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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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