Insights of Random Mismatch
I have explained the basic concept of "Random Mismatch" , Current Mirror Offset and
Differential Pair Offset in the previous three blogs. In this blog, I will be explaining
how degeneration can help us to improve the accuracy of Current
Mirrors. I shall explain the concept through a problem statement this
time.
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
$g_m=\frac{\partial I_D}{\partial V_{GS}}=\beta \left ( V_{GS}-V_t
\right )=\frac{2I_D}{\left ( V_{GS}-V_t \right
)}=\frac{2I_D}{V_{ov}}$
Problem Statement:
1. Design a 1:1 NMOS current mirror (Fig.1) for 10% accuracy (6 sigma) of
$I_{d}$
given,
$I_{ref}=5 \mu A$
$\mu _nC_{ox}=30\mu A/V^2$
$A_{Vt}=3mV\mu m$
$A_{\beta }=2\%\mu m$
$A_{R }=2\%\mu m$
$Minimum V_{ov}=100mV$
Calculate $V_{out1,min}$
2. Improve upto 2% accuracy (6 sigma) of $I_{d}$
(i) by increasing the L
only (Fig. 2) and keep the same W as of Fig.1. Calculate
$V_{out2,min}$
(ii) by inserting
degeneration resistor only (Fig. 3) and keep the same W and L as
of Fig.1. Calculate $V_{out3,min}$
Solution:1
$I_D=\frac{\mu _nC_{ox}\frac{W}{L} }{2}V_{ov}^{2}$
$\frac{W}{L}=\frac{2I_D}{\mu _nC_{ox}V_{ov}^2}\Rightarrow
\frac{W}{L}=33.3$ .....(1)
From the previous blog we know, the current mismatch in a current mirror will be:
$\frac{\sigma _{\Delta I_D }}{I_D}=\frac{g_m}{I_D}\sigma_{\Delta
V_t}$
$\Rightarrow \frac{\sigma_ {\Delta
I_D}}{I_D}=\frac{\sqrt{\frac{2\mu
_nC_{ox}A_{Vt}^2}{I_{ref}}}}{L}$
$\Rightarrow L=\frac{\sqrt{\frac{2\mu
_nC_{ox}A_{Vt}^2}{I_{ref}}}}{\frac{\sigma_{ \Delta
I_D}}{I_D}}$ ......(2)
Given 6-sigma accuracy is 10%$\Rightarrow 6\times {\frac{\sigma_ { \Delta I_D}}{I_D}}=10\%
=0.1$
From (2), $L=0.62\mu m\approx 0.6\mu m$
Using (1) and (2), $W=33.3\times 0.62\mu m\approx 20.8\mu m$
$M1:20.8\mu m/0.6\mu m$
$M2:20.8\mu m/0.6\mu m$
$V_{out,min1}=V_{ov1}=100mV$
Solution:2.(i)
From the previous blog we know, the current mismatch in a current
mirror will be:
$\frac{\sigma _{\Delta I_D
}}{I_D}=\frac{g_m}{I_D}\sigma_{\Delta V_t}$
$\Rightarrow \frac{\sigma_ {\Delta
I_D}}{I_D}=\frac{\sqrt{\frac{2\mu
_nC_{ox}A_{Vt}^2}{I_{ref}}}}{mL}$
$\Rightarrow mL=\frac{\sqrt{\frac{2\mu
_nC_{ox}A_{Vt}^2}{I_{ref}}}}{\frac{\sigma_{
\Delta I_D}}{I_D}}$ ......(3)
Given 6-sigma accuracy is 10%$\Rightarrow 6\times {\frac{\sigma_ { \Delta
I_D}}{I_D}}=2\% =0.02$
From (3), $mL=3.12\mu m\approx 3.1\mu m$
$W\approx 20.8\mu m$
$M3:20.8\mu m/3.1\mu m$
$M4:20.8\mu m/3.1\mu m$
$V_{out,min2}=V_{ov2}=\sqrt
{\frac{2I_D}{\mu_nC_{ox}\frac{W}{mL}}}=224mV$
$V_{out,min2}=V_{ov2}=\sqrt
{\frac{2I_D}{\mu_nC_{ox}\frac{W}{mL}}}=\sqrt{m}V_{ov1}$
Solution:2.(ii)
$I_D=\frac{\beta }{2}\left ( V_{G}-I_DR-V_t \right )^{2}$
