Question 9.11: Designing a Wilson current source (a) Design the Wilson curr...

I_{ C 2}=5 \mu A , \text { and } V_{ T }=26 mV.

(a) From Eq. (9.120),

I_{ O }=I_{ C 2}-I_{ R }\left[\frac{\left(2+\beta_{ F }\right) \beta_{ F }}{\beta_{ F }^{2}+2 \beta_{ F }+2}\right]-I_{ R }\left[1-\frac{2}{\beta_{ F }^{2}+2 \beta_{ F }+2}\right]                                  (9.120)

5 \mu A =I_{ R }\left[1-\frac{2}{\left(100^{2}+2 \times 100+2\right)}\right]

so I_{ R } \approx 5 \mu A . The reference current is

I_{ R }=\frac{V_{ CC }-V_{ BE 1}-V_{ BE 2}}{R_{1}}

That is,

 5 \mu A =\frac{30-0.7-0.7}{R_{1}}

which gives R_{1}=5.72 M \Omega.

(b) From Eq. (9.118),

I_{ C 3}=I_{ C 2} \frac{1+\beta_{ F }}{2+\beta_{ F }}                                 (9.118)

I_{ C 3}=I_{ C 1}=\frac{5 \mu A \times(1+100)}{2+100}=4.95 \mu A

Then  r_{ ol }=r_{ o 3}=\frac{V_{ A }}{I_{ C 1}}=\frac{150}{4.95 \mu}=30.3 M \Omega

r_{ o 2}=\frac{V_{ A }}{I_{ C 2}}=\frac{150}{5 \mu}=30 M \Omega

Since 1 / g_{ m 1}=1 / g_{ m 3}=V_{ T } / I_{ C 1}=26 mV / 4.95 \mu A =5.25 k \Omega,

g_{ m 1}=g_{ m 3}=190.4 \mu A / V

Since 1 / g_{ m 2}=26 mV / 5 \mu A =5.2 k \Omega,

g_{ m 2}=192.3 \mu A / V

Then    r_{\pi 1}=r_{\pi 3}=\frac{\beta_{ F }}{g_{ m 1}}=100 \times 5.25 k =525 k \Omega

r_{\pi 2}=100 \times 5.2 k =520 k \Omega

From Eq. (9.124),

R_{ e }^{\prime}=\frac{v_{1}}{i}=\frac{r_{\pi 2}+\left(r_{ ol } \| R_{1}\right)}{1+\left(r_{ ol } \| R_{1}\right) g_{ m 1}}                           (9.124)

\begin{aligned}R_{ e }^{\prime} &=\frac{520 k \Omega+(30.3 M \Omega \| 5.72 M \Omega)}{[1+(30.3 M \Omega \| 5.72 M \Omega) \times 190.4 \mu A / V ]} \\&=5.81 k \Omega\end{aligned}

From Eq. (9.123),

R_{3}=r_{\pi 1}\left\|r_{\pi 3}\right\| \frac{1}{g_{ m 3}} \| r_{ o 3} \approx \frac{1}{g_{ m 3}}                              (9.123)

R_{3}=525 k \|525 k \| 5.25 k \| 30 M \approx 5.15 k \Omega

From Eq. (9.126),

R_{ o } \approx r_{ o 2}\left[1+\frac{g_{ m 2} r_{\pi 2}}{R_{ e }^{\prime}}\left(R_{3} \| R_{ e }^{\prime}\right)\right]                                    (9.126)

R_{ o }=(30 M )\left[1+\frac{192.3 \mu \times 520 k }{5.81 k } \times 5.15 k \| 5.81 k \right]=1.44 G \Omega

and      V_{\text {Th }}=1.44 G \times 5 \mu=7.2 kV

NOTE: If we use Eq. (9.127),

\begin{aligned}R_{ o } & \approx r_{ o 2}\left(1+\frac{g_{ m 2} r_{\pi 2}}{2}\right) \\& \approx r_{ o 2}\left(1+\frac{\beta_{ F }}{2}\right)\end{aligned}                            (9.127)

R_{ o } \approx(30 M )\left(1+\frac{100}{2}\right)=1.53 G \Omega

(c) The Wilson current source for PSpice simulation is shown in Fig. 9.31. We will use parametric sweep for the model parameter βF of the transistors. The voltage source V_{ y } acts as an ammeter for the output current.

The results of simulation (.TF analysis) are as follows. (The hand calculations are shown in parentheses.)

For \beta_{ F }=100, the simulation gives

VOLTAGE SOURCE CURRENTS

NAME                     CURRENT
VCC                         -5.022E-06

**** BIPOLAR JUNCTION TRANSISTORS

**** SMALL-SIGNAL CHARACTERISTICS

For \beta_{ F }=400, the simulation gives

NOTE: R_{ o } changes from 1.602 E +09 \Omega\left(\text { for } \beta_{ F }=100\right) \text { to } 5.441 E +09 \Omega\left(\text { for } \beta_{ F }=400 \Omega\right) and depends on \beta_{ F } , as expected.