likely 1M ohms according to its function in the circuit, no matter the remnants of the burnt colour bands. It may be calculated by trial and error, by selecting a value that will give a HI on the opto output - with 50% mains input. It could even be 1M ohm as suggested, beccause of the very sensitive opto Darlington output. Remember the output should go HI during brownout ...
Could you figure it out by looking at the other end of the connection? I mean, if the signal disappears into a micro controller, you're SOL. But if, for instance, the photo transistor charges a capacitor which then triggers a relay or something like that, it's probably a soft start circuit. And then you know that the exact value isn't mission critical, it just has to light the LED when mains is applied.
I'm in team 4X5 1% also. 475 making the most sense. Can't see any colour left though on that scorched band. Also 312 divided by 66 (you're probe looked roughly 2/3 along), times 100 is 472.
Richard, I am guessing that you don't have a schematic for this amp, so I your cautionary approach is excellent. I like two of the suggestions: 1) Test the remaining half of the resistor to get an approximate estimate of its original value and 2) Be prepared to try succesively lower values until you get one to give around 12v, which would be within the working voltage of the electrolytic you had to replace.
That circuit will *only* limit the DC output voltage if there is a load. Without a load the output capacitor would charge to the peak of the AC input voltage, ignoring the fact it would detonate long before reaching that voltage. That means that the resistor in series with the IRED of the coupler simply cannot be very high. The 4N32 has a current transfer ratio specified at 500% (x5) minimum with no maximum specified. It wouldn't surprise me in the slightest if the maximum were at least 2500% That would make if a very poor choice for brown-out detect because the threshold would be extremely poorly defined. Adding a zener in series with the input of the optocoupler would improve the threshold sharpness dramatically. I have seen yellow used instead of gold as a tolerance band on resistors. The gold bands often contain metallic particles which are bad news in some circumstances. Any that I've seen like that have been high-voltage parts and simply look far higher in quality than the part in question. It looks like a low-end bargain basement part. There is absolutely no reason to use a 1% resistor from a performance standpoint. The whole circuit is just *way* too sloppy. However, the manufacture may have used a 1% part because it has a higher power rating than a 5% part of the same physical size (metal film vs carbon film) or simply because they use 1% for everything to reduce variety of inventory. 475 ohms is my bet. Metal film are better in audio because they produce lower excess noise (i.e. noise beyond "Johnson" noise) than carbon.
With a 4 and a 5 band showing the only value I can think of with those values in will be from the E96 series, which would match with the brown band as a 1% tolerance resistor, and would be a 475 ohm resistor.
@@LearnElectronicsRepair Your feeble attempt to probe into the guts of the resistor was half-assed. The guy is correct. Measure the good half and multiply by 2.
The circuit is detecting the presence of mains independently of the output of the switch mode power supply. Why it's necessary to do that is unclear, but brownout detection is a possibility. More likely though is that its an anti-thump circuit. the speaker will be disconnected before the output of the power supply goes down. With a 470uF capacitor in there its going to be very slow though, it will probably take tens of cycles to detect that the mains has gone. The reason you were getting 28 volts across the 470uF capacitor is that the only thing limiting the volts would normally be the current through the opto. With a 1M resistor in there, the voltage rises above the rating of the capacitor and only stops rising when the capacitor leaks a good few mA. As others have said, the resistor will be 475 ohms although this is absolutely not critical. The current out of the capacitive dropper (and through the opto) will be more or less constant regardless of the value of the resistor, within limits of course. The only thing the actual value will affect is the voltage the 470uF capacitor settles at.
yellow brown green black red.. i would call that circuit "anti Click" because it turns the amp off, faster than the caps discarge. in theory you could bridge that for testing.
475 ohms it is. But what is concerning is the voltage that had to be over this resistor to burn it. As it is a 1W resistor, V = SQRT( P x R ) = SQRT( 1 x 475 ) = 21,794V. Add the voltage over the optocoupler LED and there had to be at least 23V over the 16V capacitor. Probably much better to have a Zener diode over the capacitor to limit the voltage. I would also recommend using a higher rated capacitor. For example 25V capacitor and a 12V Zener.
I was thinking it would be used to mute the output to stop any pops or bangs on power up, but I would have thought there would have been a better way to do it.
Hey Richard, Best wishes for 2025! I would say the burned resistor is 45 Ohm 1%. The 470nf capacitor gives an impedance of approx. 6772 Ohms at 50Hz + the 1K resistor is a total impedance of 7772 Ohms at 50Hz resulting in a maximum current of approx. 30mA at 240VAC input. The 45 Ohm resistor is just to further limit the current to the LED in the opto-coupler. According to the datasheet of the 4N32 the LED can handle 50mA at a voltage drop of approx. 1.2V.
I see it as a 5-band resistor: yellow(4)-gray(8)-green(5)-black(0)-brown(4), which is a 485 ohm 1% tolerance resistor. Reading it the other direction, it would be a 10500M Ohm , 4% resistor, which doesn't make sense.
@@sx64man I believe some are seeing the gray band as being black. This could have to do with their particular monitor and how it represents colors, contrast and brightness. I'm watching the video on a 55" LG C1 OLED TV, so I believe my colors are accurate.
This is how we did this in the past: scratch off the ceramic on the outside of the resistor as close to the middle of the body as possible then measure from the lead end to the exposed conductive sputter coat on the outside of the resistor ceramic body. Double your reading, then puck the closest preferred value. If always worked well...
Just lower the resistor until you get about 12V on the cap. I would do this experimental because the LED behave non-linear, so it's kinda hard to calculate.
475 is a standard value for 1% resistors. The cap voltage calue of 16 and a max current through the LED of 20ma puts a limit on the resistor value to 800 ohms. so for a sane voltage across the electrolytic, 475 ohms seems reasonable. The electrolytic cap most likely failed open rather than shorted as if it was shorted there wouldn't be any voltage to burn up the resistor. The capacitive dropper/1K resistor ahead of the bridge wants to push about 30mA and with 30ma through the 475 ohm resistor it would dissipate .45W which is more than the 1/4W of the resistor, so it would burn up.
I go with the others that there probably is another colour ring between yellow and green, whatever the colour might be. Other than a brown out detection, could it also be a delay circuit to prevent the loud plop when switching on the power?
485Ω ±1% Resistor. 5 Bands: Yellow, Grey, Green, Black, Brown. I can clearly see the Grey Band. It has a straight line border, which would not be if it were just a burn mark.
Hi Richard, The 470nF capacitor in an AC circuit should be considered as a resistance, or more accurately a reactance given by the formula 1/(2pi FC). At 50Hz the reactance would around 6770 ohms. With the mains peak voltage of 300V, the short circuit current across the bridge rectifier, would be around 45mA, which is within the spec of the opto-isolator. If the resistor (R2) was 100ohms (which is insignificant with regards to the current) the voltage drop would be 4.5V plus 3V for the LED gives you 7.5V across the electrolytic, which is at least in the right ball park for the 16V capacitor. Is it possible that the resistor bands are therefore brown-black-brown? However, I am unable to offer a purpose for this circuit as a whole, to detect brownouts, the 470uF electrolytic may exhibit a significant delay in its response.