$\Rightarrow \Delta I_D=\frac{\Delta \beta }{2}\left ( V_{G}-I_DR-V_t
\right )^{2}-\Delta V_t\frac{\beta }{2}2\left ( V_{G}-I_DR-V_t \right
)$
$\hspace{2.5cm} -R\Delta I_{D}\frac{\beta}{2}2\left (V_{G}-I_DR-V_t
\right)-I_D\Delta R\frac{\beta}{2}2\left (V_{G}-I_DR-V_t \right)$
$\Rightarrow \frac{\Delta I_D}{I_D}=\frac{\frac{\Delta \beta }{2}\left (
V_{G}-I_DR-V_t \right )^{2}}{\frac{\beta }{2}\left ( V_{G}-I_DR-V_t \right
)^{2}}-\frac{\Delta V_t\frac{\beta }{2}2\left ( V_{G}-I_DR-V_t \right
)}{\frac{\beta }{2}\left ( V_{G}-I_DR-V_t \right )^{2}}-\frac{R\Delta
I_{D}\frac{\beta}{2}2\left (V_{G}-I_DR-V_t \right)}{I_D}$
$\hspace{2.5cm} -\frac{I_D\Delta R\frac{\beta}{2}2\left
(V_{G}-I_DR-V_t \right)}{\frac{\beta }{2}\left ( V_{G}-I_DR-V_t \right
)^{2}}$
$\Rightarrow \frac{\Delta I_D}{I_D}\left [1+\beta\left(
V_{GS}-V_t\right )R \right ]=\frac{\Delta \beta }{\beta }-\frac{2\Delta
V_t}{\left ( V_{GS}-V_t \right )}-\frac{2I_D\Delta R}{\left ( V_{GS}-V_t
\right )}$
$\Rightarrow \frac{\Delta I_D}{I_D} \left (1+g_mR \right )=\frac{\Delta
\beta }{\beta }-\frac{g_m}{I_D}\Delta V_t-g_mR\frac{\Delta R}{R}$
$\Rightarrow \frac{\sigma _{\Delta I_D }}{I_D}=\frac{1}{\left (1+g_mR
\right )}\sqrt{\left ( \frac{\sigma _{\Delta \beta }}{\beta } \right
)^2+\left ( \frac{g_m}{I_D}\sigma_{ \Delta V_t } \right )^2+\left (
g_mR\frac{\sigma _{\Delta R }}{R } \right )^2}$ ......(4)
Neglecting $\beta$ and $R$ mismatch, (I have already proved that we can neglect the $\beta$ mismatch in
blog2. Did neglecting $R$ mismatch matter? We shall come back to this point in
the end)
$\frac{\sigma _{\Delta I_D }}{I_D}=\frac{1}{\left(1+g_mR
\right)}\frac{g_m}{I_D}\sigma_{\Delta V_t}$ ......(5)
Given 6-sigma accuracy is 10%$\Rightarrow 6\times {\frac{\sigma_ { \Delta I_D}}{I_D}}=2\%
=0.02$
$\Rightarrow \frac{\sigma_ {\Delta
I_D}}{I_D}=\frac{1}{\left(1+g_mR\right)}\frac{\sqrt{\frac{2\mu
_nC_{ox}A_{Vt}^2}{I_{ref}}}}{L}$
$\Rightarrow 1+g_mR=\frac{\sqrt{\frac{2\mu
_nC_{ox}A_{Vt}^2}{I_{ref}}}}{\frac{\sigma_{ \Delta
I_D}}{I_D}} \frac{1}{L}$
Given $L=0.62\mu m$ and $W=20.8\mu m$,
$g_m=\frac{2I_D}{V_{ov}}=100 \mu S$
$1+g_mR=5$
$\Rightarrow g_mR=4$
$\Rightarrow R=\frac{4}{g_m}=40 k\Omega$
To generalize,
$1+g_mR=\frac{mL}{L}$
$\Rightarrow g_mR=m-1$
$\Rightarrow R=\frac{m-1}{g_m}$
From (4), $L=0.62\mu m\approx 0.6\mu m$
$W\approx 20.8\mu m$
$R=40k \Omega$
$M5:20.8\mu m/0.6\mu m$
$M6:20.8\mu m/0.6\mu m$
$R=40k\Omega$
$V_{out,min3}=I_DR+V_{ov3}=300mV$
To generalize,
$V_{out,min3}=I_DR+V_{ov3}=I_D \frac{\left( m-1
\right)}{g_m}+V_{ov1}=\left( m-1
\right)\frac{V_{ov1}}{2}+V_{ov1}=\frac{\left( m+1
\right)}{2}V_{ov1}$
Summary:
Let's summarize what we have learnt:
1. So, at first we have a simple current mirror whose accuracy
