Question B.4: Analyzing a circuit with a current-dependent current source ...

Using KCL at node 1, we get

I_{ E }=I_{ B }+I_{ C }=I_{ B }+\beta_{ F } I_{ B }=\left(1+\beta_{ F }\right) I_{ B }                                             (B.11)

Using KVL around loop I, we get

V_{ Th }=R_{ Th } I_{ B }+r_{\pi} I_{ B }+R_{ E } I_{ E }=R_{ Th } I_{ B }+r_{\pi} I_{ B }+R_{ E }\left(1+\beta_{ F }\right) I_{ B }

which gives I_{ B } as

\begin{aligned}I_{ B } &=\frac{V_{ Th }}{R_{ Th }+r_{\pi}+R_{ E }\left(1+\beta_{ F }\right)} \\&=\frac{5 V }{15 k \Omega+1 k \Omega+500 \Omega \times(1+100)}=75.19 \mu A\end{aligned}                            (B.12)

Current I_{ C } \text {, which is dependent only on } I_{ B } , can be found from

\begin{aligned}I_{ C } &=\beta_{ F } I_{ B }=\frac{\beta_{ F } V_{ Th }}{R_{ Th }+R_{\pi}+R_{ E }\left(1+\beta_{ F }\right)} \\&=\frac{100 \times 5 V }{15 k \Omega+1 k \Omega+500 \Omega \times(1+100)}=7519 \mu A\end{aligned}                                  (B.13)

Then

\begin{aligned}&I_{ E }=I_{ B }+I_{ C }=75.19 \mu A +7519 \mu A =7594 \mu A \\&V_{ C }=V_{ CC }-I_{ C } R_{C}=30 V -7519 \mu A \times 2 k \Omega=14.96 V\end{aligned}

Voltage V_{ E } can be found from

V_{ E }=R_{ E } I_{ E }=R_{ E }\left(I_{ C }+I_{ B }\right)=R_{ E }\left(\beta_{ F } I_{ B }+I_{ B }\right)=R_{ E } I_{ B }\left(1+\beta_{ F }\right)

Since I_{ C }=\beta_{ F } I_{ B } , we get

\left.V_{ E }=R_{ E } I_{ B }\left(1+\beta_{ F }\right)=R_{ E } I_{ C }\left[\left(1+\beta_{ F }\right) / \beta_{ F }\right)\right]

Thus, R_{ E } \text { offers a resistance of } R_{ E }\left(1+\beta_{ F }\right) \text { to current } I_{ B } in loop I and a resistance of R_{ E }\left(1+\beta_{ F }\right) \beta_{ F } to current I_{ C } \text { in loop II. Therefore, } R_{ E } can be split, or “reflected,” into loop I and loop II by adjusting its value such that V_{ E }  is preserved on loop I and loop II. This arrangement is shown in Fig. B.5.

NOTE: The current-dependent current source \beta_{ F } I_{ B } is often represented by a voltage-dependent voltage source g_{ m } V_{1} \text { such that } \beta_{ F } I_{ B }=\beta_{ F } V_{1} / r_{\pi}=\left(\beta_{ F } / r_{\pi}\right) V_{1} \cdot g_{ m } , which is known as the transconductance, is related to \beta_{ F } \text { by } g_{ m }=\beta_{ F } / r_{\pi} . Therefore, by substituting \beta_{ F }=g_{ m } r_{\pi} , we can apply the above equations for a voltage-dependent voltage source that is often used as the small-signal model of bipolar and field-effect transistors.