As a minor correction to my explanation, you cannot just add the 1K to the 6770 ohms reactance of the capacitor to give you the total to calculate the current in the circuit, since the calculation is quite complex. I chose to ingnore the resistor, since there is a 90 degree shift phase in current due to the capacitance, in series with the 0 degree phase shift in current from the resistor. So the addition of the 1K resistor does not really change basic circuit operation. The resistor R2 does not act as a load resistor, because the 470nF capacitor is the dominant reactance/resistance in the circuit. The purpose of R2 is to generate a suitable voltage across the electrolytic to smooth out the ripple.
I have seen speaker protection circuits that (amongst other things) directly monitor the AC supply and isolate the speakers as soon as they detect AC power disconnect - could this circuit be doing something similar ?
@@budgetmerch You're right. Just looked on Digikey. I'm very surprised that a manufacturer would produce resistors outside the known E series values. 🤷♂
I can clearly see a violet band just beside the burnt mark. So it's a 475 Ohm resistor, 1% tolerance. Assuming we have 12V on the rectifier output, the current across optocoupler's LED would be in order of 25mA , wich I think complies with the operation characteristics. Furthermore, as the voltage goes lower, the lower is the current. We could expect 9V or even less.
Is it ACTUALLY a 5-band resistor? Looks like the burn is between a green and black band. That black is just too prefect to be a burn mark (appearance at the end of the video)
There is no voltage limit with the 470uF capacitor. The circuit is based solely on the current across R2 and the optocoupler being so large that the voltage is pulled down on the 470uF. Use a 470uF capacitor with a higher voltage rating and insert a Zener diode ! A capacitor dropper does not limit the voltage. If no current flows, the 470uF capacitor has 325V! Only the current ist limited !
Agreed, yellow, black, green, black, brown. 405 @ 1%. Can you explain the effect of incoming voltage variance on the capacitative dropper. Would it have that much of an effect. Could this be a 110v defense circuit, preventing the amp from starting with an insufficient voltage?
Since so many posts were way off track, I add my wisdom. Ah, well, the answer is glaringly obvious, dear electrically-inclined friends! You see, this isn't your standard resistor-oh no! This is, in fact, a "Holy Resistor of Antioch" from the legendary E-Lore series. The mysterious burn mark is not a band but a divine scorch left by the Resistor Knight when he valiantly fought off the Voltage Spike of Doom. Its value? Clearly 4.7 Mythical Ohms (MOhms), which can only be interpreted in the presence of the Sacred Current Flow. As for its purpose, it doesn't merely control bias or voltage-this resistor's role was to protect the amp from the ferocious "Thump of Noisy Displeasure" emitted by the Great Capacitor Kraken. Sadly, it perished in the line of duty. To solve your dilemma, you must replace it not with another resistor, but with a coconut-based capacitor. Simply attach it with a sparrow feather (African or European, your choice), utter the words "Ni!" three times, and your amp shall live again. But beware of the dreaded Killer Voltage Bunnies-they lurk near the heat sinks. And remember, always bring a shrubbery.
Start with a calculation, what would the AC current limit be - that is what the current through the 470 nF capacitor at 230 V, 50 Hz pass through. Assume at first for easy calculations that the bridge is a SHORT circuit and no inrush limiting resistor. You know/remember the formula for impedance of the capacitor at any sinusoidal frequency and voltage? Well, the current would be limited to 34 mA with nominal component tolerance (which is at least +/- 10% and more likely +/- 20%). Then you can try, what voltage drops different resistors (R2) would develop at 30-some mA current. 100 ohms would drop only 3 V, hardly good. 1k would drop 30 V - too much. 500 ohms would drop 15 V. Closer. What are the reasonable resistances in E96 series? Meaning between 400 and 500 ohms… The only one ending in “5” is 475 ohms. That matches quite well with reasonable voltages and currents, doesn’t it? The missing color must be purple!
Off the top of my head, 470 ish Ohms, with around 5V across the cap, assuming 7mA being drawn by the LED with 1.5V forward drop. Personally (only my opinion), I would have replaced the opto and not left its base floating, in what could be a noisy environment. Those pesky EU types, with their reversible mains (7:18).
Resistor Value The original R2 resistor was likely 1 kOhm (Brown, Black, Red in a 4-band code or Brown, Black, Black, Brown, Yellow in 5-band). The purpose of R2 is to limit the current through the optocoupler and help stabilize the voltage across the capacitor. Using a higher value, like 1 MOhm, significantly reduces the current flow, resulting in minimal load on the capacitive dropper and causing the voltage across the capacitor to rise. Capacitive Dropper Dynamics The circuit uses a 470 nF capacitor as a capacitive dropper to limit AC current. The reactance (X_C) of this capacitor can be calculated using the formula: X_C = 1 / (2 * pi * f * C) Substituting f = 50 Hz and C = 470 nF: X_C = 1 / (2 * 3.14159 * 50 * 470 * 10^-9) X_C = 6782 Ohms This limits the AC current to: I_dropper = V_AC / X_C I_dropper = 240 / 6782 I_dropper = 35.4 mA Voltage Across the 470 uF Capacitor The 470 uF capacitor is responsible for stabilizing the DC voltage after rectification. Under normal conditions, with R2 at 1 kOhm, the current through the optocoupler can be calculated as: I_opt = (V_cap - V_f) / R2 Assuming V_cap = 12 V and the forward voltage of the optocoupler (V_f) = 1.2 V: I_opt = (12 - 1.2) / 1000 I_opt = 10.8 mA The remaining current (around 24.6 mA) charges the capacitor and compensates for leakage. However, with an incorrect R2 value (e.g., 1 MOhm), the current through the optocoupler is negligible, leaving nearly the entire 35.4 mA to charge the capacitor. Without sufficient current draw, the capacitor voltage increases beyond its 16 V rating. When this happens, the capacitor starts leaking, pulling the voltage down to approximately 28 V. This leakage current can be estimated as: I_leak = V / X_C I_leak = 28 / 6782 I_leak = 4.13 mA This leakage current is not enough to stabilize the voltage at a safe level, leading to capacitor degradation and potential failure. DC Voltage Stabilization Under normal operation, the circuit stabilizes when the current supplied by the capacitive dropper matches the current drawn by the optocoupler and any leakage. For a 1 kOhm R2, the capacitor voltage stabilizes around 10-12 V, as the current is properly distributed. Resistor Power Rating The R2 resistor should be rated to handle the power it dissipates. For a 1 kOhm resistor with 10.8 mA current: P = I^2 * R P = (0.0108)^2 * 1000 P = 0.1166 W To ensure reliability, a 0.5 W resistor is recommended.
My analogy is, there is a 1k in series with the input, so the circuit will limit at a voltage where the burnt resistor will pass 250 mA dropping Vin - 1V on the led. Seeing it's a 1/8W resistor, max V on burnt resistor is 250mV so I say it's a 2 Ohm with a discoloured multiplier ring.
