Saturday 27 February 2016

BJT Differential Amplifier Common Mode & Differential Mode Gain



I have a few questions about how to derive the differential gain and common mode gains:


schematic


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).


Thanks in advance !



Answer



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).



enter image description here


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.


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