In a push-pull setup, is the total output current limited to 2x the bias current, or can it exceed that? If it can, why? And why can't a single-ended output stage not do likewise?
Nelson wrote a really nice paper that may answer some of these questions for you. Link below. Here is a specific quote of interest:
"A limitation of single-ended Class A operation is that the
peak output current is limited to the value of the bias current. By comparison, pushpull Class A can deliver twice the bias current as peak output in Class A, and generally
much more than that in Class AB"
https://www.firstwatt.com/pdf/art_leave_classa.pdf
"A limitation of single-ended Class A operation is that the
peak output current is limited to the value of the bias current. By comparison, pushpull Class A can deliver twice the bias current as peak output in Class A, and generally
much more than that in Class AB"
https://www.firstwatt.com/pdf/art_leave_classa.pdf
I read that article, but the “why” is still unclear.
It makes sense that push-pull is limited to twice the bias current in Class A; twice the stages as single-ended, twice the bias before it goes “klunk”.
I suppose it also makes sense that single-ended cannot extend past the edge of Class A because a single output stage cannot operate in Class B.
Still though, where does the push-pull amplifier get the extra current to deliver once it has entered Class AB operation? Can it take amps from the “shut down” channels bias? If so, does that imply a power limit of 4x bias current?
It makes sense that push-pull is limited to twice the bias current in Class A; twice the stages as single-ended, twice the bias before it goes “klunk”.
I suppose it also makes sense that single-ended cannot extend past the edge of Class A because a single output stage cannot operate in Class B.
Still though, where does the push-pull amplifier get the extra current to deliver once it has entered Class AB operation? Can it take amps from the “shut down” channels bias? If so, does that imply a power limit of 4x bias current?
SE can go more than 2x bias but the opposite half of the waveform will clip because it cannot go less than 0amps. So at the point where the SE amp goes >2x bias you are above the amps max linear power output. So it can but it is of no use.
Imagine you have a single ended amp, with 100mA idle current.
Driven by signal, current can go up or down.
The problem lies in the "down" half: it can NOT go lower than zero (Physics Law, a Tube or NPN transistor can´t pass negative current) , so negative current swing is 100mA tops , period (from 100mA to 0 mA)
Do we agree so far?
Now we go the other way: to make the other exactly symmetrical sinewave half, current will need to go up to 200mA.
So far so good.
What if we raise input signal?
Positive half will rise beyond 200mA, say 250-300-350mA, as much as tube or transistor can give ... but negative half will clip/distort big time because it can not go lower than 0.
So you can "increase" output, sort of, but heavily distorted, so useless.
Now to Push Pull amp.
Suppose it also has 100mA idle current per output device (just to simplify the example).
Top transistor can be driven to 200mA ... and beyond, way beyond, say 300 -400 - 500 mA , whatever it can supply, no clipping.
Ok, and what about the other half?
It is handled by *another* transistor, which can supply as much "extra" current as the top one.
But ... but ... I want the negative half .... can it supply negative current? How come?
Good eye 😉
Don´t Physics Laws apply?
Yes they do, but we cheat. 😱
Ummmm, call it we "design" 😉 , looks nicer 🙂
Bottom tube/transistor again can only swing positive, BUT we invert phase so the "negative area current" can be now successfully supplied by a Physics allowed Positive device.
How do we do it?
We take part of drive signal , invert its phase ("phase inverter"), drive a positive swing only device (none other available) and then invert again its output so Positive swing now becomes needed Negative swing.
Output transformer not only does that, but also *combines* two signal halves into a single full cycle one.
Boy, aren´t we Humans clever? 😀 😀
Driven by signal, current can go up or down.
The problem lies in the "down" half: it can NOT go lower than zero (Physics Law, a Tube or NPN transistor can´t pass negative current) , so negative current swing is 100mA tops , period (from 100mA to 0 mA)
Do we agree so far?
Now we go the other way: to make the other exactly symmetrical sinewave half, current will need to go up to 200mA.
So far so good.
What if we raise input signal?