560Ω to 1kΩ. I think since we have a max 16V cap, we will have about 12V after the rectifier. The 4N32 needs 1.2-1.4v / 10-20mA to light up the LED. So R = (12V-1.2V)/10mA = 1080Ω. Since this is the max resistance to turn on the LED, I'd choose the next smaller standard value, which is 1kΩ. Power consumption is also easy: (12V-1.2V)*0.01A = 0.108W , so 1/4W "standard" would be sufficient. To be safe if the DC voltage is lower than 12V, as the LED's typical current is up to 20mA, it would be safe to half that resistance, 560Ω would be the next standard value. With that the usable DC voltage range would be around 7-13 V, which should be quite sufficient..
HERE IS A DIFFERENT VIEW So this is a circuit that provides a signal to the Amplifier fro the speaker so the Amp knows the correct speaker is connected to it before the Amp is turned on. It is part of the amplifier protection / start up circuit to ensure the amp will see the correct impedance before the power transistors are turned on. The resistor is a 1M ohm 5% (gold band) as this circuit does not need to draw much current. The tolerance band looks yellow because the heat has caused it to dis-colour (I have seen this happen before on a blown power resistor). I think the capacitor, being under the required voltage (16V against the 28V you measured) has over time spewed out its corrosive electrolyte that has in turn eaten through the resistor casing and caused the resistor to short out and blow. You will note the resistor has blown at the point where the electrolyte is.
Could it be a 5 band colour resistor (e.g. Brown, Black, Green, Black, Yellow) 105 Ohm 4% (circa 5V out) or (Yellow, Black, Green, Black, Brown) 405 Ohms 1% (circa 20V out) - depending on total load.
Just wanted to say I love the videos was wondering if you were ever going to do a follow up with the Arduino for beginners and how would anybody go about purchasing some of the hardware you find don’t know how shipping would be to the US to Florida though
450 Ohms @1% would be what it looks like but that's not a standard value. Neither is 405 for a 5 band with black. Violet 2nd on a 5 band would make it 475 Ohms. Would they spin custom resistors for an OTS prosumer speaker? Could be an out of spec. 390 and they marked it 405 to get the 1%! Curious either way. I think anything within the +/- 400 Ohm range would get it to work as it's just running the LED at a cut-off limit assuming it is for low power or more likely peak power on protection. Just checked Mouser and you can order 405 ohm resistors with reasonable lead times and prices.
I was doing this same thing last night, except it was a little SMD resistor, and it was vaporized and there was nothing left but a burnt mark on the PCB. And a quarter of the transistor had exploded. Fuseable resistor looked like it got a bit warm but still intact. 🙄. I took a guess at fixing it, plugged it in, and half of the remainder of the board exoloded and nearly took my hand with it.
1 watt? Eh? There's a 1W resistor at R1 but R2 is definitely only 1/4W. A 1W resistor is much larger than the one he removed.... same as R1 on this board. Maybe next time just don't say anything if you are clueless lol
Some of the soldered pads on the PCB have obviously been cleaned ( _maybe_ just soldered with no-clean solder, but even that leaves visible residue on big joints if they were done with cored solder and an iron). The pads for the bridge rectifier have not been cleaned. Neither have those for the 1k resistor. That very strongly suggests things have been replaced or at least removed for testing somewhere along the line. Whodunit?
Might be a mains detector circuit for an UPS like behavior and switchover to battery operated mode. Net the cheapest solution and would certainly be a bit complicated, but yeah, have seen worse things.
As you quite rightly say "Yellow is not used as a tolerance band " so we know straight away how to read it so '4' I'm sure the violet iis there so 475 Ω 1% ?
his is a color-coded resistor. To determine its resistance value, we'll interpret the color bands, and for the wattage, we'll assess the physical size of the resistor. Interpreting the Color Bands The resistor has the following colors: Brown (1) Black (0) Green (multiplier = 1 0 5 10 5 ) Gold (tolerance = ±5%) This corresponds to: Resistance = ( 10 ) × 1 0 5 = 1 , 000 , 000
Ω
(1 MΩ) Resistance=(10)×10 5 =1,000,000Ω(1 MΩ) With a tolerance of ±5%. Wattage Estimation From the image, this resistor appears to be a physically large, rounded resistor, often indicative of higher power ratings. This size is commonly associated with a 2-watt or 5-watt resistor. For a precise wattage rating, the manufacturer's specifications or a comparison to similar resistors is needed.
What you haven't accounted for is this R is blown and the perhaps the reason for that is it's a replacement by someone who didn't really care as much as you. So perhaps not best replying on what that resister is at all and find the right one.
My initial thought before you did the partial resistance was 4500 ohms. The maximum current for the opto is 60mA, but if it is a low voltage detector it needs to be just turned on. 12/4500 gives 2.66mA which is probably about right. Yellow, green, black =450 plus 1 extra zero for brown. No band for tolerance. But 4500 is too specialist, why would it not be 4700 since it can't be that critical? Doesn't make sense. Second thought is 1K ohms. Brown and black are correct but the heat has changed the colour of the next 2 bands, the red has turned green and the gold or brown has turned yellow. Probably brown, black, red, brown = 1000 ohms, 1%. That gives about 12mA through the opto which also makes sense. BTW, I think the purpose is to tell the amp that power is out although the main part of the amp is still being fed by the large capacitors. It allows for graceful shutdown.
Maybe the purpose is a soft-start (or delayed start). Amp is not active while the SMPS gets settled. With the inrush limiting resistor on the AC side, maybe voltage on the DC side "slowly" increases until it's high enough to turn on the Darlington pair, and enable the amp. In that time the SMPS has started up and is running the rails ok, eliminating pops etc the amp might give if that delayed start up was not there.
@CollinBaillie I think, as Richard said, it would know the power is on because the power is on. It would not necessarily immediately know that the power was off. Worth considering though.
Having the colours burnt its hard to say Yellow or Gold at the end band (leaning towards yellow) Either way though Yellow +/-4% & Gold +/-%5 tolerances. So being what looks like a 5 Band resistor & taking into account what its being used for. I'd opt for 105 Ohms (Brown-Black-Green-Black-Yellow or Gold) 🤔
I think it should be read from the yellow end, as there's a distinct bigger gap between the brown & black, indicating that brown is the tolerance. Also, 105 Ohms only exist in the 2% or tighter tolerance range. I think the band next to the burn mark could possibly be a discoloured violet.
bit of a bad design, there should be a high value resistor across the 470nF cap to stop high voltages being see on the mains plug when disconnected. This can be fatal when working at hight, your working on a ladder and you unplug the amp contact is made with the mains pins and you get DC at a few hundred volts. The current is limited by the impedance of the capacitor, Xc = 1/(2 x Pi x f x C). this provides an impedance that causes a current to flow that is out of phase of the supply voltage. So dissipates very little power. The capacitor just provides a current limit, the output voltage is defined by whatever is connected the other side of the bridge. in this case the voltage is limited by the diode in the optocoupler and the burnt out resistor. so there will only be a few volts across the 470u cap. if the diode goes open circuit then it will not limit the voltage which will increase to the mains voltage. The 470U 16V cap will try to keep the voltage lower by getting very hot before going bang. That's a problem with this type of power supply, with nothing limiting the voltage on the output, the output voltage will rise to the rectified mains voltage. You need to calculate the value of the capacitor (Xc) and use that as part of the voltage divider network.
The bridge rectifier and everything behind is a non-linear circuit. The standard calculation don't work that well. You could either simulate the thing are do some experiments.
