solid state long tail pair input

[lowell]

I'm trying to once and for all wrap my brain around how this circuit functions in your typical solid state amplifier. Assume 2 pnp bjt's sharing an emitter resistor that is connected to the positive rail(typical). And assume that the signal to the feedback input is in phase with the LTP's input. I believe this is also called a differential amp, and that it amplifies the DIFFERENCE between it's inputs. This is what I don't understand. Say the input voltage is 2v p-p and that the PA output is 10v p-p. Then we set the feedback up with a 10:1 voltage divider. (I know that there is DC feedback AND AC feedback, but lets ignore DC for now). So we are feeding back 1v p-p. Got me? Sorry a picture would be best but I don't have access to a program or scanner. Does this mean that the differential input will produce 1v p-p? (difference between inputs is 2v/1v=1v.

So here's what I foresee happening here:

When the feedback input goes UP, so does the voltage on its emitter (and in turn same for the input transistor's emitter), THUS turning on the INPUT transistor MORE, THUS reducing the positive excersion on the input transistors collector. Right? In so many words, an upgoing signal on the PA output, reduces upgoing signal on the INPUT transistor's collector, thus reducing gain?

I know I'm probably missing a LOT of details here forgive me.

[Alan0354]

I think you get the idea wrong. You should think of it as an opamp. In fact the field schematics in MOSFET power amp thread are just simple opamp in discrete form. This is the simplified drawing:



Q1 and Q2 is the PNP differential pair you want, assume you have +V and -V as supply. Assume input is reference to ground. AND assume Q1 and Q2 are matched pair to minimize offset voltage.

With no signal, tail current I is (+V-0.7V)/R where R is the resistor from emitter to +V. The bottom is called current mirror. When there is no signal and the Q1 and Q2 are match, current I is split between Q1 and Q2 and the collector current of each is 1/2I. Ic1goes through the diode and the resistor and set up a voltage. That voltage drive the base of the bottom NPN and cause a current. If the Vbe of the diode and the NPN is match and the bottom two resistor is the same. The collector current of the NPN is equal to 1/2I also because the same voltage is drop across both resistor at the bottom.

So under idle condition, the collector current from Q2 is absorbed completely by the NPN. So the voltage at the collector is 0V. But if you start driving a +ve voltage at the base of Q1 ( +ve input of opamp), before the rest of the circuit can respond, the voltage at the base of Q2 is still 0V, you have 1V of differential voltage across the base of Q1 and Q2. So the moment you put +1V to the base of Q1, Vbe of Q1 decrease( as emitter of Q2 prevent it from going up), Current of Q1 decrease and Q2 increase. The current of the current mirror decrease, so the current between the collector of Q2 and the NPN no longer equal, more current source by Q2 than the NPN can sink, so the junction start to rise up.

Now let's complete the opamp. I just put a generic AMP following the diff pair. The AMP have more non inverted gain. So the output start going up. You can see there is a voltage divider of 1/10, so if the output goes up 1V, the base of Q2 only goes up 0.1V. Now you look at the differential pair, Q1 and Q2 still not balance, there is still 0.9V across the two inputs. So Q2 still draw more current than the NPN. The voltage at the collector of Q2 still keep going up. This process will not stop until the output almost reach 10V, the base of Q2 almost goes up to +1V, then the differential pair almost balance, the collector current between Q1 and Q2 start to equal to each other again, the voltage at the collector of Q2 stop going up. So you get almost +10V.

Notice I say almost 10V? This is because to get output to go 10V, you need some input voltage. Say if the forward gain from +ve input at Q1 to the output is 1000, so in order to get 10V, you need 0.01V across the differential input. So if you put 1V at the base of Q1, you really get about 9.99V instead of 10V, because you need 0.01V across the differential pair to generate 10V. This small voltage that is needed to generate the output is called ERROR VOLTAGE. That's the reason the opamp are made to have very high open loop gain....like 10EE6, so the the error voltage is approx 0.

