(a) Write KVL around the left part of the circuit in Figure 8-46, Solve for $I_{B}$ ,
(b) Determine the current from the dependent current source. This current is $I_{C}$ .
(c) Calculate the output voltage and the power in the load resistor $R_{L}$ .
(d) Compare the power determined in (c) with the power delivered to the load if the load resistor were connected directly to the Thevenin circuit.

Question

(a) Write KVL around the left part of the circuit in Figure 8-46, Solve for [latex]I_{B}[/latex],
(b) Determine the current from the dependent current source. This current is [latex]I_{C}[/latex].
(c) Calculate the output voltage and the power in the load resistor [latex]R_{L}[/latex].
(d) Compare the power determined in (c) with the power delivered to the load if the load resistor were connected directly to the Thevenin circuit.

Accepted Answer

(a)[latex]V_{TH}- R_{TH}I_{B}- 0.7 \ V = 0[/latex]

[latex]I_{B}= \frac{V_{TH}- 0.7 \ V}{R_{TH}}= \frac{1.6 \ V - 0.7 \ V}{6.8 \ k\Omega }= 132 \ \mu A[/latex]

(b) [latex]I_{c}= β I_{B} = 200(132 \ μA) = 26.5 \ mA [/latex]

(c) [latex] V_{OUT} = I_{C} R_{L}= (26.5 \ mA)(470 \ Ω) = 12.4 \ V [/latex]

[latex]P_{L}= \frac{V^{2}_{OUT} }{R_{L}}= \frac{(12.4 \ V)^{2}}{470 \ \Omega } = 327 \ mW[/latex]

(d) [latex]P_{L} = I^{2}_{B} R_{L}= (132 \ μA)^{2}(470 \ Ω) = 8.19 μW [/latex]

The power in the load resistor is 327 mW/8.19 μW = 39.927 times greater than the power that the Thevenin input circuit could deliver to the same load. This illustrates that the transistor can operate as a power amplifier.

Question 8.13: (a) Write KVL around the left part of the circuit in Figure ...

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