Positive half will rise beyond 200mA, say 250-300-350mA, as much as tube or transistor can give ... but negative half will clip/distort big time because it can not go lower than 0.
So you can "increase" output, sort of, but heavily distorted, so useless.
Now to Push Pull amp.
Suppose it also has 100mA idle current per output device (just to simplify the example).
Top transistor can be driven to 200mA ... and beyond, way beyond, say 300 -400 - 500 mA , whatever it can supply, no clipping.
Ok, and what about the other half?
It is handled by *another* transistor, which can supply as much "extra" current as the top one.
But ... but ... I want the negative half .... can it supply negative current? How come?
Good eye 😉
Don´t Physics Laws apply?
Yes they do, but we cheat. 😱
Ummmm, call it we "design" 😉 , looks nicer 🙂
Bottom tube/transistor again can only swing positive, BUT we invert phase so the "negative area current" can be now successfully supplied by a Physics allowed Positive device.
How do we do it?
We take part of drive signal , invert its phase ("phase inverter"), drive a positive swing only device (none other available) and then invert again its output so Positive swing now becomes needed Negative swing.
Output transformer not only does that, but also *combines* two signal halves into a single full cycle one.
Boy, aren´t we Humans clever? 😀 😀
Bias current (Quiescent current, Iq) is simply the standing current applied to the transistors to keep them turned on (conducting). In a push-pull circuit, the class A operating mode is maintained so long as both transistors (both halves) are turned-on, regardless if the wave is positive or negative going. The transition to class B happens when the input signal reaches a level high enough to pull/push the output transistors far enough positive or negative that it exceeds the standing current (Iq) of the opposing side transistor and it shuts off. The limits of the current seen by the load are more a product of the voltage rail limits, transistor ratings and power supply.
ZM will correct me if I’m wrong 😀
ZM will correct me if I’m wrong 😀
JMFahey: That was fantastic, thank you!
Cody: Let me see if I understand, you can correct me off I'm off base:
At the initial state of a hypothetical amplifier where the signal can vary +-1v, the signal is 0v, Transistor Positive has 100mA and Transistor Negative has 100mA.
As the signal approaches 1v, tP approaches 200mA and tN approaches 0mA. As the signal approaches -1v, tN approaches 200mA and tP approaches 0mV. If the signal exceeds 1v, tP could exceed 200mA, as long as the transistor and power supply were able to provide that much current. tN would see negative amps and shut off. The circuit is now operating in Class B.
--
Regarding single-ended (and more to JMFahey's comment):
If the single-ended transistor operating at 100mA when v=0, 200mA when v=1 and 0mA when v=-1? If so, wouldn't that imply that the amplifier would be delivering no power at the maximum negative point in the wave? That seems... undesirable.
Cody: Let me see if I understand, you can correct me off I'm off base:
At the initial state of a hypothetical amplifier where the signal can vary +-1v, the signal is 0v, Transistor Positive has 100mA and Transistor Negative has 100mA.
As the signal approaches 1v, tP approaches 200mA and tN approaches 0mA. As the signal approaches -1v, tN approaches 200mA and tP approaches 0mV. If the signal exceeds 1v, tP could exceed 200mA, as long as the transistor and power supply were able to provide that much current. tN would see negative amps and shut off. The circuit is now operating in Class B.
--
Regarding single-ended (and more to JMFahey's comment):
If the single-ended transistor operating at 100mA when v=0, 200mA when v=1 and 0mA when v=-1? If so, wouldn't that imply that the amplifier would be delivering no power at the maximum negative point in the wave? That seems... undesirable.
I guess the x2 bias current value is an approximation and not a "mathematically" value and may depend if output stage is mosfet or bipolar devices?
what's wrong with ................ the books?
in short - if something is not clear reading article linked in post #2 and wiki article Power amplifier classes - Wikipedia......... it'll never be clear
in short - if something is not clear reading article linked in post #2 and wiki article Power amplifier classes - Wikipedia......... it'll never be clear
I read both of those articles. Neither one answered the questions in my post.
