Designing a simple modified basic current source
(a) Design the modified basic current source in Fig. 9.27(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_{ BE 3}=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 .
I_{ O }=I_{ C 2}=5 \mu A , V_{ BE 1}=V_{ BE 2}=V_{ BE 3}=V_{ CE 1}=0.7 V , \text { and } V_{ A }=150 V .
(a) From Eq. (9.92),
I_{ O }=I_{ C 2}=\frac{I_{ R }}{1+2 /\left(\beta_{ F }^{2}+\beta_{ F }\right)} (9.92)
I_{ O }=I_{ C 2}=\frac{I_{ R }}{1+2 /\left(\beta_{ F }^{2}+\beta_{ F }\right)}
which, for I_{ O }=5 \mu A , \text { gives } I_{ R }=(5 \mu A )\left[1+2 /\left(\beta_{ F }^{2}+\beta_{ F }\right)\right]=5 \mu A . From Eq. (9.93),
I_{ R }=\frac{V_{ CC }-V_{ BE 1}-V_{ BE 3}}{R_{1}} (9.93)
R_{1}=\frac{V_{ CC }-V_{ BE 1}-V_{ BE 3}}{I_{ R }}=\frac{30-0.7-0.7}{5 \mu A }
which gives the required value of the resistor as R_{1}=5.72 M \Omega .
(b) From Eq. (9.94),
R_{ o }=\frac{v_{ x }}{i_{ x }}=r_{ o 2}=\frac{V_{ A }}{I_{ C 2}} (9.94)
R_{ o }=\frac{V_{ A }}{I_{ C 2}}=\frac{150}{5 \mu A }=30 M \Omega
From Eq. (9.95),
V_{ Th }=I_{ O } R_{ o }=I_{ C 2} R_{ o }=I_{ C 2} \frac{V_{ A }}{I_{ C 2}}=V_{ A } (9.95)
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+\left(V_{ BE 1}+V_{ BE 3}\right) / V_{ A }}=\frac{1+20 / 150}{1+(0.7+0.7) / 150}=1.123
Neglecting I_{ B 3} \text {, we have } I_{ C 1} \approx I_{ O }=5 \mu A \text {. Thus, } I_{ C 2}=1.123 \times I_{ C 1}=1.123 \times 5 \mu A =5.62 \mu A , which agrees, with a degree of error, with the desired value of I_{ O }=5 \mu A .