# 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_{\text{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 $$\R_3\$$) I get the voltage gain as follows

$$(V_{\text{in1}} - 0.6)-(V_{\text{in2}}-0.6) / 2 \times (R_E+r_e) = 0-V_\text{OUT} / R_C$$

$$V_\text{OUT} / V_{\text{diff}} = -\frac{R_C}{2 \times R_E+r_e}$$

but in the AoE they have $$G_\text{diff} = \frac{R_C}{2 \times R_E+r_e}$$

What happened to the (-) sign?

Common Mode Gain:

Common Mode signal is $$(V_\text{in1} + V_\text{in2})/2 = V_\text{in2}$$

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

$$(V_\text{IN2} - 0.6)/ (R_E + r_e + 2 R_3) = 0-V_\text{OUT} / R_C$$

This is as far as I get - I don't see how I can get rid of the 0.6 V to get the right answer of $$-R_C/(2 R_3+R_E)$$

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.