This is the process how a closed loop feedback works. The differential pair is the first part of the opamp. Hope this help. I show the equivalent circuit in opamp form at the lower right. If you look at the solid state power amp, they are just opamp. If you look at it this way, it really become very simple. This is very different from the tube power amp. You don't really draw load line etc. Solid state amps are in a way simpler.

[Steve Conner]

It works the same as the long-tailed PI in a tube amp.

The two transistors talk to each other through their emitters. Applying a signal to either base will cause a differential signal at both collectors, just like the tube amp PI. The first transistor functions as a common-emitter amplifier, but it also passes the signal on to the second one through its emitter, which amplifies it as a common-base stage.

When signals are applied to both bases at once, both collectors give an output that represents the difference between the two inputs. One normal, one inverted, like the old tube amp PI. In most solid-state amps, only one of the collector outputs is used, so you never get to scope the other one, but they are always equal and opposite.

The circuit has quite a lot of gain and is quite nonlinear. A 26mV difference between the two bases will unbalance the two collector currents in a 10:1 ratio, and that is practically about the biggest signal it can handle. When used in a solid-state amp, the global NFB should maintain much less than 26mV of error signal at all times.

Why do I call it error signal? In a solid-state amp, the long-tailed pair is sensing the difference between what the amp output ought to be, and what it actually is.

[Enzo]

This may or may not help you. It is how I try to make transistors intuitive to myself.

Because I learned about transistors really after having learned about logic circuits, to internalize transistors I still think of them in terms of logic - or highs and lows. And the way I conceptualize a transistor circuit is this. If you pull a transistor's base towards its collector and away from its emitter, it turns it on, which pulls its collector towards the emitter. Pull the base towards the emitter and you turn the transistor more off, and collector relaxes back towards its supply.

So in a typical common emitter gain stage, with positive supply on the collector, the more I pull the base towards the collector - in other words the more positive I make the base with respect to emitter - the more the transistor conducts, which pulls the collector voltage down. It inverts.


SO in your case of PNP with emitter to positive, then the collector must be the more negative end, and so pulling the base down - making it less positive than it is now - towards the collector turns it on, and that pulls the collector voltage up.


So reading your diffy pair, PNP with common emitters to V+, I'd think the more negative I make the base of the left one (the input), the more positive that collector becomes, and yes, the emitter follows the base downwards. The emitters going down pulls them closer to the base on the other side (the feedback side), which turns that side more off. That in turn makes the right side collector fall away from its emitter, the voltage ther goes more negative.

Yes, in a working amp, there are influences all working together - input AND feedback signals, but to conceptualize how the feedback side works or how the input side works, I tend to imagine the other side held steady. So if the feedback signal at some instant goes more positive - base towards emitter - then the emitters follow, and that makes the input side base more negative to those emitters than it was, thus turning the left transistor on harder, and pulling the left collector up more positive. And the collector on the feedback side would then go more negative too. Thus differential operation.


SO my basic pull the base towards the collector idea is how I read one of those little control ciruits where tree or four transistors turn on a fan or a power-up mute or whatever. So if this one pulls up, then that one pulls down, which pulls thius one up, which turns on the fan. I go through just such a sequence in my head. It isn;t how they teach transistors, but it is just my way to read through the sequence of a circuit.


I never built one, but in my head I envision a transistor like a lever in a visual aid. One end is your base, the other end is your collector, and pivot point is the emitter. (Assuming V+ at top and NPN) SO pushing up on the base end of the lever makes the collector end drop lower and vice versa. Corresponds to voltages Emitter follower? Then the collector is bolted to V+, and pushing up on the base end will make the pivot (emitter) go up as well - following. OH this plan ignores current and gain and stuff, it just serves to conceptualize the signal flow.

[Steve Conner]

The Art Of Electronics explains the amplifying action of transistors in terms of a little guy called "Transistor Man".

