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Design the circuit in Fig. P6.34 to establish IC = 0.2 mA and VC = 0.5 V. The transistor
Chapter 6, Problem 6.34(choose chapter or problem)
Design the circuit in Fig. P6.34 to establish \(I_{c}=0.2 \ \text{mA}\) and \(V_{c}=0.5 \text{V}\). The transistor exhibits \(v_{BE}\) of \(0.8 \text{V}\) at \(i_{c}= 1 \ \text{mA}\), and \(\beta=100\)
Questions & Answers
QUESTION:
Design the circuit in Fig. P6.34 to establish \(I_{c}=0.2 \ \text{mA}\) and \(V_{c}=0.5 \text{V}\). The transistor exhibits \(v_{BE}\) of \(0.8 \text{V}\) at \(i_{c}= 1 \ \text{mA}\), and \(\beta=100\)
Step 1 of 6
The data is given as \({I_C} = 0.2 \times {10^{ - 3}}\;{\rm{A}}\), \({V_C} = 0.5\;{\rm{V}}\), \({v_{BE}} = 0.8\;{\rm{V}}\), \({i_C} = 1 \times {10^{ - 3}}\;{\rm{A}}\), \(\beta = 100\) and let’s assume \({V_T} = 25 \times {10^{ - 3}}\;{\rm{V}}\).
The modified circuit diagram is shown in Figure 1.
Step 2 of 6
Determine the value of base-emitter voltage as,
\({V_{BE}} = {v_{BE}} + {V_T}\ln \left( {\frac{{{I_C}}}{{{i_c}}}} \right) \)
Substitute the values in the above equation, and we get,
\({V_{BE}} = 0.8 + 20 \times {10^{ - 3}}\ln \left( {\frac{{0.2 \times {{10}^{ - 3}}}}{{1 \times {{10}^{ - 3}}}}} \right)\)
\({V_{BE}} = 0.8 - 0.40