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I'm confused about one current-based explanation for why the active current mirror restores a differential pair's gain by letting each differential input reinforce the other.

The explanation (from Razavi) is like this:

The output voltage rises by means of two mechanisms: M2 draws less current from X to ground and M4 pushes a greater current from VDD to X.

My confusion is that while I can see that a drop in potential at F means M4 wants to source more current and the drop at M2's gate means that it wants to sink less current, don't these fight against each other (at least without a load attached to the output?)

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"Active load" is an abstract circuit concept but it can be explained in a simple intuitive way by more elementary electrical concepts as voltage divider, potentiometer, variable resistor, etc...

So, M2 and M4 can be considered as two variable "resistors" R2 and R4 in series forming a variable "voltage divider"... or a sensitive "potentiometer" - Fig. 1. It is controlled in a differential manner so that when R2 increases, R4 decreases and vice versa (the resistance crossfades). As a result, the total resistance R2 + R4 stays constant, the common current I = Vdd/(R2 + R4) does not change... but the output voltage vigorously changes; hence the high gain.

Potentiometer_analogy

Fig. 1. An old picture of mine about the same topic.

Like any voltage divider configuration, this stage has a voltage output that does not need a load current; on the contrary, it "likes" to work without load (open circuit). So any "current-based explanations" (including these in Razavi's book) are absurd when introducing the main idea of the standalone active load stage. Their place is later, when the next stage is connected, a load current is consumed and the active load stage does not work at "ideal" load conditions.

I have explained many times this extremely simple but clever circuit idea to my students in such a simple way based on their intuition and common sense... and they have always grasped it instantly. I introduce it in the very beginning of my course when talking about passive resistor circuits.

See also my more detailed answer and the discussion about the exotic current feedback amplifier (its output stage is based on the same idea).

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    \$\begingroup\$ Imho, this is a far better answer because it correctly simplifies what's actually happening. Transistors are voltage controlled resistors. Even if the current wasn't constant due to the mirroring, the same effect occurs. Alter even one resistor in the divider and you'll alter the output voltage; simple as that. \$\endgroup\$
    – horta
    Mar 2 at 16:36
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    \$\begingroup\$ Slight correction: In some applications, the transistor can be used as a (dynamic) resistor. \$\endgroup\$
    – LvW
    Mar 2 at 17:51
  • \$\begingroup\$ @horta, Exactly! Simply put, clear and unambiguous! What prevents people from expressing themselves in this way? The transistor is a variable resistor... so the network of a transistor + collector resistor is a variable voltage divider with one varying resistance, the other is static… in the simple active load stage, the other resistance is dynamic (self-varying) so the network is a dynamic voltage divider... in the more sophisticated (OP's) current-mirror active-load stage, both resistances are differentially varying and dynamic... so this network is a fully dynamic voltage divider... \$\endgroup\$ Mar 3 at 11:25
  • \$\begingroup\$ @LvW, I agree with your "correction". Really, when the input base-emitter voltage is constant, the transistor output collector-emitter part behaves as a "constant-current dynamic resistor". The simple active load is implemented in this way. In the OP's circuit, the active load is additionally controlled from the side of the input in an opposite direction. Thus we obtain a fully symmetric pair of two differentially controlled dynamic resistors. I rememver that, in the 70's, my teacher on Amplifier devices called this pair of two confronting transistors "controlled active load"... \$\endgroup\$ Mar 3 at 11:41
  • \$\begingroup\$ Let me say, that there is something that prevents me from seing the BJT simply as a resistor. I am afraid, such a view will put every newcomer on a wrong track. What happens when you change the collector resistor? Will the current change (as for each "normal" voltage divider?). No - of course not. Does such a view help to better understand the transistor principle? I have severe doubts. (Einstein: Explain everything as simple as possible - but not simpler!) \$\endgroup\$
    – LvW
    Mar 3 at 13:25
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Even with no external load, the output resistance of the differential pair is finite due to channel length modulation. The drain current in saturation depends on a drain-source voltage drop: \$I_D\$ ~ \$(V_{GS} - V_{th})^2(1 + \lambda V_{DS})\$.

UPDATE emphasizing the use of superposition principle

In the course notes (EE215A) of Samueli School of Engineering, UCLA, the lecture on Differential Amplifiers, Prof. Razavi derives the voltage gain of differential pair with active load as a product of the stage transconductance and the output resistance, \$A_V = -G_mR_{out}\$. The derivation proceeds in three steps. First, the transconductance is computed:

Vin0out

\$G_m = -I_{out}/(V_{in+}-V_{in-})\$. The voltages at differential inputs are \$±V_{in}\$, the output is grounded via a measuring voltage source (\$0V\$). Notice and remember the output current \$I_{out}\$.

