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" and Current Mirror Offset in the previous two blogs. In this blog, I will be explaining how to quantify the offset in a diff pair and how to meet a certain offset spec in a diff pair by sizing the transistors properly with help of mismatch calculations.
Offset Derivation of Diff Pair:
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 $\beta =\mu C_{ox}\frac{W}{L}$
$\Rightarrow \Delta I_D=\frac{\Delta \beta }{2}\left ( V_{GS}-V_t \right )^{2}-\Delta V_t\frac{\beta }{2}2\left ( V_{GS}-V_t \right )+\Delta V_{GS}\frac{\beta}{2}2\left (V_{GS}-V_t \right)$
Constant current so, $\Delta I_{D}=0$
$0=\frac{\frac{\Delta \beta }{2}\left ( V_{GS}-V_t \right )^{2}}{\frac{\beta }{2}2\left ( V_{GS}-V_t \right )}-\Delta V_t+\Delta V_{GS}$
$\Delta V_{GS}=-\frac{\Delta \beta }{\beta }\frac{\left ( V_{GS}-V_t \right )}{2}+\Delta V_t$
$\Delta V_{GS}=-\frac{\Delta \beta }{\beta }\frac{I_D}{g_m}+\Delta V_t$
$\Rightarrow \sigma _{\Delta V_{GS} }=\sqrt{\left ( \frac{\sigma _{\Delta \beta }}{\beta }\frac{I_D}{g_m} \right )^2+\left ( \sigma_{ \Delta V_t } \right )^2}$ ......(1)
Neglecting $\beta $ mismatch, (In the previous blog, I have proved that we can easily neglect $\beta$ mismatch with respect to $V_t$ mismatch while doing the analysis. Still, I will prove once again in Problem1)
$\sigma _{\Delta V_{GS} }=\sigma_{\Delta V_t}$ ......(2)
Let's jump into a practical problem to understand the significance of the above analysis. In industry, diff pair and current mirror are probably the most used circuits in any analog block. So, it's very important to know how to design a diff pair and current mirror very accurately to meet the specs. Now this example can be demonstrated in 2 ways.
- Analysis: You will be given the sizes of the current mirror transistors and you have to find the accuracy of the current mirror.
- Synthesis: You have to design the current mirror to meet a given accuracy spec.
So, I shall take 2 problems. Problem 1 will be an Analysis of a diff pair and Problem 2 will be a Synthesis/Design problem where I will merge the concepts of a diff pair and current mirror both!
Problem Statement 1:
Calculate Input Referred Offset $V_{io}$ (6-sigma) of the given NMOS diff pair
given,
- $I_{tail}=5 \mu A$
- $\mu _nC_{ox}=15\mu A/V^2$
- $A_{Vt}=15mV\mu m$
- $A_{\beta }=2\%\mu m$
- $\left (\frac{W}{L}\right )_{1,2}=\frac{40 \mu m}{1 \mu m}$
Solution:
$I_D=\frac{I_{tail}}{2}=2.5\mu A$
$g_{m1,2}=\sqrt{2I_D\mu_nC_{ox}\left (\frac{W}{L} \right )_{1,2}}=54.77\mu S$
From the previous blog we know, the threshold mismatch $\Delta V_t$ and $\beta$ mismatch between two MOSFETs of same sizes will be:
$\sigma _{\Delta V_t}=\frac{A_{Vt}}{\sqrt{WL}}=3.87mV$
$\frac{\sigma _{\Delta \beta }}{\beta}=\frac{A_{\beta}}{\sqrt{WL}}=3.16mV$
$\Rightarrow \frac{\sigma _{\Delta \beta }}{\beta}\frac{I_D}{g_m}=\frac{A_{\beta}}{\sqrt{WL}}\frac{I_D}{g_m}=0.14mV$
From (1), $\sigma _{\Delta V_{GS} }=\sqrt{\left ( \frac{\sigma _{\Delta \beta }}{\beta }\frac{I_D}{g_m} \right )^2+\left ( \sigma_{ \Delta V_t } \right )^2}=\sqrt{\left (0.14mV\right)^2+\left (3.87mV\right)^2}=3.87mV$
$\therefore V_{io}=6\sigma_{\Delta V_{GS}}=23.23mV$
It is clear from the above example that we can safely ignore the $\beta$ mismatch with respect to $V_t$ mismatch. In the next example, we shall proceed with only $V_t$ mismatch.