Both explain the fundamental operating nature of a class A circuit, which I understand. When my knowledge falls short is the applied electronics of these circuits, not the theory.
I think my understanding of a push-pull stage is correct, as described in my last post. If it's not, please let me know.
Where I'm more confused is with single-ended.
JFFahey said "Positive half will rise beyond 200mA, say 250-300-350mA, as much as tube or transistor can give ... but negative half will clip/distort big time because it can not go lower than 0."
That would imply power approaches 0 as the signal becomes more negative. There's something missing in my understanding of the way the circuit controls the driver, because if power was 0, shouldn't it mean that the is no power going to the driver?
In a push-pull circuit, the negative side increases in amperage when the signal goes more negative, and the positive side increases i amperage when the signal goes more positive. How can the amperage in a single-ended stage increase with positive voltage and decrease with negative?
Single ended «increases» at either the positive or negative rail, depending on whether it is a single ended N or P channel circuit. Not sure if «increase» is the right word. It is linear up to bias current, beyond that not much to go on, as I read it. But I’m a GH, so it is not all I fully or even halfway understand, or need to understand. Mighty keeps me continuosly challenged just beyond the point of my understanding, Papastyle, so as to slowly force me to into self education.
it is explained, but you're missing catch in articles to ...... catch it
maybe video could help
try this one ( I didn't check it thoroughly, hope guy knows what he's talking)
Understanding Amplifier Class A & Class AB - YouTube
maybe video could help
try this one ( I didn't check it thoroughly, hope guy knows what he's talking)
Understanding Amplifier Class A & Class AB - YouTube
Hehe, that's how I feel I'm doing with this hobby. I kinda take away 80% of everything and have to keep looping around clarifying things. It's one reason I ask so many questions. Like, in each of the articles ZM keeps linking me, I understand 80-95% of it, but I'm missing a few small but key points :/
That's why I ask specific questions, to get specific answers, without having to reread and rewatch videos and articles that mostly repeat information I already understand
I tried in a better way than I could explain 🙂
elbow grease is what is giving you knowledge, brain is of secondary importance here
your "specific question" is not good enough ...... in many cases lack of knowledge is preventing you of asking right question
been there, done that , not once - tried to do things by puzzle approach then realized that without efforts involved in comprehensive study of subject I'm not going anywhere
one hour more invested in beginning will spare you of many hours later, and quite possible even plenty of $ wasted in ignorant mistakes
again, been there, done that, not once
elbow grease is what is giving you knowledge, brain is of secondary importance here
your "specific question" is not good enough ...... in many cases lack of knowledge is preventing you of asking right question
been there, done that , not once - tried to do things by puzzle approach then realized that without efforts involved in comprehensive study of subject I'm not going anywhere
one hour more invested in beginning will spare you of many hours later, and quite possible even plenty of $ wasted in ignorant mistakes
again, been there, done that, not once
OK, short way
say that you have dual rails, each being 100Vdc , so called +/-100Vdc
voltage envelope on disposal is 200V, so hypothetically you can have 200Vpp voltage sine to your load
say that you have two stages, one is SE, second is PP, both having 1A standing current (Iq)
now , SE:
current envelope , it doesn't matter how big rails are , is defined by Iq
major fact is that you imagine current envelope while still being linear, meaning that positive and negative half of cycle are same
you can imagine two halves as result of two opposite behavior of transistor - one is when it is conducting more and other is when it is conducting less
while - when we open transistor fully , it will conduct as much rail(s) can give to load ( load/speaker being most hungry here) there is opposite situation :
- how much transistor can conduct less ? amount is exactly the same as value of Iq
so, in quiescent state, there is 1A of Iq, and thus half of current envelope is 1A, simply because difference between 1A and 0A is 1A
that's why current envelope ( linear one) is twice Iq , for SE stage .......... 1A down to 0A, 1A up to 2A
