BJT Differential Amplifier Common Mode & Differential Mode Gain

Question

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 $$-R_C/(2 R_3+R_E)$$

Answer 1

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.