An NPN bipolar junction transistor circuit is configured with voltage-divider biasing from… — Armed Services Vocational Aptitude Battery (ASVAB)

An NPN bipolar junction transistor circuit is configured with voltage-divider biasing from a single $15\text{ V}$ DC power supply. The voltage divider consists of an upper resistor $R_1 = 20\text{ k}\Omega$ connected to the positive supply and a lower resistor $R_2 = 10\text{ k}\Omega$ connected to ground. The emitter terminal is connected to ground through an emitter resistor $R_E = 1\text{ k}\Omega$ . Assuming a standard silicon base-to-emitter forward voltage drop of $V_{BE} = 0.7\text{ V}$ and negligible base current, what is the approximate DC emitter current ( $I_E$ ) flowing through the transistor?

A
$5.0\text{ mA}$
$4.3\text{ mA}$ Answer
C
$5.7\text{ mA}$
D
$10.0\text{ mA}$

Answer

4.3\text{ mA}

In an NPN voltage-divider biased circuit, the base voltage relative to ground is determined by the divider ratio:

V_B = 15\text{ V} \times (10\text{ k}\Omega / 30\text{ k}\Omega) = 5.0\text{ V}

. Because the base-emitter junction is forward-biased, it drops

0.7\text{ V}

, establishing an emitter voltage of

V_E = 5.0\text{ V} - 0.7\text{ V} = 4.3\text{ V}

. Dividing this emitter voltage by the

1\text{ k}\Omega

emitter resistor yields an emitter current of

4.3\text{ mA}

Step-by-Step Solution

Calculate the DC voltage at the base terminal (

V_B

) using the voltage divider rule

V_B = 15\text{ V} \times \frac{10\text{ k}\Omega}{20\text{ k}\Omega + 10\text{ k}\Omega} = 15\text{ V} \times \frac{1}{3} = 5.0\text{ V}

The base voltage is established by the ratio of the lower resistor to the total resistance across the supply when base current loading is negligible.

Determine the DC voltage at the emitter terminal (

V_E

) by accounting for the base-emitter forward voltage drop

V_E = V_B - V_{BE} = 5.0\text{ V} - 0.7\text{ V} = 4.3\text{ V}

In an active NPN transistor, the base-emitter junction is forward-biased, dropping approximately 0.7 V from base to emitter.

Apply Ohm's law across the emitter resistor (

R_E

) to calculate the emitter current (

I_E

)

I_E = \frac{V_E}{R_E} = \frac{4.3\text{ V}}{1\text{ k}\Omega} = 4.3\text{ mA}

The emitter current equals the voltage developed across the emitter resistor divided by its resistance.

Key Concept

Transistor Voltage-Divider DC Biasing and Emitter Current Calculation

Estimated Time:1m 30s

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