# Designing a common-emitter amplifier using BJT

Discussion in 'Analog & Mixed-Signal Design' started by AlternatedCurrent, Oct 1, 2018.

1. ### AlternatedCurrent Thread Starter New Member

Oct 1, 2018
6
0
Hi guys, I am learning, so, if you will spank me, do it softly...

I am trying to design a common-emitter amplifier just for the whole purpose of learning.

Can you guys see if the calculations I have done are correct and help me calculate the input/output coupling capacitors?

The circuit will be like this:

I have chosen the following values: Vcc = 18, Ic = 8mA, transistor 2N2222A.

hfe for that current is 225, according to this: http://web.mit.edu/6.101/www/reference/2N2222A.pdf

I have chosen Ve to be 1V.

Vce minimum is chosen to be 2V. I see it is not good to have a Vce lower than that.

If Vce minimum = 2V, them Vc minimum = Vcemin + Ve = 2 + 1 = 3V

So, the output may swing from 3 to 18 V, right?

The middle of this range is 10.5V. So I have designed Vc to be biased ad 10.5V.

For maximum output,

=================
CALCULATING RE
=================
ie = ic + ib
ie = 8mA + 8mA/225
ie = 8.0355 mA

if Ve = 1
and Ve = Re.ie
then

Re = 1/8.0355mA = 124.44 Ω

=================
CALCULATING RC
=================
If Vc is set at 10.5V and Vcc is 18V,
Rc should produce a voltage drop of

18 - 10.5 = 7.5V

so,

Vrc = 7.5 = Rc.ic
7.5 = Rc(8mA)
Rc = 937.5 Ω

=================
CALCULATING R1 and R2
=================
To make the circuit stable I choose i1 to be 10% of iC
so,
i1 = 10% of 8ma = 800 uA.

if Ve = 1 and Vbe = 0.7V, then Vb = 1.7V
__________________
i2 = i1 - ib

R2.i2 = Vb
R2 (i1 - ib) = Vb
R2 = Vb/(i1 - ib)

R2 = 1.7/(800uA - 35.55uA)
R2 = 2223 Ω
_________________

R1 = (Vcc - Vb)/i1
R1 = (18 - 1.7)/800uA
R1 = 20,375 Ω
________________

=================
CALCULATING Cin
=================
First I calculate the input impedance that, as far as I know is equal to

Zin = R1 // R2 // (rπ + Re(hfe + 1))

rπ = kT/qib = 26mV/35.55uA
rπ = 727.09 Ω

So,
Zin = 20,375 // 2223 // 727.09
Zin = 1874 Ω

_____________

As far As I have researched, Cin is calculated to have a reactance equal to 2/3 of Zin at the cut frequency. Suppose 20Hz.

Xc = 2/3 of Zin = 1249 Ω

So

Cin = 1/2πfXc
Cin = 1/2π(20)(1249)
Cin = 6.36uF

___________________

Now I don't know how to calculate Cout. Suppose the collector will drive a 600Ω load (high impedance headphone).

___________________

Two questions:

1) are my calculations correct
2) how do I calculate Zout and consequently Cout, to be connected at the collector?

Thanks.

2. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,693
1,321
Only without the load resistance, this is true.
As soon as you connect the load resistance, the voltage swing will change

The positive swing can not be larger than :

Vo_+max = (Vcc - Vc) * RL/(Rc + RL) = Icq *Rc||RL = 7.5V * 600Ω/(600Ω + 937Ω) = 2.9V

And the negative swing

Vc_sat ≈ Vcc * (Re||RL)/(Rc + Re||RL) + Vc* (Rc||Re)/(RL + Rc||Re) ≈ 3.4V

Vo_-max = Vc_sat - Vc = 3.4V - 10.5V = -7.1V

I do not check the math. But it does not look bad.

And the next thing that you should do is to calculate the Q-point with the values you just selected.

Xc_out < RL at 20Hz

For example

Xc = 0.5 * RL = 300Ω

C = 0.16/(20Hz *300Ω) ≈ 22μF

Last edited: Oct 1, 2018

Mar 10, 2018
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4. ### Audioguru Expert

Dec 20, 2007
10,896
1,245
The schematic does not show.
The hFE of a 2N2222A "might" be 225 but is a range of numbers because some have less gain and others have more gain. A circuit must be designed to work properly with any number of hFE in its datasheet.

