JNW_SV_SKY130A

JNW_SV

A translinear gain cellcircuit is a wonderful thing. It’s not that common these days, so allow me to do an explantion.

You can find details in Analog Integrated Circuit Design

Consider the circuit below. The important transitors are XM1, XM2, XM3 and XM4. Call the bias current in XM1 and XM2 $I_1$ and $I_2$ in XM3 and XM4.

The transistors are all biased in weak inversion, so the VGS of XM1 and XM2 will be

\[I_{D1} = I_0 e^{(V_{GS1}-V_TH)/nU_T} = I_0 e^{V_{TH}/nU_T}e^{V_{GS1}/nU_T} = A e^{V_{GS1}/B}\]

and

\[V_{GS1} = B \ln (I_{D1}/A)\]

Assume that the gate is at $ 0 V$ it’s not really at $0 V$, can see from the schematic, but it makes the analysis easier.

\[V_{S1} = -B \ln (I_{D1}/A)\]

where A and B are constants.

Consider a small current difference into IIP and IIN where $$IIP = i_1$ and $IIN=-i_1$, since the gate is shared, the differential voltage on source will be

\[\Delta V = V_{S2} - V_{S1} = -B \ln\left(\frac{I_1-i_1}{A}\right) + B \ln \left(\frac{I_1+i_1}{A}\right)\]

which can be simplified to

\[\Delta V = B \ln \left( \frac{I_1 + i_1 }{ I_1 - i_1 } \right)\]

Now consider the differential pair XM3 and XM4. The voltgage $V-$ must be given by

\[V+ - V_{GS3} + V_{GS4} = V-\]

or

\[\Delta V = B \ln \frac{I_3}{I_4}\]

or

\[\frac{I_3}{I_4} = e^{\Delta V/B}\]

It must be the case that

\[2I_2 = I_3 + I_4\]

As the current in the tail current source is the sum of the differential pair currents, as such, we can compute

\[2I_2 = I_4 e^{\Delta V/B} + I_4\]

inserting for $\Delta V$ we get

\[2 I_2 = I_4 \frac{I_1 + i_1 }{ I_1 - i_1 } + I_4\]

or

\[I_4 = \frac{2 I_2}{\frac{I_1 + i_1 }{ I_1 - i_1 } + 1}\]

multiply over and under with $I_1 - i_1 $ and we get

\[I_4 = \frac{2 I_2(I_1 - i_1)}{I_1 - i_1 + I_1 + i_1} = I_2 - \frac{I_2}{I_1} i_1\]

Computing for $I_3$ and taking $i_o = I_4 - I_3$ one gets (provided tounge straight and all those $-$ checks)

\[i_o = \frac{I_2}{I_1} i_1\]

The input current is a linear scaling with the current difference between $I_2$ and $I_1$, which I find a remarkable result.

I remember learning about translinear circuits 25 years ago, and I found it beautiful that one could use the exponential relationship of bipolars, and mosfets in subthreshold to create linear gain.

In the circuit below the $I_1$ is a mirrored version of the bias current, while $I_2$ comes from a current DAC.

JNWSW_CM

Current DAC. There is always an certain current out from the DAC to keep the gain cell alive.