Mate this video is brilliant. I have found electrionics I have found to be my least favorite class due to just only doing calculations and not understanding the problems that these circuits solve. Seeing this type of problem solving and putting the theory into practice is what engineering is all about, I wish courses had your style of teaching electronics! Keep up the good work :)
You are a geniously excellent teacher. Seen this class A-B config so many times, and more or less got it intuitively. But somehow you manage to put it in a way (simply...) that suddenly the penny dropped and I now fully understand how that diode-thing actually works and does what I have known it does without truly feeling it. Same goes for the inclusion of the opamp, wow.
Thanks dude! You are the first to make me grasp this. I've built paint-by-numbers projects, even a tasty Class-A amp but I didn't have a Scooby Doo how it really worked... Now I can make an AC motor driver out of some filtered squares :)
Ive binge watched a bunch of your videos and now I feel dumb... I literally thought all this time you were talking about powering an actual fan. now it makes more sense.
I have not the slightest idea who this man is. But he is an excellent educatainer or entercator. Only now I know why resistors' resistance must be as low as rationally and logically possible to avoid noise. Many people and articles talk about this phenomenon but none actually explain the reason why resistors must be small enough but not overly small that resistors become a wire.
Push-pull would work great with power mosfets yep, it's common. I don't understand your other question: Are you saying you're trying to work with a 100V supply? I'd just say look at spec sheets to find parts rated for it.
10:27 as for the feedback, I'm not sure, but I have a feeling that the output will still be the same as the input signal even if there is no feedback. I know it sounds spooky. I mean having the op amp as an usual voltage follower and keep the BJTs handle the output on their own.
I know this is an old post, but for anyone reading this, the feedback most certainly is necessary to keep the signal the same as the input. Otherwise your input will immediately be driven to [+/-] rail voltage because of differential amplification and the extremely high gain of the op amp.
This is the second time I ended up here by searching you tube because I sort of got it - but wanted someone to explain it more logically. I just built a push pull TIP31/32 on an audio amp, it worked but the diode biasing thing messes my head up. I think last time I was spewy about the cuts making the speech relentless and unpausing. Seems to be a few breaths added this time but - space is beautiful. Imagine if your whiteboard was just full of lines and had no space between things ... like you had to squeeze everthing into the tiny corner and ignore the rest of the potential ... and it was just unpleasant and cramped ... something like that. Your channel will, with various editing improvements ... become huge I'm sure as you have the pulse of what people don't get and need to understand.
Super. just to clarify myself a bit, 17:35 with this configuration will the load be getting a voltage of + or - thus a digital-like or logic-like voltage (I mean just the rail voltage), or will it be getting the (supposingly) analog voltage of the input (the most left) which is varying and smaller than the rail voltage ? If it follows the input voltage, this means that the voltage across collector-emitter is not 0 in general.
This video is a little confusing because I wasn't using the "circuit ground" symbols yet. The solid-circle negative is circuit ground, and the dashed-circle negative is negative rail. This is an analog output: The op-amp is configured in non-inverting unity-gain mode: The signal is on the non-inverting input, and the output (lso connected to the load) is on the inverting input. The op-amp is going to try to change its own output until its non-inverting and inverting inputs are effectively equal, so the load will be receiving the same voltage as the signal.
THANKS FOR THE GREAT VIDEO!!! At about timestamp 2:10 you state that if we are only using a digital driver (I'm using a 555 in the astable mode) this single transistor is all we need to reverse the high current. What I don't understand is that when the output of the digital driver is low "0", the transistor cuts off (or does it?) which would result in NO current flow. Am I missing something? I want to connect the 555 so that it sources AND sinks the higher current. It's output goes from a low of "0v" to a high of "5v -15v". THANKS MUCH,
I didn't say it would reverse the high current: The point of the examples I give before I get to the actual push-pull one is to illustrate the problems I'm trying to solve with the push-pull, which are to have bidirectional current and to have power amplification (meaning the signal is not driving the load). In the configuration at the two-minute mark, if driven with a binary signal, it's basically "open-collector" with no pull-up resistor, so it will turn on and off a load with power amplification. I was just trying to say that if all you need is power amplification and all you need to do is turn it on and off (for example, a light or a fixed-speed DC fan), then that would work and you don't need something fancier.
