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3. {25 points) Figure 1 given below shows a common source amplifier that is biased using a current mirror. Assume that $R_B$ and $C_B$ have infinite values. $L = 120nm$ for all transistors, $W_1 = 3\mu m$, $V_{tn} = 0.5V$, $V_{DD} = 1.5V$ $\mu_n C_{ox} = 200\mu A/V^2$, $I_{REF} = 100\mu A$, $R_L = 1k\Omega$, $\gamma = 0$ $V_{DD}$ $I_{REF}$ $V_{DD}$ $R_L$ $I_{REF}$ $V_B \circ$ $M_3$ $V_o$ $R_B$ $M_1$ $M_2$ $C_B$ $V_i$ $R_B$ $M_1$ $M_2$ $C_B$ $V_i$ Figure 1 Figure 2 a) In Figure 1, assuming $\lambda = 0$, find $W_2$ so that the amplifier has a gain of $A_v =$ $v_o/v_i = -5$. What is $I_{D2}$ in this case? b) In Figure 1, keep $W_2$ same as in part (a), but now let $\lambda = 0.2V^{-1}$. What is the new drain current $I_{D2}$? What is the new $A_v$? Explain the source of mismatch with your results from part (a). Hint: You need to re-calculate $V_{GS1}$. c) In order to more accurately control $I_{D2}$, a cascode device is added, as shown in Figure 2. Find the DC bias voltage $V_B$ that will make $I_{D2}$ match what was found in part (a) exactly. Assume $(W/L)_3 = (W/L)_2$ and $\lambda = 0.2V^{-1}$ still.

          3. {25 points) Figure 1 given below shows a common source amplifier that is
biased using a current mirror. Assume that $R_B$ and $C_B$ have infinite values.
$L = 120nm$ for all transistors, $W_1 = 3\mu m$, $V_{tn} = 0.5V$, $V_{DD} = 1.5V$
$\mu_n C_{ox} = 200\mu A/V^2$, $I_{REF} = 100\mu A$, $R_L = 1k\Omega$, $\gamma = 0$
$V_{DD}$
$I_{REF}$
$V_{DD}$
$R_L$
$I_{REF}$
$V_B \circ$
$M_3$
$V_o$
$R_B$
$M_1$
$M_2$
$C_B$
$V_i$
$R_B$
$M_1$
$M_2$
$C_B$
$V_i$
Figure 1
Figure 2
a) In Figure 1, assuming $\lambda = 0$, find $W_2$ so that the amplifier has a gain of $A_v =$
$v_o/v_i = -5$. What is $I_{D2}$ in this case?
b) In Figure 1, keep $W_2$ same as in part (a), but now let $\lambda = 0.2V^{-1}$. What is the
new drain current $I_{D2}$? What is the new $A_v$? Explain the source of mismatch with
your results from part (a). Hint: You need to re-calculate $V_{GS1}$.
c) In order to more accurately control $I_{D2}$, a cascode device is added, as shown in
Figure 2. Find the DC bias voltage $V_B$ that will make $I_{D2}$ match what was found in
part (a) exactly. Assume $(W/L)_3 = (W/L)_2$ and $\lambda = 0.2V^{-1}$ still.

3. 25 points) Figure 1 given below shows a common source amplifier that is
biased using a current mirror. Assume that RB and CB have infinite values.
L = 120nm for all transistors, W1 = 3μ m, Vtn = 0.5V, VDD = 1.5V
Cox = 200μ A/V^2, IREF = 100μ A, RL = 1kΩ, γ = 0
VDD
IREF
VDD
RL
IREF
VB ∘
M3
Vo
RB
M1
M2
CB
Vi
RB
M1
M2
CB
Vi
Figure 1
Figure 2
a) In Figure 1, assuming λ = 0, find W2 so that the amplifier has a gain of Av =
vo/vi = -5. What is ID2 in this case?
b) In Figure 1, keep W2 same as in part (a), but now let λ = 0.2V^-1. What is the
new drain current ID2? What is the new Av? Explain the source of mismatch with
your results from part (a). Hint: You need to re-calculate VGS1.
c) In order to more accurately control ID2, a cascode device is added, as shown in
Figure 2. Find the DC bias voltage VB that will make ID2 match what was found in
part (a) exactly. Assume (W/L)3 = (W/L)2 and λ = 0.2V^-1 still.

Added by Todd M.

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