Voltage Divider CE amplifier

fireback

Newbie
Hello i have this question and was wondering if someone could help me,
Lets say we are given a simple CE amplifier(with voltage divider on base and Rc, Re resistances)
If we are given Ic=3mA , Vcc=20V , β=100, Rth=10kΩ, Vce=10V, how can we calculate the values for R1 and R2?
Also, how can we calculate them if we are not givven the equivalent Rth value?

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Audioguru

Super Moderator
Staff member
For a place to start from, turn on the transistor sufficiently so that output voltage centers around a level about halfway between supply V and ground. Current changes through the 'totem pole ' creates voltage changes (the totem pole consisting of RL, RE and the transistor).

Notice bias V needs to be about 0.6 V greater than V at emitter leg.

With hardware it's practical to substitute a potentiometer for R1 R2. Say 100 kohm.

Akanimo

Usually, I use I2 to be between 5 to 10 times Ib.
R2 = (Ve+Vbe)/I2.
R1 = (Vcc-V2)/(Ib+I2)

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Audioguru

Akanimo, you did not explain what I2 is. The current in R1 and R2?
You also did not explain why is the current fairly high? Because the transistor hFE has a wide range?

Akanimo

Akanimo, you did not explain what I2 is. The current in R1 and R2?
You also did not explain why is the current fairly high? Because the transistor hFE has a wide range?
Oh! I believed it would be obvious. I2 is the current flowing through R2. The current through R1 is (I2+Ib). I2 is selected to be much greater than Ib for bias stability purposes from instability which could result from a lot of factors including noise and so on.

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Audioguru

The current through R1 is (I2+Ib). I2 is selected to be much greater than Ib for bias stability purposes from instability which could result from a lot of factors including noise, input signal and so on.
No. The high current required in R2 is not affected by those AC things, it is DC and is because a transistor has a wide range of DC hFE. You do not want not enough base voltage and current for a transistor that has low hFE.

Akanimo

No. The high current required in R2 is not affected by those AC things, it is DC and is because a transistor has a wide range of DC hFE. You do not want not enough base voltage and current for a transistor that has low hFE.
Okay. I've never seen the hFE connection to I2 before, and I've not read it fro somewhere. It just hasn't occurred to me.

I probably would need to ask a few questions just to clarify:

- considering the range of hFE specified in the datasheet for a particular transistor (we can use bc547b as an example), what value of better do you use, the lowest, midpoint or highest? Why?

- You mentioned base voltage in connection with hFE and I2. Please could you explain with some detail how they connect?

FvM

Super Moderator
Staff member
I agree with Akanimo. A base voltage divider current of five- to tenfold base current is a standard design rule taught in second year of electronics engineering study.

Above might be of help.

Regards, Dana.

Audioguru

Your example of a BC547B transistor is unusual because it has its hFE selected. The BC547 has an hFE from 110 to 800 but the BC547B has its hFE selected to be from 200 to 450. The BC547A has low hFE and the BC547C has a high hFE.

A transistor with a low hFE loads down and reduces the voltage set by the base voltage divider resistors. Then if you decide to make current in the voltage divider 1/10th the mid hFE of a transistor then you should calculate that a transistor with minimum hFE will work properly.

FvM

Super Moderator
Staff member
You can say 5 times maximal Ib and save most of the verbose explanations. It's however still a rule of thumb. In a more accurate design procedure, you'd specify the acceptable bias point shift over hFE variation.