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Identifying +ve and -ive terminals of differential pair.

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itsfebin

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Hi,

I have a basic question. Suppose I have a differential pair circuit. How can I distinguish -ve and the +ive inputs?

thanks,
Febin
 

Hi,

I had assumed +ve and -Ve signals to a diff-pair could be connected to the gates of any of the 2 mos devices, since they are identical and matched. Can you please tell me how one of the terminals can then be considered +ive or -ve w.r.t the other?

In a 2 stage opamp, the o/p of the 1st stage is taken from the collector of one transistor in diff-pair. What is the gate terminal of this particular transistor from whose collector provides the i/p to 2nd stage? +ive or -ve?

Thank you once again.
 

if you apply a signal between + and - inputs ,
you should see an amplified output with the same phase.(no phase shift).

apply same by reversing to + and - i/ps , you should see an inverted(or 180) phase shifted output.

(or) apply blindly input signal .
see o/p.
reverse input .
see o/p.
the one with nophase shiftis+and - inputs.

srizbf
10thmay2010
 

itsfebin said:
Hi,
In a 2 stage opamp, the o/p of the 1st stage is taken from the collector of one transistor in diff-pair. What is the gate terminal of this particular transistor from whose collector provides the i/p to 2nd stage? +ive or -ve?
Thank you once again.
Click to expand...

The output signal at the collector (drain) of a transistor referenced to its own base (gate) is always out of phase by 180 deg. What do you think: should this be called positive or negative?
 

Hi LvW,

alrt. So, this particular gate from which the i/p to 2nd stage is taken, is the -ve terminal. Am I right?

Thank you very much.
 

itsfebin said:
Hi LvW,

alrt. So, this particular gate from which the i/p to 2nd stage is taken, is the -ve terminal. Am I right?
Click to expand...

Yes, of course - as far as the output voltage is concerned.
You can imagine that the differential pair acts as a common emitter (T1)/common base (T2) resp. common source/common gate combination. Thus, you have two transistors T1, T2 and two outputs (input at T1):
*output voltage T1 (common source): 180 deg out of phase
*output voltage T2 (common base with input at the emitter node): in phase with the input.
Finally, because of perfect symmetry you can ground the input of T1 and use T2 instead. Now, you have the same situation with an exchange of the role of T1 and T2 .
 

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