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I'm trying to understand how a differential amplifier rejects common mode noise and amplifies the differential/normal mode noise.

For that I use the following circuit topology and simulation:

enter image description here

When Vin+ and Vin- are differential Vout increases as follows:

enter image description here

When Vin+ and Vin- are same Vout doesn't move much as below:

enter image description here

In case of common mode inputs, the increase in Vout is less than the increase in Vin. So common mode voltages get attenuated or we can say the gain is negative.

In case of differential inputs however, the increase in Vout is much more than the increase in Vin. So differential mode voltages are amplified. Hence means positive gain.

I can see these as a result in simulations.

But I don't know what causes this logically. I mean current source is causing all of these but how? Why common mode inputs cause almost no difference in Vout but differential mode voltages have big gain? Is there a way to tell what is going on in this circuit in transistor level by a step by step manner to explain why are the results are like these?

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  • \$\begingroup\$ The current source isn't "causing all this." You could replace it with a resistor (which isn't a great current source, but works after a fashion.) The BJTs are the key here. Assuming they are matched (BCM62 or BCM61, for example), then an input differential between the bases of 60 mV will cause about a 10:1 difference in the collector currents. A difference of 0 will split the collector currents equally, 1:1. In all cases the sum of the collector currents (plus the minor base currents) will equal your current sink's value. \$\endgroup\$
    – jonk
    Dec 17, 2017 at 23:52
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    \$\begingroup\$ Why will the "emitter voltage" be constant in differential case, and will increase in the common-mode case? \$\endgroup\$
    – floppy380
    Dec 18, 2017 at 2:32
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    \$\begingroup\$ and how does resistor can act as a current source? \$\endgroup\$
    – floppy380
    Dec 18, 2017 at 3:17
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    \$\begingroup\$ Attenuation is not negative gain. It's gain magnitude less than 1. \$\endgroup\$
    – Mike
    Dec 18, 2017 at 5:19
  • \$\begingroup\$ Major key to grasping diff amps: abandon hfe! Transistors are voltage-in, current out, see Ebers-Moll equation. The BJT is a transconductance element, like any vacuum tube or FET. Notice tiny differential voltage as input, versus enormous diff current output. Then, 10K converts output current to volts. \$\endgroup\$
    – wbeaty
    Dec 18, 2017 at 5:29

1 Answer 1

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If the circuit is symmetric (both components and excitations), then the current from the current source will be divided equally between the left and the right branches. Therefore, even though Vin+ and Vin- both increase, the current through Q1 will stay approximately constant at 0.5mA. This requires that the emitter voltage increases to keep the base-emitter voltage approximately constant.

If the excitations are instead differential, then the current through one transistor will increase and the current through the other will decrease, so that the sum is always equal to 1mA. This results in a large change in Vout.

EDIT:

Suppose that Vin+ increases and Vin- decreases. The current through the BJT on the left will therefore increase, and the current through the BJT on the left will decrease. Due to symmetry, it turns out that the emitter voltage will be constant with differential excitations. Then we can find $$\Delta I_C = g_m \Delta V_{b}.$$ On the left, this change in collector current will be positive, and on the left negative. The gain of the amplifier is then (roughly): $$ A_v = g_m R_1,$$ since the output voltage will change by the product of the resistor and the change in collector current.

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    \$\begingroup\$ I think I understand your first paragraph and "emitter voltage increases to keep the base-emitter voltage approximately constant"... But for the second part I couldn't understand clearly. How does current source act when inputs are differential and causes large gain? \$\endgroup\$
    – floppy380
    Dec 17, 2017 at 23:34
  • \$\begingroup\$ Are you familiar with small-signal analysis (aka incremental analysis)? \$\endgroup\$
    – andars
    Dec 17, 2017 at 23:39
  • \$\begingroup\$ Yes kind of. In small-signal analysis the nonlinear behaviour becomes linear because of small change in Vbe. \$\endgroup\$
    – floppy380
    Dec 17, 2017 at 23:42
  • \$\begingroup\$ See my edit, and let me know if it is still unclear. \$\endgroup\$
    – andars
    Dec 17, 2017 at 23:54
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    \$\begingroup\$ Why will the "emitter voltage" be constant in differential case, and will increase in the common-mode case? \$\endgroup\$
    – floppy380
    Dec 18, 2017 at 0:02

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