Sorry I am new to this but I have a couple questions. So at the very end the op amp is a unity gain so it will try to match the output to be exactly the input. So does that mean there is no amplification? I have a hunch that this might just mean there is no VOLTAGE gain but there is a current gain right? If so how do you calculate the current gain from this circuit? If I added resistors to the feed back system of the op amp would I be able to get both a current AND a voltage gain?
Very good video even for me who is a novice in electronics. A question i have however, is what if my signal source is a square wave, do i need to have PNP connected to the negative voltage generated from a split supply or any other way?
This question doesn't entirely make sense. It depends on what you're trying to do with your square wave. If you're strictly dealing with digital electronics and you're using the op amp as a buffer then you most certainly do not want your system capable of delivering negative voltage to your circuit. Digital circuitry generally does not handle negative voltage well at all.
Hi there. Your video is quite informative. Im trying to drive a mosfet gate(4000pf). I had two option first one is to buy a gate driver ic which is not available here and other option is to make a push pull transistor circuit which i tried a lot of time and remained unsuccessful. I have Used bd139 bd140 transistor pair. But i dont know why the output stays high like 9v and when i apply positive pulse then otput just fall a little bit 8.85v. Which is absolutely not a good way to drive mosfet gate. Could you help me to make the circuitry or provide me some schematic which would work fine
I put this together in a simulator and the output voltage was exactly the same as an input one (isn't the way OA works?). To get some gain I had to add a voltage divider to feed a part of the signal to an inverting input.
So the signal's one lead goes into the OpAmp + and its ther leg goes directly into the circuit ground and no resistor. Is this correct? Also the load is not hooked up to the positive rail, it is hooked up between the two Transistor?
Power inputs on the op amp go to +/- voltages, respectively. The positive reference pin is attached to your input signal. Both the load and your negative reference pin are attached in the center between the two transistors.
Please do tell me if anything I explained was still confusing or you'd just like more clarification. I can always make more videos with more detail or going over specific things.
@@simplyput2796 Hello :) Well, for example 1:08 , I dont understand why - from load is attached to the collector ? This would need to me some extra explanation. So where the (-) from the emiter goes than ? Your learning is amazing but I find in your videos a lot of information presented as obvious (that are not for a beginner) and you quickly jump from one fact to the other to there other and summarizing "simple as that " :D . That's why I think this is not "simply put" for even not a relatively beginner. Absolutely I don't want to criticize ,however the pace and " obviousness " of some facts is for someone that would not need to have it "simply put" :-)
Well, I'm a bus driver learning electronics from nothing more than Google and Wikipedia, so I can assure you this is definitely at a "beginner" level. :P The arrangement at the beginning does have the load connected to the collector: It's an "open-collector" arrangement which I have done a video on. And the (-) is the 0V "reference voltage", which will usually be the negative end of your power supply. I'm jumping through the different arrangements in that video quickly because I'm not trying to actually teach them as arrangements you should use: I'm explaining the problems with them. That's why I spend a lot more time on the final one, the one I'm saying is the proper solution.
parapos thanks man! I learn from many sources and more and more things I can combine and understand. But it is my passion for last 3 years :) it is such a broad topic 😄
How does the positive input signal steer the base of the NON transistor and the negative input signal steer the base of the PNP transistor, given such a placement of the biasing diodes?
Probably the best explanation of amplifier design principles I have seen.
Absolute gold.
Great video! One of the best videos explaining how push pull output stage works
You're doing great work and I hope you keep going.
Dude its magical
Mate this video is brilliant. I have found electrionics I have found to be my least favorite class due to just only doing calculations and not understanding the problems that these circuits solve. Seeing this type of problem solving and putting the theory into practice is what engineering is all about, I wish courses had your style of teaching electronics! Keep up the good work :)
Brilliantly explained
amazing explanation. Thank you!
This is amazing explanation. Thanks a lot. You are terrific.
You are a geniously excellent teacher. Seen this class A-B config so many times, and more or less got it intuitively. But somehow you manage to put it in a way (simply...) that suddenly the penny dropped and I now fully understand how that diode-thing actually works and does what I have known it does without truly feeling it. Same goes for the inclusion of the opamp, wow.
Man, wonderful lesson.
Very good instructive video. Excellent explanation. Thanks much.
Beautiful instructive content !
your videos are so good!
thank you for sharing your knowledge, this looks all so easy and obvious for you :) You explain really well.
Brilliant!
Great explanation! Bravo!
very good explantion !
Great tutorial!
You're a wonderful teacher! I've learned things from you that I didn't learn
In my engineering classes in college. Please keep up the wonderful work.
