Insights of Random Mismatch Part-2

        Insights of Random Mismatch

"Random Mismatch" is a tricky but very important topic with respect to Analog Design. As a learner, (still I am a learner, jokes apart!) I never found the crisp one shot material related to "Random Mismatch" in the internet. Of course you will find some piecewise information (with some boring mathematics without any application) about this topic but you can't find any material where the real application of mismatch has been demonstrated with respect to analog design viewpoint. In reality, you have to meet a given accuracy spec (say 1% or 5%), and that's where random mismatch calculations come into picture to decide the W and L of the MOSFETs in your circuit. I will try to show some demonstrations in this blog. 

I have explained the basic concept of "Random Mismatch" in the previous blog. In this blog, I will be explaining how to quantify the mismatch in a current mirror and how to meet a certain accuracy spec in a current mirror by sizing the transistors properly with help of mismatch calculations.

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Offset Derivation of Current Mirror:

The following I-V equation for a MOSFET in saturation is used:

 
$I_D=\frac{\beta }{2}\left ( V_{GS}-V_t \right )^{2}$ where $\beta =\mu C_{ox}\frac{W}{L}$

and $g_m=\frac{\partial I_D}{\partial V_{GS}}=\beta \left ( V_{GS}-V_t \right )=\frac{2I_D}{\left ( V_{GS}-V_t \right )}$

$\Rightarrow \Delta I_D=\frac{\Delta \beta }{2}\left ( V_{GS}-V_t \right )^{2}-\Delta V_t\frac{\beta }{2}2\left ( V_{GS}-V_t \right )+\Delta V_{GS}\frac{\beta}{2}2\left (V_{GS}-V_t \right)$

$\Delta V_{GS}=0$ for a Current Mirror

$\Rightarrow \frac{\Delta I_D}{I_D}=\frac{\frac{\Delta \beta }{2}\left ( V_{GS}-V_t \right )^{2}}{\frac{\beta }{2}\left ( V_{GS}-V_t \right )^{2}}-\frac{\Delta V_t\frac{\beta }{2}2\left ( V_{GS}-V_t \right )}{\frac{\beta }{2}\left ( V_{GS}-V_t \right )^{2}}$

$\Rightarrow \frac{\Delta I_D}{I_D}=\frac{\Delta \beta }{\beta }-\frac{2\Delta V_t}{\left ( V_{GS}-V_t \right )}$

$\Rightarrow \frac{\Delta I_D}{I_D}=\frac{\Delta \beta }{\beta }-\frac{g_m}{I_D}\Delta V_t$

$\Rightarrow \frac{\sigma _{\Delta I_D }}{I_D}=\sqrt{\left ( \frac{\sigma _{\Delta \beta }}{\beta } \right )^2+\left ( \frac{g_m}{I_D}\sigma_{ \Delta V_t } \right )^2}$ ......(1)

Neglecting $\beta $ mismatch, (Did neglecting $\beta $ mismatch matter? We shall come back to this point in the end)

$\frac{\sigma _{\Delta I_D }}{I_D}=\frac{g_m}{I_D}\sigma_{\Delta V_t}$ ......(2)

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Let's jump into a practical problem to understand the significance of the above analysis. In industry, current mirror is probably the most used circuit in any analog block. So, it's very important to know how to design a current mirror very accurately. Now this example can be demonstrated in 2 ways.

1. Analysis: You will be given the sizes of the current mirror transistors and you have to find the accuracy of the current mirror.
2. Synthesis: You have to design the current mirror to meet a given accuracy spec. 

No. 2 is more common and practical scenario. So, I will go ahead with a synthesis problem instead of an analysis problem.

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Problem Statement:

Design a 1:50 PMOS current mirror for 1% accuracy (6 sigma) of $I_{d}$ 

given, 

$I_{ref}=1 \mu A$ 

$\mu _pC_{ox}=30\mu A/V^2$

$A_{Vt}=3mV\mu m$

$A_{\beta }=2\%\mu m$

$Minimum V_{ov}=250mV$

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Solution:

$I_D=\frac{\mu _pC_{ox}\frac{NW}{L} }{2}V_{ov}^{2}$

$\frac{NW}{L}=\frac{2I_D}{\mu _pC_{ox}V_{ov}^2}\Rightarrow \frac{W}{L}=1.07$ .....(3)

From the previous blog we know, the threshold mismatch $\Delta V_t$ between two MOSFETs of sizes $\left (W_1/L_1  \right )$ and $\left (W_2/L_2  \right )$ will be:

$\sigma _{\Delta V_t}=\frac{A_{Vt}}{\sqrt{2}}\sqrt{\frac{1}{W_1L_1}+\frac{1}{W_2L_2}}$

$\Rightarrow \sigma _{\Delta V_t}=\frac{A_{Vt}}{\sqrt{2}}\sqrt{\frac{1}{WL}+\frac{1}{NWL}}=\frac{A_{Vt}}{\sqrt{2WL}}\sqrt{1+\frac{1}{N}}$