Problem Statement 2:
Design a diff pair with active current mirror load with input-referred offset (6-sigma) of 3mV
given,
- $I_{tail}=2 \mu A$
- $\mu _nC_{ox}=150\mu A/V^2$
- $\mu _pC_{ox}=40\mu A/V^2$
- $A_{Vtn}=8.5mV\mu m$
- $A_{Vtp}=4mV\mu m$
- $g_{mp}=40\mu S$
- $\left (V_{ov}\right )_n=250mV$
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Solution:
$I_D=\frac{I_{tail}}{2}=1\mu A$
$g_{mn}=\frac{2I_D}{V_{ovn}}=8\mu S$
Given, $V_{io}=6\sigma_{\Delta V_{GS}}=3mV \Rightarrow \sigma_{\Delta V_{GS}}=0.5mV \Rightarrow \sigma_{\Delta V_{GS}}^2=0.25\mu V^2$
Assuming half of this variance from diff pair and rest half is from current mirror.
$\sigma_{\Delta V_{GS}}^2 \left(diff pair\right)=0.125\mu V^2 \Rightarrow \sigma_{\Delta V_{GS}}\left(diff pair\right)=0.353 V$
$\sigma_{\Delta V_{GS}}^2\left(Current Mirror\right)=0.125\mu V^2 \Rightarrow \sigma_{\Delta V_{GS}}\left(Current Mirror\right)=0.353 V$
Diff Pair:
$\sigma_{\Delta V_{GS}}\left(Diff pair\right)=\sigma_{\Delta V_{tp}}=\frac{A_{Vtp}}{\sqrt{WL_p}}=0.353 V$
$\Rightarrow \left(WL\right)_p=128 {\left(\mu m\right)}^2$......(3)
$\left({\frac{W}{L}}\right)_p=\frac{{g_{mp}}^2}{2I_D \mu_PC_{ox}}$
$\Rightarrow \left({\frac{W}{L}}\right)_p=20$......(4)
Solving (3) and (4), $\left({\frac{W}{L}}\right)_p=\frac{50.5\mu m}{2.5\mu m}$
Current Mirror:
$\sigma_{\Delta V_{GS}}\left(Current Mirror\right)=\frac{\sigma_{\Delta I_D}}{g_{mp}}=\frac{\sigma_{\Delta I_D}}{I_D}\ast \frac{I_D}{g_{mp}}=\frac{g_{mn}}{I_D} \sigma_{\Delta V_{tn}} \ast \frac{I_D}{g_{mp}}$
$\sigma_{\Delta V_{GS}}\left(Current Mirror\right)=\frac{g_{mn}}{g_{mp}} \sigma_{\Delta V_{tn}}=\frac{g_{mn}}{g_{mp}} \frac{A_{Vtn}}{\sqrt{WL_n}}=0.353V $
$\Rightarrow \left(WL\right)_n=23.1 {\left(\mu m\right)}^2$......(5)
$\left({\frac{W}{L}}\right)_n=\frac{{g_{mn}}^2}{2I_D \mu_nC_{ox}}$
$\Rightarrow \left({\frac{W}{L}}\right)_n=0.213$......(6)
Solving (5) and (6), $\left({\frac{W}{L}}\right)_n=\frac{2.2\mu m}{10.4\mu m}$
$M1:50.5\mu m/2.5\mu m$
$M2:50.5\mu m/2.5\mu m$
$M3:2.2\mu m/10.4\mu m$
$M4:2.2\mu m/10.4\mu m$
Conclusion:
So, this is the way to design diff pair and current mirror with the help of "Random Mismatch". The take away is, we can easily neglect $\beta$ mismatch with respect to $V_t$ mismatch while doing the analysis.
So, as promised I have shown some frequently used practical examples 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.
