So I've done this is the schematic i made, I know we have to go CC CE CC but im not sure what values i should make the resistors and capacitors.Please give it a try.
We know the input and output are AC coupled. So draw capacitors at input and output. The input cap can be small because the impedance >20k. The output cap is large because the load is 600 ohms. Note it must pass 20hz. Do you know how to choose the capacitor?
I know you can not use a op-amp. But you can start out with a op-amp and get the gain of 50, the input capacitor and output capacitor parts working. Then go beck and design a simple 3 stage op-amp circuit using transistors.
There is a comment about 3 stages. Input, gain, output.
The output stage needs to pull almost to 0V and almost to 10V.
Do some part if this and come back and ask questions.
Supply=10V. 2N3904 current gain larger than 100.input resistance no smaller than 20K
Just a quick question for you...
We went of the input resistance very briefly so I dont really remember it. The impedance at the input of the transistor would be R1//R2 correct? So we would have 1.6k. If you increase both R1 and R2 to 40k you would have 20k as the valuse so i can put anything higher.Did you study this in class?
Supply=10V. 2N3904 current gain larger than 100.
Set Q1 base=2V (I just picked a number)
Just to keep the math simple R2=2k, R1=8k total = 10k. Do you see how that works?
R3 will change things a little. If R3=1k and the gain is 100 then the Base will look like 100k to ground. (Ask questions)
100k parallel with R2=2k. So the 100k will not change things much.
Do some math: what is the impedance at the input of the amp? Tell me how do you do this.
View attachment 214238
How will you change the values of R1...R3 so you have a value larger than 20K. You can pick any number. 40k,100k, 25k...
Now to find the value of C1:
Some where you have a formula for F=1/(something)*(something)*R*C. Find that formula in your book.
F= must be smaller than 20hz. So pick a number. 10hz
R= the input impedance of the input state. (larger than 20k)
C=? That is the unknown
If you get really lost; run the simulation. Do only this one transistor circuit. (frequency sweep from 10hz to 100khz) See what happens. Change the value of C1 and see what happens to the frequency response at the low end.
This is the first time doing something like this so I am kind of lost. How would i go about making 2 transistors share the gain. ThanksJust a quick question for you...
Why do you have only one stage with gain when you have three transistors to work with?
The two voltage followers do not provide voltage gain and from the spec's you need at least 50 with no load and nearly that with load, so that leaves the entire gain of 50 tasked to that one transistor Q2.
A single transistor that has to handle a gain of 50 all by itself is going to produce more distortion as a byproduct than if two transistors share the gain If two transistors share the gain then each transistor stage only has to provide a gain of approximately 7, which is much much less than 50. That means less overall distortion. For two transistors that share gain equally the gain for each transistor is the square root of the entire gain. The square root of 50 is about 7.07 which is close to 7 or if you like you can use 7.1 to be sure.
So i would advise to design something that uses two of the transistors to share the gain.
Oh, well do you know how to design a one transistor amplifier with a given gain?This is the first time doing something like this so I am kind of lost. How would i go about making 2 transistors share the gain. Thanks
Hello again,If im only given the gain i would not know how to do it but if im given other parameters i might know how to do it. Also i have a question how do you get the quiescent current drawn from the power supply
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by Luke James
by Luke James
by Jake Hertz