5. ### AlternatedCurrent Thread Starter New Member

Oct 1, 2018
6
0
How should I do that?

6. ### AlternatedCurrent Thread Starter New Member

Oct 1, 2018
6
0

are you telling me that with the load the swing will go from 2.9V to -7.1V? How can that be if the power supply is not symmetrical?

7. ### Audioguru Expert

Dec 20, 2007
10,896
1,245
When you learn about transistors then you will design a circuit so that it works with any passing transistor.

8. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,693
1,321
Yes, exactly. This is what I was trying to tell you.
I was talking about the output voltage swing across the load resistance (600Ω) after signal pass the output capacitor.

For the positive half cycle of an input signal the transistor conducts more current. This lowers the collector voltage.
Therefore the output capacitor is discharging through the transistor and the load resistance.
And the current through the load resistance is flowing from the lower part of a load resistor to the upper part.
And this is why we have a negative voltage across the load resistance. Because now the current through the load resistor is flowing in the opposite direction.
On the other hand for a negative half cycle at the input, the transistor collector current decreases.
So, the voltage at the transistor collector increases. And the output capacitor is now charging from Vcc via Rc resistor.
This circuit diagram tries to show the situation.

And the voltage at collector will swing from 13.4V to 3.4V which corresponds to the voltage swing at the load resistance from +2.9V to -7.1V.

In my opinion, you don't have to do anything. Because you are lucky by picking the voltage divider current 0.1*Ic = 800μA.
Because "normally" we are picking the lowest possible Hfe value from the datasheet. And we are using this value in our calculations.

9. ### Audioguru Expert

Dec 20, 2007
10,896
1,245
Where can you buy a 2N2222A transistor that has an hFE of 225? Nowhere. You get whatever they have and they will not test a few hundred to maybe find one. You must design the circuit so that it works properly with any 2N2222A transistor that passes spec's.

10. ### AlternatedCurrent Thread Starter New Member

Oct 1, 2018
6
0
OK, but in theory, a voltage divider biased circuit is not so dependent of hFE like other kinds of design and this amplifier is being designed for the whole purpose of learning.

thanks

Dec 20, 2007
10,896
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Oct 1, 2018
6
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13. ### shteii01 AAC Fanatic!

Feb 19, 2010
4,519
717
Based on input capacitor. I am going to guess that application is audio. So everyone can assume the load is 8 ohm.

14. ### AlternatedCurrent Thread Starter New Member

Oct 1, 2018
6
0
not exactly... the load is 600 ohms, like I said in the example (= high impedance headphone)

15. ### Audioguru Expert

Dec 20, 2007
10,896
1,245
You posted your schematic on another website which might be why it failed on your first post. It shows on your last post but is missing an output coupling capacitor and is missing all parts values.

On this forum site we attach schematics HERE, not somewhere else. See "Upload a file" at the bottom of making a post.

16. ### TANDBERGEREN Active Member

Jan 20, 2014
90
5
Many designs in Audio is very spesific in terms of what Hfe the transistors used have.
Look at the LIA-amplifier design, for instance. There You will get in trouble trying to put in "just any transistor" regarding to Hfe.
Also in many other designs we consider the transistors Hfe closely, and on purpose selects sortings as "A", "B" or "C" for the actual design.

AlternatedCurrent is pointing at the fact that he is learning, and I think You should treat him accordingly.

hobbyist likes this.
17. ### Audioguru Expert

Dec 20, 2007
10,896
1,245
Well, hFE is DC current gain. A transistor like a 2N2222A at 8mA has an hFE from 100 to 310. A very simple circuit designed for this transistor with a measured hFE value of 225 will be cutoff is the hFE is only 100 and will be saturated if the hFE is 300, then your circuit will not work. Make it work by using DC negative feedback. The emitter resistor provides AC and DC negative feedback but maybe not enough. Calculate the Q-point when the hFE is 100 and 300.