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[SOLVED] Voltage Divider Biased CE Amplifier Design

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b.heuju

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The question is as below:

Design a voltage divider type dc biased CE amplifier to obtain B (beta) independent biasing. Use appropriate guidelines to support your design. Given parameters are: Vcc = 12 VDC, Ic = 2 mA, B (beta) = 150.

From what I know, we can use 1/3 rule or 10/1 rule. For this question the answer required is the value of resistance for Rc, Re, and voltage divider resistors. I quite don't understand the application of the guidelines.
 

You will have to be more explicit about what guidelines this class is expected to follow. I don't know of any commonly accepted guidelines for such a question. As a wild guess, I will assume that the "1/3 rule" means the +12 volts is divided equally into thirds, so that Ve=4v and Vc = 8v. If Ic = 2ma., that means Rc = 2K. And if Beta=150, that means Ie must represent the collector current (2ma) plus 1/150 of the collector current coming from the base. Therefore Ie = 2.01333 ma. So if Ve is to be +4 volts, Re must be 1987 Ohms.

As for what resistors to use in the voltage divider that biases the base, there is no one unique answer. Any reasonably small value resistors will provide a bias voltage that is relatively unaffected by Beta. The lower the voltage divider resistance, the more resistant the bias voltage will be. But very low resistance values have their own problems, such as drawing more power supply current and loading down the input signal applied to the base. So the choice of voltage divider resistors is a compromise, taking these other factors into account. These other factors can be different for different applications, depending on what is important.

I think the "10/1" rule means the voltage divider resistors are chosen so that they carry 10 times as much current as the base does. By following this guideline you can be assured that the bias voltage never varies by more than 10% over all reasonable values of Beta. If that is what you want to do, and if Ic=2ma, then Ib = 2ma/150 = 13.3 uamps. Assuming that Vbe=0.7 volts, we need Vb= 4.7v. So all that remains is to pick voltage divider resistors that carry Ib x 10 = 133 uamps, and produce a voltage of 4.7 v.
 

    V

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Tunelabguy - I agree to all of your explanations. However, I was (and still I am) not sure if the questioner knows the meaning of these "rules".
 

Thanks.
I hooked up with a book yesterday: Microelectronic Circuits, Adel S. Sedra and Kenneth C. Smith. According to this book as a rule of thumb, Vbb = 1/3 Vcc, Vcb = 1/3 Vcc, IcRc = 1/3 Vcc. And as for 10/1 I suppose it means current through VDB resistors should be 10 Ie to 0.1 Ie. I think, Tunelabguy also means the same.
But since I need to design Beta independent why is Beta used in your case. Although I think its not a problem. Designing the circuit can be done in any way, do not need to stick to any specific rule. So I suppose what Tunelabguy says is also correct. Thanks. If there is any other way with logical explanation then I would be happy to hear it.
 

Thanks.
But since I need to design Beta independent why is Beta used in your case.

I think, a design that is independent on beta is impossible.
The only way is to reduce the beta sensitivity. And this means: strong negative feedback.
 
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    b.heuju

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But since I need to design Beta independent why is Beta used in your case.
You will notice that even though I used Beta in my calculations, the effect of Beta was very small. If Beta were infinite, Re would have come out to be 2K, the same as Rc. As for the use of Beta in calculating the resistance of the voltage divider resistors, the only role played by Beta was again for an approximation. The amount of current taken by the base depends on Beta. A very high Beta means very little current is needed in the base. Whatever that current is, the rule of 10/1 (I suppose) means that the current through the voltage divider resistors should be 10 times as great as the base current. So we at least need an approximate Beta to pick those resistors. Once we pick those resistors, we can then change Beta a little and the bias voltage will not change (very much). But if you change Beta by a very large amount, the bias voltage will change more. So no design is perfectly independent of Beta. The best we can do is to make it reasonably independent of Beta. By using the rule of 10/1 designed around Beta=150, we can be assured the bias voltage will not change by more than 10% for values of Beta running from 75 all the way up to infinity. That's pretty good, even if it is not perfect.
 
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    b.heuju

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Thanks I get the idea now.
 

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