This was a great video. I actually knew most of this already, but you put it all together in a way that gives me a better grasp of how all the parts relate to one another. In particular, I'd never seen the formula for ripple voltage related to current draw, frequency, and capacitance. I knew greater current draw resulted in higher ripple, but I didn't know the mathematical relation.
Just plain great! Oh, so good! You pulled together several elements for me and cleaned up my understanding of the subject. I am so happy and grateful. Thank you, again, for all your effort to share with us and teach us. You are just such a special, generous person! I know that you do this, because you enjoy the hobby and sharing. But, another element that you can add to your satisfaction is that by educating us, you are also protecting us from getting hurt... As in electrocuted... There is no telling how many lives you have literally saved. Good on you!
I got a BS in Elec Eng Tech in 1989. Granted we were focused more on using microprocessors and programming them in assembly language. However, we had to do circuit analysis using differential equations and all that. Anyway, It was really all just a bunch of math and how to actually build something like this was secondary. So after 30 years of writing software and retiring I am now getting back to the electronics side of things.
Thanks. I have some filter capacitor considerations to make/understand to convert a 220VAC 50Hz inverter to function at 120VAC 60Hz. This helps me tremendously to see what is happening how to adjust or use the filter capacitors after the bridge.
Good info. I read online that one way to roughly determine the load capacity of an unmarked / unknown power transformer for a tube amp is to load the B+ until the voltage drops by about 10% and that is likely a nominal load. Not sure if that's a good rule of thumb or not. Would be great to see one of your detailed videos on this.
Just information I need I have just in rolled in electronics II class in community college. One of my class projects is building a headphone amplifier kit with a separate external power supply. www.pmillett.com/nuhybrid.html Using a vacuum tube technology that was just developed in Japan a couple years ago. Using the ( Korg Nutube). The design is by Peter Millett based off his original design that used low-voltage vacuum tubes called “ Millet Hybrid “ which has a lineage going all the way back to the days of their original (Starving Student Headphone Amplifier) I was going over his circuit board design to figure out what capacitors I could enlarge for a slightly better bass response. Also building my own linear power supply and try to clean up and provide the cleanest DC voltage I possibly can building myself.
You can use a simulator and experiment ...you will notice big differences in PS set-up and how to fine tune all values of the components constituting the power supply for your own project/amplifier etc , and adjust for your personal needs . Go to the site of Duncan and upload the simulator , it's free . Your ears will tell you if you got it right .....and confirm by measurement if you wish to check..
Mr carlsons lab was doing a restore on an old vacuum tube piece of equipment... I don't remember which one it was but from my understanding it used a filter choke on the power supply and when he added a capacitor before the choke, the voltage output climbed dramatically... I think that he might have been working on a tube amplifier because I vaguely remember that he was able to make it output much more power than it was actually being rated for... But the lesson was really about getting rid of voltage ripple and if someone tried to add a capacitor before the filter choke, they would be raising the voltage output way beyond the voltage ratings of components such as capacitors further down the line and you will end up with exploded capacitors without realizing what you have done... It's a pity that he didn't go deeper into detail about this topic because it is a trap for many people... Although this concept could potentially be useful if you needed a little more voltage for your circuit while the transformer available to you doesn't have the required voltage but has current to spare... Say something like a low voltage transformer from an old Ark welder....
Once you are below 1% distortion an amp is an amp is an amp is an amp...Features you need and enough power for your speakers and room. Done deal. Love the dsp in that one.
Yes Tony, you do an excellent job of explaining that beginners like me can understand. Keep it up. Perhaps one subject you can consider doing a video is good power supply design for audio amplifiers with attention to headroom. For example, though it costs more will putting a beefy power supply into a smallish 50W output high current amplifier to give it at least 3dB of headroom, can this be considered an ideal approach for the best sonics and dynamics?
A bridge has 2 times the diode voltage drop over a center tap arrangement. Moreover, voltage drop changes with load, it is not a constant .7V for diodes. Always consult the data sheet of the rectifier used.
hello thanks for video , very technical and informative and made it clear many unknown things for me. Will bulged ( swelled on top) big capacitor (10000 uF) on first step of linear power supply( bridge diodes), cause voltage drop in output?
Hi Tony I have a new Cayin integrated tube amplifier A-300b Mk II. When I turn off I hear a soft thump through my speakers. Some say normal but another owner with same amp and similar speaker efficiency has no thump. What could be wrong with my amp? Thanks!
