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4Phreal

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## Homework Statement

Consider a 2N3904 NPN transistor in a circuit. Treat the lightbulb as a 100 ohm resistor. For different values of R, what is the current through the lightbulb? Enter two digits rounded to the nearest 0.1 amps.

If R=500 ohms, the current through the lightbulb is:

If R=5 kohms, the current through the lightbulb is:

If R=50 kohms, the current through the lightbulb is:

http://i.imgur.com/mbk0cKK.jpg

## Homework Equations

Ic = βIb

## The Attempt at a Solution

For the 500 ohm resistor:

Ib = (10V - 0.6V) / 500 ohms=0.0188A

Ic=β*0.0188A=100*0.0188A=1.9A

0.6V is the common value used for silicon transistors

It appears beta for the 2N3904 NPN is 100. In other words, Ic/Ib = 100

For the 5k and 50k ohm resistors, in order for the circuit to have any current, the potential difference between base and emitter has to be large enough to activate the transistor. For the 2N3904 NPN, there must be a 0.2 mA current for it to be amplified.

**4. Other relevant information**

The answer to this question using a 2k ohm resistor is 0.1 A. No idea how they got that answer

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