Author Topic: Why would you swap a NPN for a PNP in this circuit?  (Read 1382 times)

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Offline Evy

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Why would you swap a NPN for a PNP in this circuit?
« on: August 14, 2017, 08:23:58 pm »
Negative feedback, 2 stage voltage amplifier:

Signal graph:


Biasing realization:

(ignore the current mirror in the top left corner above T1)

I am talking about T3 in this case which was a NPN in the signal graph and is now a PNP in the biased version of the amplifier.

In my book they mention that its done to  "adress the voltage level difference" which I didn't understand. Could that be the difference in voltage between VCC and the base of the transistor. Right now it seems to be 0.6V and they mentioned that it could be negative if they didn't use a PNP but I don't understand what that would accomplish. I don't see the point with the swap.

Could anyone please help me understand why you would swap two transistors like that?
« Last Edit: August 14, 2017, 08:30:31 pm by Evy »
 

Offline w2aew

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Re: Why would you swap a NPN for a PNP in this circuit?
« Reply #1 on: August 15, 2017, 03:28:23 am »
In both cases, T3 is a common emitter amplifier, providing a second stage of voltage gain in the circuit. 

Personally, I'd ignore the "signal graph" schematic - that'a a bizarre way to learn about circuit operation - just my opinion.

In the biased version, a PNP common emitter amp is used because the voltage at the active load (unlabeled transistor current source) conveniently provides the a good bias voltage for the base of T3.  If a NPN was used here, it would be a common collector amplifier and would not provide any voltage gain (it would be a unity gain buffer).
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Offline T3sl4co1l

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Re: Why would you swap a NPN for a PNP in this circuit?
« Reply #2 on: August 15, 2017, 04:07:43 pm »
If it were NPN --

How would you do it?

The NPN base must be slightly above its emitter voltage.  The NPN base must also be above the diff pair collector voltage.  Thus the NPN emitter needs to sit on an elevated supply to begin with.

The NPN collector must be above its emitter voltage as well.  So the output must be very high up, indeed.

You cannot make a DC coupled, unity gain follower with such a configuration: the voltages are in completely the wrong ranges!

Suppose you wanted to cascade two of these stages (it happens often in analog signal paths, cascading several op-amp circuits, or amplifiers in general).  Where do the voltages need to be, now?

Going back to the P/N transformation I mentioned earlier: it can be applied to any single component as well.  Reflect on the benefits / challenges this affords the DC voltages and currents in the circuit. ;)

Tim
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Offline IanMacdonald

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Re: Why would you swap a NPN for a PNP in this circuit?
« Reply #3 on: August 15, 2017, 05:49:49 pm »
Basically the output of a long-tailed pair (T1/T2) only has a limited range of voltage swing, since the emitters are at a substantial part of the supply voltage, and T1 collector cannot go below its emitter voltage.  So if you used an NPN emitter follower for T3 the output swing would be rather limited.

Using a PNP means that its collector can swing over the full supply voltage range.
 

Offline T3sl4co1l

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Re: Why would you swap a NPN for a PNP in this circuit?
« Reply #4 on: August 15, 2017, 07:55:44 pm »
Basically the output of a long-tailed pair (T1/T2) only has a limited range of voltage swing, since the emitters are at a substantial part of the supply voltage, and T1 collector cannot go below its emitter voltage.  So if you used an NPN emitter follower for T3 the output swing would be rather limited.

Using a PNP means that its collector can swing over the full supply voltage range.

Oh, and yes, subbing in an NPN, adjusted for polarity, gets you an emitter follower, not a common-emitter stage -- very different overall circuit!

Tim
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