EEVblog 1482 - Mains Capacitor Zener Regulator Circuit

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  • Опубликовано: 26 ноя 2024
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  • @derekloudon8731
    @derekloudon8731 2 года назад +20

    Hi Dave, these brief 'explanation' videos are a very useful resource for 'faded memory' refreshers (I'm a long retired engineer). Thank you ☺️

  • @padmasaripalli4325
    @padmasaripalli4325 2 дня назад

    Great explanation of how zener based Derivation of DC from AC mains works. It saved a lot of angst for me. I changed caps and zeners to no avail. now I see the problem is with capacitor on AC input

  • @KeritechElectronics
    @KeritechElectronics 2 года назад +24

    The Zener regulator makes perfect sense with the series resistance impedance on the AC side and low current loads.
    Good that you give the safety warnings, they're absolutely essential with this type of circuits.
    BTW it looks like I'm starting to teach people electronics.
    Ttoday at our local hackerspace I had to explain the FULL BRIDGE RECTIFIER's principle of operation to one guy who has a university degree in IT but was not taught the basics of EE.
    He had a pair of IN-12B Nixie tubes that he wanted to use in a project, only didn't know how to wire them... poor thing. I breadboarded a simple 200VDC PSU with two power transformers connected back to back for galvanic separation from mains, and taught him how to control these nixies using transistors.

  • @TheCod3r
    @TheCod3r 2 года назад +8

    Love these whiteboard videos Dave. By far the best series for me

  • @petehiggins33
    @petehiggins33 2 года назад +52

    This capacitance degradation in 'Polly put the kettle on' capacitors is a well known phenomena but doesn't seem to be acknowledged by the manufacturers. I've had several small electrical devices like coffee machines and mains timers fail due to this effect. It would be interesting to set up a test to count the number of breakdown events occurring and the rate of capacitance reduction. I suspect it's mainly a problem in 230V countries where many areas, like mine (and BigClive's), still operate at 240V - 250V or maybe it's a function of how clean the mains supply is. It doesn't appear to be a problem on DC supplies, spacecraft use these capacitors to filter the main DC bus and they work happily for a couple of decades. Having said that they are voltage derated by at least 50%. I would do the test myself but I don't have an oscilloscope which would be necessary to establish the size of the current pulses.

    • @EEVblog
      @EEVblog  2 года назад +88

      I could potentially do that, as I have a mains surge generator. I can run it continuously and get capacitance data.

    • @teardowndan5364
      @teardowndan5364 2 года назад +8

      I suspect voltage surge events large enough to cause breakdown in X-caps and damage proportional to the amount of energy involved are relatively common if you don't have a whole-home surge protector wired into your breaker box or snapped directly onto its bus bars as required in some of the newest electrical codes. Whenever something on your local distribution transformer shorts to ground/neutral/phase-phase and a breaker breaks 100+A fault current, all of the energy stored in stray inductance along the way has to go somewhere. Stuff triggering reclosers and other distribution equipment can cause transient nastiness too.

    • @jimw7916
      @jimw7916 2 года назад +1

      MASON?

    • @Basement-Science
      @Basement-Science 2 года назад +13

      @@EEVblog please do that. Also test some higher-voltage rated caps and test it with and without a MOV across it.

    • @WacKEDmaN
      @WacKEDmaN 2 года назад +7

      yup..id be interested to see that too please Dave

  • @Brfff
    @Brfff 2 года назад +2

    I wish we had your RUclips videos back during my university engineering days in the 90s - thanks!

  • @scottpelletier1370
    @scottpelletier1370 2 года назад +5

    Very cool. I had an LED bulb go out so I smashed the power supply out of it to see how it worked. This pretty much looks exactly like what I saw... Simple rectifier spitting 50 to 60 volts DC out to run a panel of 50 little LEDs. Great explainer!

  • @wurl
    @wurl 2 года назад +3

    Hi Dave. Actually I think that the cuircuit with the two zener-diodes in series is a quite smart solution.
    Just imagine the case of a 3.3V linear regulator hanging on the 24V rail; that would mean, in addition to the relay current comes the control logic current: a cheap uC + temp sensor + some LEDs, might sum up to additional 15mA; meaning you need twice the input capacitance and have twice the power drop on the 24V zener.
    Compared to that, the shown configuration provides the current for the 3.3V rail almost for free.
    I guess, the relay coil version is selected to match the control logic current consumption, in order to get the total component cost down.
    BTW: greetings from Germany. I like your channel a lot.

  • @tlhIngan
    @tlhIngan 2 года назад +6

    While capacitive droppers make poor power factors it really helps the power company. Most load anywhere is generally inductive - think electric motors used for practically everything in industry and residential. Fans used in ventilation, or motors driving compressors in HVAC and refrigeration or motors pumping liquids and such. Thus power companies know the power factor is generally inductive and they have to add capacitive reactance to balance it. There are capacitor banks on distribution lines and at substations to compensate for the inductive loads. You using something with a capacitive dropper helps bring the inductive load down a bit. Though modern LED lights are back to being resistive loads as they are using linear regulation driver ICs. But since the LED array drops about 90% of the line voltage the linear regulator doesn't waste much power. So I wouldn't worry too much about capacitive droppers at all.

