Common emitter amplifier simulation - 1 mV to 100 mV

Hi,

Why not simply using an Opamp?

If you want to discuss about your circuit, please give a link to the site where the schematic and description is from.

Klaus

joniengr said:

but I am not getting true signal at the output.

Click to expand...

What does it mean? Did you made the same configuration on test board and get different results?

You are biasing base probably too high. So, even with 1V of Vce i doubt that this BJT is in forward linear region (because of Vcb)

Look at the collector voltage (2.1V) to see that it is much too low.
A simple calculation shows that the transistor biasing has it saturated and turned on very hard so that it cannot amplify a signal.

CE Amplifier calculator for biasing -


Also you have not stipulated desired Zin or Zout for the CE amp ?


Regards, Dana.

My circuit has a voltage gain of 115 times, not 100 times. It needs a 2.1 ohm resistor between the emitter and R4/C2 for a gain of 100 times.

Hi,

Now I get a gain of 27. Here is the simulation. Which components I need to change to get 100 gain.

Hi,

Now I get gain of 100.
But the problem is that I have a load of 1 K ohm. If I connect capacitor in series to block the DC and than connect to the load, I don't get the gain of 100.

Hi,

The gain reduces to 45 if I connect a capacitor and a load resistor.

Hi,

you ask rather basic questions. Thus there are many thousad of internet sources who answer them. Documents, internet pages, turorials, online calculators, even videos.
At the same time you don´t tell us which documents/tutorials you read to find the solution on your own.
(even if you follow danadakk´s link you get an online caclulator to play around)
--> please give links

--> Do those simple internet seach "common emitter gain setting" on your own. Show us your effort.

--> please refer to the links, tell us your ideas and your cacluations. Ask more detailed questions.

Mind: "Helping" in a forum has to do with motiviation. The more effort you show, the more you motivate us ... the more and the more detailed answers you get.


Klaus

By overloading the original circuit by adding a 1k load makes it a different circuit and since you do not see why then I am gone. Good bye.

Your 1 uF Xc is 159 ohms at 1 Khz, so you have a V divider with 1 K load of
~ .86.

But thats not the whole story, as pointed out by others here.


Regards, Dana.

Hi,

danadakk said:

Your 1 uF Xc is 159 ohms at 1 Khz, so you have a V divider with 1 K load of
~ .86.

Click to expand...

How do you come to 0.86?

If you just have a 1uF capacitor and a 1k load (igonoring th other circuitry)... then you get 0.9876 times the input voltage at the resistor.

If you add the 1uF + 1k to the existing 1k2 then you reduce the gain to about 0.452 times than before (@R load)

Klaus

KlausST said:

Hi,


How do you come to 0.86?

If you just have a 1uF capacitor and a 1k load (igonoring th other circuitry)... then you get 0.9876 times the input voltage at the resistor.

If you add the 1uF + 1k to the existing 1k2 then you reduce the gain to about 0.452 times than before (@R load)

Klaus

Click to expand...

My error, did not compute classic Ztotal. .98 is correct. Back to school for me.....


Regards, Dana.