Now available as a T-shirt.
Transistor Man Shirt [Mens Large] ID: 843 - $19.95 : Adafruit Industries, Unique & fun DIY electronics and kits

The original passage from the book.
The Art of Electronics - Paul Horowitz, Winfield Hill - Google Books

[Alan0354]

The Art Of Electronics explains the amplifying action of transistors in terms of a little guy called "Transistor Man".

Now available as a T-shirt.
Transistor Man Shirt [Mens Large] ID: 843 - $19.95 : Adafruit Industries, Unique & fun DIY electronics and kits

The original passage from the book.
The Art of Electronics - Paul Horowitz, Winfield Hill - Google Books

I have not read this book, and I don't want to be disrespectful. This is a post mentioned about this book in post #15. My suggestion is the book I mentioned in #13. My user name is Yungman over there.

[lowell]

"Yes, in a working amp, there are influences all working together - input AND feedback signals, but to conceptualize how the feedback side works or how the input side works, I tend to imagine the other side held steady. So if the feedback signal at some instant goes more positive - base towards emitter - then the emitters follow, and that makes the input side base more negative to those emitters than it was, thus turning the left transistor on harder, and pulling the left collector up more positive. And the collector on the feedback side would then go more negative too. Thus differential operation."

Enzo, exactly what I was trying to say in my initial post. Cool! And, probably due to your many helpful posts, I also envision transistors doing what you describe. It's like the base is a see saw. Weigh the base DOWN and the other end (collector) goes UP, and vice versa.

So now I see it as simple as this. The feedback input (Vbe) of a diff pair changes the voltage on its emitter, and in turn, the emitters (plural) of the PAIR (they share Re). This in turn affects the Vbe of the input transistor and therefore affects how ON it is. Finally, this affects its output (Vc).

[Alan0354]

The good old long tail tube PI is differential pair, there is no difference conceptual wise. It works the same way. The key to understand is the differential pair share the same current, changing the differential voltage ( voltage between Q1 and Q2 in my drawing) just steer the current from one transistor to the other. One requirement of designing a differential pair with resistor R to set the current I is you need to have the voltage across the R much much bigger than the differential voltage. Or else, the I change as you can see in my drawing. Say if you hav +V only +10V and R is 1K. I is 9.3mA. But if you put 1V at the input, the output goes to 10V, the voltage at Q2 is 1V. Now the voltage drop across R is 8.3V. So I decrease about 10%. This will cause common mode problem and power supply rejection problem. Usually people use current source to provide the tail current, not a resistor. So the current I remain constant even when the voltage at the emitter of Q1 and Q2 changes. This will improve common mode rejection. and power supply rejection.

It is good to look at it in a very simplistic way, but I still think you should read more into it as there is more to the differential pair. It is one very important circuit for very good reason. Take a look at this article:

http://opencourseware.kfupm.edu.sa/c...Handout_5b.pdf

Since you are here, you must be into tubes. This is an explanation of the long tail PI, which is differential pair.

http://www.aikenamps.com/LongTailPair.htm

Read this, you understand this, you understand BJT differential pair. I am learning more about tubes, I found it very easy to relate transistor to tubes. They work pretty much the same. I am a lot more familiar with transistors, I just think of tubes are transistor with different characteristics. This holds true for MOSFET and JFET or any other exotic transistors. Find their similarities and there difference and work with them.

For example the NPN BJT is very similar with beam power tube characteristic, both have high output resistance. If you look at the collector curve vs the plate curve, they almost look the same. Even though collector curve is Ic vs Ib, but you can set up using base voltage and it will be just like beam power tubes. The difference is the bias voltage. BJT need 0.7V, Beam tubes need -30 to 50V. You get the drift. You design circuit almost the same. Just concentrate on their difference and work around those, you'll be learning in no time.

In a way, tubes are a little more complicated. You don't have complimentary pair like PNP vs NPN. The output impedance of tubes are lower and always get in the way. Tubes conduct grid current when positive. You can design very low output impedance with transistors and totally bypass all the load line stuff. Because of the low output impedance, you get rid of the OT all together. You adjust the class of amplifier by adjusting the idle current. It is easier in every aspect.