  • @tomasbergh
    @tomasbergh 2 года назад +6

    Finally some nerdy electrovideo! By the way, the zener needs to be able to handle the power of 24v 16mA when the relay is not activated. Hope you are doing fine Dave! Thanks! /Tomas

    • @kennmossman8701
      @kennmossman8701 2 года назад +1

      rather use a single 24V 1W Zener, you can use two 12V 400mW Zeners in series

  • @TheHuesSciTech
    @TheHuesSciTech 2 года назад +12

    Worth calling out that the strange-looking stacking of 24V and 3V3 kinda allows the microcontroller to be powered "for free"-ish in some vague sense. I.e., a 240V AC mains + dropper cap = a high compliance, practically-fixed-current AC current source; if you were to run the MCU off a LDO off the 24V, then the current for the MCU and the current for the relay coil would be added up. But with those elements in series, it's only the compliance voltage of the current source that needs to be increased, and since it's already at 240V, it's all good to go.

    • @adama7752
      @adama7752 2 года назад

      You should lookup the LDO's efficiency. Not great

    • @robertbackhaus8911
      @robertbackhaus8911 2 года назад +2

      Another thing I like, and it comes to the low current - relays need more current to turn on, but a lot less to hold them on. So that 24 volt supply backed up with the capacitor can provide all the current needed to activate the relay, but then the voltage can drop away as it likes, and the current with it, and still hold the relay on.
      With the relay coil and 24v zener providing current for the digital circuits, you don't have to worry about a voltage drop causing the micro to reset.
      We can see how well the circuit worked, with that capacitor dropping to less than half its value before it caused a problem.

  • @CaspaB
    @CaspaB 2 года назад +4

    Thanks Dave. I've been perusing the comments and realise how much learning can be got from quite simple circuits such as this.

    • @bertblankenstein3738
      @bertblankenstein3738 2 года назад

      You might also like Big Clive.

    • @CaspaB
      @CaspaB 2 года назад +1

      @@bertblankenstein3738 Big Clive is ok but doesn't have the theoretical background IMHO.

  • @MrDoneboy
    @MrDoneboy 2 года назад +1

    Great explanation, Dave about the loss of capacitance, not creating enough voltage, which caused the problems further down the circuit. Hope that made sense.

  • @trelligan42
    @trelligan42 2 года назад +12

    Plugging in a capacitive load into the electrical grid is not a problem. Yes, there is (reactive?) power that the grid needs to supply and then accept back each cycle; this produces non-imaginary current flowing through wires that heats them up and has to be made up for by the generators.
    But this is capacitive reactance, and is 180° in phase from inductive reactance. The (US) national grid operates on a power factor of 0.8, which is inductive. Most of the loads on the grid are inductive; transformers and motors mostly. That means the generators have to supply power to build up the magnetic field in all this inductance, which heats wires &etc.
    I once worked for a company with many large assembly lines. They had many large motors each and we used heaters with some inductance themselves. In order to prevent extra fees because we were inductive-loading the power factor out of tolerance, we had a Capacitor Farm. All those magnetic fields growing and shrinking in our motors were feeding power into our own capacitors locally, so the generators saw a normal load and the long-distance wires to the power company could stay cool.
    So if you plug a capacitive load into your house receptacle, rejoice in knowing that you're just balancing out a bit of your refrigerator's inductive load, and making those generators a bit less loaded.

    • @axelBr1
      @axelBr1 2 года назад +7

      Actually the power company will love you having these capacitive loads, not a problem at all, because as you say, most loads on the grid are inductive.

    • @matthewmaxwell-burton4549
      @matthewmaxwell-burton4549 2 года назад +9

      They won't like you if you add a cap behind a diode bridge, you're filling the mains with current harmonics. Which is even worse.

    • @quetzal4042
      @quetzal4042 2 года назад

      @@matthewmaxwell-burton4549 But Dave is dead wrong when he says "the generator still has to provide that 3.7 watts of power", isn't he? It has to provide the current, and there will be some loss in the lines, but only a small fraction of the 3.7 watts. If it were just a capacitor directly connected to the generator, there would be no net transfer of power, right?

    • @matthewmaxwell-burton4549
      @matthewmaxwell-burton4549 2 года назад

      @@quetzal4042 power in is always equal to power out. Even if we are talking about reactive power, it has to come from somewhere. Reactive power is a bummer because it provides no work and only heats wires. A cap cause the V to lag behind I. So just by adding a cap, the gen will have to supple reactive power.

  • @kiltrash1
    @kiltrash1 2 года назад +1

    In some of these types of circuit the zener only limits the maximum supply voltage if the load drops, rather than to regulate the supply voltage. For example, the zener in our fan controller limits the supply rail to +12V whereas the normal running voltage is about 5½V.

    • @robegatt
      @robegatt 2 года назад

      Like in the reverse polarity circuits to protect the mosfet gate (someone uses just the zener though)

