# Homework Help: BJT simple Transistor problem

1. Nov 16, 2013

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

We have a transistor with RB, RE and A LED being a GaAs Light-emitting Diode (LED with a forward voltage of approximately 1.5V.

Values for RB= 2K, Vs-offset=5, Vs-amplitude=4, V-frequency= 1KHz, let β=100

what is the value of the emitter resistor for which the LED sinusoidal current will
have an offset level of I=27 mA?

Use two methods to create such a current:
Method A: Keep RE constant (say 200Ω) and vary VOFF of the VSIN source.
Method B: Keep VOFF constant (say 5V) and vary RE.

In each case, once you set the DC current to 27mA, check the sinusoidal collector current
iC(t), and make sure that it doesn’t get clipped. If it does, reduce VAMPL of the VSIN
source, from 4V to a lower level, but as large as you can make it to keep the LED
currents from clipping

2. Relevant equations

3. The attempt at a solution

By I offset do they mean the value of I current (in collector)?

If it is then I used the formula for current in emitter

Ie= (Vin-Vbe)/ (Re- Rb/β+1)

plugging in values and solving for Re I get => Re= 139.46Ω.

I guess this is more of a PSPICe problem but I just want to make sure if by the current offset they meant the value of the current.

2. Nov 17, 2013

### Staff: Mentor

Most likely. The LED is the collector load, is it?

Check that formula.

3. Nov 17, 2013

### rude man

4. Nov 17, 2013

Here's a better view of the HW problem.

I copied that exact same diagram on PSPICE.
Made a Bias point analysis and got these Voltage and Current values.

http://postimg.org/image/bcry7ph7l/ [Broken]

Now to solve the problem. I'm not really sure what to do but I used the formula on "Hint".

I use the formula for the emmiter current:
IE= (Vin- VBE)/ (RE + RB/β+1)

Do I have to just solve in Method 1 plug in values and solve for Vin, keeping RE constant?

And in Method 2 solve for RE and plug in values keeping VOFF constant?

Last edited by a moderator: May 6, 2017
5. Nov 17, 2013

### rude man

Yes.

In both cases you are gunning for Ie = 27 mA.

PS I would just say Ie = (V_OFF - 0.7)/R_E.

Last edited: Nov 17, 2013
6. Nov 17, 2013

### rude man

Forget Rb/beta +1. RE = (5-0.7)/0.027 ohms.

7. Nov 17, 2013