# Current through a lightbulb after flowing through transistor

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1. May 13, 2015

### 4Phreal

1. The problem statement, all variables and given/known data
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

2. Relevant equations
Ic = βIb

3. 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

Last edited: May 13, 2015
2. May 13, 2015

### Hesch

You must find some characteristics for the transistor ( Icollector vs. Ibase characteristics ).

It's very hard to find ( I've tried ) because the transistor is typical used as a switching transistor, where of the mentioned characteristics are not of interest.

3. May 13, 2015

### 4Phreal

What kind of characteristics?

For the 2N3904 NPN transistor, when the base is more positive than the emitter by 0.6 volts (for the silicon devices) current will flow from the base to the emitter. When this base current IB is sufficiently large, the transistor is “turned on”. A current of 1 mA or larger will turn on the 2N3904 NPN transistor. As mentioned in my original post, the beta value of amplification here between IB and IC is 100. What other characteristics am I needing?

4. May 13, 2015

### Hesch

Something like this:

Plot two points ( A and B ) and draw a load line. Calculate Ib and you can read the Q-point, determining Ic.

( The figure is not for a 2N3904 ).

5. May 13, 2015

### 4Phreal

Something like this?

I kind of doubt that I have not been given sufficient information to answer the question without finding some obscure characteristics. But I also clearly have no idea what I'm doing, so my doubts don't hold much weight

6. May 13, 2015

### Hesch

I cannot read the base currents in the figure, but yes, that's it.

The characteristics are not "obscure"

7. May 13, 2015

### 4Phreal

Oh sorry, the graph is for IB = 10 µA, 20 µA, 30 µA, ... , 80 µA.

8. May 13, 2015

### Hesch

The highest resistor value in #1 is 50 kΩ → Ib = 188 μA.

That's a problem.

9. May 13, 2015

### 4Phreal

It's a problem because we don't have that data point on the graph? I'm almost certain the answer for the 50 kΩ is 0.0

10. May 14, 2015

### Hesch

Are the curves in #5 for a 2N3904 ?

( Curves differ for different types of transistors ).

11. May 14, 2015

### Andrew Priest

0.1, 0.1, 0.0. I assume you're in Physics 140 at BYU? All you do for this problem is see if the transistor is on or not. To do this, the current going into the transistor from the R side has to be greater than 1 mA. If it is, then the current through the light bulb=9.8/100, or 0.1 in his class.