@kriswillems5661 the cap is just an impedance defined by the cap value and the mains frequency. Normally you would see something like a zener diode to define the output voltage. In this case its just the burnt resistor and the Vfd of the optocoupler diode. Once the resistor started to burn and fail the next limiting factor for the voltage is the 470u capacitor, which then popped its clogs as the voltage increased. If you don't pull any current then the dc voltage will be rectified mains. It all just relies on the impedance of the 470nf cap. You could replace the cap with a resistor that matches the Xc value and it would still work. But the current in the resistor would be inphase with the supply voltage so the resistor would get very hot very quickly. This circuit only works due to the capacitor shifting the phase angle of the current, thus very little power is generated in the cap.
Yeah, lack of the bleed resistor shows a bit of a lazy design, and a supply like this is only useful in cheap LED stuff or as it seems here a simple, isolated way of verifying the presence of a supply ? or as a mute because the mains is passed through so it's not protecting anything as for the value, shall go and look what I think but my eyes are not what they used to be.... 475 Ω 1%
would been easier and cheaper to use a small 16 volt transforner and a bridge rectifier.. with the transformer you already have the isolation and that is enough to power the led... they try to be fancy where they do not have to be !!!
@@LearnElectronicsRepair (1 MΩ) With a tolerance of ±5%. Wattage Estimation From the image, this resistor appears to be a physically large, rounded resistor, often indicative of higher power ratings. This size is commonly associated with a 2-watt or 5-watt resistor. For a precise wattage rating, the manufacturer's specifications or a comparison to similar resistors is needed.
@@ioannisbalouktsis962 The body of the resistor is physically too small to carry two or five watts! Difficult to determine the exact size of that board without actually being there but that's a 1/4 watt resistor at most.
Horrible circuit, this is wrong example of circuit which made in principle "mute" or "delay speaker connection/disconnection"... I know well these circuits trom some old AV receivers, but in right state with all what is necessary not as this. There from my point of view is missing fuse or fusible resistor, 1K resistor in case of failure made nice burner, as next is missing paralel resistor for 470n capacitor and last problem is that on DC side isn´t zener diode or load resistor which made more stable DC voltage. As load use only in series diode in optocoupler with resistor is wrong. This is nice example of circuit with less parts as is possible, but will be best really use only one diode instead directifier (irony). In combination with wooden box nice to have it at home, isn´t there marked telephone number for fire rescue? or 112? THX🤮
*As this is a challenge, and to give you guys time to think about the answer I won't reply here to all your comments here say if you are right or wrong. But I WILL publish another video in a couple of days or so discussing your answers and revealing the correct answer, and why*
I don't know a lot about electronics, but while trying to fix my own stuff I've ran across resistors that don't have a tolerance band. So that would make this resistor a 4,500 Ohm resistor. I think anyways. The yellow would be 4, then green is 5, then black is 0, then your brown would be another zero added on. I think that's how the numbers would work. I just figured the value with the last band being the multiplier instead of the tolerance. I have no idea how to plug that value into a formula to see if it would actually function with that value. All I know is that I've seen those type of resistors that just don't have a tolerance band at all, don't know why. Hopefully this helps maybe. If not, I'm sorry, I hope you can figure it out soon. 🙂
This is causing quite a stir! 🤣 Looking forward to the follow-up video.
That must be 475 ohms. This is a common value with metal film 1% resistors. The purple ring (number 7) has changed color due to the high temperature and it is seen as black. So, a resistor of 470 ohm / 0.25 -0.5W does the job. Also, the resistor was damaged first, then the electrolytic capacitor, because the voltage is higher if there is no load.
Yellow Gray Green Black Brown. - 485 Ohms 1%. Still and odd value but the grey could have originally been violet which is 475 ohms !
Look carefully at the burned section you can clearly see the band.
@@TheCrakkle There must be a leakage on the 470nf capacitor, no way to burn that resistor with the expected current of around 33 ma. (or maybe the dark glue...)
You can see that there's a bigger gap between the brown & black, indicating that the brown band is the tolerance (1%). As pointed out by others, there's another band where the burn mark is. It's quite obvious, as it has a straight edge, unlike a burn mark which is uneven. It looks like a black band, but that would give a value that doesn't exist in the E96 series. I can only think that it's a discoloured violet, giving a value of 475 Ohms.
I suspect it's not detecting brownout but is there to ensure the amplifier output doesn't suddenly shift when mains is applied, causing speaker thump - a soft power-on. The optocoupler output probably disables part of the output stages until the rest of the circuit has stabilised after power-up.
I'm going to suggest 475 Ohms. As others have pointed out already, there appears to be a purple band under the charring. Great video for working things out from first principle.
475 Ohms from the E96 table makes absolutely sense.
My first go to to get a rough idea was what you did at the end… the test would have been about 2/3’s the actual value, so likely a burnt off band. Give the 1%, the only 4_5 will be 475 so it’s burnt off the violet second band.
The opto-isolator will have a current specification, you can calculate the resistor value based on that.
Or, use a potentiometer and simply adjust for maximum smoke.
Awesome.
I approve of using magic smoke for working out the resistor value.
I agree with 5 band with 475R value. I think function is to prevent switch on thump, possibly switch off thump controlling amp bias or loudspeaker relay. Given capacitor tolerances I would have opted for higher voltage on the electrolytic. Also, a resistor across the electrolytic/opto diode would have provided more consistent turn off by the opto. I also wondered if all four bridge rectifier diodes are ok, failure could result in much higher voltage across the electrolytic but I guess it would have exploded in that case!
That's a 5-band resistor. The burned area hides another band. Scratch it slightly then look through microscope and see if you can get the real color of it. Pretty sure it has 5 bands.
Yeah I thought that as well.
Richard, I think you're on the right track. I did a quick simulation, and a resistor in the 400 ohm range will limit the voltage across the 470uF output capacitor to around 12VDC. And yeah, it looks like it's a 5 band resistor with a value in the 400 ohm range. And somewhat suprisingly if you increase it even to around 1k ohms the output capacitor voltage (assuming a 310 volt peak 50Hz utility voltage) will already be up around 26VDC. So yeah, around 400 ohms makes sense, and limits the optocoupler input amps to around 35mA. So I'm guessing a voltage transient (lightning strike?) got into the circuit, and since the same voltage is across both the output cap and resistor they both exploded. I suppose that's what you get when you save a few $$ by using a cheap capacitor dropper rather than a transformer. But the bigger question I have is what does this circuit do? If my simulation is right, it looks like you need to get the input AC down around 20 volts to turn off the optocoupler. So why didn't they just use a simple AC relay to detect loss of the AC input? Or maybe they're using this circuit to somehow limit audio transients?
observe band spacing closely when dealing with burnt resistors to avoid such mishaps, tolerance band usually has distinct spacing with value bands being equidistant or close to it
405 Ohms.... I see another band on the resister under the burnt parts..
Yellow Black green black brown
I was just going to post 405 ohms - noticed the black band giving Yellow Black Green Black Brown, so 405Ω @ 1% tolerance. Makes sense to me anyway. 👍
but...there is no such value. Not even in the E192 series (0.5% & tighter).
