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I am reading The Art of Electronics.

I found a related question here.

I understand that differential gain of the amplifier is $$G_{\text{diff}}=\frac{R_c}{2(r'+R_e)}$$ but how?

OK, using hybrid-pi model. Does that mean that AC current can flow from emitter to base? Doesn't that mean that the emitter will get more positive and base negative or is this small signal change negligible compared to biasing voltages?

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  • \$\begingroup\$ Take a look here electronics.stackexchange.com/questions/609271/… \$\endgroup\$
    – G36
    Jul 3, 2022 at 17:59
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    \$\begingroup\$ And yes, the small-signal change is negligible compared to biasing voltages. For the diff amp, Vin = Vbe1+ Vbe2. This means that as Vbe1 increases the Vbe2 must decrease by the same amount. For example, if Vbe1 changes from |0.60V| to |0.61V| the Vbe2 will drop by |0.01V| from |0.6V| to |0.59V| And any differential input voltage larger than 100mV will cut-off one of the transistors. electronics.stackexchange.com/questions/337939/… \$\endgroup\$
    – G36
    Jul 3, 2022 at 18:39

3 Answers 3

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Differential mode Gain

In hybrid-pi model, we just omit the dc offset from all parameters in circuit and think in a different way and solve the circuit very easily, however deep in mind, we shouldn't forget about the biasing currents and voltages. So if in h-pi model a current equal to \$i\$ is running through emitter to base, it means that in our actual circuit the current through base is \$I_{bias}-i\$, and yes, \$i\$ is so much smaller than \$I_{bias}\$ that the direction of current through base doesn't change.

So if you're asking that how the two transistors are influencing eachother to maintain the emitter voltage constant, the answer is, "the small current which is due to small imbalance in inputs comes out from one emitter and goes into the other emitter, and the overall current trough R3 doesn't change and stays constant even if there is small imbalance in the inputs, So the emitters' voltage stays constant". Remember that the direction of main current out of emitters, is outwards, because the amount of so called small current is smaller than bias currents.

Note: be careful that single ended gain is half of differential gain. $$\frac{Vo_2-Vo_1}{Vin_2-Vin_1} = 2\frac{Vo_2}{Vin_2-Vin_1} $$

The answer for differential mode ends here.If you are looking for so called "common-mode" gain, where two input voltages are equal, read below, IF NOT SKIP THE REST of the answer.


Common mode Gain

Since our excitation is symmetrical, the two symmetric sides of our circuit will function exactly similar.

As a first step, we can change \$R_e\$ with two parallel resistors with value \$2R_e\$. (Note that the equivalent value of two parallel \$2R_e\$s, is \$R_e\$)

Now here is the important trick. Because of our symmetrical inputs and symmetrical circuit, the voltage of emitter resistors will be identical even if there is no wire between emitters of BJTs. So we can easily omitt the middle connection between two sides as depicted below: common-mode equivalent circuit

And at the end, we have a simple common-emitter gain stage, with emitter degeneration resistor \$2R_e\$. Gain is simply $$ \frac{V_o}{V_{in}} =- \frac{R_c}{r_m + 2R_e}, r_m = g_m^{-1}=\frac{V_{thermal}}{I_c}$$

where \$I_c\$ is bias current passing through collector.

In the comments you asked:

If voltage doesn't change at the emitter resistor and current resistor junction, then how can they influence each other?

well in common-mode input, emitter voltage changes and if there is any resistors between ground and emitter, we can state that the current of that resistor will also change according to emitter voltage.

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  • \$\begingroup\$ The question was about "Gdiff" - and, therefore, I assume that the questioner was not referring to the common mode amplification. \$\endgroup\$
    – LvW
    Jul 3, 2022 at 12:50
  • \$\begingroup\$ @LvW thanks, I will edit my answer ASAP. \$\endgroup\$
    – MaNegah
    Jul 3, 2022 at 12:59
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I think you speak about the unsymmetrical gain - that means: Signal input at one base node while second base node is grounded.

In this case, the most simple way to find the gain is to consider the circuit as a common emitter stage with signal feedback.

Neglecting the very large current source resistance in the common emitter lag, the feedback resistance for the first stage (Q1) is Rf=Re+Re+(1/g) with 1/g=re (input resistance at the emitter node of Q2). This assumes that both transistors have equal currents Ic1=Ic2 with g1=g2=g.

Therefore, the gain is (classical feedback scheme): Gdiff=-gRc/(1+g*Rf).

After inserting the above expression for Rf and substituting g=1/r´ you arrive at the given gain formula Gdiff.

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  • \$\begingroup\$ I have the feeling that OP is talking about the differential ("symmetrical gain") gain. \$\endgroup\$ Jul 1, 2022 at 13:43
  • \$\begingroup\$ @LvW Yes, I was talking about symetrical gain. I understand the situation when one BJT is grounded, but I dont't understand symetrical situation. If voltage doesn't change at the emitter resistor and current resistor junction, then how can they influence each other? Aren't they simply two common emitter amplifiers with gain -Rc/(r+Re) each? \$\endgroup\$
    – TheUnknown
    Jul 1, 2022 at 13:51
  • \$\begingroup\$ In case of symm. input signals (Vin1=-Vin2), we have the same gain expression when the output signal at one collector only (NOT the difference) is referenced to the differential input Vin=Vin1-Vin2=2Vin1. Answering your question - yes, the expression as given in the last line of your comment is corrent, when the output is referenced to ONE input only (Vin1 or Vin2). \$\endgroup\$
    – LvW
    Jul 1, 2022 at 16:26
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Does that mean that AC current can flow from emitter to base?

The sum of the bias base current and AC current flows through the base-emitter junction from the base to emitter, according to the conventional current direction. So the base current does not change its direction but only its magnitude.

Doesn't that mean that the emitter will get more positive and base negative or is this small signal change negligible compared to biasing voltages?

In differential mode, when the input voltages change with the same rate but in different directions, the emitter voltage will stay unchanged (as though it is grounded) since the two transistors oppose to each other.

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