  • @billharris6886
    @billharris6886 2 года назад

    Hey Dave, I have seen these crude line capacitor dropper power supplies used since the early 1990's. I worked for a company that used these circuits in LED bulbs but, without the surge resistor. I modified the circuit by inserting a 100 ohm pulse rated resistor (US, 120 Vac), which improved reliability and power factor. These circuits are very vulnerable to voltage spikes and outputs from triac dimmers and squarewave/modified power inverters. The circuit you showed has a worst case surge current of 6.6 amps, assuming a clean sinewave. That particular circuit was likely designed by an engineer, then redesigned by management to further lower costs, then redesigned again by manufacturing who would only allow 10 reels of parts to be used on their pick & place machine. If the surge resistor was pulse rated, the redesigns probably changed it to a 1206 thick film chip resistor to reduce cost.
    Of course, who cares about ripple on the regulated supply lines or voltage drift over temperature. These zener voltages are horribly unstable over temp. Also, placing a filter capacitor in parallel with a zener feed from a rectifier, can never be free of ripple (I wonder how the microcontroller reacts to that?).
    I am assuming this circuit was powering a digital thermostat for a low cost space heater? With this circuit, it no longer replicates the mechanical thermostat it is replacing because, it will no longer run off DC (assuming these was no fan being used).
    This circuit has horrible power factor as you mentioned, although would be insignificant when the heating element is energized. The power factor of the circuit is usually overlooked due to the low power draw but, becomes a problem due to the amount of power line harmonics it generates and resulting power loss at the power plant, which is roughly 7 times the power as measured at the device itself.
    The power factor problem was recognized in the 1990's and since then, home appliances drawing greater than 50 watts employ some type of power factor correction. The worst offenders now are induction motors but, they have been redesigned as well such that the worst power factor I have witnessed lately is 0.85. Additionally, most all electronics use an off-line-switcher instead of a step-down 50/60 Hz power transformer. So, the mains are far less inductive these days.
    If your house is powered from a simple 2 or 4 pole generator, the power factor can quickly consume available generating capacity. Let's assume you are charging a solar battery bank from your generator. The difference in power factor between battery chargers that have 0.5 versus 0.9 is huge, with the 0.5 needing almost twice the generator capacity.

  • @Nabilphysics
    @Nabilphysics 2 года назад +2

    this kind of video suits you 100%

  • @WacKEDmaN
    @WacKEDmaN 2 года назад +10

    Excrement!...erm Excellent! :P
    youre starting to step into BigClives world with this one Dave!...always good to have the technical background tho!
    speaking of Zappy!.. ive got a cheap coffee grinder.. ive copped multiple zaps off it!.. it obviously doesnt have any bleeder resistors across the capacitor...ive started to short the plug prongs on a metal frame every time i pull it from the wall

  • @StevenHodder
    @StevenHodder 2 года назад +1

    The grid delivering the power is largely inductive, so power factor correction usually means adding capacitance. Except for very long EHV lines where there is a huge amount of distributed capacitance and series inductors need to be added.

    • @vogonjelc
      @vogonjelc 2 года назад

      Yep some old factories had capacitor banks for power factor correction. But in this case those 3.7 w were passed away in form of reactive power

    • @StevenHodder
      @StevenHodder 2 года назад +2

      Technically 3.7 VAr were provided into the grid by the capacitor dropper ;)

    • @vogonjelc
      @vogonjelc 2 года назад

      @@StevenHodder except it should be in parallel to the load if my memory serves me right. And this is in series.

  • @christopherjackson2157
    @christopherjackson2157 2 года назад

    I find the whiteboard videos really helpful. I don't have any ee background I just like to tinker with and fix things as a hobby so I lack a lot of the fundamentals.

  • @MyProjectBoxChannel
    @MyProjectBoxChannel 2 года назад +1

    big Clive done a similar failure mode video. There was a capacitor in a cap drop circuit that lost capacitance over time. I think it was the power supply for the clock inside a oven. there was also a relay circuit, the relay wouldn't operate due to a cap dropper losing its capacitance over time.

  • @arvydasn2278
    @arvydasn2278 2 года назад +2

    Great timing with this content. I was just looking into my failed mains ac power meter. It seems to use this to power dc circuit and voltage is way to low.
    So it seems you just pointed me to the culprit (ac capacitor). All other possible sections were testing as good.

  • @alainderauglaudre4722
    @alainderauglaudre4722 2 года назад +5

    What happens when the NPN transistor is turned on on time 16:28 ? Which one of the two transistor will blow up first ? Did you miss a limiting current resistor between the PNP and the NPN ?

    • @eDoc2020
      @eDoc2020 2 года назад +1

      I'm sure there was a resistor.

  • @atkelar
    @atkelar 2 года назад +1

    I came across a resistor based version of this in the standby circuit of the Revox power amp restoration I recently completed. It gets way too toasty for my taste, so I disconnected it; besides, the optocoupler that turns on the mains is probably shot too. I replaced the original with a higher value to cut down the current, but it still was too hot even after short periods of powering it. Note: I'm convinced the zener would have properly conducted, since the original value was also supposed to work across 110V and I have 240V here, so double R should not be an issue for the current...

  • @CaspaB
    @CaspaB 2 года назад +4

    Hi Dave.
    A trick with this circuit is to select the upper electrolytic value so that when fully charged it will reliably close the relay, but set the quiescent current (set by the 0.22 UF cap) to a value which will still hold the relay in. With the spec sheet you showed, the relay should hold in easily at half the voltage and the cap could be reduced to 0.1 UF, so long as the upper electrolytic cap was sufficiently charged up by the time the "start" signal arrived.

    • @ferrumignis
      @ferrumignis 2 года назад +3

      That will minimise power dissipation in the relay, but will make the circuit prone to failing with little degradation of the X class cap.

    • @brainndamage
      @brainndamage 2 года назад +1

      That's one way, that lowers the reactive and real power draw, but the real power is still drawn even when the relay is on, dissipated by the zener.
      Another method is to keep the cap at 220n (enough current to turn on the relay) but short out the 24v supply with a current-limited transistor. That way the real power will go to practically 0 when the relay is off, and the zener stays cool. The reactive power will still be drawn however. This is easy with this stacked zener arrangement, since the logic will still get power if the relay supply is shorted. It is possible even in the single-rail version, just that you still need to keep the relay driver (disconnect the relay to turn it off) but add a transistor voltage clamp circuit that clamps the supply voltage to whatever your logic regulator needs minimum (could use a LDO to make it very close to the actual logic voltage)

    • @CaspaB
      @CaspaB 2 года назад

      @@ferrumignis Agreed.