Hey Tony, know of a small simple soft start circuit? You made me glance over my home made bench power supply and since I have a huge filter cap in it.I need to toss in a soft start circuit.
You could build one, but I found them in kit form as well as assembled for a few dollars on eBay. Just search "soft start module" or "time delay on relay" and you'll see a whole bunch of them. Thanks for watching!
At car stereo with heavy amps in it, we used for 13,8VDC big caps to protect battery and generator, 1F caps, which are huge, these were nuts to be honest, and some guys placed them themselves, which caused electric fires, with bad isolated connections, and use of too thin cables, to provide all the power to the stereo. I have here a 21mm² cable that is just pulverized on the inside. Because of the high currents.
Thanks for your good tutorial I have two questions to ask: 1) How do I limit the out put voltage from 16volts to 12volts when my load is a 12volt load 5amps 2) The value of the smoothing capacitor I calculated is too large how do I reduce the capacitor value with 1volt ripple appearing on the load
The best way to limit the voltage is with a regulator IC, such as the 7812. If the circuit is more than 1 amp, you will need to add a current pass transistor. Just search 7805 pass transistor and you should find more info on that. If you use a resistor, the voltage drop will vary with changes in current load, so that would not be a good idea. As for the capacitor value, you can calculate it with this formula: For 60 Hz mains: C = I (8.3/V) For 50 Hz mains: C = I (10/V) Where: C is capacitance in microfarad, I is the amount of current the power supply will deliver and V is the amount of ripple voltage you desire. 8.3 and 10 are the respective time periods for 1/2 cycle of the mains voltage in milliseconds for either 60 Hz or 50Hz mains. For a 5 amp power supply: C = 5(8.3/1) = 5(8.3) = 41 microfarad. In your example, 1 volt of ripple would be very high, as you only have a 12 volt power supply. To reduce ripple to 10 millivolts, you would need 5(8.3/.01) = 4150 microfarad. Hope that helps
@@xraytonyb Thank you and I really appreciate God bless you One more question; a situation where there is no regulator will the peak voltage produce after rectification can it destroy the load
Thanks so much Tony, I learnt a lot. I know this may be a stretch, but I would be grateful if you could point me in the right direction; would replacing two 6800uF filter caps in an old Kenwood 5700 with two 10,000uF capacitors be pushing it too far? Would replacing each with 2 in parallel change anything? Thank you very much.
You roughly double the capacity, hence doubling the time it takes to charge them up. Remember that an empty capacitor presents a dead short to your rectifier as well as to the transformer. I don't know if the 5700 has a softstart, but if it is voltage controlled, it will make no difference at all. Both the transformer and the rectifier may survive, but you put a lot of stress on both of them on power up. Why you wanna do that anyway? Is it humming or too noisy? Maybe the capacitors are just leaking to much dc due to age. But then change them with the original value. Just my two cents.
@@agrimm61 thank you 😊 Going with original values. I guess I was being greedy and not knowing enough thought hey I can buy double the value for a couple of dollars more!
Lovin the videos!!! ThankZ - quick question I’m semi new to electronic but have been building stuff for many years but I’m now trying to expand on my understanding. I have a nice multimeter and just picked up a Rigol DS1052e I’m trying to design a power supply from the ground up and overall learn more. Is there any place you can point me to further understand. The video was great by the way THANKZ AGAIN
Hi Tony Thanks for the refresher info, you are very easy to learn from !! Just a small suggestion do you think it may be better to connect mains voltages up using a connecting block that Big Clive uses,it is a lot safer and teacher's new starters a safe way to connect up Sorry for my suggestion it is not mean't to insult you Thanks for the great video
I actually have clear heat shrink tubing over both connections to insulate them. It's a bit hard to see on the camera. I thought about getting a quicktest a while back, but they are very expensive here in the states and that would be another big thing on my bench. They are really cool, though! Thanks for watching!
