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Is there any reason why the gain of a single ended input BJT differential amplifier, with one Rc, is a good approximation of the value of Adm for a common mode input.

The simulation shows the frequency response for a single ended input and the circuit below shows th common mode input circuit

Single ended input frequency reesponse

Common mode input circuit

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  • \$\begingroup\$ At the risk of stating the obvious, those inputs aren't common mode. What are you really trying to do here? \$\endgroup\$
    – user16324
    Jun 7, 2020 at 15:33
  • \$\begingroup\$ Also, there is not a single-ended input here... \$\endgroup\$
    – Circuit fantasist
    Jun 7, 2020 at 15:38
  • \$\begingroup\$ How is the circuit not common mode? I thought this meant when there are two inputs to the diff amplifier \$\endgroup\$
    – S123l2000
    Jun 7, 2020 at 15:41
  • \$\begingroup\$ @Circuitfantasist The single ended circuit did not have the AC voltage source on the left hand side \$\endgroup\$
    – S123l2000
    Jun 7, 2020 at 15:42
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    \$\begingroup\$ Common-mode means the inputs are equal. Or at least that their average is nonzero. \$\endgroup\$
    – Hearth
    Jun 7, 2020 at 16:11

1 Answer 1

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The single ended circuit did not have the AC voltage source on the left hand side

When you apply no AC signal to Q1's base, the shared emitter connection is still largely held at a small signal value and the gain of the right-side input to the left-side output is the same formula as the full differential mode amplifier gain: -

$$\dfrac{R_C}{2\cdot r_E}$$

The "2" factor in the above equation is because we are applying a differential signal. This means that the voltage gain is halved because each base only receives 50% of the full differential signal.

When applying a single ended signal to a differential amplifier the gain is twice as high but given we might only be applying half the full differential signal, it becomes the same formula.

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  • \$\begingroup\$ Thankyou very much! \$\endgroup\$
    – S123l2000
    Jun 7, 2020 at 15:57
  • \$\begingroup\$ @Andy aka....when the "shared emitter connection is still largely held at circa 0 volts AC....." and there is no ac signal to the base of Q1 - how can the left side output carry any amplified signal voltage? \$\endgroup\$
    – LvW
    Jun 7, 2020 at 16:23
  • \$\begingroup\$ @LvW well it can't - I used the term "largely" because I know that the AC signal voltage seen at that point is still quite small (but not zero). Maybe I should be clearer. \$\endgroup\$
    – Andy aka
    Jun 7, 2020 at 16:26
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    \$\begingroup\$ Well - I think, the answer is quite simple: Seen from the most right side (from the base of Q2) both transistors form a two-stage common-collector - common base amplifier. And the gain of such an amplifier is (1/2)gm*Rc because the large RE is shunted with the small input resistance of the common-base amp: (1(gm)||RE is set equal to 1/gm. Hence, the common emitter point carries the input signal for Q1, \$\endgroup\$
    – LvW
    Jun 7, 2020 at 21:51
  • \$\begingroup\$ @LvW While I agree with your reasoning, really speaking there's always a certain resistance on the base of both the inputs (for example the signal source resistance), hence the common base amplifier assumption is not always well satisfied...even if this base resistance is often very low and coluld be disregarded in the calculations... \$\endgroup\$
    – barrow
    Jun 10, 2020 at 21:44

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