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Ver simple question about CE amplifier DC biasing !!!

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ahmad_abdulghany

ahmad_abdulghany

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Hello...

i want to ask a very BASIC quesion about Amplifier DC biasing...
assume that i'm designing a CE amplifier, with base biased by a potential divider from Vcc with Resistances R1 and R2...
Which is better, to increase R1 and R2 such that Ib is comparable or greater than current passing through them? OR decrease their values such that i can assume that I1≈I2 (e.g. I1=10Ib)??? Why?

Assume the next figure for illustration...
**broken link removed**​

Thanks alot in advance,
Ahmad,
 

The R1//R2 have interest in:

1-input resistance saw by signal source.
2-series resistance of DC voltage bias.

Ideal:
For 1 R1//R2=∞
For 2 R1//R2 finite, because we need current (mA) with specific DC voltage source.


We need to make a compromise between 1&2 to reach our fullfilements.

R1//R2>>rb=VT/IB


First choice IB then choice R1&R2. R1&R2 have importance in dependence of IB(β) like RE.
 
I usually start off assigning a DC voltage that I want across RE. 1 volt is probably ok--it will give a fair amount of negative feedback to the bias circuit. I also decide on the quiescent collector current desired.

You then know how much quiescent current is flowing in RE. Assume that the base current is Ib = Ire/β. This tells you how much base current has to flow. I would use a typical value for β from the transistor data sheet.

I then make the current flowing thru the R1/R2 divider to be around 4X Ib. That way the base voltage point is fairly "stiff", irregardless of if Ib changes over temperature.
 
Well, I understood it,
Simply, to apply the potential divider on Vb to be equal to Vccx[R2/(R1+R2)], current flowing through R1 and R2 should be almost the same, that's why we choose it much greater than Ib (say 10 times Ib), otherwise, Thevinen Theorem should be applied to calculate it..

that's all..
anyway, thank you every body for contribution,
Ahmad,
 

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