Second, the output resistance is computed:

0inVout

\$R_{out} = V_{out}/(-I_{out})\$. The voltages at differential inputs are zero, the output is grounded via a measuring voltage source. The testing voltage \$V_{out}\$ at the output is selected in such a way as to provide an output current \$-I_{out}\$. The output current here is equal to the output current of the first (transconductance) step, but flows in the opposite direction. The small-signal model circuit is linear, and you can set the output current to any value by scaling the testing voltage \$V_{out}\$ of this circuit.

Finally, we apply the superposition principle to our linear small-signal model circuit and sum up the voltages and currents obtained in the transconductance and output-resistance measurements:

VinVout

As a result of this summation, the voltages at differential inputs are \$\{V_{in}, -V_{in}\}\$, the output voltage is \$V_{out}\$, and the output current is zero (\$I_{out}-I_{out}\$). The output current is zero, we can cut off the output path to ground. In the end we arrive at the solution for our small-signal model circuit where the input is \$\{V_{in}, -V_{in}\}\$ and the output is \$V_{out}\$ with a zero output current. The zero output current means that the circuit output is open, and we can calculate the voltage gain for the small signal model of an unloaded differential pair with active load: \$A_v = V_{out}/(V_{in+}-V_{in-})\$; \$V_{in+}-V_{in-}\$ is given; \$V_{out}\$ is calculated at the second step.

Other explanations of the diffpair voltage gain are possible, having their upsides and downsides. The explanation from Razavi's textbook and lectures is made in the mainstream of EE approaches to circuit analysis. It can be readily used for calculations by hand; implemented in software applications; it gives in one batch all three parameters of a two-port representation of differential pair -- transconductance, output resistance and voltage gain.

TL;DR)

At the moment I have no copy at hand of "Design of Analog CMOS Integrated Circuits", so I am looking at a copy of Prof. Razavi's lecture slides (pp. 11-12). The graph he draws for large-signal behavior indicates that he considers an "intrinsic" gain of the diffpair with active load: there is no external load in the circuit.

The small-signal behavior analysis on the same pages supports this observation: the small-signal model shows that output resistances of diffpair transistors are finite and there are no external load (no path to a "next stage"). The only shortcoming of the small-signal circuit diagram is how the ratio of voltages at the diffpair outputs is drawn: with a simple active load, the voltage at the drain node of M1/M2 is much smaller than the voltage at the M3 & M4 drains (output voltage). The circuit with a simple active load converts a differential input to a single-ended output.

The small-signal derivation of diffpair's transconductance, output resistance and voltage gain is straightforward and almost omitted in the lecture. In my answer, I use the MNA method to derive the formulas.

smallsignalmeasurement

To verify the symbolic calculation with the help of a SPICE simulation, I assign the values to the components in this circuit. Using the circuit linearity, the signal voltage is set to one volt for convenience of comparing symbolic and numerical calculation figures.

Write down KCL equations for calculating the voltages (w.r.t. the ground) at the circuit nodes: \$V_1\$ is a voltage at the M1/M3 drain node, \$V_P\$ is a voltage at the node P (for Pair, the connected sources of M1 & M2 diff pair transistors), \$V_{out}\$ is a voltage at the M2/M4 drain node. $$ g_{m3}V_1 + {\frac {V_1} {r_{O3}}} = g_{m1}(V_{in}-V_P) + {\frac {V_P - V_1} {r_{O1}}} \tag 1 $$ $$ g_{m3}V_1 + {\frac {V_1} {r_{O3}}} = -g_{m2}(-V_{in}-V_P) + {\frac {V_{out} - V_P} {r_{O2}}} \tag 2 $$ where the symbols \$g_{m1}=g_{m2}=g_m, r_{O1}=r_{O2}=r_O\$ are the identical transconductances and output resistances of M1, M2; \$r_d = {r_3||(1/g_{O3})}\$ is the resistance of the diode-connected M3. To measure a transconductance, the output must be (AC-)shorted, \$V_{out}=0\$. From the equation (2) $$ V_P = {\frac {-g_mV_{in}+V_1/r_d} {g_m - 1/r_O}} $$ Substitute \$V_P\$ into (1) $$ V_1 = {\frac {g_{m}r_O} {2r_O+r_d}}·(V_{in}-(-V_{in})) \\ I_{out} = V_1/r_d + g_{m4}V_1 $$ \$I_{out}\$ is the output current; it is a sum of the current flowing from the left branch of the circuit, equal to \$V_1/r_d\$, and the current from the current source \$g_{m4}\$, equal to \$g_{m4}V_1\$. There is no current through \$r_{O4}\$, because the voltage drop across \$r_{O4}\$ is zero when measuring transconductance (\$V_{out} = 0\$).