Very good instructive video.....Thanks
Thanks dude! You are the first to make me grasp this. I've built paint-by-numbers projects, even a tasty Class-A amp but I didn't have a Scooby Doo how it really worked...
Now I can make an AC motor driver out of some filtered squares :)
This guys knows how big of an idiot I am, and I appreciate him for it.
Thx for this very informative vid
Ive binge watched a bunch of your videos and now I feel dumb... I literally thought all this time you were talking about powering an actual fan. now it makes more sense.
Awesome videos
Everything is Hunkydory...Best explanation of Push-Pull on UA-cam
Thanks for the video =)
Very nice
great explanation, as usual !!! I'm looking forward for the 555-timer video. it seems they can be used in many different applications.
That series is coming soon! Only one more video before that.
Wow man,you know what you are doing nice 👍,,,,,,,South Africa 🇿🇦🇿🇦
I have not the slightest idea who this man is. But he is an excellent educatainer or entercator.
Only now I know why resistors' resistance must be as low as rationally and logically possible to avoid noise. Many people and articles talk about this phenomenon but none actually explain the reason why resistors must be small enough but not overly small that resistors become a wire.
good
I've enjoyed these videos. Haven't seen new ones but hopefully he's doing well.
Great video. Would this circuit also work with mosfets? And do you need for several 100V resistors instead of an opamp?
Push-pull would work great with power mosfets yep, it's common. I don't understand your other question: Are you saying you're trying to work with a 100V supply? I'd just say look at spec sheets to find parts rated for it.
Great video. Can this be done with a single-supply voltage (vs split)?
Yes. It does.
That "headphones" are something special!.... are they some special edition?? You should patent them!
I haven't even realized until half of the vid that there is no speaker on the headphone lol
10:27 as for the feedback, I'm not sure, but I have a feeling that the output will still be the same as the input signal even if there is no feedback. I know it sounds spooky. I mean having the op amp as an usual voltage follower and keep the BJTs handle the output on their own.
I know this is an old post, but for anyone reading this, the feedback most certainly is necessary to keep the signal the same as the input. Otherwise your input will immediately be driven to [+/-] rail voltage because of differential amplification and the extremely high gain of the op amp.
What is the minimum value of the DC biasing current in class A output stage? Explain you answer.
This is the second time I ended up here by searching you tube because I sort of got it - but wanted someone to explain it more logically. I just built a push pull TIP31/32 on an audio amp, it worked but the diode biasing thing messes my head up. I think last time I was spewy about the cuts making the speech relentless and unpausing. Seems to be a few breaths added this time but - space is beautiful. Imagine if your whiteboard was just full of lines and had no space between things ... like you had to squeeze everthing into the tiny corner and ignore the rest of the potential ... and it was just unpleasant and cramped ... something like that. Your channel will, with various editing improvements ... become huge I'm sure as you have the pulse of what people don't get and need to understand.
Super. just to clarify myself a bit, 17:35 with this configuration will the load be getting a voltage of + or - thus a digital-like or logic-like voltage (I mean just the rail voltage), or will it be getting the (supposingly) analog voltage of the input (the most left) which is varying and smaller than the rail voltage ?
If it follows the input voltage, this means that the voltage across collector-emitter is not 0 in general.
This video is a little confusing because I wasn't using the "circuit ground" symbols yet. The solid-circle negative is circuit ground, and the dashed-circle negative is negative rail.
This is an analog output: The op-amp is configured in non-inverting unity-gain mode: The signal is on the non-inverting input, and the output (lso connected to the load) is on the inverting input. The op-amp is going to try to change its own output until its non-inverting and inverting inputs are effectively equal, so the load will be receiving the same voltage as the signal.
THANKS FOR THE GREAT VIDEO!!! At about timestamp 2:10 you state that if we are only using a digital driver (I'm using a 555 in the astable mode) this single transistor is all we need to reverse the high current. What I don't understand is that when the output of the digital driver is low "0", the transistor cuts off (or does it?) which would result in NO current flow. Am I missing something? I want to connect the 555 so that it sources AND sinks the higher current. It's output goes from a low of "0v" to a high of "5v -15v".