    • @CaspaB
      @CaspaB 2 года назад +1

      @@brainndamage Yes! I like your thinking. No-one anticipates shorting a power supply could possibly reduce power consumption.

  • @admirerofclassicalelectron2858
    @admirerofclassicalelectron2858 2 года назад

    Very interesting video. And a wise choice of a RPN calculator, 'cause there are no problems here with implied multiplication whatsoever. I like the good old HP 41C.

    • @CaspaB
      @CaspaB 2 года назад

      And the HP35. I bought one in 1973. Cost me a week's pay.

  • @jensschroder8214
    @jensschroder8214 2 года назад +5

    In capacitor power supplies with relays, a relay with the highest possible voltage and lowest current is often used.
    It is often common for the zener diode to get hot when the relay is off because the current is then heating up the diode. If the relay is on, the diode is often very cold because the power is then in the relay.
    The series connection is clever because the 16mA is available anyway, an additional 3.3 or 5 volts are irrelevant for the current.
    The disadvantage of the capacitor power supply is that 16mA always flow. Yes, always!

  • @ChristophLaimer
    @ChristophLaimer 2 года назад +3

    I think the explanation about the Zener-Diodes and the degrading dropper capacitor is a bit awkward: those diodes are mainly needed to limit the voltage for the transistors, in particular if the the relay coil is turned off. For a longer lasting circuit, the dropper capacitor probably should be 300nF, and then the Zener is also "eating" the surplus energy. It's true, that the Zener are getting redundant, when the dropper capacitor degrades too much over time.

  • @RGD2k
    @RGD2k 2 года назад +4

    The current capacity is not that dodgy: Having the rails in series means the current gets to be used twice: That same ~15mA is always available for the micro regardless of whether the relay is on or off: It's in series, not parallel.

  • @olivierconet7995
    @olivierconet7995 2 года назад

    This HP41C calcularor 😍, typical of a good engineer

  • @veganath
    @veganath 2 года назад +1

    Wow! That I found this in it's entirety intriguing speaks volumes of you terrific delivery, thanks Dave

  • @uwezimmermann5427
    @uwezimmermann5427 2 года назад

    Today me and a colleague also had a failure in a capacitive dropper circuit - but it was not the capacitor. After the diode bridge there was a simple linear regulator with a 30 V zener-diode, a 10 kΩ resistor to limit the zener current and an NPN-transistor. This intention of the supply was to keep the coil of a 24 Vdc relay energized - pumping 29 V into the coil, why?
    However, after about 25 years in duty both the transistor and the zener diode had failed which I diagnosed in the end from the about 0V over the zener diode and 0.4 V base-emitter voltage of the transistor.

  • @MrDoneboy
    @MrDoneboy 2 года назад +4

    "Come a Gutsa". Gotta love Dave's dielect! Lol.

    • @WacKEDmaN
      @WacKEDmaN 2 года назад +1

      ..you could say its dielectric! 😜

  • @stelmo502
    @stelmo502 2 года назад

    ALWAYS wanted to know about these type of circuts. THANKS

  • @alicangul2603
    @alicangul2603 2 года назад

    Dave, correct me if I'm wrong but the 3.5 "W" (VA) is not supplied by the grid, it is supplied by the power factor correction panel at the facility main distribution center. For example, installed alongside transformer LV output panels.
    If reactive power consumption of your generator's load increases, you do not push more water, steam etc. to your generator. You change the field excitation current.

    • @robertbackhaus8911
      @robertbackhaus8911 2 года назад

      Your supplier has to provide that VA, whether it is supplied by locally, or even, at one time near here, with a smallish hydro plant being run out of phase to the wider grid. But as another person said, generally the load on the grid is net inductive from motor and transformers, so adding capacitive loads like this helps.

  • @axelBr1
    @axelBr1 2 года назад +2

    "Without the load you get nothing", apart from 240V RMS, on the outlet of the rectifier presumably? I came across a "reactive voltage dropper" circuit a few years ago when I tried repairing a paper shredder. Being a naive mostly home taught electrical engineer, I was expecting to find a transformer and voltage regulator circuit to drive the LED indicators and optical paper sensor, but all I found was a capacitor and resistor. After a quick Google I found out what they were doing and why. And promptly decided that poking around a circuit board that wasn't isolated from 240V wasn't going to do my life expectancy much good. Did consider buying a plug-in earth leakage trip so I didn't trip my entire flat when the inevitable touching of the live line happened.

    • @Broken_Yugo
      @Broken_Yugo 2 года назад +2

      I consider a good quality 5ma RCD/GFCI, tested often, to be mandatory when working on any opened up mains referenced gear, even if you don't intend to do any live probing.

  • @gusmartin6053
    @gusmartin6053 2 года назад +2

    If you take the capacitor apart and unroll the layers, I wonder if you could see the damage? What would that look like under the microscope?

    • @ferrumignis
      @ferrumignis 2 года назад +1

      They are encapsulated in epoxy, I think it would be very difficult to get apart without destroying the evidence.

  • @michaelhawthorne8696
    @michaelhawthorne8696 2 года назад

    Nice video Dave
    13:20 This Power Factor business means you're getting your circuit to work but the energy supplier has to carry the cost of the use of your circuit.
    Conventional meters in the house can't measure the Apparent power and so you actually save some money.
    The suppliers want everyone to go to Smart meters which can....
    I wonder why !!