Thanks for the video I am trying to understand this concept. Can anyone help me I am trying to understand this concept because I am having an interference issue on a brushless gimbal. I have recently connected a 5.8ghz transmitter to transmit the video signal meanwhile I control the gimbals brushless motors using a 2.4ghz receiver. The brushless gimbal and both transmitters are connected to the same power source and whenever I turn on the video transmitter it causes the brushless gimbal to have erratic responses to the 2.4ghz input commands. I tried to solve this by using a 12v Battery Eliminator Circuit or BEC which has its own capacitors built in to try to clean up the power but I still get the same issue and I want to add a capacitor to help with the interference. I am not sure how to work out the type to add. Should I just add a 10V 100uF or will I need to go up to a 25V but I would have thought a 25V would store up power and then release it when it gets to 25V? Just thought that a 10V 100uF would end up being too low for a 12V device. Any help will be great thanks.
The voltage rating on a capacitor has nothing to do with storage capacity. It is the amount of voltage in the circuit that can damage the capacitor. A 25v 100 mf capacitor is the same as a 600v 100mf capacitor. The only difference is that you can use the 600v cap to filter a high voltage supply. Why not just make all capacitors 600v? Size and money. You just need to pick a voltage rating on the cap slightly higher than the highest voltage it is going to encounter.
i am working on Recapping a Technics SA-8000x it has two 4700uf caps in the filter what do you think the safe ratting i can go up too? 5,600 or just stick with 4700uf?
Sure this video isn't from yesterday - still it's great to listen. What gave me a bit to think was at 41:25. Isn't there an error in C formula? If there is known equation where I=C*dV/dt then C * dV = I * dt and finally C=(I*dt)/dV (compare it with one shown on the right: C=I/f*V) I - power the circuit is rated for, dV - ripple, dt - 60 Hz half-cycle = 1000/(60*2), Since author assumed 1 Amp and 1 V, equation gave good result. Change amps and volts to something more exotic and compare: (0,3A * 8.3ms)/2.5V is not 0,3A/(8.3ms * 2.5V) EDIT: 46:22 Hmm. Answer(?) You come up with 0.001xxx F and mine comes up with 1.xxx mF.
It just depends on the application. Generally, the transformer itself provides isolation, as long as the primary and secondary are not physically connected (i.e. it's NOT an autotransformer). This is why we call transformerless radios "hot chassis" radios. One side of the mains is directly connected to the chassis, causing a possible shock hazard. The isolation transformer provides isolation between the mains and chassis, making it safer to service. In a radio WITH a power transformer, the power transformer provides isolation in the same way, making the isolation transformer unnecessary. Grounding the chassis won't hurt and could ensure that the chassis is at the same earth potential as any other grounded equipment that could come in contact with it. That said, it is not necessary, as long as the mains leads are only connected to the primary of the transformer and NOT the chassis or any other part of the device. Hope that helps. Thanks for watching!
Someday we'll try to get to that. In the meantime, check out RUclips channel DiodeGoneWild. He is my absolute favorite for videos on switching power supplies.
At 8:50 or so you claim the transformer outputs the right output voltage with 110V in, but that's not the case, its rated to output 12.6V out for 120V in at _full load_, ie 3A rms. Your experiment with the 47 ohm resistor was only a 10% load. Normally you'd expect about 10% more output than rated for no-load, since typically the windings drop about 5% each at full load for practical transformer designs, so about 10% drop in voltage total at full load.
It is *not* a filter capacitor. It is not filtering any frequencies. It's a smoothing capacitor because it's smoothing out the peaks to give a flatter voltage waveform.
Really? And all this time I thought it was a low pass filter that was affected by frequency, capacitance and the load placed across it. Thanks for the education ;)
The equation C = I/(f Vr) shows that to reduce Vr to zero would require an infinitely large capacitor, i.e. as Vr -> 0, C -> infinity; interesting. So, you can never have zero ripple unless you use a regulator, again, interesting.
ruclips.net/video/jMkD3QxmN2I/видео.html isnt the diode positive on the transformer side , the stripe is negative from the transformer side ,off course positive from load side but if you put positive on stripe it wont work , dont know what you were talking there , got me confused ,diode has no polarity , in classical terms so who knows but off course stripe goes away from the positive transformer side to make it conduct, transformer is positive side , so the part of diode that is connected to positive transformer is positive part of diode , and that is not Stripe anode , its cathode , like you do and show, but say something else .
Does anybody recall Ohm's law written like I=U/R ? (U instead of V) I learned it like that in high school, 45 years ago... I can imagine for just one explanation: in physics V stands for speed (in Latin: velocitas) But I am far from sure that this is correct. Edit: I did some research. Indeed, U is correct, (from the Latin word: 'Urgere", which means to push or to press) This makes sense, voltage 'pushes' the current through the resistor.