@@Manticore1960 Yes there is, definitely. 🙂Who actually has them available in stock in small quantities could be another matter, but that value is available. Vishay, for example. Not a common value, granted, but seemingly dumb values like that can be purchased. There is a black band there.
@@Manticore1960 yes there is such a value.
Digikey has them in stock from 3 different manufacturers.
Yes - you can get a 405 ohms resistor - but it will be meaningless to specify such a special component in this application.
The resistor was most likely 1M ohms according to its function in the circuit, no matter the remnants of the burnt colour bands.
It may be calculated by trial and error, by selecting a value that will give a HI on the opto output - with 50% mains input.
It could even be 1M ohm as suggested, beccause of the very sensitive opto Darlington output.
Remember the output should go HI during brownout ...
Could you figure it out by looking at the other end of the connection? I mean, if the signal disappears into a micro controller, you're SOL. But if, for instance, the photo transistor charges a capacitor which then triggers a relay or something like that, it's probably a soft start circuit. And then you know that the exact value isn't mission critical, it just has to light the LED when mains is applied.
I'm in team 4X5 1% also. 475 making the most sense. Can't see any colour left though on that scorched band. Also 312 divided by 66 (you're probe looked roughly 2/3 along), times 100 is 472.
Richard, I am guessing that you don't have a schematic for this amp, so I your cautionary approach is excellent.
I like two of the suggestions:
1) Test the remaining half of the resistor to get an approximate estimate of its original value and
2) Be prepared to try succesively lower values until you get one to give around 12v, which would be within the working voltage of the electrolytic you had to replace.
On the last picture of the video I thought I can see a 5th color band under the burnt section.
Maybe do up a schematic of the amp board where the output side of the opto-isolator is connected. That might be more indicative as to its purpose.
it looks like a 5 band resistor on camera.. Brown, Black, Green, then either Black or Burnt Violate and last Yellow?. Good luck
That circuit will *only* limit the DC output voltage if there is a load. Without a load the output capacitor would charge to the peak of the AC input voltage, ignoring the fact it would detonate long before reaching that voltage. That means that the resistor in series with the IRED of the coupler simply cannot be very high.
The 4N32 has a current transfer ratio specified at 500% (x5) minimum with no maximum specified. It wouldn't surprise me in the slightest if the maximum were at least 2500% That would make if a very poor choice for brown-out detect because the threshold would be extremely poorly defined. Adding a zener in series with the input of the optocoupler would improve the threshold sharpness dramatically.
I have seen yellow used instead of gold as a tolerance band on resistors. The gold bands often contain metallic particles which are bad news in some circumstances. Any that I've seen like that have been high-voltage parts and simply look far higher in quality than the part in question. It looks like a low-end bargain basement part.
There is absolutely no reason to use a 1% resistor from a performance standpoint. The whole circuit is just *way* too sloppy. However, the manufacture may have used a 1% part because it has a higher power rating than a 5% part of the same physical size (metal film vs carbon film) or simply because they use 1% for everything to reduce variety of inventory. 475 ohms is my bet. Metal film are better in audio because they produce lower excess noise (i.e. noise beyond "Johnson" noise) than carbon.
With a 4 and a 5 band showing the only value I can think of with those values in will be from the E96 series, which would match with the brown band as a 1% tolerance resistor, and would be a 475 ohm resistor.
1% resistors often have 4 bands so yellow, x, green, black plus gold for 1%. So I suggest the missing band (x) = violet. So 475 ohms.
Gently scratch it at the 1/2 way mark and measure from unburnt side then x2 .....
Now what did I say about watching the whole video? it seems like you didn't want to see all the clues 😵💫
@@LearnElectronicsRepair Your feeble attempt to probe into the guts of the resistor was half-assed. The guy is correct. Measure the good half and multiply by 2.
@@LearnElectronicsRepair The probe was way to the right of half way.
@@g4z-kb7ct Ouch! 🤣
@@g4z-kb7ct OK I can take the criticism, it was constructive 😉
The circuit is detecting the presence of mains independently of the output of the switch mode power supply. Why it's necessary to do that is unclear, but brownout detection is a possibility. More likely though is that its an anti-thump circuit. the speaker will be disconnected before the output of the power supply goes down. With a 470uF capacitor in there its going to be very slow though, it will probably take tens of cycles to detect that the mains has gone.
The reason you were getting 28 volts across the 470uF capacitor is that the only thing limiting the volts would normally be the current through the opto. With a 1M resistor in there, the voltage rises above the rating of the capacitor and only stops rising when the capacitor leaks a good few mA. As others have said, the resistor will be 475 ohms although this is absolutely not critical. The current out of the capacitive dropper (and through the opto) will be more or less constant regardless of the value of the resistor, within limits of course. The only thing the actual value will affect is the voltage the 470uF capacitor settles at.
I am going to say that is a 5 band resistor that is 105 ohm with the gold band indicating 5% tolerance as there is a black band under the black glue.
yellow brown green black red.. i would call that circuit "anti Click" because it turns the amp off, faster than the caps discarge. in theory you could bridge that for testing.
475 ohms it is. But what is concerning is the voltage that had to be over this resistor to burn it. As it is a 1W resistor, V = SQRT( P x R ) = SQRT( 1 x 475 ) = 21,794V. Add the voltage over the optocoupler LED and there had to be at least 23V over the 16V capacitor.
Probably much better to have a Zener diode over the capacitor to limit the voltage. I would also recommend using a higher rated capacitor. For example 25V capacitor and a 12V Zener.
I was thinking it would be used to mute the output to stop any pops or bangs on power up, but I would have thought there would have been a better way to do it.
Hey Richard,
Best wishes for 2025!
I would say the burned resistor is 45 Ohm 1%. The 470nf capacitor gives an impedance of approx. 6772 Ohms at 50Hz + the 1K resistor is a total impedance of 7772 Ohms at 50Hz resulting in a maximum current of approx. 30mA at 240VAC input. The 45 Ohm resistor is just to further limit the current to the LED in the opto-coupler. According to the datasheet of the 4N32 the LED can handle 50mA at a voltage drop of approx. 1.2V.
I see it as a 5-band resistor: yellow(4)-gray(8)-green(5)-black(0)-brown(4), which is a 485 ohm 1% tolerance resistor. Reading it the other direction, it would be a 10500M Ohm , 4% resistor, which doesn't make sense.
that's what I came up with also 485
@@sx64man I believe some are seeing the gray band as being black. This could have to do with their particular monitor and how it represents colors, contrast and brightness. I'm watching the video on a 55" LG C1 OLED TV, so I believe my colors are accurate.
This is how we did this in the past: scratch off the ceramic on the outside of the resistor as close to the middle of the body as possible then measure from the lead end to the exposed conductive sputter coat on the outside of the resistor ceramic body. Double your reading, then puck the closest preferred value. If always worked well...
Just lower the resistor until you get about 12V on the cap. I would do this experimental because the LED behave non-linear, so it's kinda hard to calculate.
475 is a standard value for 1% resistors. The cap voltage calue of 16 and a max current through the LED of 20ma puts a limit on the resistor value to 800 ohms. so for a sane voltage across the electrolytic, 475 ohms seems reasonable. The electrolytic cap most likely failed open rather than shorted as if it was shorted there wouldn't be any voltage to burn up the resistor. The capacitive dropper/1K resistor ahead of the bridge wants to push about 30mA and with 30ma through the 475 ohm resistor it would dissipate .45W which is more than the 1/4W of the resistor, so it would burn up.
