Common Emitter Amp Design

Audioguru

Joined Dec 20, 2007
11,248
I said that you should let the voltage across the emitter resistor be about 1/10th the voltage across the collector resistor. You calculated that the collector resistor is 9V/6mA= 1.5k then the total of the emitter resistors is 150 ohms. The collector can swing up 18V - 9V= 9V and it can swing down 9V - 0.9V= 8.1V WHICH IS NOT SYMMETRICAL. That is why I said to let the quiescent collector voltage be 9.5V so the output swing is up 8.5V and down 8.6V which is almost symmetrical which is required.

Why are you now calculating a collector resistor with a current of 120mA?? That is the current when the transistor has a maximum saturation voltage of 0.4V that DOES NOT MATTER in your circuit. When it is saturated then its base current should be 1/10th the collector current that also DOES NOT MATTER in your circuit since your circuit never saturates. You are supposed to use a quiescent collector current of 6mA, not saturated at 120mA.

Rc+Re=18V/9mA is not correct because the resistors should have about 9.5V across them (not 18V) and the current is supposed to be 6mA, not 9mA. Let the quiescent collector voltage be 9.5V. Then Rc has 18V - 9.5V= 8.5V across it and with 6mA then the value of Rc is 8.5V/6mA= 1417 ohms, use 1.5k. The total of the emitter resistors is 0.9V/6mA= 150 ohms.

I did not use a load line, I simply used common sense to let the output swing be symmetrical.
 

Audioguru

Joined Dec 20, 2007
11,248
Select β = 200 and calculate Ib.
At a collector current of 10mA, a 2N2222 transistor has a minimum β of 75 and is not spec'd for typical or maximum so let β be 100.

Split Re for gain of 40.
No.
The requirement is an input of 0.2V peak and an output of 8V peak-to-peak so the voltage gain with a load should be 20. Calculate for a gain of about 23 without a load.
 

Thread Starter

embi

Joined Oct 7, 2012
20
Rc=(Vcc-Vce-Ve)/Ic

Vcc=18V, Vce=8V, Ve=(1/10)Vcc=1.8V, Ic=6mA

Rc=1.36kΩ

Re=Ve/Ic=1.8/6m=300Ω

R2=(1/10)β*Re=1/10*100*300=3kΩ

Vb=(Vcc*R2)/(R1+R2) -- (1)

Vb=Vbe+Ve, Vbe=0.7V, Ve=1.8V
Vb=2.5V

from (1)
R1=R2(Vcc-Vb)/Vb
R1=18.6k

so, R1=18.6kΩ, R2=3kΩ, Rc=1.36kΩ, Re=300Ω

Is this correct?
But how do I find the values for the capacitors and the resistor at the output?
 

Audioguru

Joined Dec 20, 2007
11,248
Rc=(Vcc-Vce-Ve)/Ic

Vcc=18V, Vce=8V, Ve=(1/10)Vcc=1.8V, Ic=6mA

Rc=1.36kΩ

Re=Ve/Ic=1.8/6m=300Ω

R2=(1/10)β*Re=1/10*100*300=3kΩ

Vb=(Vcc*R2)/(R1+R2) -- (1)

Vb=Vbe+Ve, Vbe=0.7V, Ve=1.8V
Vb=2.5V

from (1)
R1=R2(Vcc-Vb)/Vb
R1=18.6k

so, R1=18.6kΩ, R2=3kΩ, Rc=1.36kΩ, Re=300Ω
Is this correct?
No.
The voltage gain will be much too high at about 180. The emitter needs an unbypassed resistor in series to add some negative feedback which reduces the gain and reduces the distortion.
Make a new schematic so you can see everything together.

18.6k is not a standard value. Use 18k then the base voltage and currents are a little different.

[/quote]But how do I find the values for the capacitors and the resistor at the output?[/QUOTE]
The capacitor values are calculated for the lowest frequency you want (1khz).
The output resistor is simple to calculate and loads the output which reduces the output amplitude to set the gain at exactly 20.
 