THANKS MUCH,
I didn't say it would reverse the high current: The point of the examples I give before I get to the actual push-pull one is to illustrate the problems I'm trying to solve with the push-pull, which are to have bidirectional current and to have power amplification (meaning the signal is not driving the load). In the configuration at the two-minute mark, if driven with a binary signal, it's basically "open-collector" with no pull-up resistor, so it will turn on and off a load with power amplification. I was just trying to say that if all you need is power amplification and all you need to do is turn it on and off (for example, a light or a fixed-speed DC fan), then that would work and you don't need something fancier.
Sorry I am new to this but I have a couple questions. So at the very end the op amp is a unity gain so it will try to match the output to be exactly the input. So does that mean there is no amplification?
I have a hunch that this might just mean there is no VOLTAGE gain but there is a current gain right? If so how do you calculate the current gain from this circuit?
If I added resistors to the feed back system of the op amp would I be able to get both a current AND a voltage gain?
Very good video even for me who is a novice in electronics. A question i have however, is what if my signal source is a square wave, do i need to have PNP connected to the negative voltage generated from a split supply or any other way?
This question doesn't entirely make sense. It depends on what you're trying to do with your square wave. If you're strictly dealing with digital electronics and you're using the op amp as a buffer then you most certainly do not want your system capable of delivering negative voltage to your circuit. Digital circuitry generally does not handle negative voltage well at all.
Hi there. Your video is quite informative. Im trying to drive a mosfet gate(4000pf). I had two option first one is to buy a gate driver ic which is not available here and other option is to make a push pull transistor circuit which i tried a lot of time and remained unsuccessful. I have
Used bd139 bd140 transistor pair. But i dont know why the output stays high like 9v and when i apply positive pulse then otput just fall a little bit 8.85v. Which is absolutely not a good way to drive mosfet gate. Could you help me to make the circuitry or provide me some schematic which would work fine
I put this together in a simulator and the output voltage was exactly the same as an input one (isn't the way OA works?). To get some gain I had to add a voltage divider to feed a part of the signal to an inverting input.
Excellent video and very interesting. I have to ask though, was that a Home Improvements reference? More Power argh argh argh argh. 😁
Please mentor me I live in the middle of nowhere no family and I'm trying to learn this on my own it's slow going on my own
So the signal's one lead goes into the OpAmp + and its ther leg goes directly into the circuit ground and no resistor. Is this correct?
Also the load is not hooked up to the positive rail, it is hooked up between the two Transistor?
Power inputs on the op amp go to +/- voltages, respectively. The positive reference pin is attached to your input signal. Both the load and your negative reference pin are attached in the center between the two transistors.
great stuff but it is not an easy stuff. definitely at least intermediate level :) I will come back one day :)
Please do tell me if anything I explained was still confusing or you'd just like more clarification. I can always make more videos with more detail or going over specific things.
@@simplyput2796 Hello :) Well, for example 1:08 , I dont understand why - from load is attached to the collector ? This would need to me some extra explanation. So where the (-) from the emiter goes than ? Your learning is amazing but I find in your videos a lot of information presented as obvious (that are not for a beginner) and you quickly jump from one fact to the other to there other and summarizing "simple as that " :D . That's why I think this is not "simply put" for even not a relatively beginner. Absolutely I don't want to criticize ,however the pace and " obviousness " of some facts is for someone that would not need to have it "simply put" :-)
Well, I'm a bus driver learning electronics from nothing more than Google and Wikipedia, so I can assure you this is definitely at a "beginner" level. :P The arrangement at the beginning does have the load connected to the collector: It's an "open-collector" arrangement which I have done a video on. And the (-) is the 0V "reference voltage", which will usually be the negative end of your power supply. I'm jumping through the different arrangements in that video quickly because I'm not trying to actually teach them as arrangements you should use: I'm explaining the problems with them. That's why I spend a lot more time on the final one, the one I'm saying is the proper solution.
@@jatza07 you should watch all videos from the beginning to get a better understanding.
i know it's a big task, but worth it.
parapos thanks man! I learn from many sources and more and more things I can combine and understand. But it is my passion for last 3 years :) it is such a broad topic 😄
This guy really is a treasure. Hopefully he will return some day.
BTW: PLEASE make that ground symbol...
I did! I've been using it in my newer videos. It's pretty sweet, now that I've gotten used to it.
very well but too much trim in video make me feel bad!!
Is it possible to share your email?
If you want to discuss things off of UA-cam, use the Discord link in the description.
did the sea moose eat the sea moss or what it a sea mouse?
How does the positive input signal steer the base of the NON transistor and the negative input signal steer the base of the PNP transistor, given such a placement of the biasing diodes?