I go with the others that there probably is another colour ring between yellow and green, whatever the colour might be. Other than a brown out detection, could it also be a delay circuit to prevent the loud plop when switching on the power?
485Ω ±1% Resistor. 5 Bands: Yellow, Grey, Green, Black, Brown. I can clearly see the Grey Band. It has a straight line border, which would not be if it were just a burn mark.
I too suggest it is a 475 Ohm resistor giving 33mA at 16v and 25mA at 12 v. About half the maximum forward current of the device
as well as brown-outs it would tell the amp if it was inadvertently plugged into a 115V supply.
Hi Richard, The 470nF capacitor in an AC circuit should be considered as a resistance, or more accurately a reactance given by the formula 1/(2pi FC). At 50Hz the reactance would around 6770 ohms.
With the mains peak voltage of 300V, the short circuit current across the bridge rectifier, would be around 45mA, which is within the spec of the opto-isolator. If the resistor (R2) was 100ohms (which is insignificant with regards to the current) the voltage drop would be 4.5V plus 3V for the LED gives you 7.5V across the electrolytic, which is at least in the right ball park for the 16V capacitor. Is it possible that the resistor bands are therefore brown-black-brown?
However, I am unable to offer a purpose for this circuit as a whole, to detect brownouts, the 470uF electrolytic may exhibit a significant delay in its response.
As a minor correction to my explanation, you cannot just add the 1K to the 6770 ohms reactance of the capacitor to give you the total to calculate the current in the circuit, since the calculation is quite complex. I chose to ingnore the resistor, since there is a 90 degree shift phase in current due to the capacitance, in series with the 0 degree phase shift in current from the resistor.
So the addition of the 1K resistor does not really change basic circuit operation. The resistor R2 does not act as a load resistor, because the 470nF capacitor is the dominant reactance/resistance in the circuit. The purpose of R2 is to generate a suitable voltage across the electrolytic to smooth out the ripple.
I think it is a 475 ohm resistor.
I have seen speaker protection circuits that (amongst other things) directly monitor the AC supply and isolate the speakers as soon as they detect AC power disconnect - could this circuit be doing something similar ?
475 Ohms 1% as this is the only value in E-series with 4x5 0.7 Watts
405Ω @1% resistors are available and I'm pretty sure that's what it should be.
@@budgetmerch In the 1% (E96) series, there's nothing between 402Ω & 412Ω. Even in the E192 series (0.5%) 405Ω doesn't exist. There's 402Ω then 407Ω.
@@Manticore1960 We'll find out soon, right here on Richard's channel! 😀
@@budgetmerch had a quick look at digikey and the 405 Ohms were either MIL-SPEC or 0.1% which makes it seem unlikely
@@budgetmerch You're right. Just looked on Digikey. I'm very surprised that a manufacturer would produce resistors outside the known E series values. 🤷♂
I can clearly see a violet band just beside the burnt mark. So it's a 475 Ohm resistor, 1% tolerance. Assuming we have 12V on the rectifier output, the current across optocoupler's LED would be in order of 25mA , wich I think complies with the operation characteristics. Furthermore, as the voltage goes lower, the lower is the current. We could expect 9V or even less.
Is it ACTUALLY a 5-band resistor? Looks like the burn is between a green and black band. That black is just too prefect to be a burn mark (appearance at the end of the video)
There is no voltage limit with the 470uF capacitor. The circuit is based solely on the current across R2 and the optocoupler being so large that the voltage is pulled down on the 470uF.
Use a 470uF capacitor with a higher voltage rating and insert a Zener diode !
A capacitor dropper does not limit the voltage. If no current flows, the 470uF capacitor has 325V!
Only the current ist limited !
Agreed, yellow, black, green, black, brown. 405 @ 1%.
Can you explain the effect of incoming voltage variance on the capacitative dropper. Would it have that much of an effect. Could this be a 110v defense circuit, preventing the amp from starting with an insufficient voltage?
Speakers usually come in pairs. Look inside the other speaker
Since so many posts were way off track, I add my wisdom.
Ah, well, the answer is glaringly obvious, dear electrically-inclined friends! You see, this isn't your standard resistor-oh no! This is, in fact, a "Holy Resistor of Antioch" from the legendary E-Lore series. The mysterious burn mark is not a band but a divine scorch left by the Resistor Knight when he valiantly fought off the Voltage Spike of Doom.
Its value? Clearly 4.7 Mythical Ohms (MOhms), which can only be interpreted in the presence of the Sacred Current Flow. As for its purpose, it doesn't merely control bias or voltage-this resistor's role was to protect the amp from the ferocious "Thump of Noisy Displeasure" emitted by the Great Capacitor Kraken. Sadly, it perished in the line of duty.
To solve your dilemma, you must replace it not with another resistor, but with a coconut-based capacitor. Simply attach it with a sparrow feather (African or European, your choice), utter the words "Ni!" three times, and your amp shall live again. But beware of the dreaded Killer Voltage Bunnies-they lurk near the heat sinks.
And remember, always bring a shrubbery.
Start with a calculation, what would the AC current limit be - that is what the current through the 470 nF capacitor at 230 V, 50 Hz pass through. Assume at first for easy calculations that the bridge is a SHORT circuit and no inrush limiting resistor. You know/remember the formula for impedance of the capacitor at any sinusoidal frequency and voltage? Well, the current would be limited to 34 mA with nominal component tolerance (which is at least +/- 10% and more likely +/- 20%). Then you can try, what voltage drops different resistors (R2) would develop at 30-some mA current. 100 ohms would drop only 3 V, hardly good. 1k would drop 30 V - too much. 500 ohms would drop 15 V. Closer. What are the reasonable resistances in E96 series? Meaning between 400 and 500 ohms… The only one ending in “5” is 475 ohms. That matches quite well with reasonable voltages and currents, doesn’t it? The missing color must be purple!
Off the top of my head, 470 ish Ohms, with around 5V across the cap, assuming 7mA being drawn by the LED with 1.5V forward drop. Personally (only my opinion), I would have replaced the opto and not left its base floating, in what could be a noisy environment. Those pesky EU types, with their reversible mains (7:18).
Resistor Value
The original R2 resistor was likely 1 kOhm (Brown, Black, Red in a 4-band code or Brown, Black, Black, Brown, Yellow in 5-band). The purpose of R2 is to limit the current through the optocoupler and help stabilize the voltage across the capacitor. Using a higher value, like 1 MOhm, significantly reduces the current flow, resulting in minimal load on the capacitive dropper and causing the voltage across the capacitor to rise.
