# BJT Differential Amplifier Common Mode & Differential Mode Gain

I have a few questions about how to derive the differential gain and common mode gains: simulate this circuit – Schematic created using CircuitLab

Differential Gain:

Taken from Art of Electronics - "Imagine a symmetrical input signal wiggle in which input 1 rises by v_in (a small signal variation) and input 2 drops by the same amount As long as both transistors stay in the active region, point A remains fixed."

I don't follow how A would be fixed?

However taking it to be true (no current through R3) I get the voltage gain as follows

(V_in1 - 0.6)-(V_in2-0.6) / 2*(R_E+r_e) = 0-V_OUT / R_C

V_OUT / Vdiff = -R_C/(2*R_E+r_e)

but in the AoE they have G_dff = R_C / 2*(R_E+r_e) what happened to the (-) sign?

Common Mode Gain:

Common Mode signal = (V_in1 + V_in2)/2 = V_in2

Following the suggestion to split the pair into 2 sections (I'm looking at section on the right)

(V_IN2 - 0.6)/ (R_E + r_e + 2*R3) = 0-V_OUT / R_C

This is as far as I get - I don't see how I can get rid of the 0.6V to get the right answer of -R_C/(2*R3+R_E).

They simply do AC small-signal analysis.

So you can skip $V_{BE}$ if you do AC analysis.

The $r_e$ resistance "represents" the change in $V_{BE}$.

$\Delta V_{BE} = i_e\cdot r_e$

As for the voltage at point $A$.

This voltage remains fixed due to the fact that we are again dealing with symmetrical AC signal (no AC current through R3) and "Imagine a symmetrical input signal wiggle in which input 1 rises by $V_{IN}$(a small signal variation) and input 2 drops by the same amount".

For example, $V_{IN}$ if will increase $I_{E1}$ current from $1mA$ to let as say $1.2mA$ (due to $V_{be1}$ increase) and $I_{E2}$ will decrease by the same amount from $1mA$ to $0.8mA$

$$ΔIe1 = 1.2mA - 1mA = 0.2mA$$

$$ΔIe2 = 0.8mA - 1mA = - 0.2mA$$

So the AC current sum of the emitters currents $Iee = ΔIe1+ΔIe2$ will be equal to $0A$.

Because the AC component of a $Ie1$ and $Ie2$ are equal in magnitude but 180° out of phase.

This means that $Iee$ current is constant, no AC component. Hence the potage at point $A$ remains fixed.

(1.2mA + 0.8mA = 2mA = constant). As for this "minus" sign in the gain equation. We usually omit this "minus" sign because we know what this "minus" sign represents/means. This "minus" only informs us that the output voltage is the 180-degree phase shift with respect to the input voltage.