Strangely enough, when I went to school, we learned "E=IR" (E was short for electromotive force, measured in volts) . Sometimes it was "V=IR", were V was just the first letter of the word Volt. When I started repairing X-Ray equipment, a lot of the European-built equipment (i.e. Philips & Siemens) would refer to voltage as "U" in the service documents, and still do today. For instance, "UH" would mean "heater voltage" (as heater was another word for the filament in a vacuum tube). Here in the U.S. you see all three letters used. Welcome to the land of nonstandard standards!
LOL! Fortunately, you guys in the US don't talk about 3/8 of a volt or 11/16 of an ohm. No offense, but I wonder if and when the US will stop using all kinds of body parts to measure distances ...
1959Berre In germany it is still U to this day, V for voltage has never been used since obviously we don't call it voltage but "Spannung" meaning tension in english. Just the unit is Volt.
Lester Barrett V or Volt is just the unit. In many languages the word for voltage does not start with a V, just like in german. so we use the already existing U for voltage. E is energy nowadays. And while E is for energy the unit you use would be J for Joule. Same thing as V and U. Or I for current but the unit is Amperes.
There is an error at 24:35 - 27:18. You mention that (in a half wave rectified sine wave) with a peak voltage of 16 volts one should expect an RMS of 0.707*16 = 11.3 V but this is NOT correct. For a HALF wave rectified signal the RMS voltage is actually half the peak voltage and not 0.707 times the peak voltage. Thus the correct RMS in this case is actually 8 volts. Now when your DMM measures AC voltage it puts a capacitor in series eliminating any DC offset which is why the AC reading (alone) is not accurate. However when a periodic signal has a DC offset the RMS value of that signal is the square root of the sum of the squares of the DC reading plus the AC reading. In this case you have a DC value of 5.18 V and an AC value of 6.42 volts. Taking the square root of the sum of the squares of these two values gives on 8.25 volts in agreement with the predicted RMS (8 volts above). Similarly with the center tapped full wave rectifier at 29:13 You have a peak voltage of 8 volts. SInce this is a full wave rectified signal the RMS value actually is 0.707 times the peak voltage and thus the expected RMS is 8*0.707 = 5.66 volts. Now your measured DC was 4.85 volts and AC was 2.69 volts. Taking the square root of the sum of the squares of these two values gives one a measured 5.55 volts, once again in agreement with the calculated result.
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Thank you for the video Tony B.
Takes me back 30+ years. Great little refresher considering I have forgotten pretty much all of my electrical theory.
This was a great video. I actually knew most of this already, but you put it all together in a way that gives me a better grasp of how all the parts relate to one another. In particular, I'd never seen the formula for ripple voltage related to current draw, frequency, and capacitance. I knew greater current draw resulted in higher ripple, but I didn't know the mathematical relation.
Just plain great! Oh, so good! You pulled together several elements for me and cleaned up my understanding of the subject. I am so happy and grateful. Thank you, again, for all your effort to share with us and teach us. You are just such a special, generous person! I know that you do this, because you enjoy the hobby and sharing. But, another element that you can add to your satisfaction is that by educating us, you are also protecting us from getting hurt... As in electrocuted... There is no telling how many lives you have literally saved. Good on you!
After surviving engineering school many years ago this is the first time I totally understand electronics circuits. After $100k in student debt.
I got a BS in Elec Eng Tech in 1989. Granted we were focused more on using microprocessors and programming them in assembly language. However, we had to do circuit analysis using differential equations and all that. Anyway, It was really all just a bunch of math and how to actually build something like this was secondary. So after 30 years of writing software and retiring I am now getting back to the electronics side of things.
Technicians sometimes make the best teachers or someone with the passion of the tech.
One of the BEST channels on the WHOLE RUclips.
Greeat video - perhaps you could cover the black art of bypass capacitors in power supplies.
Excellent explanation and demonstration; thanks for the formula!
thanks much Tony, exactly what I needed, especially hearing again how inrush current can kill a circuit was timely
This helps me understand the basics for my electronics lab on voltage regulators. Thank you!
