Figure 9.1.1
For the differential cascode amplifier in Fig. 9.1.1, all the transistors have the same threshold voltage $\left|V_{t}\right|=0.5 \mathrm{~V}$ and are operating at the same overdrive voltage. For the purpose of calculating dc quantities, neglect the Early effect. The current source $I$ requires a minimum dc voltage $V_{C S}=0.25 \mathrm{~V}$ for its proper operation.
(a) With $v_{G 1}=v_{G 2}=0 \mathrm{~V}$, find:
i. The bias current in each of the eight transistors.
ii. $\left|V_{O V}\right|$ for each of the eight transistors. iii. $g_{m}$ for each transistor.
iv. $r_{O}$ and $A_{0}$ for each transistor assuming $\left|V_{A}\right|=$ $5 \mathrm{~V}$.
v. The dc voltage at the drains of $Q_{1}$ and $Q_{2}$.
(b) Find the input common-mode range.
(c) With $v_{G 1}=v_{i d} / 2$ and $v_{G 2}=-v_{i d} / 2$, where $v_{i d} \ll 2 V_{O V}$, give the differential half-circuit and use it to determine the differential gain $v_{o d} / v_{i d}$.

Question

Quizplus · Accepted Answer

The Answer is $Figure 9.1.1 (a) (i) I_{D}=\frac{I}{2}=25 \mu \mathrm{A} (ii) Since for Q_{7} and Q_{8} , V_{S G}=2.5-1.75=0.75 \mathrm{~V} and since all transistors have the same \left|V_{t} ight| and \left|V_{O V} ight| , then \left|V_{O V} ight|=0.75-0.5=0.25 \mathrm{~V} (iii) g_{m}=\frac{2 I_{D}}{\left|V_{O V} ight|}=\frac{2 imes 0.025}{0.25}=0.2 \mathrm{~mA} / \mathrm{V} (iv) \begin{aligned} r_{o} & =\frac{\left|V_{A} ight|}{I_{D}}=\frac{5}{0.025}=200 \mathrm{k} \Omega \ A_{0} & =g_{m} r_{o}=0.2 imes 200=40 \end{aligned} (v) V_{D 1}=V_{D 2}=0.75-V_{G S 3,4}=0 \mathrm{~V} (b) \begin{aligned} v_{I C M \max } & =V_{D 1}+V_{t}=0+0.5=0.5 \mathrm{~V} \ v_{I C M \min } & =-V_{S S}+V_{C S}+V_{G S 1,2} \ & =-2.5+0.25+0.75=-1.5 \mathrm{~V} \end{aligned} Thus, (c) Figure 9.1.2 From the differential half-circuit in Fig. 9.1.2, we find \begin{aligned} \frac{v_{o d}}{v_{i d}} & =g_{m 1} R_{O} \ & =g_{m} imes \frac{1}{2}\left(g_{m} r_{o} ight) r_{o} \ & =\frac{1}{2}\left(g_{m} r_{o} ight)^{2} \ & =\frac{1}{2} A_{0}^{2}=\frac{1}{2} imes 40^{2}=800 \mathrm{~V} / \mathrm{V} \end{aligned}$

Figure 9.1.1
(a)
(i)

I_{D}=\frac{I}{2}=25 \mu \mathrm{A}

(ii) Since for

Q_{7}

and

Q_{8}

,

V_{S G}=2.5-1.75=0.75 \mathrm{~V}

and since all transistors have the same

\left|V_{t} ight|

and

\left|V_{O V} ight|

, then

\left|V_{O V} ight|=0.75-0.5=0.25 \mathrm{~V}

(iii)

g_{m}=\frac{2 I_{D}}{\left|V_{O V} ight|}=\frac{2 imes 0.025}{0.25}=0.2 \mathrm{~mA} / \mathrm{V}

(iv)

\begin{aligned}r_{o} & =\frac{\left|V_{A} ight|}{I_{D}}=\frac{5}{0.025}=200 \mathrm{k} \Omega \A_{0} & =g_{m} r_{o}=0.2 imes 200=40\end{aligned}

(v)

V_{D 1}=V_{D 2}=0.75-V_{G S 3,4}=0 \mathrm{~V}

(b)

\begin{aligned}v_{I C M \max } & =V_{D 1}+V_{t}=0+0.5=0.5 \mathrm{~V} \v_{I C M \min } & =-V_{S S}+V_{C S}+V_{G S 1,2} \& =-2.5+0.25+0.75=-1.5 \mathrm{~V}\end{aligned}

Thus,
(c)
$Figure 9.1.1 (a) (i) I_{D}=\frac{I}{2}=25 \mu \mathrm{A} (ii) Since for Q_{7} and Q_{8} , V_{S G}=2.5-1.75=0.75 \mathrm{~V} and since all transistors have the same \left|V_{t} ight| and \left|V_{O V} ight| , then \left|V_{O V} ight|=0.75-0.5=0.25 \mathrm{~V} (iii) g_{m}=\frac{2 I_{D}}{\left|V_{O V} ight|}=\frac{2 imes 0.025}{0.25}=0.2 \mathrm{~mA} / \mathrm{V} (iv) \begin{aligned} r_{o} & =\frac{\left|V_{A} ight|}{I_{D}}=\frac{5}{0.025}=200 \mathrm{k} \Omega \ A_{0} & =g_{m} r_{o}=0.2 imes 200=40 \end{aligned} (v) V_{D 1}=V_{D 2}=0.75-V_{G S 3,4}=0 \mathrm{~V} (b) \begin{aligned} v_{I C M \max } & =V_{D 1}+V_{t}=0+0.5=0.5 \mathrm{~V} \ v_{I C M \min } & =-V_{S S}+V_{C S}+V_{G S 1,2} \ & =-2.5+0.25+0.75=-1.5 \mathrm{~V} \end{aligned} Thus, (c) Figure 9.1.2 From the differential half-circuit in Fig. 9.1.2, we find \begin{aligned} \frac{v_{o d}}{v_{i d}} & =g_{m 1} R_{O} \ & =g_{m} imes \frac{1}{2}\left(g_{m} r_{o} ight) r_{o} \ & =\frac{1}{2}\left(g_{m} r_{o} ight)^{2} \ & =\frac{1}{2} A_{0}^{2}=\frac{1}{2} imes 40^{2}=800 \mathrm{~V} / \mathrm{V} \end{aligned}$

Figure 9.1.2
From the differential half-circuit in Fig. 9.1.2, we find

\begin{aligned}\frac{v_{o d}}{v_{i d}} & =g_{m 1} R_{O} \& =g_{m} imes \frac{1}{2}\left(g_{m} r_{o} ight) r_{o} \& =\frac{1}{2}\left(g_{m} r_{o} ight)^{2} \& =\frac{1}{2} A_{0}^{2}=\frac{1}{2} imes 40^{2}=800 \mathrm{~V} / \mathrm{V}\end{aligned}

Figure 911 for the Differential Cascode Amplifier in Fig ∣Vt∣=0.5 V\left|V_{t}\right|=0.5 \mathrm{~V}∣Vt​∣=0.5 V

Figure 911
for the Differential Cascode Amplifier in Fig $\left|V_{t}\right|=0.5 \mathrm{~V}$