Question 3.13: For the BJT circuit in Fig. 3.43 β = 150 and V BE = 0.7 V . ......
For the BJT circuit in Fig. 3.43 β = 150 and V_{BE} = 0.7 V . Find v_{o} .
1. Define. The circuit is clearly defined and the problem is clearly stated. There appear to be no additional questions that need to be asked.
2. Present. We are to determine the output voltage of the circuit shown in Fig. 3.43. The circuit contains an ideal transistor with β= 150 and V_{BE} = 0.7 V .
3. Alternative. We can use mesh analysis to solve for v_{o} We can replace the transistor with its equivalent circuit and use nodal analysis. We can try both approaches and use them to check each other. As a third check, we can use the equivalent circuit and solve it using PSpice.
4. Attempt
■ METHOD 1 Working with Fig. 3.44(a), we start with the first loop.
-2 + 100 kI_{1} + 200k(I_{1} – I_{2} ) = 0 or 3I_{1} – 2I_{2} = 2 × 10^{-5} (3.13.1)
Now for loop 2.
200k(I_{2} – I_{1} ) + V_{BE} = 0 or -2I_{1} + 2I_{2} = -0.7 × 10^{-5} (3.13.2)
Since we have two equations and two unknowns, we can solve for I_{1} and I_{2} . Adding Eq. (3.13.1) to (3.13.2) we get;
I_{1} = 1.3 × 10^{_5} A and I_{2} = (-0.7 + 2.6 ) 10^{-5}/2 = 9.5 μASince I_{3} = -150 I_{2} = – 1.425 m A, we can now solve for using loop 3:
-v_{o} + ^{1}kI_{3} + 16 = 0 or v_{o} = -1.425 + 16 = 14.575 V■ METHOD 2 Replacing the transistor with its equivalent circuit produces the circuit shown in Fig. 3.44(b). We can now use nodal analysis to solve for v_{o}
At node number 1: V_{1} = 0.7 V
At node number 2 we have:
150 I_{B} + (v_{o} – 16 ) /1k = 0 or
v_{o} = 16 – 150 × 10³ × 9.5 × 10^{-6} = 14.575 V
5. Evaluate. The answers check, but to further check we can use PSpice (Method 3), which gives us the solution shown in Fig. 3.44(c).
6. Satisfactory? Clearly, we have obtained the desired answer with a very high confidence level. We can now present our work as a solution to the problem.