was 10%
2. Our objective was to improve its accuracy up to 2%,
means we want its accuracy 5X better. Now this can be done in
two ways:
(i) Fig. (2) We know that the accuracy of the current mirror is
inversely proportional with L (W has nothing to do with the
accuracy). So if want 5X better accuracy, we simply have to
increase the L by 5X and that's exactly what the math also
tells us. But if we keep the W same, that means we are
decreasing the (W/L) ratio by 5 times, which indeed will
increase the headroom requirement by $\sqrt{5}X$
(ii) Fig. (3) Also we saw, by degenerating the current mirror, we
can improve the accuracy by a factor of $\left(1+g_mR\right)$.
So, instead of increasing the L by 5X, we choose the value of R,
such that $1+g_mR=5$, keeping the W and L same. Now, in this
case, as (W/L) is not changed, the overdrive will not change but
one extra IR drop will be added which will increase the headroom
requirement. Also, due to the mismatch between the resistors,
some additional mismatch will be added which may degrade the
accuracy, but by choosing resistor with sufficient W and L, we
can minimize the effect of R mismatch.
3. Now as circuit designer, we have to understand the trade
offs. So basically, if your IP has very low VDD (less headroom),
probably you will prefer to increase L to improve accuracy. On
the other hand, if your IP has tight area constraints, you may
end up with degeneration to improve the accuracy of your current
mirror.
The below table will summarize the above statements.
|
|
$\frac{W}{L}, R=0$
Fig. (1) |
$\frac{W}{mL}, R=0$
Fig. (2) |
$\frac{W}{L}, R=\frac{m-1}{g_m}$
Fig. (3) |
|
Accuracy |
$X\%$ |
$\frac{X}{m}\%$ |
$\frac{X}{m}\%$ |
|
Vout,min |
$V_{ov}$ |
$\sqrt{m}{V_{ov}}$ |
$\frac{\left(m+1\right)}{2}V_{ov}$ |
|
Area |
$2*WL$ |
$2*mWL$ |
$2*WL+2*\left(WL\right)_R$ |
NOTE: $m>1$
Bonus Assignment:
Aren't you curious, how to quantify the effect of R mismatch?
I won't solve this, but I shall give you hints how to
calculate it.
Let's take the Fig. (3). For the ease of calculation, let's
take the area of resistor is same as of the mosfet i.e. 20.8um
x 3.1um.
$\frac{\sigma _{\Delta R }}{R
}=\frac{A_R}{\left(WL\right)_R}$
$\frac{\sigma _{\Delta I_D
}}{I_D}=\frac{1}{\left (1+g_mR \right )}\sqrt{\left (
\frac{\sigma _{\Delta \beta }}{\beta } \right )^2+\left (
\frac{g_m}{I_D}\sigma_{ \Delta V_t } \right )^2+\left (
g_mR\frac{\sigma _{\Delta R }}{R } \right )^2}$
Use these two equations, and calculate the extra error
introduced by the R mismatch (Ignore the $\beta$ mismatch). Do
let me know in the comment section. Do you think this error is
significant enough?
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