# Cascaded RC Coupled BJT Amplifier

Discussion in 'The Projects Forum' started by lozadarod, Feb 21, 2013.

Nov 3, 2012
10
0
Hello guys,

Our professor asked us to design a amplifier then print it in PCB. My amplifier is a Cascaded RC Coupled BJT. So I came up to these values (in the figure). My problem is that the Vout that I measured with my voltage gain (w/ load = 3401) is different from the Vout from multisim which is 63.7. Is it because of the large Vin I inputted that's why the formulas Av1 = - (Rc||Zin2 / re) and Av2 = - Rc / re are invalid? or something else?

And also, another problem is that my Vout is distorted when the Vin is 100mVpkpk, I know I should lower the Vin for it to be not distorted but in our laboratory the minimum Vin we can inputted from the function generator is 80-100mVpkpk. Any idea what I should do?

Tnx for the help guys!!!

Note: The Beta is 133.4630818 if you will ask.

2. ### patricktoday Member

Feb 12, 2013
157
42
The problem is that this amplifer is amplifying way too much for a 9V supply and that's why your output is a square wave rather than a sine wave as in the input. Because of this, dividing your output by the input isn't a valid gain calculation.

You are biasing so that there's 1 volt at both emitters so you will have 1ma of DC current across each transistor. This works out to a re of about 25 ohms. If you want to reduce the gain to a more manageable level and provide some ability to allow higher input voltages (such as from your lab supply), you could insert a resistor in series with each of your emitter bypass caps. About 270 ohms each should get you in range. Then you can replace [(270 || 1k) + 25] in place of re in your formulas. Raising the resistors' values will reduce the gain and therefore allow higher input voltage without clipping the signal.

Also, your AV2 calc should include the load resistor along with the collector resistor so that should be (Rc || 10K).

3. ### crutschow Expert

Mar 14, 2008
13,505
3,376
To reduce the output voltage from the function generator you can use a resistive voltage divider.

Nov 3, 2012
10
0
@patricktoday
O_O tried it and it works!! Tnx so much sir!
BTW, where should I replaced [(270 || 1k) + 25] in my formula?

@crutschow
What is a resistive voltage divider? Sry for the noob question

Tnx guys!! I learned a lot

5. ### crutschow Expert

Mar 14, 2008
13,505
3,376

Nov 3, 2012
10
0
Oh ok sir, so what I should do to that? I'm confused I already using a voltage divider in my circuit. Sry again >.<

7. ### patricktoday Member

Feb 12, 2013
157
42
That's great the emitter resistors helped. You listed these as your gain formulas:
Av1 = - (Rc||Zin2 / re) and Av2 = - Rc / re

I was suggesting you replace "re" with the new calculation: [(270 || 1k) + 25]

But Av2 should include the load resistor in parallel.

So let's say NewRe = [(270 || 1k) + 25]

Av2 = (3.6k || 10k) / NewRe
Av1 = (3.6k || 10k || 2.2k || {transistor 2 base impedance}) / NewRe
and
A_total = Av1 * Av2

8. ### crutschow Expert

Mar 14, 2008
13,505
3,376
I'm confused also. Your originally stated the output of you function generator was too high. The voltage divider is to reduce that voltage.

Nov 3, 2012
10
0
@patricktoday
Tnx! I got it. It's almost the same with the multisim. BTW sir, can u explain to me what is the use of 270ohms in the circuit. And, why the Vout became not distorted?

@crutschow
Hahaha, ok I got it now. Tnx! XD

Also guys, when I increase the Vin @ 200mVpkpk the signal became distorted and its voltage gain value became too far with my calculated voltage gain value. Why it is? Tnx again

10. ### patricktoday Member

Feb 12, 2013
157
42
The reason that it was distorted in the first place was that your total gain X your p-p source voltage was higher than your supply, 9V. You put an emitter bypass cap in place which effectively changed the emitter-side resistance from 1k to about 25 ohms. That increased the gain factor greatly. You could build the circuit with _only_ the 1k resistor and it would work but the gain would be very low. So adding a resistor in series with the bypass capacitor creates a "happy medium" where you can get a gain somewhere between the two extremes.

Try adding a second oscilloscope trace to the emitter of your second transistor and look at both channels using the DC option on the scope. I bet somewhere between 100mV p-p and 200mV p-p on your source you'll see that the emitter voltage swing "bumps against" the downward swing of your collector voltage. Try changing the source to different values to see this.

When you build an amplifier circuit it's always going to have a maximum input voltage and exceeding that is always going to cause distortion.

Nov 3, 2012
10
0
@patricktoday
Oh, ok I got it now!! Super Tnx!! I know now how to explain it to my prof. In my latest configuration, my voltage gain here is 53.05 and in the multisim the voltage gain is about 49.95 (@ Vin = 100mVpkpk) almost the same.

"When you build an amplifier circuit it's always going to have a maximum input voltage and exceeding that is always going to cause distortion."

So sir how will I know the maximum input voltage in my circuit? Is there any way to know it? Tnx!

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12. ### patricktoday Member

Feb 12, 2013
157
42
That's great! Glad to hear that.

Well, just think about math for a minute, not electronics. Your supply is 9V, you are going to lose, say, 1V off the top and the bottom due to the transistors' Vbe drop for one thing, so you have 7 volts max for your output swing. Your gain is about 50. So what times 50 makes 7?