Thread Starter

embi

Joined Oct 7, 2012
20
Were the formulas I used right?

The voltage gain will be much too high at about 180. The emitter needs an unbypassed resistor in series to add some negative feedback which reduces the gain and reduces the distortion.
what do u mean at 180? and how do i calculate the value for the unbypassed resistor? (what is the formula for it)


18.6k is not a standard value. Use 18k then the base voltage and currents are a little different.
I dont think that matters for this problem. We just use a simulator. (microcap)

The output resistor is simple to calculate and loads the output which reduces the output amplitude to set the gain at exactly 20.
what is the formula I use for this?

To calculate the capacitor values I used Xc=1/10R for the bypass capacitor
but what value for impedance do I use for the other 2 capacitors?

Thank you
 
Last edited:

Audioguru

Joined Dec 20, 2007
11,248
The voltage gain of a transistor is simply the collector resistor value (parallel with the load resistance) divided by the unbypassed emitter resistor value (in series with the transistor's internal emitter resistance). Your latest circuit had the entire emitter resistor bypassed so the voltage gain was maximum about 180 or more.

The requirement is a voltage gain of exactly 20.

The cutoff frequency where the level is reduced by -3dB (times 0.707) is when a coupling capacitor's value is 1 divided by 2 x pi x R. You have 3 capacitors so each one should have a cutoff frequency that is 1/3rd what is needed.
 

Thread Starter

embi

Joined Oct 7, 2012
20
So,
20=[Rc//Rl]/[Re]
Rc=1.36k
but what is Rl and Re? And the internal emitter resistance?


I don't understand what you mean for the capacitors?
for the bypass cap:
the resistor value is 300
so:
X=[1]/[2*pi*f*C]
C=[1/30]/[2*pi*1000]
C=5.31uF

is that right? if not what is the right formula?
 

Audioguru

Joined Dec 20, 2007
11,248
So,
20=[Rc//Rl]/[Re]
Rc=1.36k
but what is Rl and Re? And the internal emitter resistance?
Almost.
The voltage gain is the total collector resistance (Rc parallel to the load resistance) divided by the total emitter resistance (RE in series with the internal emitter resistance.

Every tutorial about transistors says the internal emitter resistance formula (26mV/emitter current in Amps).

I don't understand what you mean for the capacitors?
for the bypass cap:
the resistor value is 300
so:
X=[1]/[2*pi*f*C]
C=[1/30]/[2*pi*1000]
C=5.31uF

is that right? if not what is the right formula?
No.
The reactance (AC resistance) of a capacitor is 1 divided by (2 x pi x f x C).
Then the cutoff frequency is 1 divided by (2 x pi x R x C).

The lowest f in your project is 1kHz so each of the 3 capacitors must have a cutoff frequency of 1kHz/3= 333Hz.
Then C= 1 divided by (2 x pi x 333Hz x 300 ohms)= 1.6uF. Then calculate the other capacitor values.
 

Thread Starter

embi

Joined Oct 7, 2012
20
Vcc=18V, Vce=8V, Ve=(1/10)Vcc=1.8V, Ic=6mA

Rc=(Vcc-Vce-Ve)/Ic
Rc=1.36kΩ

Re=Ve/Ic=1.8/6m=300Ω

R2=(1/10)β*Re=1/10*100*300=3kΩ


Vb=Vbe+Ve, Vbe=0.7V, Ve=1.8V
Vb=2.5V

Vb/R2=(VCC-VB)/R1–0.006/β
R1=17350Ω

so, R1=17.35kΩ, R2=3kΩ, Rc=1.36kΩ, Re=300Ω

I then split Re into r_E and R_E, aiming for a gain of 20:
r_E=1.36kΩ/20=68Ω
r_e=68Ω, rE=232Ω