  • @jeffreylewis145
    @jeffreylewis145 2 года назад +1

    I never concern myself with capacitors loading the grid, figuring that all of the inductive loads on the grid more than compensate. So the capacitors are benefiting the grid. Do I have this right?

  • @Richardincancale
    @Richardincancale 2 года назад +22

    Actually in this case where it’s driving a heater, using a 5 watt resistor to drop the voltage would actually just contribute to the heating!

    • @robegatt
      @robegatt 2 года назад +6

      Lol, just replace the whole thing with a big resistor and you're good to go.

    • @westelaudio943
      @westelaudio943 2 года назад +7

      @@robegatt
      Weren't old school electric heaters just that... Long resistor wire with several taps, some bimetallic magic to shut it off when too hot, and an AC shaded pole fan.
      I wonder why a friggin' heater even needs a regulated PSU and MCUs more powerful than ENIAC in it.

    • @robegatt
      @robegatt 2 года назад

      @@westelaudio943 lol they are still selling those, I have a couple of them, very cheap stuff and they actually DO what they're made for! So I fully agree!

    • @bertblankenstein3738
      @bertblankenstein3738 2 года назад

      @@westelaudio943 in my Uni days we had some serious resistors capable of kWs in a lab. Three of those, three phase power and man that delivered.

  • @nicholasvalentine4725
    @nicholasvalentine4725 2 года назад

    Interesting circuit break down, thanks!

  • @cmuller1441
    @cmuller1441 2 года назад +2

    9:50 Isn't it a bit strange to do "order of magnitude" calculation using 3 digits of precision for the resistor (1.45 *10^4) ? Mains voltage only has 2 digits (2.4 *10^2). It would have been nice to round that to 15k for example.

    • @CaspaB
      @CaspaB 2 года назад +1

      Adding extra (unnecessary) precision helps in understanding the maths.
      For example in coil inductance calcs (very imprecise with iron or ferrite cores), pi could be rounded to 3 which is only in error by 5% but readers may not know where it came from.
      Using 3.1416 clears any confusion.

    • @volodymyrzakolodyazhny
      @volodymyrzakolodyazhny 2 года назад

      Mains voltage has three digits - '2', '4' and '0'.

  • @BezosAutomaticEye
    @BezosAutomaticEye 2 года назад

    4:05 thank you. just prior after watching the other video I spent 5 minutes googleing Australian slang. Thought I was missing something.

  • @Brian_Of_Melbourne
    @Brian_Of_Melbourne 2 года назад

    Perhaps a followup to this would be to model the circuit with one of the many SPICE type simulators? It would then be possible to see the phase of the input current wrt the voltage, to look at the effect of having the relay on or off (on current drawn and its phase). It would also be possible to demonstrate the effect(s) of using a resistor rather than a capacitor.

  • @PaulSteMarie
    @PaulSteMarie 2 года назад +1

    An HP-41C? Your taste in calculators just stepped up several notches.

    • @4L3XG
      @4L3XG 2 года назад

      Actually it stepped up exponentially! Nice calc.

  • @namenotshown9277
    @namenotshown9277 2 года назад

    Damn this guy is good, excellent explanation.

  • @piotrludorowski9529
    @piotrludorowski9529 2 года назад

    Yes we find it interesting. Thanks Dave :)

  • @Gaugolon
    @Gaugolon 2 года назад

    Thanks for the excellent explanation... I guess that is also the way YAMAHA build their standby power supply in their home cinema amplifiers... btw. also a 220nF capacitor ^^

  • @johnsonlam
    @johnsonlam 2 года назад

    Thank you very much, another fun learning video.

  • @aerobaticant
    @aerobaticant 2 года назад

    Love the calculators making cameos on your whiteboard videos. I'm always surprised at how many of them I have myself, including the HP-41 here. :-)

  • @gudenau
    @gudenau 2 года назад +4

    I mean, it's an electrical heater. A resistor just keeps a low level heater running. Say it's a feature!

  • @buckstarchaser2376
    @buckstarchaser2376 2 года назад

    Assuming this circuit is out of a coiled-wire resistive heater, with or without a fan, wouldn't the heater's primary operation effectively offset the capacitive line reactance at the plug?

  • @minus3dbintheteens60
    @minus3dbintheteens60 2 года назад

    I'll have to look for your power factor content, even though I get what it is, I'm yet to feel like I fully understand it, I definitely don't feel like I understand how the PF correction circuits work.

  • @topherteardowns4679
    @topherteardowns4679 2 года назад +1

    Is there a particular reason they use 24v relays instead of 12v? If they are pinching pennies, wouldnt it save them twice over. (Able to use a lower value cap), etc..

    • @EEVblog
      @EEVblog  2 года назад +1

      Hard to say. Maybe they use it on other designs and are just reusing stock parts.

    • @slademcthornbody9060
      @slademcthornbody9060 2 года назад +5

      12v relay may require higher current draw (bigger capacitor)

    • @robegatt
      @robegatt 2 года назад

      @@EEVblog exactly, 24v are way more common in industrial applications.

    • @Broken_Yugo
      @Broken_Yugo 2 года назад +2

      Probably the usual answer in consumer electronics design, it happens to be cheapest solution. I would assume because higher voltage coils are less common until you hit mains level, which are larger, more costly, more costly driver transistor, etc. Lower voltage is going to need more current for the same switching action (and a larger cap and zeners to match).

