Designing a simple basic current source
(a) Design the basic current source in Fig. 9.25(a) to give an output current of I_{ O }=5 \mu A . The transistor parameters are \beta_{ F }=100, V_{ CC }=30 V , V_{ BE 1}=V_{ BE 2}=V_{ CE 1}=0.7 V , \text { and } V_{ A }=150 V .
(b) Calculate the output resistance R_{ o }, Thevenin’s equivalent voltage V_{ Th } , and the collector current ratio if V_{ CE 2}=20 V .
V_{ BE 1}=V_{ BE 2}=V_{ CE 1}=0.7 V , \text { and } V_{ A }=150 V.
(a) From Eq. (9.83),
I_{ C 1}=I_{ C 2}=\frac{I_{ R }}{1+2 / \beta_{ F }} (9.83)
I_{ O }=I_{ C 2}=I_{ C 1}=\frac{I_{ R }}{1+2 / \beta_{ F }}
which, for I_{ O }=5 \mu A , \text { gives } I_{ R }=(5 \mu A )\left(1+2 / \beta_{ F }\right)=5.1 \mu A . From Eq. (9.84),
\frac{V_{ CC }-V_{ BE 1}}{R_{1}} \times \frac{1}{1+2 / \beta_{ F }} (9.84)
R_{1}=\frac{V_{ CC }-V_{ BE 1}}{I_{ R }}=\frac{30 V -0.7 V }{5.1 \mu A }=5.75 M \Omega
(b) From Eq. (9.87),
R_{ o }=\frac{v_{ x }}{i_{ x }}=r_{ o 2}=\frac{V_{ A }}{I_{ C 2}} (9.87)
R_{ o }=\frac{V_{ A }}{I_{ C 2}}=\frac{150 V }{5 \mu A }=30 M \Omega
From Eq. (9.88),
V_{ Th }=I_{ O } R_{ o }=I_{ C 2} R_{ o }=I_{ C 2} \frac{V_{ A }}{I_{ C 2}}=V_{ A } (9.88)
V_{ Th }=V_{ A }=150 V
From Eq. (9.86),
\frac{I_{ C 2}}{I_{ C 1}}=\frac{1+V_{ CE 2} / V_{ A }}{1+V_{ CE 1} / V_{ A }} (9.86)
\frac{I_{ C 2}}{I_{ C 1}}=\frac{1+V_{ CE 2} / V_{ A }}{1+V_{ CE 1} / V_{ A }}=\frac{1+20 / 150}{1+0.7 / 150}=1.128
Thus, I_{ C 2}=1.128 \times I_{ C 1}=1.128 \times 5 \mu A =5.64 \mu A , which agrees, with a degree of error, with the desired value of I_{ O }=5 \mu A .