# Self-Biasing Transistor Amplifier

Discussion in 'General Electronics Chat' started by baseball07, Sep 20, 2008.

1. ### baseball07 Thread Starter Active Member

Apr 24, 2007
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Hello everyone. I would like to design a self biasing transistor amplifier such as the one attached, however, I need help in choosing resistors RL and RB. I am not exactly sure how to choose resistors in a transistor amplifier to keep it in the active region and find the quiescent point, so if someone can explain that to me I woudl appreciate it.

How can I maximize Vout? Would Vout be equal to Vbe + RB(base current + input current)?

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2. ### blocco a spirale AAC Fanatic!

Jun 18, 2008
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This isn't a good way to bias a transistor as it is device dependent. The resistor values would have to be selected for each transistor.

Apr 24, 2007
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4. ### Wendy Moderator

Mar 24, 2008
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By self biasing you are talking that particular configuration? There are better ways, why are you working on this one? You have a negative feedback going, which will mess somewhat with the gain calculations.

5. ### Audioguru New Member

Dec 20, 2007
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The transistor's minimum hFE is 100 and its typical hFE is from 125 to 225 depending on its collector current.

Then use Ohm's Law and simple arithmatic to calculate the value of the two resistors.
It should be driven from a low source impedance and have a fairly high load impedance.

6. ### silvrstring Active Member

Mar 27, 2008
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baseball07,

I am assuming you must be doing this for a class. Everyone's comments are right; there are better ways to build a BJT amplifier circuit. However, you need to build a self-biased BJT amp. So let's look at it.

First, you need to choose your resistors--so you will have a good starting point.

For this circuit,
Vcc = (Rc*Ic) + (Rb*Ib) + Vbe.

You want Vce ≈ 1/2 Vcc. Then Vrc will also ≈ 1/2 Vcc. So pick some Rc value and determine the current needed to produce a Vrc drop of 1/2 Vcc. This will be Ic.

Then Rb = (Vcc - Rc*Ic - .7) / Ib.

As far as I know, you've done all you can mathematically at this point. From here on, I hope you have some parts in front of you (or an electronics simulation program); you can increase the values of Rb to decrease the Hfe gain and increase your Vce to 1/2 Vcc.

The problem with this circuit is that it is temperature dependent, and Beta dependent. To boot, Beta changes with Vce but Vce depends on Beta? I think that would be a topic for someone who understands device physics.

Whatever, Just start with the math, and feel your way from there. You should be able to get a good bit of gain. If you experience clipping, just reduce your input signal or increase your DC supply voltage. Hope this helps.

Last edited: Sep 20, 2008
7. ### baseball07 Thread Starter Active Member

Apr 24, 2007
39
0
Ok. I thought self-biasing was better because it was more stable. What is the best BJT configuration for a phoyodetector?

Silvrstring, thank you got your help with the calculations.

8. ### Audioguru New Member

Dec 20, 2007
9,411
896
An opamp is a much better amplifier than a transistor.
It has a very high input resistance, a very low output resistance and much more gain. It has a low distortion.

There are a few types of photo-detectors. LDR, photo-diode and photo-transistor.
Which type do you have?

9. ### Wendy Moderator

Mar 24, 2008
20,772
2,540
This assumes the transistor isn't the photodetector. The classic 4 resistor common emitter design is both temperature stabilized and (relatively) beta insensitive.

10. ### baseball07 Thread Starter Active Member

Apr 24, 2007
39
0
Audioguru -

I have a Si photodiode PIN type. I tried an opamp but the output didn't have enough current to drive the rest of my circuit.

Thanks for the circuit Bill. Is this circuit a function of voltage since its has a divider network?