From (2), $\frac{\sigma _{\Delta I_D }}{I_D}=\frac{g_m}{I_D}\sigma_{\Delta V_t}$

$\Rightarrow \frac{\sigma _{\Delta I_D }}{I_D}=\frac{\sqrt{2I_D\mu _pC_{ox}\frac{NW}{L}}}{I_D}\times \frac{A_{Vt}}{\sqrt{2WL}}\sqrt{1+\frac{1}{N}}$

$\Rightarrow \frac{\sigma_ { \Delta I_D }}{I_D}=\frac{\sqrt{2NI_{ref}\mu _pC_{ox}\frac{NW}{L}}}{NI_{ref}}\times \frac{A_{Vt}}{\sqrt{2WL}}\sqrt{1+\frac{1}{N}}$

$\Rightarrow \frac{\sigma_ {\Delta I_D}}{I_D}=\frac{\sqrt{\frac{\mu _pC_{ox}A_{Vt}^2}{I_{ref}}\left ( 1+\frac{1}{N} \right )}}{L}$

$\Rightarrow L=\frac{\sqrt{\frac{\mu _pC_{ox}A_{Vt}^2}{I_{ref}}\left ( 1+\frac{1}{N} \right )}}{\frac{\sigma_{ \Delta I_D}}{I_D}}$ ......(4)

Given 6-sigma accuracy is 1%$\Rightarrow 6\times {\frac{\sigma_ { \Delta I_D}}{I_D}}=1\% =0.01$

From (4), $L=9.96\mu m\approx 10\mu m$

Using (3) and (4), $W=1.07\times 10\mu m\approx 10.7\mu m$

$M1:10.7\mu m/10\mu m$

$M2:50*(10.7\mu m)/10\mu m$

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Now, let's do the same analysis but this time we shall also take $\beta $ mismatch into account.

From the previous blog we know, the $\beta $ mismatch between two MOSFETs of sizes $\left (W_1/L_1  \right )$ and $\left (W_2/L_2  \right )$ will be:

$\frac{\sigma _{\Delta \beta }}{\beta }=\frac{A_{\beta }}{\sqrt{2}}\sqrt{\frac{1}{W_1L_1}+\frac{1}{W_2L2}}$

$\Rightarrow \frac{\sigma _{\Delta \beta }}{\beta }=\frac{A_{\beta}}{\sqrt{2}}\sqrt{\frac{1}{WL}+\frac{1}{NWL}}=\frac{A_{\beta}}{\sqrt{2WL}}\sqrt{1+\frac{1}{N}}=\frac{A_{\beta}}{\sqrt{2\frac{W}{L}L^2}}\sqrt{1+\frac{1}{N}}$ .....(5)

Putting (2) & (5) into (1),

$\frac{\sigma_ { \Delta I_D }}{I_D}=\sqrt{\left \{ \frac{A_{\beta}}{\sqrt{2\frac{W}{L}L^2}}\sqrt{1+\frac{1}{N}} \right \}^2+\left \{ \frac{A_{Vt}\sqrt{\left ( 1+\frac{1}{N} \right )\frac{\mu _pC_{ox}}{I_{ref}}}}{L} \right \}^2}$

$\Rightarrow L=\frac{\sqrt{\left ( \frac{\mu _pC_{ox}A_{Vt}^2}{I_{ref}}+\frac{A_{\beta}^2}{2\frac{W}{L}} \right )\left ( 1+\frac{1}{N} \right )}}{\frac{\sigma_ { \Delta I_D }}{I_D}}$ .....(6)

Given 6-sigma accuracy is 1%$\Rightarrow 6\times {\frac{\sigma \left ( \Delta I_D \right )}{I_D}}=1\% =0.01$

Using (6), $L=12.96\mu m\approx 13\mu m$

Using (3) and (6), $W=1.07\times 13\mu m\approx 13.8\mu m$

$M1:13.8\mu m/13\mu m$

$M2:50*(13.8\mu m)/13\mu m$

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Conclusion:

So, there is a small change in size after taking $\beta $ mismatch into consideration but this complicates the equations. That's why, we generally neglect the $\beta $ mismatch with respect to $V_{th}$ mismatch.

So, as promised I have shown a frequently used practical example of random mismatch. Possibly you have never seen such detailed analysis regarding "Random Mismatch". Believe me, I have invested a lot of time to derive these equations because neither these are available in books nor in the internet. Also, after deriving, I have cross checked the results by simulations to ensure I am not passing any wrong information to the readers. So, if you appreciate this blog, in the comment section please let us know, does this content/blog add some value to you? If you have any suggestion/feedback for us to make this blog better, you are highly appreciated. In the next blog, I will come with the same analysis for a diff pair.

Declaimer: As these are my own derived equations, if anywhere I have done any mistake, please let me know. I am always ready to accept my mistake and learn through that mistake.