**************
with PP stage , A class of same is defined again by same principle (how much less?!?) and A class envelope is same for SE and PP, but PP is able to transfer in B class on both sides, so B class envelope is defined with Current envelope , which is (one) rail value /divided with R load ............ then doubled (having two rails)
so, this is probably best I can do with words ....... and if that is better to you than watching nice vid, your choice
say that you have dual rails, each being 100Vdc , so called +/-100Vdc
voltage envelope on disposal is 200V, so hypothetically you can have 200Vpp voltage sine to your load
say that you have two stages, one is SE, second is PP, both having 1A standing current (Iq)
now , SE:
current envelope , it doesn't matter how big rails are , is defined by Iq
major fact is that you imagine current envelope while still being linear, meaning that positive and negative half of cycle are same
you can imagine two halves as result of two opposite behavior of transistor - one is when it is conducting more and other is when it is conducting less
while - when we open transistor fully , it will conduct as much rail(s) can give to load ( load/speaker being most hungry here) there is opposite situation :
- how much transistor can conduct less ? amount is exactly the same as value of Iq
so, in quiescent state, there is 1A of Iq, and thus half of current envelope is 1A, simply because difference between 1A and 0A is 1A
that's why current envelope ( linear one) is twice Iq , for SE stage .......... 1A down to 0A, 1A up to 2A
**************
with PP stage , A class of same is defined again by same principle (how much less?!?) and A class envelope is same for SE and PP, but PP is able to transfer in B class on both sides, so B class envelope is defined with Current envelope , which is (one) rail value /divided with R load ............ then doubled (having two rails)
so, this is probably best I can do with words ....... and if that is better to you than watching nice vid, your choice
Although apparently the longest path, it´s better to start with the basics and work your way up step by step.
If you ask "step 5" questions, but have not mastered 1 to 4, each answer will be right but will not match/join others, so must be remembered by heart.
If you learn from the ground up, you will answer yourself because you will have the tools to do so.
It ends up being the shortest and most useful path.
If you ask "step 5" questions, but have not mastered 1 to 4, each answer will be right but will not match/join others, so must be remembered by heart.
If you learn from the ground up, you will answer yourself because you will have the tools to do so.
It ends up being the shortest and most useful path.
I've always done best by running ahead toward something I'm interested in, then figuring out what I tripped over on the way. 😛
ZF: I think I get it:
Transistor can only go up, not down. Bias boosts ground floor, but ground is still absolute while sky is infinite.
In SE, signal can only vary +/- bias point, because if + goes higher than - can, - end will crash into the ground.
In PP, same principle applies, but *both* sides are going up. So + side can go up as high as it's dreams and silicone can carry it, then comes back down to earth, and - side goes as high as it's dreams and silicone can take it. Neither size is going "down", they both go "up", then the two sides of the wave are strapped onto each others backs and go off down the yellow brick road to the driver.
Now maybe my explanation of my understanding confuses *you*, but I do think I have it. Both SE and PP could send the same amount of power out the back, assuming the same devices, but SE would clip off half the waveform as soon as it exceeded 2x bias.
--
(Also, ZF, that video was/is very good. I'm half-way through on 2x. So, thanks, gotta give this round to you 😉 😛 🙂 )
ZF: I think I get it:
Transistor can only go up, not down. Bias boosts ground floor, but ground is still absolute while sky is infinite.
In SE, signal can only vary +/- bias point, because if + goes higher than - can, - end will crash into the ground.
In PP, same principle applies, but *both* sides are going up. So + side can go up as high as it's dreams and silicone can carry it, then comes back down to earth, and - side goes as high as it's dreams and silicone can take it. Neither size is going "down", they both go "up", then the two sides of the wave are strapped onto each others backs and go off down the yellow brick road to the driver.
Now maybe my explanation of my understanding confuses *you*, but I do think I have it. Both SE and PP could send the same amount of power out the back, assuming the same devices, but SE would clip off half the waveform as soon as it exceeded 2x bias.
--
(Also, ZF, that video was/is very good. I'm half-way through on 2x. So, thanks, gotta give this round to you 😉 😛 🙂 )