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I am trying to solve various differential-amplifier homework problems for which I have the solution to study.

For the simplicity of most problems we are able to neglect base-currents. Moreover, it is to be assumed that the transistors are perfectly matched with a Beta = 100 for both Q1 and Q2.

However, for the example shown below, if we neglect any additional base current, by deduction, are we also able to remove (short out) the emitter-resistance as well? It appears that is what the solution manual did, though, there is no explanation.

Given Schematic:

schematic

simulate this circuit – Schematic created using CircuitLab

Where the derived AC-equivalent is:

schematic

simulate this circuit

And where I derived the expression for the differential gain Ad as:

enter image description here

But the solution somehow derived the following equation: For which I could only surmise that in the AC-analysis, considering one-half of the amplifier, since they are identical mirrors, the common-emitter node gets grounded?

enter image description here

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2 Answers 2

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Consider a single emitter follower - the output impedance at the emitter is low (usually just a few ohms). This is because the emitter "follows" the base voltage when the BJT is properly biased.

Tying two emitters together means that both emitters present just a few ohms to each other. It follows from this that you can regard the emitter junction as being a fairly hard AC signal point and can, when ignoring DC levels be regarded as ground.

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  • \$\begingroup\$ Simply to reinforce my own understanding, what do you mean when you say a "fairly hard" AC signal point? Do you mean pretty strong AC signal with no DC-offset? \$\endgroup\$
    – Matt
    Commented Oct 1, 2018 at 17:11
  • \$\begingroup\$ I mean that the common node shared by the two emitters appears to be produced from a voltage source with low series impedance. \$\endgroup\$
    – Andy aka
    Commented Oct 1, 2018 at 18:35
  • \$\begingroup\$ @Matt If you are finished with this question and have a satisfactory answer you should close the question down by formally accepting an answer. If you are still confused or need clarification then leave a comment. You might also consider taking a look at your other questions - they have not had closure yet. What did you resolve to do with the Zigbee radio question? \$\endgroup\$
    – Andy aka
    Commented May 24, 2020 at 15:15
  • \$\begingroup\$ Sorry - but the answer is wrong. If the answer would be correct, there would be - for example - no output signal at the collector of Q2 for a single input at the base of Q1. In fact, both output nodes show a signal because the whole combination works as a common-collector-common-base combination. Only for symmetrical operation (Vin1=-Vin2) the common emitter node is fixed and can be considered as ac-grounded. \$\endgroup\$
    – LvW
    Commented Oct 14, 2021 at 13:36
  • \$\begingroup\$ @LvW the original question does contain the formula V2 = -V1 hence there is an implicit assumption that the amplifier is operated symmetrically. My answer is restricted to what was asked (and directed) by the OP. \$\endgroup\$
    – Andy aka
    Commented Oct 14, 2021 at 13:45
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There is only one single case for which the common emitter node can be regarded as grounded (for ac):

Fully differential application with Vin1=V1 and at the same time Vin2=-V1

And this is exactly the condition as shown in the circuit. For all other cases (one input grounded and, in particular, for common mode signals) the emitter resistor plays an important role and must not be considered as grounded.

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