How to Design a Simple Transistor Amplifier with Given Data?

[Mutaja]

Homework Statement

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

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

Homework Equations

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 V_A. Where is V_A 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 I_E = 0.5mA and V_C = 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 (V_BC)

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.

 

[Mutaja]

Can't see any edit button, but if it's there, I'm sorry about this double post.

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.

 

[berkeman]

Yes, that is the correct approach. :-)

 

[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 I_B = I_C. If this is wrong, please try to explain why.

I_B = V_SIG divided by R_SIG -> V_SIG = 0.5mA *2.5kΩ
V_SIG 1.25V

I_C = 100*I_B = 50mA -> current divider between R_L and R_C.

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

Is any of this correct?

Any input or guidance will be greatly appreciated.

 

[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.

 

[NascentOxygen]

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

Yes, assume V_BE = 0.7v.

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

Surely you mean I_C = β.I_B?

 

[rude man]

II get this:

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

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

Forget R_L. This is a dc setup computation.
1. Compute V_E. Assume V_B = 0 (accurate to 10 mV). Assume V_BE = 0.7V.
2. Compute R_E to give 0.5 mA.
3. Compute R_C to give V_C = +5.0V.