Nice video Tony it does us all good to go back to basics now and again......................................Berni
Fantastic. This answered several of my questions. Thank you
Thanks. I have some filter capacitor considerations to make/understand to convert a 220VAC 50Hz inverter to function at 120VAC 60Hz. This helps me tremendously to see what is happening how to adjust or use the filter capacitors after the bridge.
Another great video Tony, thank you so much !!!
Great video, Tony!
Thank you for a great and simple info
Thanks you for making this video. Amazing help.
Good info. I read online that one way to roughly determine the load capacity of an unmarked / unknown power transformer for a tube amp is to load the B+ until the voltage drops by about 10% and that is likely a nominal load. Not sure if that's a good rule of thumb or not. Would be great to see one of your detailed videos on this.
really nice test equipment compared to my hobby stuff 😎 thanks , great info
Great video Tony!
Very nice presentation. Thanks you very much.
thanks sir, it helped me a lot while making power supply using 7805.
Great Video for beginners, I had to figure this all out the hard way. Lol
Just information I need I have just in rolled in electronics II class in community college. One of my class projects is building a headphone amplifier kit with a separate external power supply.
www.pmillett.com/nuhybrid.html
Using a vacuum tube technology that was just developed in Japan a couple years ago. Using the ( Korg Nutube). The design is by Peter Millett based off his original design that used low-voltage vacuum tubes called “ Millet Hybrid “ which has a lineage going all the way back to the days of their original (Starving Student Headphone Amplifier) I was going over his circuit board design to figure out what capacitors I could enlarge for a slightly better bass response. Also building my own linear power supply and try to clean up and provide the cleanest DC voltage I possibly can building myself.
You can use a simulator and experiment ...you will notice big differences in PS set-up and how to fine tune all values of the components constituting the power supply for your own project/amplifier etc , and adjust for your personal needs .
Go to the site of Duncan and upload the simulator , it's free .
Your ears will tell you if you got it right .....and confirm by measurement if you wish to check..
thank you so much for making this video, it really helped me out!!
Mr carlsons lab was doing a restore on an old vacuum tube piece of equipment... I don't remember which one it was but from my understanding it used a filter choke on the power supply and when he added a capacitor before the choke, the voltage output climbed dramatically... I think that he might have been working on a tube amplifier because I vaguely remember that he was able to make it output much more power than it was actually being rated for... But the lesson was really about getting rid of voltage ripple and if someone tried to add a capacitor before the filter choke, they would be raising the voltage output way beyond the voltage ratings of components such as capacitors further down the line and you will end up with exploded capacitors without realizing what you have done... It's a pity that he didn't go deeper into detail about this topic because it is a trap for many people... Although this concept could potentially be useful if you needed a little more voltage for your circuit while the transformer available to you doesn't have the required voltage but has current to spare... Say something like a low voltage transformer from an old Ark welder....
Once you are below 1% distortion an amp is an amp is an amp is an amp...Features you need and enough power for your speakers and room. Done deal. Love the dsp in that one.
Thanks a lot for this very much helpful video.
Thanks! Great tutorial. Could you do a tutorial on bleeder resistor circuits?
Thanks for refreashing our knowledge. =)
Keep making great videos!
26:59 - How would the Fluke be thrown off? Doesn't it measure true RMS or is that only applied to 'AC volts'?
True rms only applies to ac voltage. DC voltage does not vary so there would be no rms
Awesome Video as usual, Thanks!
Peace love and harmony from here. Great video
Yes Tony, you do an excellent job of explaining that beginners like me can understand. Keep it up.
Perhaps one subject you can consider doing a video is good power supply design for audio amplifiers with attention to headroom.
For example, though it costs more will putting a beefy power supply into a smallish 50W output high current amplifier to give it at least 3dB of headroom, can this be considered an ideal approach for the best sonics and dynamics?
I hope to be getting into some of this in my next series.
Exciting please let me know.
A bridge has 2 times the diode voltage drop over a center tap arrangement. Moreover, voltage drop changes with load, it is not a constant .7V for diodes. Always consult the data sheet of the rectifier used.
hello thanks for video , very technical and informative and made it clear many unknown things for me. Will bulged ( swelled on top) big capacitor (10000 uF) on first step of linear power supply( bridge diodes), cause voltage drop in output?
I know those spliced wires are protected with clear shrink wrap, but on video they still look scar, LOL. Keep those videos coming.
