# Basic Class A amplifiers

Discussion in 'Homework Help' started by tm2019, Mar 28, 2011.

1. ### tm2019 Thread Starter New Member

Mar 28, 2011
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0
Im having real trouble getting my head around class A transistor amplifiers.

I have to design a class A amp with a gain of 20db with a wide bandwidth.

I have discovered a formula for the gain , which is Rc / Re ( Collecter resistor , and emitter resistor).

The problem is there are alot more aspects of the curcuit, which I don't quite understand. And I'm thinking that the gain formula is definetly not the only calculation I have to make.

Does anyone have some links , and / or a brief explanation I need to get my head round this!

The DC and AC aspects are confusing, and my lecturer is not the best at explaining things!

Cheers

Tom

2. ### tm2019 Thread Starter New Member

Mar 28, 2011
6
0
I have managed to construct a curcuit that has a gain of 10.

My questions now:

What is the purpose of the DC potetiometer?

Why is a coupling capacitor used on the emitter?

3. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
Are you able to post your circuit schematic?

4. ### tm2019 Thread Starter New Member

Mar 28, 2011
6
0
I have posted the curcuit on multisim ?

I showed the curcuit to my lecturer (R2) should be 20K. He said that the collector current will be too small.

So I just lower the ratios of R4 and R3.

What sort of current is needed for this curcuit its a superheterdyne reciever. This is supposed to be the 1st stage of amplification from the atenna.

5. ### hobbyist Distinguished Member

Aug 10, 2008
773
62
Hi,

Please Put your circuit on your computer screen, then hit the key "print screen" then it can be copied over into MS Paint, and saved as a jpg. file.

It will then show up on the forum.

6. ### tm2019 Thread Starter New Member

Mar 28, 2011
6
0
Im I being stupid or can you not see the curcuit i have posted in post 2 ?

Dec 26, 2010
2,147
300
Well, I can't see it.

8. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
783
tm2019

I can't see the schematic either.

You should be able to see it in your post as you review the thread posts. If you can't see it, then neither can anyone else.

Did you use the attachments tool (the one with the paper clip symbol) on the tool bar above the text window?

9. ### tm2019 Thread Starter New Member

Mar 28, 2011
6
0
Oops looks like this forum doesnt work with linked images.

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10. ### hobbyist Distinguished Member

Aug 10, 2008
773
62
Hi,

For a class A amp. it is good to have around half the supply voltage across, R4.

If you use the value you chose, then this would be one way to get the values of the other resistors.

First the 100K is a very large resistor for this application, so that's why I can use this as an example, because through this example, you could then use these equations to design for a more feasable output resistance.

VC = VCC/2

VC / RC = IC

so 6v. / 100K = 60uV.

IC x R3 = VE = 0.6v.

Vbe = 0.7v for a silicon transistor.

so VE + Veb = VB = 1.3v.

For now choose R2 to be around 10 x > than R3 = 100K
Now ID is the divider current that will flow through R1 and R2
so ID = VB / R2 = 1.3v. / 100K = 13uA.

R1 = (VCC - VB) / ID = [(12 - 1.3v) / 13uA.] = ~823K

These values are very outrageous, because the collector current is very low, due to a very high collector resistor, but use this method to design a class A amp, with a collector resistor of around 1K to 10K and see what you get with your simulation.

11. ### tm2019 Thread Starter New Member

Mar 28, 2011
6
0
Ok im confused now

how does this curcuit have a gain of 10 ? It dosent follow the rule rc / re ?!?

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