Capacitive Dropper Dynamics
The circuit uses a 470 nF capacitor as a capacitive dropper to limit AC current. The reactance (X_C) of this capacitor can be calculated using the formula:
X_C = 1 / (2 * pi * f * C)
Substituting f = 50 Hz and C = 470 nF:
X_C = 1 / (2 * 3.14159 * 50 * 470 * 10^-9)
X_C = 6782 Ohms
This limits the AC current to:
I_dropper = V_AC / X_C
I_dropper = 240 / 6782
I_dropper = 35.4 mA
Voltage Across the 470 uF Capacitor
The 470 uF capacitor is responsible for stabilizing the DC voltage after rectification. Under normal conditions, with R2 at 1 kOhm, the current through the optocoupler can be calculated as:
I_opt = (V_cap - V_f) / R2
Assuming V_cap = 12 V and the forward voltage of the optocoupler (V_f) = 1.2 V:
I_opt = (12 - 1.2) / 1000
I_opt = 10.8 mA
The remaining current (around 24.6 mA) charges the capacitor and compensates for leakage.
However, with an incorrect R2 value (e.g., 1 MOhm), the current through the optocoupler is negligible, leaving nearly the entire 35.4 mA to charge the capacitor. Without sufficient current draw, the capacitor voltage increases beyond its 16 V rating. When this happens, the capacitor starts leaking, pulling the voltage down to approximately 28 V. This leakage current can be estimated as:
I_leak = V / X_C
I_leak = 28 / 6782
I_leak = 4.13 mA
This leakage current is not enough to stabilize the voltage at a safe level, leading to capacitor degradation and potential failure.
DC Voltage Stabilization
Under normal operation, the circuit stabilizes when the current supplied by the capacitive dropper matches the current drawn by the optocoupler and any leakage. For a 1 kOhm R2, the capacitor voltage stabilizes around 10-12 V, as the current is properly distributed.
Resistor Power Rating
The R2 resistor should be rated to handle the power it dissipates. For a 1 kOhm resistor with 10.8 mA current:
P = I^2 * R
P = (0.0108)^2 * 1000
P = 0.1166 W
To ensure reliability, a 0.5 W resistor is recommended.
My analogy is, there is a 1k in series with the input, so the circuit will limit at a voltage where the burnt resistor will pass 250 mA dropping Vin - 1V on the led.
Seeing it's a 1/8W resistor, max V on burnt resistor is 250mV so I say it's a 2 Ohm with a discoloured multiplier ring.
560Ω to 1kΩ.
I think since we have a max 16V cap, we will have about 12V after the rectifier. The 4N32 needs 1.2-1.4v / 10-20mA to light up the LED. So R = (12V-1.2V)/10mA = 1080Ω. Since this is the max resistance to turn on the LED, I'd choose the next smaller standard value, which is 1kΩ. Power consumption is also easy: (12V-1.2V)*0.01A = 0.108W , so 1/4W "standard" would be sufficient.
To be safe if the DC voltage is lower than 12V, as the LED's typical current is up to 20mA, it would be safe to half that resistance, 560Ω would be the next standard value. With that the usable DC voltage range would be around 7-13 V, which should be quite sufficient..
HERE IS A DIFFERENT VIEW
So this is a circuit that provides a signal to the Amplifier fro the speaker so the Amp knows the correct speaker is connected to it before the Amp is turned on. It is part of the amplifier protection / start up circuit to ensure the amp will see the correct impedance before the power transistors are turned on. The resistor is a 1M ohm 5% (gold band) as this circuit does not need to draw much current. The tolerance band looks yellow because the heat has caused it to dis-colour (I have seen this happen before on a blown power resistor). I think the capacitor, being under the required voltage (16V against the 28V you measured) has over time spewed out its corrosive electrolyte that has in turn eaten through the resistor casing and caused the resistor to short out and blow. You will note the resistor has blown at the point where the electrolyte is.
Could it be that the LED is becoming intermittent, allowing the voltage to rise, which is why the capacitor started to bulge?
Could it be a 5 band colour resistor (e.g. Brown, Black, Green, Black, Yellow) 105 Ohm 4% (circa 5V out) or (Yellow, Black, Green, Black, Brown) 405 Ohms 1% (circa 20V out) - depending on total load.
Just wanted to say I love the videos was wondering if you were ever going to do a follow up with the Arduino for beginners and how would anybody go about purchasing some of the hardware you find don’t know how shipping would be to the US to Florida though
The circuit does not need a precision resistor. 1M, The yellow might be from over heating.
450 Ohms @1% would be what it looks like but that's not a standard value. Neither is 405 for a 5 band with black. Violet 2nd on a 5 band would make it 475 Ohms. Would they spin custom resistors for an OTS prosumer speaker? Could be an out of spec. 390 and they marked it 405 to get the 1%! Curious either way. I think anything within the +/- 400 Ohm range would get it to work as it's just running the LED at a cut-off limit assuming it is for low power or more likely peak power on protection.
Just checked Mouser and you can order 405 ohm resistors with reasonable lead times and prices.
I was doing this same thing last night, except it was a little SMD resistor, and it was vaporized and there was nothing left but a burnt mark on the PCB. And a quarter of the transistor had exploded. Fuseable resistor looked like it got a bit warm but still intact. 🙄. I took a guess at fixing it, plugged it in, and half of the remainder of the board exoloded and nearly took my hand with it.
Yellow Gray Green Black Brown. the Gray probably was Violet.
Look carefully at the burned section.
I say it's a 1 watt 450 ohm resistor. Why did it blow though?
@@kriswillems5661 Half it's coating is missing, that doesn't happen without serious internal damage.
1 watt? Eh? There's a 1W resistor at R1 but R2 is definitely only 1/4W. A 1W resistor is much larger than the one he removed.... same as R1 on this board. Maybe next time just don't say anything if you are clueless lol
Is there a black band between the yellow and green band. The burn mark looks a bit too straight in places to me
Metal film 1% resistor 475 Ohm.
Some of the soldered pads on the PCB have obviously been cleaned ( _maybe_ just soldered with no-clean solder, but even that leaves visible residue on big joints if they were done with cored solder and an iron). The pads for the bridge rectifier have not been cleaned. Neither have those for the 1k resistor. That very strongly suggests things have been replaced or at least removed for testing somewhere along the line. Whodunit?
I think it is a 1k 5% (Brown black red gold) it has turned green from the heat
Might be a mains detector circuit for an UPS like behavior and switchover to battery operated mode. Net the cheapest solution and would certainly be a bit complicated, but yeah, have seen worse things.
As you quite rightly say "Yellow is not used as a tolerance band " so we know straight away how to read it so '4' I'm sure the violet iis there so 475 Ω 1% ?
yellow grey green black 485 ohm
Curious why would brown-out needed to be detected and not over-voltage?
Heya, nice this with some time to thing about this. I would say 450 ohms just by looking at the ring colours
It appears the resistor just has some of the resin on it.
In that case then it must be intended to read open circuit
@ I didn’t see it measured end to end. However the fact you could poke a needle into the side requires a hole which is a result of localized burning.
his is a color-coded resistor. To determine its resistance value, we'll interpret the color bands, and for the wattage, we'll assess the physical size of the resistor.
Interpreting the Color Bands
The resistor has the following colors:
Brown (1)
Black (0)
Green (multiplier =
1
0
5
10
5
)
Gold (tolerance = ±5%)
This corresponds to:
Resistance
=
(
10
)
×
1
0
5
=
1
,
000
,
000
Ω
(1 MΩ)
Resistance=(10)×10
5
=1,000,000Ω(1 MΩ) With a tolerance of ±5%.