Hi Tony I have a new Cayin integrated tube amplifier A-300b Mk II. When I turn off I hear a soft thump through my speakers. Some say normal but another owner with same amp and similar speaker efficiency has no thump. What could be wrong with my amp? Thanks!
Hey Tony, know of a small simple soft start circuit? You made me glance over my home made bench power supply and since I have a huge filter cap in it.I need to toss in a soft start circuit.
You could build one, but I found them in kit form as well as assembled for a few dollars on eBay. Just search "soft start module" or "time delay on relay" and you'll see a whole bunch of them. Thanks for watching!
At car stereo with heavy amps in it, we used for 13,8VDC big caps to protect battery and generator, 1F caps, which are huge, these were nuts to be honest, and some guys placed them themselves, which caused electric fires, with bad isolated connections, and use of too thin cables, to provide all the power to the stereo. I have here a 21mm² cable that is just pulverized on the inside. Because of the high currents.
Thanks for your good tutorial
I have two questions to ask:
1) How do I limit the out put voltage from 16volts to 12volts when my load is a 12volt load 5amps
2) The value of the smoothing capacitor I calculated is too large how do I reduce the capacitor value with 1volt ripple appearing on the load
The best way to limit the voltage is with a regulator IC, such as the 7812. If the circuit is more than 1 amp, you will need to add a current pass transistor. Just search 7805 pass transistor and you should find more info on that. If you use a resistor, the voltage drop will vary with changes in current load, so that would not be a good idea.
As for the capacitor value, you can calculate it with this formula:
For 60 Hz mains: C = I (8.3/V)
For 50 Hz mains: C = I (10/V)
Where: C is capacitance in microfarad, I is the amount of current the power supply will deliver and V is the amount of ripple voltage you desire. 8.3 and 10 are the respective time periods for 1/2 cycle of the mains voltage in milliseconds for either 60 Hz or 50Hz mains.
For a 5 amp power supply: C = 5(8.3/1) = 5(8.3) = 41 microfarad. In your example, 1 volt of ripple would be very high, as you only have a 12 volt power supply. To reduce ripple to 10 millivolts, you would need 5(8.3/.01) = 4150 microfarad.
Hope that helps
@@xraytonyb Thank you and I really appreciate God bless you
One more question; a situation where there is no regulator will the peak voltage produce after rectification can it destroy the load
The 10milivolt ripple, do you mean that will appear on the out put of supply or it is what the filter capacitor will remove?
From your example 8.3ms is 0.0083s hence 5(0.0083/1) = 41500microFarad
Also 5(0.0083/0.01) = 41.5microFarad kindly explain in detail please
Thanks Tony!
Very well explained.
V. Informative! Thank you!
Thanks so much Tony, I learnt a lot. I know this may be a stretch, but I would be grateful if you could point me in the right direction; would replacing two 6800uF filter caps in an old Kenwood 5700 with two 10,000uF capacitors be pushing it too far? Would replacing each with 2 in parallel change anything? Thank you very much.
You roughly double the capacity, hence doubling the time it takes to charge them up. Remember that an empty capacitor presents a dead short to your rectifier as well as to the transformer. I don't know if the 5700 has a softstart, but if it is voltage controlled, it will make no difference at all. Both the transformer and the rectifier may survive, but you put a lot of stress on both of them on power up. Why you wanna do that anyway? Is it humming or too noisy? Maybe the capacitors are just leaking to much dc due to age. But then change them with the original value. Just my two cents.
@@agrimm61 thank you 😊 Going with original values. I guess I was being greedy and not knowing enough thought hey I can buy double the value for a couple of dollars more!
Perfect, realy perfect. Thanks
Lovin the videos!!! ThankZ - quick question I’m semi new to electronic but have been building stuff for many years but I’m now trying to expand on my understanding. I have a nice multimeter and just picked up a Rigol DS1052e I’m trying to design a power supply from the ground up and overall learn more. Is there any place you can point me to further understand. The video was great by the way THANKZ AGAIN
Take a look at the series @ ruclips.net/user/katkimshowvideos
The power supply videos may not be grouped together.
cougar1861 awesome thankZ!!!
Hi Tony
Thanks for the refresher info, you are very easy to learn from !!