    • @robertbackhaus8911
      @robertbackhaus8911 2 года назад

      A balance of high enough voltage to keep the required currents down, and a low enough voltage to be able to control easily with your low power digital circuitry.

  • @martenthornberg275
    @martenthornberg275 2 года назад

    Having a power resistor waste 5 W as heat isn't necessarily a bad thing in a heater, since heat is what you are trying to create! :) (But in almost every other case it would be bad of course). In this case it would still be a problem because the zener regulator would use the same amount of current regardless if the relay is energised or not. Could they have used a power resistor to only power the relay, and some other way to power the control circuit that only needs a tiny current (one of the linear regulators mentioned in the other video maybe?). That would be more efficient and a resistor is less likely to cause problems than the capacitor.

    • @zyghom
      @zyghom 2 года назад

      yes but... if you dissipate 5W on resistor, you pay, when it is on capacitor then power plant pays ;-)

  • @ilanmagen
    @ilanmagen 2 года назад +2

    All in all they did excellent design "design to fail"

  • @CaspaB
    @CaspaB 2 года назад +2

    Hi Dave; you said PNP but you've drawn it as a PNPN.

  • @entropyachieved750
    @entropyachieved750 2 года назад +2

    Nice topic...

  • @keithking1985
    @keithking1985 2 года назад

    Nice stuff Thanks Dave 😊 👍

  • @comet1954
    @comet1954 2 года назад

    Nice calculator you have there! Just like mine....

  • @randycarter2001
    @randycarter2001 2 года назад

    The thing about that design is the microcontroller was still functioning normally even when the power supply was only working at 50% capacity.

  • @SuperSnowShow
    @SuperSnowShow 2 года назад

    Hi Dave, I have a question. Could it make a difference if I swap a polar capacitor (220nf) n that circuit against a bipolar one? Could the bipolar cap have lower ESR and mess up the circuit?

    • @kennmossman8701
      @kennmossman8701 2 года назад

      It is a non-polar cap. You can not use a polar cap

  • @keithcitizen4855
    @keithcitizen4855 2 года назад

    is it not safe to put fingers on a 5v usb lead from say an arlec plug adapter that has usb outlet ?

  • @PoiSonSonic
    @PoiSonSonic 11 месяцев назад

    Quick question. If the diode rectifier provides 240V+ and we shave off 24V and 3.3V, where does the rest go? Does it get blocked by the diodes? Does it get dissipated by them? Thanks

  • @jimmybrad156
    @jimmybrad156 Год назад

    Has anyone heard of mains supplies in Australia ever going over ~253v for a long time eg. more than half a second to say a few minutes?
    Do the street/suburb/local transformers have cut-outs to deal with this?
    I guess all the MOVs in devices would go up in smoke after a second or two, and a lot of devices would end up with damaged parts?

  • @clems6989
    @clems6989 2 года назад

    Fun video dave, this and the repair video. And quite informative ....

  • @anthonyvolkman2338
    @anthonyvolkman2338 2 года назад

    Is the rail voltage at the rectifier 27 volts or if the second Zener Diode 5 volts is it 29 volts? If so then it' may be using the Zener diodes in series with the virtual ground. That would make sense such as using 2 12 volt Zeners to make a 24 volt power source.

  • @tranxn7971
    @tranxn7971 2 года назад

    Thanks a lot Dave for this vidéo. I am not sure I understand how a 16 mA power supply can source enough current for both the relay (15 mA) and the logic part (?? mA).

    • @Gengh13
      @Gengh13 2 года назад +1

      With that zener arrangement they act like a 24+3.3v supply, that current flows through both, it's a pretty clever arrangement.

  • @donreid358
    @donreid358 2 года назад

    Nice calculator, does it still work? I was the engineer for the CPU on the first 41C model.

  • @jamesjohn2537
    @jamesjohn2537 2 года назад

    good explanation thanks

  • @GordieGii
    @GordieGii Год назад

    Couldn't one use a capacitor on each leg to get an isolated output?

  • @fredflintstone1
    @fredflintstone1 2 года назад +1

    wouldn't you have 240 x 1.4 so really 330 volts after the bridge???

    • @8bitMicroFan
      @8bitMicroFan 2 года назад +1

      Yes, the 220nF capacitor limits the current. Voltage "regulation" is done by the circuitry after the bridge rectifier. In a LED bulb, the LEDs in series act as zeners. When one of the LEDs fail, the smoothing capacitor in parallel to the LEDs will pop, if it's not rated for 350V.

    • @thisnthat3530
      @thisnthat3530 2 года назад +1

      Open circuit it would be 339V, but the zenners effectively create a voltage divider with the capacitor to limit the voltage of the output.

    • @EEVblog
      @EEVblog  2 года назад +1

      That would only be the case if the dropper resistor was after the bridge. But in this case the capacitor is the dropper resistor, and it's *before* the bridge, so it's not like havign a bridge direct across the mains. If you had no clamping zener and just the caps then yes that would be the case.

    • @robegatt
      @robegatt 2 года назад

      For a voltage divider to work you need some current flowing. The zeners or whatever load closes the circuit makes thing running.

  • @t1d100
    @t1d100 2 года назад

    Good info. Thank you.

  • @andymouse
    @andymouse 2 года назад

    I found this very informative...cheers.

    • @fredflintstone1
      @fredflintstone1 2 года назад +1

      Cheese!!!!!!!!!!!!!!!!!!!!!!

