Chapter 10, Problem 10.1EP

The circuit parameters for the two-transistor current source shown inFigure 10.2(b) are $V^{+}=3V$ , $V^{-}=-3V$ , and $R_1=47k\Omega$ . The transistor parameters are $\beta =120,V_{BE}\left(on\right)=0.7V$ , and $V_A=\infty$ . Determine $I_{REF},I_O$ , and $I_{B1}$ .(Ans. $I_{REF}=0.1128mA,I_O=0.1109mA,I_{B1}=0.9243\mu A$ )

To determine

The value of $I_{\text{REF}}$ , $I_0$ and $I_{B1}$ .

Answer to Problem 10.1EP

The values are:

  $I_{\text{REF}}=0.1128\text{mA}$

  $I_0=0.1109\text{mA}$

  $I_{B1}=0.9242\text{μA}$

Explanation of Solution

Given:

The given circuit is shown below.

  

  $\begin{array}{l}{V}^{-}=-3\text{V}\\ V^{+}=3\text{V}\end{array}$

  $R_1=47\text{k}\Omega$

  $\beta =120$

  $V_{\text{BE}}(\text{on})=0.7\text{V}$

Calculation:

  $V_A=\infty$ (Ideal case)

Apply KVL in input Loop

  $V^{+}-I_{\text{REF}}{R}_1-V_{\text{BE}}(\text{on})-V^{-}=0(1)$

From equation,

  $\begin{array}{c}{I}_{\text{REF}}=\frac{V^{+}-V_{BE}(on)-V^{-}}{R_1}\\ I_{\text{REF}}=\frac{3-0.7-(-3)}{47\times 1000}\\ =0.1128\text{mA}\end{array}$

From figure,

  $\begin{array}{c}{I}_0=I_c=\frac{Iref}{1+\frac{2}{\beta }}\\ I_0=\frac{0.1128\times {10}^{-3}}{1+\frac{2}{120}}\\ =0.1109\text{mA}\end{array}$

Also, both transistors are identical ( $I_{B1}=I_{B2}$ ). So,

  $\begin{array}{l}\frac{2I_c}{\beta }=I_{B1}+I_{B2}\\ \frac{2I_c}{\beta }=2I_{B1}\end{array}$

So,

  $I_{B1}=\frac{I_c}{\beta }$

  $\begin{array}{c}{I}_{B1}=\frac{0.1109\times {10}^{-3}}{120}\\ =0.9245\text{μA}\\ \approx \text{0}\text{.923μA}\end{array}$