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This is a snipet from the book Design of Analog CMOS Integrated Circuits (page 112).

This is analysis of differential pair.

For \$g_{m1} = g_{m2} = g_m\$, reduces to

$$\begin{equation} (V_X − V_Y ) = −g_m R_D V_{in1} \tag{4.21} \end{equation}$$

By virtue of symmetry, the effect of Vin2 at X and Y is identical to that of Vin1 except for a change in the polarities: \$ (V_X − V_Y ) = g_m R_D V_{in2} \tag{4.22} \$

Adding the two sides of (4.21) and (4.22) to perform superposition, we have $$\frac{V_X - V_Y}{V_{in1}-V_{in2}} = -g_m R_D$$

Well, adding two sides together will give us the following. I have no idea why the book give us the equation above.

$$2(V_X - V_Y) = g_m R_D (V_{in2}-V_{in1})$$

$$\Rightarrow \frac{V_X - V_Y}{V_{in1}-V_{in2}} = -g_m R_D/2$$

Velvet
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kile
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3 Answers3

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If you think about one side of a differential amplifier, Vout= -gm•RD•Vin. The two sides at opposite polarity mean you end up with double the gain, but with a differential amplifier, you split the input as ±Vin/2 on each side, meaning it cancels out and you get the same gain you'd get for the equivalent half circuit.

Edit: for a concrete example, imagine an amplifier with gm•RL=10 and Vin=10mV AC. Vin1 would equal +5mV, and Vin 2=-5mV. VX=-gm•RL•Vin1=50mV, and similarly VY=-50mV. Your differential output signal is VX-VY, which is 100mV. That gives you a total gain of 100mV/10mV=10, which is just gm•RL.

mapplejacks
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  • Do you mean $V_{in1}$ and $V_{in2}$ is halved? – kile Nov 30 '23 at 15:42
  • @kile I mean that in a single ended amplifier, if you had a DC bias VDC and then a 10mV peak-to-peak Vin signal, the differential equivalent would be having VDC+Vin/2 on one side and VDC-Vin/2 on the other. It's the same amplitude of input signal, but in the single ended case it's all together, and in the differential case it's split in two. But since this is true with the output as well, you add back the two differential halves of the output to get the full signal. – mapplejacks Nov 30 '23 at 16:11
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The book specifically uses notation for $$(V_X-V_Y) |due\ to\ vin1$$ and $$(V_X-V_Y) |due\ to\ vin2$$ When it combines the two by superposition, it then specifically notes $$(V_X-V_Y)tot$$ So, I don't think it is literally adding $$(V_X-V_Y) + (V_X-V_Y)$$

instead half of the total $$(V_X-V_Y)$$ comes from the contribution due to vin1 and the other half comes from the contribution to due to vin2. It might have been clearer to say $$(V_X-V_Y)tot = (V_X-V_Y) |due\ to\ vin1 +(V_X-V_ Y) |due\ to\ vin2$$ where $$(V_X-V_Y) |due\ to\ vin1 = (V_X-V_Y) |due\ to\ vin2 = \frac{(V_X-V_Y)tot }{2}$$

pat
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1

For transistor-based differential amplifiers we have to discriminate between different operation modes: (1) Input voltages at one input only (single-in) or (2) with different polarities at both inputs at the same time (differential-in). More than that, the output voltage is available (3) at one collector (drain) or (4) as a difference between both output nodes (symmetrical-out).

For a single input (unsymmetrical case) and single output (at one drain node only) we can treat the whole circuit as a common drain-common gate combination. Because the input resistance of the common gate stage (1/gm) acts as a feedback resistor for the 1st stage, the gain G is (sign and common-mode influence not considered):

  • Single-in/single-out: G=gm*RD/(1+gm/gm)=gmRD/2
  • Therefore: Single-in/symmetrical-out: G=gmRD.
  • Differential-in (Vin1=-Vin2) and symmetrical-out: G=2gmRD
LvW
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