Just a small suggestion do you think it may be better to connect mains voltages up
using a connecting block that Big Clive uses,it is a lot safer and teacher's new starters
a safe way to connect up
Sorry for my suggestion it is not mean't to insult you
Thanks for the great video
I actually have clear heat shrink tubing over both connections to insulate them. It's a bit hard to see on the camera. I thought about getting a quicktest a while back, but they are very expensive here in the states and that would be another big thing on my bench. They are really cool, though! Thanks for watching!
Thanks for the video I am trying to understand this concept. Can anyone help me I am trying to understand this concept because I am having an interference issue on a brushless gimbal. I have recently connected a 5.8ghz transmitter to transmit the video signal meanwhile I control the gimbals brushless motors using a 2.4ghz receiver. The brushless gimbal and both transmitters are connected to the same power source and whenever I turn on the video transmitter it causes the brushless gimbal to have erratic responses to the 2.4ghz input commands. I tried to solve this by using a 12v Battery Eliminator Circuit or BEC which has its own capacitors built in to try to clean up the power but I still get the same issue and I want to add a capacitor to help with the interference. I am not sure how to work out the type to add. Should I just add a 10V 100uF or will I need to go up to a 25V but I would have thought a 25V would store up power and then release it when it gets to 25V? Just thought that a 10V 100uF would end up being too low for a 12V device. Any help will be great thanks.
The voltage rating on a capacitor has nothing to do with storage capacity. It is the amount of voltage in the circuit that can damage the capacitor. A 25v 100 mf capacitor is the same as a 600v 100mf capacitor. The only difference is that you can use the 600v cap to filter a high voltage supply. Why not just make all capacitors 600v? Size and money. You just need to pick a voltage rating on the cap slightly higher than the highest voltage it is going to encounter.
Thank you so much
The conversion factor of Vpp to Vrms is 2.828 ( square root of 2 time 2) roughly 3. So 36 volts divided by 3 would be 12 V.
great fundamentals, thanks ....
i am working on Recapping a Technics SA-8000x it has two 4700uf caps in the filter what do you think the safe ratting i can go up too? 5,600 or just stick with 4700uf?
Sure this video isn't from yesterday - still it's great to listen. What gave me a bit to think was at 41:25.
Isn't there an error in C formula? If there is known equation where I=C*dV/dt then C * dV = I * dt and finally C=(I*dt)/dV (compare it with one shown on the right: C=I/f*V)
I - power the circuit is rated for, dV - ripple, dt - 60 Hz half-cycle = 1000/(60*2),
Since author assumed 1 Amp and 1 V, equation gave good result. Change amps and volts to something more exotic and compare:
(0,3A * 8.3ms)/2.5V is not 0,3A/(8.3ms * 2.5V)
EDIT: 46:22 Hmm. Answer(?) You come up with 0.001xxx F and mine comes up with 1.xxx mF.
I think frequency should be in radians/sec, i.e. 2*pi*f, and not f (in Hz) as shown in the video.
Really great.
Would this setup be safer or more dangerous if the transformer was grounded (when not using an additional isolation transformer)?
It just depends on the application. Generally, the transformer itself provides isolation, as long as the primary and secondary are not physically connected (i.e. it's NOT an autotransformer). This is why we call transformerless radios "hot chassis" radios. One side of the mains is directly connected to the chassis, causing a possible shock hazard. The isolation transformer provides isolation between the mains and chassis, making it safer to service. In a radio WITH a power transformer, the power transformer provides isolation in the same way, making the isolation transformer unnecessary. Grounding the chassis won't hurt and could ensure that the chassis is at the same earth potential as any other grounded equipment that could come in contact with it. That said, it is not necessary, as long as the mains leads are only connected to the primary of the transformer and NOT the chassis or any other part of the device. Hope that helps. Thanks for watching!
Terima kasih cikgu(tq sir)
Very perfect teachings but sir please make a video on switch mode power supply
Someday we'll try to get to that. In the meantime, check out RUclips channel DiodeGoneWild. He is my absolute favorite for videos on switching power supplies.
totally understood this thanks
6:40 I calculate the internal impedance of that transformer to be 0.28 Ohms :) [ (14.25V - 14.16V)/0.303A] :)
Hmm... Another video topic?
At 8:50 or so you claim the transformer outputs the right output voltage with 110V in, but that's not the case, its rated to output 12.6V out for 120V in at _full load_, ie 3A rms. Your experiment with the 47 ohm resistor was only a 10% load. Normally you'd expect about 10% more output than rated for no-load, since typically the windings drop about 5% each at full load for practical transformer designs, so about 10% drop in voltage total at full load.
