# Homework Help: Simple transistor amplifier

1. Nov 3, 2014

### Mutaja

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

The circuit above shows a simple transistor amplifier with the given data:
β=100, VA = 100V, RL = 10kΩ

a) neglect the "early-effect" and design the circuit so that IE = 0.5mA and VC = 5V.

2. Relevant equations

3. The attempt at a solution

I'm familiar with the β part, it means DC gain, or collector current divided by base current. What I'm not familiar with is VA. Where is VA measured? If it's a typo, what can it be? I would ask my professor this, but he's away for the week so that makes it slightly difficult.

More to the problem itself. The DC analysis here should be very straight forward, I feel I just need some help to get started.

I need to set up an equation here so that IE = 0.5mA and VC = 5V. Do I also need to analyse the base-to-collector "junction" to see if it's a forward or reverse biased pn junction? That way I get the base-to-collector voltage (VBC)

Pardon me if I'm far off here, but I'm looking at a slightly different circuit for DC analysis.

Any help of guidance here will be greatly appreciated, thanks.

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2. Nov 3, 2014

### Mutaja

I've had some time to think about this problem now, and it seems logic to me that what I'm dealing with here is a forward biased pn junction at the base-to-emitter junction. This is because of the direction of the current through the emitter is the same direction as the current through the base.

Can I assume that the voltage across the base and emitter then is 0.7v? I have a similar problem in my lecture notes, but it doesn't clarify where the 0.7V comes from.

If I'm correct this far, I should be able to use this information to set up an equation for the current through the base, and then use that to set up an equation and calculate which numbers I need on my resistors?

Hopefully I'm on to something here.

3. Nov 3, 2014

### Staff: Mentor

Yes, that is the correct approach. :-)

4. Nov 3, 2014

### Mutaja

I'm not back home and I can spend some time solving this.

Using my example in my book, which is the following:

I get this:

I'm assuming IB = IC. If this is wrong, please try to explain why.

IB = VSIG divided by RSIG -> VSIG = 0.5mA *2.5kΩ
VSIG 1.25V

IC = 100*IB = 50mA -> current divider between RL and RC.

For VC = 5V, I'm using the current divider rule between RC and RL forcing RC to be 101.01Ω if 50mA is going through the collector, and it's divided between RC and RL. That means 0.5mA is going through RL and 49.5mA is going through RC.

Is any of this correct?

Any input or guidance will be greatly appreciated.

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5. Nov 4, 2014

### Mutaja

It's been a day or so without any replies, and I haven't gotten anywhere on my own so I'm bumping this thread.

As always, and input, tips or guidance will be greatly appreciated.

6. Nov 4, 2014

### Staff: Mentor

The circuit in your post #1 has capacitor coupling to RL so the DC current does not divide between RC and RL. You can ignore RL when designing biasing. In your design you need for Vsig to be both a source of DC (for biasing) and a small signal source (to illustrate AC amplification).

Yes, assume VBE = 0.7v.

Surely you mean IC = β.IB?

Last edited: Nov 4, 2014
7. Nov 5, 2014

### rude man

Close enough. Accurate to within 1% since β=100. The right thing to do.
(Note that I changed your IB to IE. The BJT equations are IC ~ IE = βIB if the device is biased in the active region.)

Forget RL. This is a dc setup computation.
1. Compute VE. Assume VB = 0 (accurate to 10 mV). Assume VBE = 0.7V.
2. Compute RE to give 0.5 mA.
3. Compute RC to give VC = +5.0V.