In the current-mirror circuits in Fig. 13.3.1, all transistors are operating at the same current and have the same $V_{O V}$ and the same $V_{A}$. Identify each current-mirror (i.e., give its name) and give its output resistance $R_{O}$ in terms of $I_{\mathrm{REF}}, V_{O V}$, and $V_{A}$. Also, give the minimum voltage $V_{O}$ each mirror requires to operate properly. Give $V_{O}$ in terms of $V_{t}$ and $V_{O V}$.

Question

Quizplus · Accepted Answer

The Answer is $(a) Simple or basic current mirror. \begin{aligned} R_{O} & =r_{O 2}=\frac{V_{A}}{I_{O}}=\frac{V_{A}}{I_{\mathrm{REF}}} \ V_{O \min } & =V_{O V} \end{aligned} (b) Cascode current mirror. R_{o}=\left(g_{m 4} r_{o 4} ight) r_{o 2} where \begin{aligned} g_{m 4} & =\frac{2 I_{D}}{V_{O V}}=\frac{2 I_{\mathrm{REF}}}{V_{O V}} \ r_{o 2} & =r_{o 4}=\frac{V_{A}}{I_{D}}=\frac{V_{A}}{I_{\mathrm{REF}}} \end{aligned} Thus, \begin{aligned} R_{o} & =\left(\frac{2 V_{A}}{V_{O V}} ight)\left(\frac{V_{A}}{I_{\mathrm{REF}}} ight) \ V_{O \min } & =2 V_{G S}-V_{t}=V_{t}+2 V_{O V} \end{aligned} (c) Modified Wilson current mirror. R_{O} is the same as that of the cascode mirror, R_{O}=\left(\frac{2 V_{A}}{V_{O V}} ight)\left(\frac{V_{A}}{I_{\mathrm{REF}}} ight) Also, V_{O \min } is the same as that of the cascode mirror, V_{O ext { min }}=V_{t}+2 V_{O V} (d) Wide-swing current mirror. R_{O} is the same as that of the cascode, R_{O}=\left(\frac{2 V_{A}}{V_{O V}} ight)\left(\frac{V_{A}}{I_{\mathrm{REF}}} ight) However, here V_{O \min } is \begin{aligned} V_{O \min } & =V_{G 4}-V_{t} \ & =V_{t}+2 V_{O V}-V_{t}=2 V_{O V} \end{aligned}$
$(a) Simple or basic current mirror. \begin{aligned} R_{O} & =r_{O 2}=\frac{V_{A}}{I_{O}}=\frac{V_{A}}{I_{\mathrm{REF}}} \ V_{O \min } & =V_{O V} \end{aligned} (b) Cascode current mirror. R_{o}=\left(g_{m 4} r_{o 4} ight) r_{o 2} where \begin{aligned} g_{m 4} & =\frac{2 I_{D}}{V_{O V}}=\frac{2 I_{\mathrm{REF}}}{V_{O V}} \ r_{o 2} & =r_{o 4}=\frac{V_{A}}{I_{D}}=\frac{V_{A}}{I_{\mathrm{REF}}} \end{aligned} Thus, \begin{aligned} R_{o} & =\left(\frac{2 V_{A}}{V_{O V}} ight)\left(\frac{V_{A}}{I_{\mathrm{REF}}} ight) \ V_{O \min } & =2 V_{G S}-V_{t}=V_{t}+2 V_{O V} \end{aligned} (c) Modified Wilson current mirror. R_{O} is the same as that of the cascode mirror, R_{O}=\left(\frac{2 V_{A}}{V_{O V}} ight)\left(\frac{V_{A}}{I_{\mathrm{REF}}} ight) Also, V_{O \min } is the same as that of the cascode mirror, V_{O ext { min }}=V_{t}+2 V_{O V} (d) Wide-swing current mirror. R_{O} is the same as that of the cascode, R_{O}=\left(\frac{2 V_{A}}{V_{O V}} ight)\left(\frac{V_{A}}{I_{\mathrm{REF}}} ight) However, here V_{O \min } is \begin{aligned} V_{O \min } & =V_{G 4}-V_{t} \ & =V_{t}+2 V_{O V}-V_{t}=2 V_{O V} \end{aligned}$

(a) Simple or basic current mirror.

\begin{aligned}R_{O} & =r_{O 2}=\frac{V_{A}}{I_{O}}=\frac{V_{A}}{I_{\mathrm{REF}}} \V_{O \min } & =V_{O V}\end{aligned}

(b) Cascode current mirror.

R_{o}=\left(g_{m 4} r_{o 4} ight) r_{o 2}

where

\begin{aligned}g_{m 4} & =\frac{2 I_{D}}{V_{O V}}=\frac{2 I_{\mathrm{REF}}}{V_{O V}} \r_{o 2} & =r_{o 4}=\frac{V_{A}}{I_{D}}=\frac{V_{A}}{I_{\mathrm{REF}}}\end{aligned}

Thus,

\begin{aligned}R_{o} & =\left(\frac{2 V_{A}}{V_{O V}} ight)\left(\frac{V_{A}}{I_{\mathrm{REF}}} ight) \V_{O \min } & =2 V_{G S}-V_{t}=V_{t}+2 V_{O V}\end{aligned}

(c) Modified Wilson current mirror.

R_{O}

is the same as that of the cascode mirror,

R_{O}=\left(\frac{2 V_{A}}{V_{O V}} ight)\left(\frac{V_{A}}{I_{\mathrm{REF}}} ight)

Also,

V_{O \min }

is the same as that of the cascode mirror,

V_{O ext { min }}=V_{t}+2 V_{O V}

(d) Wide-swing current mirror.

R_{O}

is the same as that of the cascode,

R_{O}=\left(\frac{2 V_{A}}{V_{O V}} ight)\left(\frac{V_{A}}{I_{\mathrm{REF}}} ight)

However, here

V_{O \min }

is

\begin{aligned}V_{O \min } & =V_{G 4}-V_{t} \& =V_{t}+2 V_{O V}-V_{t}=2 V_{O V}\end{aligned}

In the Current-Mirror Circuits in Fig VOVV_{O V}VOV​ And the Same VAV_{A}VA​

In the Current-Mirror Circuits in Fig $V_{O V}$ And the Same $V_{A}$