Re: Transistors

On a sunny day (Mon, 12 Nov 2007 13:18:52 -0800) it happened john
<conphiloso@xxxxxxxxxxx> wrote in
<1194902332.623320.152710@xxxxxxxxxxxxxxxxxxxxxxxxxxx>:

Hi,

I am new to electronics and working with transistors for the first
time. I chose 2N4401transistor from Fairchild to begin with. You can
find the data *** at http://www.fairchildsemi.com/ds/2N/2N4401.pdf

I have following questions

1. The base current Ib controls the collector current Ic. But I am
unable to find the base current from the data ***.

Well, some can be found from the graph on page 3,
'typical pulsed current gain versus collector current.'
For example at Ic 100mA at -40C the beta is about 110, so the Ib was .9 mA.

2. I am also unable to find the maximum and minimum values of Ib too.

3. Is it ok to determine Ic and then calculate Ib using Ic = beta x
Ic.

Yes, but beta depends strongly on Ic and temp, as that graph shows.

4. What should I consider to do regarding DC biasing of the transistor
before applying the AC signal to the transistor?

The idea is to apply an AC CURRENT within the operating DC bias current.


The operating point needs to be stabilised too, there are 2 main ways for common emittor:
----------------------- +
| |
R1 R3
| |-------------- out
| c
|--- b
- C1-| e
| |--- R5 ---
| | |
R2 R4 C2
| | |
------------------------------ 0V


In this case the divider R2 / R1 + R2 sets the base voltage,
and the DC emiter current is about Ur2 - 0.7 / R4
The collector current is about the same.

For AC you can see that a 1V AC signal does not only see R4,
abut als the R5 C2 parallel to R4, so for AC there is a bigger
emitter current change (for the same emittor voltage change),
resulting in a higher gain.
So R5 C2 sets the AC gain basically, together with R3.
The voltage on R3 is Ic x R3.
The input impedance is R1 parallel with R2 parallel with the
network (R4 parallel with R5 + C2) x beta.
So what is in the emittor appears beta x higher in the base.



A simpler DC biasing system could look like this:
+Ub
|
R2
|------ out
--R1-|
| |
| c
-- C1-----b
e
|
-------------------

In this case the ACinput impedance in higly non-linear
due to the be junction, but if driven from a high impedance source
that non-linear effect gets much smaller.

For max outpu swing Vce should be about 1/2 Ub,
so Ur1 is 1/2 Ub - Vbe or about 1/2Ub - 0.7.
You know IR2(=Ic) for 1.2 Ub drop, you know beta, you then know Ib.
and can calculate R1.
You could improve linearity by adding an emittor resistor.



Please advice!
John
.

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