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I'm studying the differential amplifier on different books (Razavi and Sedra-Smith).

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I've understood how this circuit works, but I have some questions. When authors analyze this circuit (2 identical mosfets or two identical BJTs with a current source below them, as shown in figure) for small signals, they always assume differential inputs, that is: v1 = -v2

Question: why is this assumption required? When this circuit is used as the first stage of an operational amplifier in a negative feedback fashion, who says that the inverting and non-inverting terminals will have v1 = -v2 ? Negative feedback says that the input differential input of the op-amp is very very small (and for a very very small differential input collector currents of the differential amplifier are linear), I don't understand then why v1 should be equal to -v2. Small signal analyses should (in my opinion) consider the case in which v1-v2 is very very small, but not necessarily v1=-v2

Thanks

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  • \$\begingroup\$ V1 can never practically equal -V2 because the biasing required on both inputs raises both V1 and V2 to be positive values. In my experience, when analyzing diff amps you set V1 and V2 to be independent but also analyze the common-mode value and the effects that it might have on the output signal. Therefore I refute your observation that analyzers always assume that V1 = -V2 both from a practical standpoint and an analytical standpoint. This makes it very difficult to answer a question based on such shaky ground. Well-done on breaking the questions down BTW but sorry to undermine this one. \$\endgroup\$ – Andy aka Aug 24 '18 at 8:57
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The background for the assumption (V1-V2) is simple: Gain calculation.

Consider the following: We have two arbritrary input voltages Vx and Vy.

These voltages can be split into two parts:

  • Common mode voltage; Vcm=(Vx+Vy)/2 ;

  • Push-pull voltage: Vpp=(Vx-Vy)/2 ;

  • Hence, it is easy to show that Vx=Vcm+Vpp and Vy=Vcm-Vpp

Now it is easy to calculate the gain:

  • Vcm is amplified with the common mode gain Acm (which is very low and approaches zero if the common dynamic emitter resistor is infinte >>> ideal current source) .

  • Vpp is amplified with the push-pull gain App which can be given immediately because Vpp is applied to both inputs - however with opposite sign. As a consequence, the collector current increase in one transistor will be compensated by a corresponding decrease in the other transistor. As a consequence, there will be no negative feedback effect - and the gain App is identical to the classical gain of an emitter stage without feedback: App=-gm*Rc.

  • Hence, the output voltages are:

(1) Vout1=AcmVcm+AppVpp=VcmAcm+gmRc(Vx-Vy)/2

(2) Vout2=AcmVcm-AppVpp=VcmAcm-gmRc(Vx-Vy)/2

  • For an ideal emitter current source, we have Acm=0 and the output voltages are:

Vout1,2=(+-)Vd*(Vx-Vy) with the differential gain Vd=-gmRc/2.

Fazit: The calculation of the combined gain is very simple if we split each of the two input voltages Vx, Vy into two parts (common mode resp. push-pull mode). Of course, the push-pull mode dominates for a high-quality diff. amplifier - and the gain for the push-pull voltages is very easy to calculate because these parts are equal in magnitude, but opposite in sign.

Note that the voltages V1, V2 as mentioned in the original question are identical to Vx and Vy, respectively: V1=Vx=Vcm+Vpp and V2=Vy=Vcm-Vpp.

However, if we consider the DIFFERENCE V1-V2 only (and assuming Acm=0) the common mode parts play no role (are not amplified) and it is sufficient to consider this difference only:

V1-V2=Vpp-(-Vpp)

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Small signal analyses should (in my opinion) consider the case in which v1-v2 is very very small, but not necessarily v1=-v2

I agree with that, however when \$V_1 \ne -V_2\$ things do get slightly more complicated as then your input signal isn't truly differential anymore as by making \$V_1 \ne -V_2\$ introduces a commonmode signal. It is easier to consider the commonmode signal separately by evaluating \$V_1 = V_2\$

Suppose we start with \$V_{in} = 0 mV\$ so \$V_1 = V_2=0 mV\$ and then we evaluate what happens when we apply a not-truly differential signal.

Say we apply \$V_1 - V_2 = 4 mV\$ but we do that by making \$V_1= 3 mV\$ and \$V_2= -1 mV\$ then \$V_1 \ne -V_2\$ is true, this signal isn't truly differential.

In the small signal model then the voltage at the common emmitter of the NPNs will be roughly in between that voltage. So when \$V_1 = V_2 = 0 mV\$ there will be 0 mV at the emitters but when \$V_1= 3 mV\$ and \$V_2= -1 mV\$ there will be +1 mV at the emitters.

That + 1mV is the common mode voltage. Suppose that current source \$I_{EE}\$ has an output impedance, then that current will change a little. So the transistors will be biased slightly differently. That makes things complex!

Since the small signal model is a linearization we can simply split that into two cases: differential signal behavior and commonmode signal behavior. And to make it truly differential we need \$V_1 = V_2\$.

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