I calculated R_L for a gain of 20:
20=(R_C||R_L/r_E)
R_L=17kΩ


I then calculated the capacitor values:
Bypass cap:
C_E=(1/(2*pi*f(min)*232))*3
C_E=2.06uF

the other 2 capacitors:
C_1:
R1//R2//Transistor input resistance:
17.35k//3k//1k=718
1/(718*C1)=2*pi*1000
C1=0.22uF

C_2:
1/(R_L*C2)=2*pi*1000
1/(17000*C2=2*pi*1000
C2=9.36nF



I got an output that comes close, but not quite exactly right to the required 4V peak output. Can you see where did I go wrong?
I attached the circuit and Vout.
 

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Audioguru

Joined Dec 20, 2007
11,248
Your gain without the transistor's internal emitter resistance is (17k//1.36k)/68= 18.52.

The internal emitter resistance is 26mV/6mA= 4.33 ohms.

Then the total gain is (17k//1.36k/(68 + 4.33)= 17.4, not 20.0 .
 

Thread Starter

embi

Joined Oct 7, 2012
20
ok, I must have gone wrong with something, because it doesnt matter how large I make the Load the gain can't reach 20.

I even tried 1GΩ and it only gives me a 3.668V peak, only about 18.8 gain.
 

Audioguru

Joined Dec 20, 2007
11,248
1.36k/((68 + 4.33)= 18.8 so you are correct.

Then replace the 68 ohms resistor with 56 ohms:
1.36k/(56 + 4.33)= 22.54 . I said to calculate the resistors for a gain of 23 with no load.
Now calculate a load resistance that will reduce the gain to 20.0 .
 

Thread Starter

embi

Joined Oct 7, 2012
20
i calculated r_E for a gain of 25, so it gave a nice 54.4Ω value for r_E
then I calculated the load to bring it down to 20.

now I have this:
The gain however is about 19.3 (7.72V peak-to-peak output)
 

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Audioguru

Joined Dec 20, 2007
11,248
Your calculation is missing the 4.33 ohms internal emitter resistance of the transistor.
From my post #31: "The internal emitter resistance is 26mV/6mA= 4.33 ohms."
Then your gain without a load is actually only 23.16, not 25.

Are you allowed to design the circuit using non-standard resistor and capacitor values?
 

Thread Starter

embi

Joined Oct 7, 2012
20
ok.
So i changed r_E to be 56Ω
then calculated the load using: ((x*1360)/(x+1360))/(56+4.33)=20
and solving for 'x'. I got 10697Ω. (yes I can use non-standard values for the resistors)
I now get a gain of 19.3275. I'm thinking this is probably as close as it gets. (mathematically)

Now I experimentally increased the Load and got the gain to 19.965 (With R_load=15kΩ)
 

Thread Starter

embi

Joined Oct 7, 2012
20
I tested the circuit for the frequency range, and the gain is smaller at low frequency <~5kHz and at 10MHz the gain is only 16.
 

Audioguru

Joined Dec 20, 2007
11,248
If you calculate so that each of the three capacitors produces a low cutoff frequency that is -3dB (2.828V peak) then all three reduce the output to -9dB (1.41V peak). Simply increase the values of the capacitors.

The transistor with gain cannot have a flat response as high as 10MHz unless you peak the high frequencies. Try adding a low value capacitor parallel to the unbypassed emitter resistor.
 

Thread Starter

embi

Joined Oct 7, 2012
20
How do I calculate the power dissipation of the transistor?

From the datasheet I found that Pmax=500mW

by power dissipation of the transistor, do they mean dissipated in the load?
I found the following equation for that:

P_L=Icq^2*R_L/2


I'm guessing I have to adjust this formula to account for the collector and load resistor in parallel?


I also have to calculate the power dissipated by each component with and without an applied input signal. To do this should I just find the current through a resistor using the simulator and apply P=RI^2?
 
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