Homework Help: Single Stage Common Emitter Amp Design

1. Nov 5, 2012

charkins

1. The problem statement, all variables and given/known data

I am designing a voltage divider common emitter amplifier given the following parameters

Av= 40 dB +/- 1dB
Zi> 1000 ohms
Zout< 5000 ohms
Stability factor = 5
Beta*re= 650 ohms
Beta minimum = 35
Beta maximum = 300
Vcc = 20 volts
Rsource = 470 ohms
Vce= 10 volts
Ve= 2 volts
We can use the approximation that Ic=Ie
My teacher has given us a "hint" saying to work backwards from the AC analysis.
I need to solve for Re, R1, R2, Rc and the respective currents.

2. Relevant equations[/b
Ic=Ie
Ib=Beta*Ic
Vb=(Vcc(R2))/(R1+R2)
Vcc-Vce-Ic(Re+Rc)=0
re=(26*10^-3)/Ic
Stability Factor=Rthevenin/Re
Av in db=20log(Vo/Vi)

3. The attempt at a solution

I am completely lost where to start with this circuit. So I have not attempted it.

2. Nov 5, 2012

aralbrec

This sounds like a discrete design using a voltage divider to bias Vb, Rc and Re to help bias and set gain, capacitor on the emitter to bypass Re for common emitter small signal gain, capacitor coupled source attached to the base and capacitor coupled load at the collector.

But it's speculation until we can see a circuit diagram.

3. Nov 6, 2012

charkins

http://tinypic.com/view.php?pic=2ew1t0l&s=6

I believe this is how the circuit should be constructed

I just cannot figure out how to solve for the missing values with the given parameters.

4. Nov 7, 2012

aralbrec

You have quite a bit of information given, some of it is ac performance and some of it is dc bias conditions. You will need to calculate those specifications in terms of circuit components, which will lead to simultaneous constraints on the component values.

The only way to find these constraints is to start writing equations. As a first step, I would look at the dc circuit and try to determine the ac transistor model parameters. You have a few explicit dc bias conditions: Vce = 10 volts, Ve = 2 volts. This also specifies the dc voltage across Rc. If you know the dc current, you have Rc and Re specified. One thing that grabs my attention is that βre = 650 ohms. This is because re depends on the bias current so you may be able to use this information to find the dc bias current. Now β is always statistically variable so this is not the best way to find the dc bias current if there is another way. The other use for βre is in computing ac input impedance since this will be the value of re reflected into the base side of the circuit (β>>1).

With the dc voltage at the emitter, you know the voltage at the base, which lets you find a ratio for R1 and R2. From there you will have to look at the ac model. The ac input impedance will place a second constraint on R1 and R2, which (given Re and βre), you will be able to solve. If you couldn't find Re before, the output impedance may place a second needed constraint on Rc, which will specify Re from dc bias conditions. Actually that may be one of the first steps to find Rc and Re.

As you can see, you need to write down the equations and keep in mind what the specifications are telling you about component constraints.

I see you've moved onto a JFET in similar configuration so maybe you've left this one already but the problem is almost the same.

Last edited: Nov 8, 2012