Designing a Widlar current source
(a) Design the Widlar current source in Fig. 9.28(a) to give I_{ O }=5 \mu A \text { and } I_{ R }=1 mA .. The parameters are V_{ CC }=30 V , V_{ BE 1}=0.7 V , V_{ T }=26 mV , V_{ A }=150 V , \text { and } \beta_{ F }=100 .
(b) Calculate the output resistance R_{ o } and Thevenin’s voltage V_{\text {Th }} \text {. Assume } V_{ T }=26 mV \text {. }
I_{ C 2}=5 \mu A , \text { and } V_{ T }=26 mV .
(a) From Eq. (9.99),
I_{ R }=\frac{V_{ CC }-V_{ BE 1}}{R_{1}} (9.99)
1 mA =\frac{30-0.7}{R_{1}}
so R_{1}=29.3 k \Omega .From Eq. (9.100),
\begin{aligned}I_{ R } &=I_{ C 1}+I_{ B 1}+I_{ B 2} \\&=I_{ C 1}\left(1+\frac{1}{\beta_{ F }}\right)+\frac{I_{ C 2}}{\beta_{ F }}\end{aligned} (9.100)
1 mA =I_{ Cl }\left(1+\frac{1}{100}\right)+\frac{5 \mu A }{100}
which gives I_{ Cl } \approx 990 \mu A . From Eq. (9.98),
V_{ T } \ln \left(\frac{I_{ C 1}}{I_{ C 2}}\right)=I_{ C 2} R_{2} (9.98)
26 mV \times \ln \left(\frac{990 \mu A }{5 \mu A }\right)=5 \mu A \times R_{2}
so R_{2}=27.5 k \Omega.
(b) r_{ o 2}=V_{ A } / I_{ C 2}=150 /(5 \mu A )=30 M \Omega . From Eq. (9.110),
r_{\pi 2}=\frac{\beta_{ F }}{g_{ m 2}}=\frac{\beta_{ F }}{I_{ C 2} / V_{ T }}=V_{ T } \frac{\beta_{ F }}{I_{ C 2}}=\beta_{ F } \frac{V_{ T }}{I_{ C 1}} \times \frac{I_{ C 1}}{I_{ C 2}}=\frac{\beta_{ F }}{g_{ m 1}} \times \frac{I_{ C 1}}{I_{ C 2}} (9.110)
r_{\pi 2}=\frac{V_{ T } \beta_{ F }}{I_{ C 2}}=26 m \times \frac{100}{5 \mu}=520 k \Omega
Also from Eq. (9.109),
\frac{1}{g_{ m 1}}=\frac{1}{I_{ C 1} / V_{ T }}=\frac{V_{ T }}{I_{ C 1}} (9.109)
g_{ m 2}=\frac{I_{ C 2}}{V_{ T }}=\frac{5 \mu A }{26 mV }=192.3 \mu A / V
From Eq. (9.112),
R_{ o } \approx r_{ o 2}\left[1+g_{ m 2}\left(R_{2} \| r_{\pi 2}\right)\right] (9.112)
R_{ o } \approx 30 M \Omega \times[1+192.3 \mu A / V \times(27.5 k \Omega \| 520 k \Omega)]=180.68 M \Omega
Using the approximation in Eq. (9.115), we have
\begin{aligned}R_{ o } & \approx r_{ o 2}\left(1+g_{ m 2} R_{2}\right) \\& \approx r_{ o 2}\left(1+\frac{I_{ C 2} R_{2}}{V_{ T }}\right)\end{aligned} (9.115)
R_{ o } \approx 30 M \Omega \times\left(1+\frac{5 \mu A \times 27.5 k \Omega}{26 mV }\right)=188.66 M \Omega
and V_{ Th }=R_{ o } I_{ C 2}=188.65 M \Omega \times 5 \mu A =943.3 V
NOTE: The Widlar current source gives a low output current at high output resistance, and Thevenin’s equivalent voltage is very high.