    • @andymouse
      @andymouse 2 года назад

      @@fredflintstone1 squeak !!!

  • @VandalIO
    @VandalIO 2 года назад

    Would you do a video on esaki or Gunn diodes

  • @dreamcat4
    @dreamcat4 2 года назад

    yeah it was an interesting and useful video dave. actually im sorry to have zoned out a bit in the middle with the specifics. but that is more due to my lack of interest due to having an inherent utter conpept for all these god awful circuits that i just find so apalling. rather than it being any fault in your educational style or presentation these. you made some great points, and also i am sure that inevitably will be coming across a great many real world examples of these. to need to actually at least identify them at least. to have a good grasp of what they are. so will need to watch this again. but more to the point: what other types of banger / substandard ac power circuits are there apart from this style? you know.... the others for which i can also harbor a similar level of contempt / that goes boom and/or kicks out a crap ton of interference. all for saving those precious bom cost dollar. it would be great to see them all. and maybe even some rankings table with columns pros and cons. one column could be interference. another column power factor. another column effeciency (and therefore also that means heat --> reliability). another column bom cost, etc. etc. a true rankings of the worst of the worst. haha

    • @dreamcat4
      @dreamcat4 2 года назад

      or perhaps another interesting angle would be to ask: whats the cheapest / most cost cutting such circuit that gets ul approval / certification. that complies with both those safety and emi requirements. but i suppose the thing is that the actual execution is also just as important as the design. i am reminded of big clive's recent video about the $1 usb power plug from the thrift store. that looked a lot like an official apple charger plug. but actually didnt have sufficient level of isolation on the pcb between the live input and 5v usb output that you touch with your hands. and the unbranded transformer was not sufficiently galvanically isolated either, when taking apart the windings no plastic sheet / seperator.
      so many diifferent ways these things can fail to meet expectations. it would be difficult to summarize into some single chart or table. so nevermind i guess
      btw in this video i watched it 2nd time over. and got almost everything now. it really comes alive and comes across as a very much more interesting subject the way you present it /explain it. only thing i didnt quite understand still was why they put the 3v3 zener in series with the 24v zener, if they could have instead put them in parallel. but perhaps the current in one zener was halping to regulate and limit (or kindda average out component variations) in the other zener. to make individual fault less likely to occur. would be my guess (of course when not knowing the specs of the zeners, perhaps the 3v3 zener has different current or something, if both were in parallel with the same shared 15k impedence capacitor). must be something or other along those lines. or for some similar reason

  • @mmaranta785
    @mmaranta785 2 года назад

    You come a gutsa!

  • @botoxpig417
    @botoxpig417 2 года назад

    Here we go. We are at school again !

  • @jimmybrad156
    @jimmybrad156 Год назад

    I guess a MOV might've saved the X2 cap losing so much capacity?

  • @Rx7man
    @Rx7man 2 года назад +1

    Considering this device is actually called a HEATER, some 3.7 watt power dissipation from a resistive dropper wouldn't worry me too much.. if it was in a light or something else then I'd be more concerned

    • @slademcthornbody9060
      @slademcthornbody9060 2 года назад +4

      It would draw 3.7 watt when switched off (with electronic timer or thermostat running) if that was the case. The relay switches on thr heating element.

    • @CaspaB
      @CaspaB 2 года назад +1

      The 3.7W would be an important consideration if the whole thing was built into a mains plug, with the high-powered heater elsewhere.

    • @Rx7man
      @Rx7man 2 года назад

      @@CaspaB it's not, check the previous video on the heater repair he references

    • @CaspaB
      @CaspaB 2 года назад

      @@Rx7man ok I didn't see the previous vid.

  • @christatler7378
    @christatler7378 2 года назад

    Dave, it’s ok to say ‘Impedance’. I lost count the times you use finger quotes to say ac resistance.

  • @Scrogan
    @Scrogan 2 года назад

    If you want to use a non-isolated MOSFET to switch rectified mains, you need a more nuanced capacitive dropper circuit.

  • @onjofilms
    @onjofilms 2 года назад

    Interesting stuff.

  • @shanesrandoms
    @shanesrandoms 2 года назад +1

    Enjoy these whiteboard videos.
    Though you weren't able to explain the whole circuit without spoilers 🤣

  • @user.A9
    @user.A9 2 года назад

    That chassis is hot as

  • @RixtronixLAB
    @RixtronixLAB 2 года назад

    Nice video, thanks :)

  • @EGL24Xx
    @EGL24Xx 2 года назад

    Don't forget, reactance is just the imaginary component of the impedance. The impedance is a complex number.

  • @urugulu1656
    @urugulu1656 2 года назад

    dave didnt YOU come a gutsa here? the zeners in series would mean that the zener voltages should add up and 24v + 3v3 or 5v would be roughly 30v and if you guesstimate the coil current from the datasheet at 30v (given are only 24v and 36v) you would get anywhere between 15mA and 10mA thus the zener current would still be edgy but not that edgy. if it really was in the order of 1mA i would expect the circuit to not even work with the good cap. assuming a coil current of 12mA at 30v (right between 15 and 10mA) there would be 4mA for the zeners which i learned to be roughly sufficent for small signal level type zeners...

    • @robertbackhaus8911
      @robertbackhaus8911 2 года назад

      Relays require more current to be engaged, but less current to reliably hold them on. So on turn on, plenty of current is provided by that smoothing capacitor to engage the relay. Then the voltage can drop away, as 10 or less mA is enough to hold the relay engaged. This gives the zener a break!
      Remember that this circuit worked fine until that capacitor lost more than half its value. So there's no problem with the circuit design, but a clear quality issue with their choice of capacitor supplier.