Wattage Estimation
From the image, this resistor appears to be a physically large, rounded resistor, often indicative of higher power ratings. This size is commonly associated with a 2-watt or 5-watt resistor. For a precise wattage rating, the manufacturer's specifications or a comparison to similar resistors is needed.
No way is that tiny resistor 2W or 5W. 🤣
That is max 1/2 W and its between 450 and 500 ohm.
What you haven't accounted for is this R is blown and the perhaps the reason for that is it's a replacement by someone who didn't really care as much as you. So perhaps not best replying on what that resister is at all and find the right one.
My initial thought before you did the partial resistance was 4500 ohms. The maximum current for the opto is 60mA, but if it is a low voltage detector it needs to be just turned on. 12/4500 gives 2.66mA which is probably about right. Yellow, green, black =450 plus 1 extra zero for brown. No band for tolerance. But 4500 is too specialist, why would it not be 4700 since it can't be that critical? Doesn't make sense.
Second thought is 1K ohms. Brown and black are correct but the heat has changed the colour of the next 2 bands, the red has turned green and the gold or brown has turned yellow. Probably brown, black, red, brown = 1000 ohms, 1%. That gives about 12mA through the opto which also makes sense.
BTW, I think the purpose is to tell the amp that power is out although the main part of the amp is still being fed by the large capacitors. It allows for graceful shutdown.
Maybe the purpose is a soft-start (or delayed start). Amp is not active while the SMPS gets settled. With the inrush limiting resistor on the AC side, maybe voltage on the DC side "slowly" increases until it's high enough to turn on the Darlington pair, and enable the amp. In that time the SMPS has started up and is running the rails ok, eliminating pops etc the amp might give if that delayed start up was not there.
@CollinBaillie I think, as Richard said, it would know the power is on because the power is on. It would not necessarily immediately know that the power was off. Worth considering though.
Having the colours burnt its hard to say Yellow or Gold at the end band (leaning towards yellow) Either way though Yellow +/-4% & Gold +/-%5 tolerances.
So being what looks like a 5 Band resistor & taking into account what its being used for. I'd opt for 105 Ohms (Brown-Black-Green-Black-Yellow or Gold) 🤔
I think it should be read from the yellow end, as there's a distinct bigger gap between the brown & black, indicating that brown is the tolerance. Also, 105 Ohms only exist in the 2% or tighter tolerance range. I think the band next to the burn mark could possibly be a discoloured violet.
its a 560 ohm resistoer 1/2 w metal film
i see the colors yellow, black, green, black and as last the tolorance brown so it's 415 ohm ( no zerrows addet) and the tolorance must be 1%
It's like you deliberately missed the banding on the burnt band. I guess a yt video allows you to pretend that you didn't see it.
It looks like a 405 ohm 1% Yellow, Black, green, Black, Brown?
405 does not exist in any E-series, it must be a 475 Ohms resistor
color code brown- black-ded
bit of a bad design, there should be a high value resistor across the 470nF cap to stop high voltages being see on the mains plug when disconnected.
This can be fatal when working at hight, your working on a ladder and you unplug the amp contact is made with the mains pins and you get DC at a few hundred volts.
The current is limited by the impedance of the capacitor, Xc = 1/(2 x Pi x f x C). this provides an impedance that causes a current to flow that is out of phase of the supply voltage. So dissipates very little power. The capacitor just provides a current limit, the output voltage is defined by whatever is connected the other side of the bridge. in this case the voltage is limited by the diode in the optocoupler and the burnt out resistor. so there will only be a few volts across the 470u cap. if the diode goes open circuit then it will not limit the voltage which will increase to the mains voltage. The 470U 16V cap will try to keep the voltage lower by getting very hot before going bang.
That's a problem with this type of power supply, with nothing limiting the voltage on the output, the output voltage will rise to the rectified mains voltage.
You need to calculate the value of the capacitor (Xc) and use that as part of the voltage divider network.
The bridge rectifier and everything behind is a non-linear circuit. The standard calculation don't work that well. You could either simulate the thing are do some experiments.
@kriswillems5661 the cap is just an impedance defined by the cap value and the mains frequency. Normally you would see something like a zener diode to define the output voltage. In this case its just the burnt resistor and the Vfd of the optocoupler diode. Once the resistor started to burn and fail the next limiting factor for the voltage is the 470u capacitor, which then popped its clogs as the voltage increased. If you don't pull any current then the dc voltage will be rectified mains. It all just relies on the impedance of the 470nf cap. You could replace the cap with a resistor that matches the Xc value and it would still work. But the current in the resistor would be inphase with the supply voltage so the resistor would get very hot very quickly. This circuit only works due to the capacitor shifting the phase angle of the current, thus very little power is generated in the cap.
Yeah, lack of the bleed resistor shows a bit of a lazy design, and a supply like this is only useful in cheap LED stuff or as it seems here a simple, isolated way of verifying the presence of a supply ? or as a mute because the mains is passed through so it's not protecting anything as for the value, shall go and look what I think but my eyes are not what they used to be.... 475 Ω 1%
@@andymouse
Eek
the gap to big between 1st colour. so i think a colour missing
would been easier and cheaper to use a small 16 volt transforner and a bridge rectifier.. with the transformer you already have the isolation and that is enough to power the led... they try to be fancy where they do not have to be !!!
475 ohm 470 ohm
Red,black,green,brown/black,gold 2050/205 ohms
Time for a baloney sandwich.
I don't think they are good for you even though with lashings of pickles they taste great, or did you mean something else !
MFR-50FTE52-475R
475 ohms?
4% 485
Photograph of this resistor to chatgpt and it will give you the value of it
If the resistor was not damaged, sure. But this is damaged beyond recognition, ChatGPT cannot do anything with this. Stop thinking that AI is magic.
give it a go and let us know
@@LearnElectronicsRepair
(1 MΩ) With a tolerance of ±5%.
Wattage Estimation
From the image, this resistor appears to be a physically large, rounded resistor, often indicative of higher power ratings. This size is commonly associated with a 2-watt or 5-watt resistor. For a precise wattage rating, the manufacturer's specifications or a comparison to similar resistors is needed.
@@ioannisbalouktsis962 The body of the resistor is physically too small to carry two or five watts! Difficult to determine the exact size of that board without actually being there but that's a 1/4 watt resistor at most.
Horrible circuit, this is wrong example of circuit which made in principle "mute" or "delay speaker connection/disconnection"... I know well these circuits trom some old AV receivers, but in right state with all what is necessary not as this. There from my point of view is missing fuse or fusible resistor, 1K resistor in case of failure made nice burner, as next is missing paralel resistor for 470n capacitor and last problem is that on DC side isn´t zener diode or load resistor which made more stable DC voltage. As load use only in series diode in optocoupler with resistor is wrong. This is nice example of circuit with less parts as is possible, but will be best really use only one diode instead directifier (irony). In combination with wooden box nice to have it at home, isn´t there marked telephone number for fire rescue? or 112? THX🤮
How about you just look at the schematic.
Don't think he has one yet but I agree at this point it would be great to see it.
la Résistance est une 1000K!!!!!Maron !!Noir!!Vert!!!et Jaune =5% the Resistance is a 1000K!!!!!Brown!!Black!!Green!!!and Yellow =5%