Circuit shows a small signal BJT audio amplifier. Design the amplifier with a small signal voltage gain of 50. You may chose an appropriate transistor.

The circuit diagram is attached herewith.

 

I got the small signal voltage gain from ac analysis as
50 = (Ib*β(Rc))Vi
Since Ib = Ic/β
50 = (Ic*Rc)/Vi
If I assume the input voltage is 0.01V
From the datasheet of BC546
I set operating point of the transistor to Ic = 10^-2 Amp, Vce = 5 and Vcc = 10
50 = (0.01*Rc)/0.01

Rc = 50 Ohms

Is this correct?

 

No, the voltage gain is equal
Av = Rc/re ≈ Rc/(26mV/Ic) ≈ 40 * Ic * Rc

 

Thank you, If i may ask, how did you find the value of re as 26mV/Ic ?

 

What small signal model do you use ?

 

Emitter model,
When I do the calculation I get β in front of the equation.
Av = β(Rc/re)

 

Beta is current gain, not voltage gain.
Your transistor circuit has no negative feedback so its voltage gain will be about 180 if the load resistance is high and it will be distorted.

 

Emitter model,
When I do the calculation I get β in front of the equation.
Av = β(Rc/re)

I've never heard about this model. But the correct voltage gain equation look like this.

Av = (Rc*β)/( re *(β+1)) ≈ Rc/re

And re is a small signal internal emitter resistance.

re = dVbe/dIe = Vt/Ie ≈ Vt/Ic ≈ 1/gm

Where
Vt is a thermal voltage equal 26mV at room temperature
http://en.wikipedia.org/wiki/Boltzmann_constant#Role_in_semiconductor_physics:_the_thermal_voltage

 

Okay thanks a lot. You have used the Base model nah?
So if I set the operating point to Vcc = 10V and Ic = 10 mA
Can I get the value for Re from below equation straight away which will also satisfy the goal of a voltage gain of 50?

By applying KVL to the output loop,

Vcc = IcRc + Vce + IeRe

To set the Q point to the middle
Vce ≈ Vcc/2

Ic ≈ Ie

Re = (5 - IcRc)/Ic

 

For such a small voltage gain you shoudl remove CE capacitor and then
Av = Rc/(RE + re). It's not good idea to have low gain amplifier with CE capacitor.

 

Note that any load resistance will be effectively in parallel with Rc so it will reduce the gain. So calculate the AC gain of 50 to be (Rc//Rl)/(RE unbyp + Re).

I think for a voltage gain as high as 50 then you need two emitter resistors in series. One is bypassed for DC stability and the other in series the the internal emitter resistance to set the AC voltage gain.

 

Then how to find R1 & R2???

 

Then how to find R1 & R2???

1) Allow the total of the two series emitter resistors to be 1/10th the resistance of the collector resistor.

2) The current in the emitter resistors is almost the same as the current in the collector resistor so use Ohm's Law to calculate the emitter voltage at the quiescent current.

3) Look at the datasheet of the transistor to see what its base-emitter voltage is at the quiescent current and add it to the emitter voltage to see the required base voltage.

4) Look at the typical hFE of the transistor on its datasheet at the quiescent current and calculate the typical base current.

5) Calculate the R1 and R2 voltage divider so that its current is 10 times the base current and makes the required base voltage.

 

It would probably take about five minutes to breadboard this thing in the lab on a proto board using a 2N2222 transistor and a box of resistors. We had a similar assignment when I was a senior back in 1978. That's how we did it, but we didn't have the "advantage" of computers......

We had to use a signal generator and an oscilloscope.