I read in sci.electronics.design that Lily Bepant <

[email protected]>

I have jumped through many equations to find that: (I'm novice, so

don't assume that these are correct, please correct wherever possible)

Rc+Re=Vbb/Ic(max)

No, this is not correct. Vbb would be the voltage of an independent base

bias supply. And you do not design every stage to run at Ic(max)!

Ic(max).. I looked up 2N3904 manual and found that Ic(max) is 200 mA

for it. With 12V Vbb, it means Rc+Re= 60 Ohms. We want a gain of 10,

thus Rc/Re should be 10, which yields to the result that Re = 60/11 =

5.4 ohms Rc = 55 ohms (rounded these).

These values seem to be weird to me. What am I doing wrong?

Using that wrong equation. Begin with Vcc - the *collector* supply

voltage = 12 V. Now, the terms of the question do not include anything

that allows you to calculate a required collector current. So you have

to choose one. With a 4.7 kohm collector load, and 11 V available once

you have dropped 1 V across the emitter resistor (assumed for the moment

it is decoupled by a parallel capacitor at signal frequencies) to get

reasonably stable d.c. operating conditions, you would get as much

output voltage swing as possible by dropping 5.5 V across the collector

resistor, which gives you a collector current of 5.5/4.7 = just under

1.2 mA. A standard value 820 ohm emitter resistor gives you 1 V on the

emitter then.

Also, what would I take B (beta) of this thing? This is a design

question, therefore I understand that I should design something

independent of the B (beta) value. But how, when

R1+R2 = [(Vbb)]/[(10Ic)/B]

or not?

Well, again, it's not Vbb but Vcc (unless you REALLY have a separate

base bias supply). What you do is to choose R1 and R2 using an average

value for beta and then see what happens to Ic with maximum and minimum

beta devices. In my book, beta min is 100 and beta max is 300 at Ic = 10

mA. I don't have a beta against Ic curve, so I'm going to assume that

the beta doesn't vary with Ic sufficiently to create a problem. The

average beta is 200.

Here, we have 12 V/(12 mA x 200)= 200 kohm. Very neat! Now there is 1 V

on the emitter and thus 1.7 V on the base. We can assume Vbe = 0.7 V

because it doesn't vary much between devices and Ic varies strongly with

it, making our d.c. stabilising circuit (emitter resistor and base

potential divider) a high-gain control loop.

So R1 + R2 = 200 kohm and R1 has 11 V across it while R2 has 1 V across

it. Strictly, it also has 0.5% less current in it, because that goes

into the base, but we won't be using high-precision resistors, so we can

neglect that. Thus R1/R2 = 11, and that's 2 equations with 2 unknowns,

R1 and R2. We get R2 = 200kohm/12 = 16.7 kohm and R1 = 183 kohm. We

would choose the standard values 180 kohm and 16 kohm.

I'll leave it to you to calculate what collector current you actually

get with 180 kohm, 16 kohm and 820 ohms, with beta values of 100, 200

and 300. For extra credit, what is Ic with a beta of 200 and the

transistor immersed in boiling water?

However, this stage has a gain of much higher than 10. The gain is

RcIe/26, with Ie in milliamps, giving a gain of 217. To get a gain of

just 10, we split the 820 ohm emitter resistor into Rc/10 = 470 ohms and

820-470 = 350 ohms (use standard value 360 ohms), and decouple (connect

a capacitor in parallel) the 360 ohms only. We could have just used the

470 ohms as the emitter resistor but that would have given only about

0.5 V on the emitter and if you do the calculations you will find that

Ic varies a lot more with beta than it does with Ve = 1 V.

I'm interested in electronics but don't have enough knowledge on it.

Thanks for your guidance.

Persevere!