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How can Gilbert Cell circuit provide linear gain variation even though ID-gm relationship is nonlinear?

As I undertstand it, the vcontrol (the voltage at the gate of M5 and M6) changes the biasing current of M5 and M6. This current are then copied by the current mirrors to M1-M4. The ID of M1-M4 define their gm which defines the gain.

But we know that ID-gm relationship is nonlinear and ID M1-M4 change pretty wide. Still I got not exactly but quite linear gain vs vcontrol.

Appreciate any hints or if you could point me to the right direction.

The x axis of the plot below is the differential control voltage that is used to vary the gain.

enter image description here enter image description here

Codelearner777
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    If you limit the input voltage and use it only in the linear region, the transfer is approximately linear. Devices that are always linear do not exist, even a resistor burns up above a certain power, so it is assumed that you're using the device in its linear region. – Bimpelrekkie Feb 18 '21 at 15:25
  • Hi, thanks for your comment. I understand that w.r.t. the input signal, i.e., the information carrying signal, the Vout vs Vin is linear (whithin specific range). I understand that one. But I do not understand why Vout vs Vcontrol is also linear even though I vary Vcontrol quite large. I might missunderstood something here. Appreciate if you could explain more. – Codelearner777 Feb 18 '21 at 15:32
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    For the Vcontrol differential pair (M5, M6), the same applies. How large we can make Vcontrol and still be in the linear part of the transfer depends on how M5 and M6 are sized. If you give M5 and M6 a **small** W/L then the linear range of Vcontrol will be much larger compared to the situation where M5 and M6 have a **large** W/L. The relations \$I_{D,M5}(V_{control})\$ and \$I_{D,M6}(V_{control})\$ will still be a tanh(x) type curve, similar to your plot. – Bimpelrekkie Feb 18 '21 at 16:00
  • If you plot the derivative you'll see it's not flat (but close). – a concerned citizen Feb 19 '21 at 10:17

2 Answers2

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What do you mean you "know that ID-gm relationship is nonlinear"?

In a MOSFET model, typically used for back-of-the-envelop calculations, $$ I_D = \frac {\mu _nC_{ox}}{2}{\frac {W}{L}}\left(V_{GS}-V_{th}\right)^{2}\left(1 + \lambda (V_{DS}-V_{DSsat})\right) $$ , and the transconductance is \$g_m = {{\partial I_D}/{\partial V_{GS}}} = {{2I_D}/(V_{GS}-V_{th})}\$. In this model and while in saturation, \$I_D\$ is a linear function of \$g_m\$, \$I_D = {\frac 1 2} {g_m \left(V_{GS}-V_{th}\right)}\$. Do you discuss a flawed design in which the diffpair and current mirror transistors work out of saturation most of the time, or are you talking about some advanced transistor models?

Nonlinearity arises because \$g_m\$ itself is a function of overdrive voltage (nonlinearity is required in a circuit which is used to multiply signals) $$ g_m = \mu _nC_{ox}{\frac {W}{L}}\left(V_{GS}-V_{th}\right)\left(1 + \lambda (V_{DS} -V_{DSsat})\right) $$. The MOSFET is a square law device, the current is proportional to a squared overdrive voltage. Let \$I_0\$ be the current through each branch of the balanced diffpair (\$V_{in}=0\$). With non-zero \$V_{in}\$, the difference of differential pair's drain currents \$i_{out}\$ can be derived from the MOSFET I-V characteristic, you can find it in textbooks or course notes or better do it yourself. To cut the long story short, I write down the formula, neglecting the channel length modulation effect (\$\lambda = 0\$): $$ i_{out} = \sqrt{2\beta I_0}V_{in}{\sqrt{1 - {\frac {\beta V_{in}^2} {8 I_0}}}} \tag{diffpair I-V} $$ where \$\beta = {\frac {\mu _nC_{ox}}{2}}{\frac {W}{L}}\$. Notice that in your circuit \$i_{out}\$ is translated to an \$V_{outv}\$ output voltage by means of the \$R_a\$ loads. Notice also that the formula is applicable (and required) only when transistors are in saturation, where the radicand is positive; the extension of the formula into all possible \$V_{in}\$ values embraces triod and cutoff operation modes and the extended \$i_{out}\$ is a monotonic function of \$V_{in}\$.

It differs from the bjt diffpair's \$\displaystyle {tanh}\$ curve, but both curves (for MOSFET and for bjt) are nonlinear at large \$V_{in}\$. For MOSFET, the criterion of linearity is \${\beta V_{in}^2} \ll {8 I_0}\$. You can also express \$I_0\$ via the device voltages according to a formula for saturation current \$I_D\$.

I write down the diffpair I-V equation once gain, this time with the voltages and currents at your M5, M6 diffpair $$ i_{SS} = \sqrt{\beta I_{SS}}V_{ctrl}{\sqrt{1 - {\frac {\beta V_{ctrl}^2} {4 I_{SS}}}}} \tag{gain control diffpair I-V} $$ \$I_{SS}\$ is a tail current through M5,M6 transistors, it corresponds to \$2I_0\$ variable of the equation for M1,M2 or M3,M4 pairs; \$i_{SS}\$ is the difference of M5 and M6 drain currents (corresponds to \$i_{out}\$ of the \$\text {diffpair I-V}\$).

You do not specify the parameters of your circuit, but from eqs. \$\text {gain control diffpair I-V}\$ and \$\text {diffpair I-V}\$ one can see that for any diff pair the \$\text {I-V}\$ linearity is controlled by the tail current. For a M5, M6 diffpair this tail current, \$I_{SS}\$, is set by a \$V_b\$ bias voltage. So, for your Gilbert cell with a variable gain network, you have two control inputs: \$V_{ctrl}\$ sets the gain ratio for M1,M2 vs. M3,M4 branches, and \$V_b\$ controls a linearity of \$V_{ctrl}\$ input. And, to be sure, there are also diff signal inputs \$V_{in+}, V_{in-}\$ in your circuit. In the Gilbert cell-based RF mixer, the \$V_{ctrl}\$ input is used on a par with the \$V_{in}\$ input as inputs for signals to be multiplied.

Summing up, the formula \$\text {diffpair I-V}\$ gives you a quantitative tool that you can use to compute and adjust the linearity of diffpair's I-V characteristics.

V.V.T
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Similar to a BJT Gilbert cell, you have a differential control voltage created to bias the current and thus gain of two differential amplifiers sharing the same differential input.

This is balanced by cross-coupled drain outputs such that the net output signal was 0 when both output diff. amps. had identical common source currents. The low gain amplifiers have a higher output impedance to achieve any voltage gain , so it was very limited , while operating in the quasi-linear region.

Yet this configuration being complementary with shared drain resistors could cancel out each other’s non-linearity as one was increasing while the other was decreasing.

This makes a very effective 4 quadrant linear multiplier in either direction from zero gain.

Tony Stewart EE75
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