  • @alklapaxida850
    @alklapaxida850 2 года назад +2

    the eternal quest for cheapness--

    • @Kirillissimus
      @Kirillissimus 2 года назад

      ... while still being as shiny and flashy as possible - a hard task with no 100% solutions.

  • @timconnors
    @timconnors 2 года назад

    I opened up the Sonoff th10 hardwired smart switch to modify it's firmware, and discovered it used this style circuit. Possibly with a switchmode regulator, but certainly with the whole circuit at some mains potential. That's ok, right? It's hardwired in and you're not touching the microcontroller. Except that there's a 3.5mm audio socket on the side to plug in a temperature probe. I then wanted to plug the waterproof temperature probe into my fishtank.
    I made another modification - plugged it directly into a 5v USB port and just buggered off the 240 part. Don't trust cheap Chinese devices without opening them up and reverse engineering them!

  • @__--JY-Moe--__
    @__--JY-Moe--__ 2 года назад

    OMG! U mean I just repaired my ''home'' appliance? wow! time 4 a Impy-dance!

  • @oschonrock
    @oschonrock 2 года назад +1

    Small correction. The generator has to provide 3.7VA not 3.7W.. Or perhaps just "the generator has provide the 'current', ie the 16mA, but this is 90 degrees out of phase, so it amounts to 'no power'" And actually quite often the "generator" doesn't have to provide that extra current at all, because there will be an inductive load somewhere more locally which just supplies (and hence offsets) the capacitive "leading" current.

    • @EEVblog
      @EEVblog  2 года назад +2

      The point is the power has to ultimately come from somewhere outside the load, there is no free lunch. I didn't want to bog the video down with the discussion on that, it's done in other videos.

    • @oschonrock
      @oschonrock 2 года назад

      @@EEVblog Fair enough, it's complicated. The current has to come from somewhere outside this device, yes, but it's probably coming from somewhere on the same circuit breaker in the same house, where there is a linear supply with a transformer (ie an inductive load).

    • @alicangul2603
      @alicangul2603 2 года назад

      Generator doesn't have to because every facility has its own capacitor bank to provide local reactive power.

    • @oschonrock
      @oschonrock 2 года назад

      @@alicangul2603 Well no. This is additional capacitive reactive power, ie leading current, which is needed. A capacitor bank can't provide that. An inductor can, and there are loads of those which are right next to the little heater/smoke detector, in the same house, or at the latest the local street transformer. In fact, these little capacitive loads are actually HELPFUL (but probably not significant) for the power company, because they are like "local power factor correction" so improve the power factor of the house (by cancelling out some of the many inductors) and thereby reduce the reactive current which the power company needs to deliver to that house, reducing required distribution wire sizes, transformer sizes, etc..

    • @alicangul2603
      @alicangul2603 2 года назад +1

      @@oschonrock Ohh that's true... I was very confused lol. Anyways, lots of reactive consumers in a premise anyways. Motors, old lamps etc. lol. This little cap doesn't need a generator to supply the negative reactive power lol!

  • @Slartibartfas042
    @Slartibartfas042 2 года назад

    WOW - 47 Ohms as inrush protecting resistor? Even when we are effectively dealing with a series circuit of an capacitor with 220nF and the electrolytics? No way that the 47 Ohms would make that much of a difference to the inrush current. It is much more likely that this 47 Ohms resistor is nothing more than just fusible resistor to break the circuit if for example the diodes in the rectifier. The 220nF cap alone would limit current down to not more than just a few Milliamps... ;-)

  • @LawpickingLocksmith
    @LawpickingLocksmith 2 года назад

    In a nutshell: Arlec is pretty cheap stuff! Not my primary choice.

  • @jdlives8992
    @jdlives8992 2 года назад

    I beg you sir to do a few videos for mains capacitor solar exposure 1850’s solar storm stuff. Super dangerous i know for the regular but ya gotta break some eggs ;-)

  • @axelBr1
    @axelBr1 2 года назад

    Disagree that the Zener was starved of voltage because the relay coil current was close to the capacitive reactance. I presume that ideally all the current would flow through the relay coil and drop the voltage below the Zener's voltage, so no current is wasted flowing through the Zener, and the purpose of the Zener is stop the outlet of the rectifier floating to 240V (RMS) when the coil is open, presumably damaging the transistor and capacitor.

  • @brumbymg
    @brumbymg 2 года назад

    I think worrying about 3.7W dissipation from a resistor in a 2kW heater is rather amusing.

  • @Graeme_Lastname
    @Graeme_Lastname 2 года назад

    G'day m8. You could have just rang the SEC and got them to change the mains frequency to 100Hz then you wouldn't 'ave to buy a new cap. Yeah, I know. Catch ya l8r m8.

  • @tomwimmenhove4652
    @tomwimmenhove4652 2 года назад +1

    You made the same mistake as in a video a while ago about power generation. The generator doesn't have to generate the power (just think about how that would work in relation to the conservation of energy. The genreator generates 3.7 watts, and it just.. disappears?). Parts of the cycle, the generator would have to push a little harder, and the next part of the cycle, energy will be delivered back to the generator, effectively pushing it forward. In a lossless system, this means there's no net power generated.

  • @theoloutlaw
    @theoloutlaw 2 года назад

    6:18 - AC 'Sthircuit' Theory?? Ahahaha!