Objective: Determine the currents and voltages in the circuit shown in Figure 5.46(b).

Assume the transistor parameters are: β = 50, $V_{B E} (on) = 0.7 V$ , and $V_{C E} (sat) = 0.2 V$ . Let $R_{C} = 1 kΩ$ and $R_{B} = 20 kΩ$ . Determine the currents and output voltage for various input conditions

Question

Objective: Determine the currents and voltages in the circuit shown in Figure 5.46(b).

Assume the transistor parameters are: β = 50, [latex]V_{B E} (on) = 0.7 V[/latex], and [latex]V_{C E} (sat) = 0.2 V[/latex]. Let [latex]R_{C} = 1 kΩ[/latex] and [latex]R_{B} = 20 kΩ[/latex]. Determine the currents and output voltage for various input conditions

Accepted Answer

The following table indicates the equations and results for this example

Condition	[latex]V_{O}[/latex]	[latex]I_{R}[/latex]	[latex]Q_{1}[/latex]	[latex]Q_{2}[/latex]
[latex]V_{1} = 0 [/latex]	5 V	0	[latex]I_{B1} = I_{C1} = 0 [/latex]	[latex]I_{B2} = I_{C2} = 0 [/latex]
[latex]V_{2} = 0 [/latex]
[latex]V_{1} = 5 V [/latex]	0.2 V	[latex] \frac{5 - 0.2}{1} = 4.8 mA[/latex]	[latex] I_{B1} = \frac{5 - 0.7}{20} = 0.215 mA[/latex]	[latex]I_{B2} = I_{C2} = 0 [/latex]
[latex]V_{2} = 0 [/latex]
			[latex]I_{C1} = I_{R} = 4.8 mA[/latex]
[latex]V_{1} = 0 [/latex]	0.2 V	4.8 mA	[latex]I_{B1} = I_{C1} = 0 [/latex]	[latex]I_{B2} = 0.215 mA[/latex]
[latex]V_{2} = 5 V [/latex]				[latex]I_{C2} = I_{R} = 4.8 mA[/latex]
[latex]V_{1} = 5 V [/latex]	0.2 V	4.8 mA	[latex]I_{B1} = 0.215 mA[/latex]	[latex]I_{B2} = 0.215 mA[/latex]
[latex]V_{2} = 5 V [/latex]			[latex]I_{C1} = \frac{I_{R}}{2} = 2.4 mA[/latex]	[latex]I_{C2} = \frac{I_{R}}{2} = 2.4 mA[/latex]

Comment: In this example, we see that whenever a transistor is conducting, the ratio of collector current to base current is always less than β. This shows that the transistor is in saturation, which occurs when either [latex]V_{1}[/latex] or [latex]V_{2}[/latex] is 5 V.

Condition	$V_{O}$	$I_{R}$	$Q_{1}$	$Q_{2}$
$V_{1} = 0$	5 V	0	$I_{B1} = I_{C1} = 0$	$I_{B2} = I_{C2} = 0$
$V_{2} = 0$
$V_{1} = 5 V$	0.2 V	$\frac{5 – 0.2}{1} = 4.8 mA$	$I_{B1} = \frac{5 – 0.7}{20} = 0.215 mA$	$I_{B2} = I_{C2} = 0$
$V_{2} = 0$
			$I_{C1} = I_{R} = 4.8 mA$
$V_{1} = 0$	0.2 V	4.8 mA	$I_{B1} = I_{C1} = 0$	$I_{B2} = 0.215 mA$
$V_{2} = 5 V$				$I_{C2} = I_{R} = 4.8 mA$
$V_{1} = 5 V$	0.2 V	4.8 mA	$I_{B1} = 0.215 mA$	$I_{B2} = 0.215 mA$
$V_{2} = 5 V$			$I_{C1} = \frac{I_{R}}{2} = 2.4 mA$	$I_{C2} = \frac{I_{R}}{2} = 2.4 mA$

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