It is *not* a filter capacitor. It is not filtering any frequencies. It's a smoothing capacitor because it's smoothing out the peaks to give a flatter voltage waveform.
Really? And all this time I thought it was a low pass filter that was affected by frequency, capacitance and the load placed across it. Thanks for the education ;)
😂
keen explaning from abt 29,00
❤
The equation C = I/(f Vr) shows that to reduce Vr to zero would require an infinitely large capacitor, i.e. as Vr -> 0, C -> infinity; interesting. So, you can never have zero ripple unless you use a regulator, again, interesting.
ruclips.net/video/jMkD3QxmN2I/видео.html isnt the diode positive on the transformer side , the stripe is negative from the transformer side ,off course positive from load side but if you put positive on stripe it wont work , dont know what you were talking there , got me confused ,diode has no polarity , in classical terms so who knows but off course stripe goes away from the positive transformer side to make it conduct, transformer is positive side , so the part of diode that is connected to positive transformer is positive part of diode , and that is not Stripe anode , its cathode , like you do and show, but say something else .
Does anybody recall Ohm's law written like I=U/R ? (U instead of V) I learned it like that in high school, 45 years ago... I can imagine for just one explanation: in physics V stands for speed (in Latin: velocitas) But I am far from sure that this is correct. Edit: I did some research. Indeed, U is correct, (from the Latin word: 'Urgere", which means to push or to press) This makes sense, voltage 'pushes' the current through the resistor.
Strangely enough, when I went to school, we learned "E=IR" (E was short for electromotive force, measured in volts) . Sometimes it was "V=IR", were V was just the first letter of the word Volt. When I started repairing X-Ray equipment, a lot of the European-built equipment (i.e. Philips & Siemens) would refer to voltage as "U" in the service documents, and still do today. For instance, "UH" would mean "heater voltage" (as heater was another word for the filament in a vacuum tube). Here in the U.S. you see all three letters used. Welcome to the land of nonstandard standards!
LOL! Fortunately, you guys in the US don't talk about 3/8 of a volt or 11/16 of an ohm. No offense, but I wonder if and when the US will stop using all kinds of body parts to measure distances ...
1959Berre In germany it is still U to this day, V for voltage has never been used since obviously we don't call it voltage but "Spannung" meaning tension in english. Just the unit is Volt.
In Belgium it always was and still is U for voltage, which in Dutch (like in German) is 'spanning'.
Lester Barrett V or Volt is just the unit. In many languages the word for voltage does not start with a V, just like in german. so we use the already existing U for voltage. E is energy nowadays. And while E is for energy the unit you use would be J for Joule. Same thing as V and U. Or I for current but the unit is Amperes.
You said "3 different...". You forgot linear power supplies with a choke input filter.
Excuse me Sir? Would you do a repair for me?
nope
is this for five year olds or something?
Sound like your comment was made by a five year old...
be nice man, it's cheap
It is for whatever you want it to be, nobody forces you to watch it.
@@duroxkilo Being nice or cheap is not how capitalism works.
There is an error at 24:35 - 27:18. You mention that (in a half wave rectified sine wave) with a peak voltage of 16 volts one should expect an RMS of 0.707*16 = 11.3 V but this is NOT correct. For a HALF wave rectified signal the RMS voltage is actually half the peak voltage and not 0.707 times the peak voltage. Thus the correct RMS in this case is actually 8 volts. Now when your DMM measures AC voltage it puts a capacitor in series eliminating any DC offset which is why the AC reading (alone) is not accurate. However when a periodic signal has a DC offset the RMS value of that signal is the square root of the sum of the squares of the DC reading plus the AC reading. In this case you have a DC value of 5.18 V and an AC value of 6.42 volts. Taking the square root of the sum of the squares of these two values gives on 8.25 volts in agreement with the predicted RMS (8 volts above). Similarly with the center tapped full wave rectifier at 29:13 You have a peak voltage of 8 volts. SInce this is a full wave rectified signal the RMS value actually is 0.707 times the peak voltage and thus the expected RMS is 8*0.707 = 5.66 volts. Now your measured DC was 4.85 volts and AC was 2.69 volts. Taking the square root of the sum of the squares of these two values gives one a measured 5.55 volts, once again in agreement with the calculated result.
Good content, Thank you