The transconductance is $$ G_m = {\frac{I_{out}} {V_{in}}} = 2g_{m}r_O{\frac {1+g_{m4}r_d} {2r_O+r_d}} \tag {Gm} $$ To calculate the output resistance, a testing voltage is applied at the output port. The small signal model is linear, and the output resistance is a ratio of an applied voltage to a current generated. The KCL equations with a zero input voltage and a testing voltage \$V_{out}\$ at the output are $$ g_{m3}V_1 + {\frac {V_1} {r_{O3}}} = g_{m1}(-V_P) + {\frac {V_P - V_1} {r_{O1}}} \tag 3 $$ $$ g_{m3}V_1 + {\frac {V_1} {r_{O3}}} = -g_{m2}(-V_P) + {\frac {V_{out} - V_P} {r_{O2}}} \tag 4 $$ Notice that when we calculate the output resistance, the voltage \$V_{out}\$ in the equation (4) is a given value that moves to the right-hand side in the process of solving the equations.

Add the equation (3) to the equation (4) to remove the terms containing \$V_P\$ $$ {\frac {V_{out} - V_1} {r_O}} = 2{\frac {V_1} {r_d}}\\ V_1 = {\frac {r_d} {2r_O + r_d}}V{out} $$

The current flowing from the left branch of the circuit into the sensing voltage source is \$I_{out}= V_1 / r_d\$. Adding to this current a current of the \$g_{m4}\$ VCCS (\$g_{m4}I_{out}r_d\$) and a current through \$r_{O4}\$ (\$V_{out}/r_{O4}\$), we arrive at $$ I_{out} = {\frac {(1+g_{m4}r_d)r_{O4} + 2r_O + r_d} {(2r_O + r_d)r_{O4}}}V_{out} \tag {Rout} $$ The output resistance is $$ R_{out} = {\frac {V_{out}} {I_{out}}} = {\frac {(2r_O + r_d)r_{O4}} {(1+g_{m4}r_d)r_{O4} + 2r_O + r_d}} $$ Finally, you can verify these tiresome symbolic calculations: run the simulation with the circuit shown above with the component values specified, the corresponding netlist is

* LTspiceXVII\Draft6.asc
rO3 N001 M3d 1Meg
B§gm3 M3d N001 I=V(M3d)/10K
Vinput Vin1 0 0
rO1 M3d P 1Meg
B§gm1 M3d P I=(V(Vin1)-V(P))/10K
rO2 Vout P 1Meg
B§gm2 Vout P I=(-V(Vin1)-V(P))/10K
rO4 0 Vout 1Meg
B§gm4 Vout 0 I=V(M3d)/10K
Vx Vout 0 1
Vmeas N001 0 0
.op
.backanno
.end

Set the values for Vinput, Vx in accordance with the text guidance, then substitute the component values into the formulas (Gm) and (Rout) and compare the results of simulation runs and manual calculations.

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  • \$\begingroup\$ Thanks for the response. I think my confusion is more that the explanation I posted claims that M4 wanting to source more current and M2 wanting to sink less current both contribute to the output node rising in potential. I would understand either one in isolation but I don't see how it works simultaneously since M4 wanting to source more current on its own should lead to a rise in output voltage and M3 sinking less should also cause the output voltage to rise, but it's a single branch, the current has to either rise or fall, not both. \$\endgroup\$
    – knzy
    Jan 14 at 5:10
  • \$\begingroup\$ Consider finite output resistance (MOSFET's channel length modulation, BJT's Early effect), and your confusion is dissolved. With infinite output resistance, the diff pair is a comparator, not an amplifier. Finite output resistance provides a linear (however narrow) range of disbalance between diff pair's input voltages. Of course, currents thru M4, M2 are identical, but the current mirror becomes less rigid \$\endgroup\$
    – V.V.T
    Jan 14 at 5:39
  • \$\begingroup\$ I think I'm still missing something. See this passage from Razavi's book for example: i.imgur.com/BDWcMZq.png This is a large signal analysis of common mode operation, and in the last sentence he says M4 current increases while M2 decreases which allows the output to rise, but in the large signal sense they're the exact same current, whatever the equivalent output impedance might be. \$\endgroup\$
    – knzy
    Jan 14 at 7:24
  • \$\begingroup\$ The explanation in Razavis book is based on currents - and therefore you need another stage which is connected at the node "Vout". This stage receives a current (as an input signal) which is the difference between the two currents. The difference should be zero without any signal input, in all other cases, this difference has a value proportional to the input signal \$\endgroup\$
    – LvW
    Jan 14 at 8:14
  • \$\begingroup\$ Or, with a nonzero parameter λ and consequently the finite transistor output resistance, (the M4 and M2 currents being undoubtedly identical), the book's explanation is still valid (even without a load), if we read this excerpt in OP's interpretation: "M4 wants to source more current and the drop at M2's gate means that it wants to sink less current". Transistors want, but cannot do it, and the only effect is the asymmetry of current mirror branches that leads to finite gain. The OP is close to correct understanding of the two mechanisms, but is reluctant to fully accept it. \$\endgroup\$
    – V.V.T
    Jan 14 at 9:13

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