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Voltage Gain of a common-emitter amplifier

  • #1
Hi,
i have a problem with the voltage-gain of a common-emitter-amplifier.
cev_398.gif

I have calculated the theoretical value of the gain given by ##G=\frac{\partial U_{a}}{\partial U_{e} } = -h_{FE} \frac{R_{c} }{R_{v} }## , which is between -13925 and -7311 for ##420\leq h_{FE} \leq 800## (given by the datasheet of the transistor )
(other important values =## U_{B}=+9V; R_{C}=4,7 kOhm; R_{V}=270 Ohm; U_{e}=0,6V*\sin(2\pi *1kHz*t)##
I have plotted ##U_{e}##and ##U_{a}##with an oscilloscope, and got ##U_{a} = +9V ## for ##U_{e}## smaller than the minimal Base-Emitter-Voltage which is 0,5V .
If ##U_{e} \geq 0.5V##, ##U_{a}## sinks to 6,5V.
I understand the qualitative process of ##U_{a}##, but not the quantitative (##U_{a}##should be much smaller, like -6000V, for ##U_{e} \geq 0.5V## ).
So why is the absolute of ##U_{e} ## so small, when it theoreticly should be much bigger.
(Btw sorry for using U instead of V for Voltage, but i already asked this question on a german physics-form and didn't want to redo the formulas.)
Thank you for your help ^^
Edit: Formulas didnt show properly, Used Transistor: BC548C
 

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Answers and Replies

  • #3
:welcome:


Which transistor? Please link the datasheet. PF, we try to always provide links to our sources.
I've attached the datasheet, its the transistor BC548C :)
 
  • #4
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The circuit you have posted lacks a biasing circuit that may allow it to operate in class A.
As depicted, it´s a class C amplifier, something used with a tuned circuit in RF applications.
Stupid question: Could it be possible some biasing resistors are missing?
 
  • #5
CWatters
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I don't think its a biasing problem. He's using it in a large signal mode or as a switch.

With the base driven hard the transistor should saturate and Vcesat is something like 0.6V in the data sheet. If it's not going below 6.5V I think something must be wrong.

Check resistor values by measuring them.

You say with "Ue > 0.5" the collector drops to 6.5V but what are you actually making Ue?. The base voltage might have to be >0.7 to stsrt turning the transistor on. What happens if Ue is say >2V?
 
  • #6
CWatters
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I have plotted ##U_{e}##and ##U_{a}##with an oscilloscope, and got ##U_{a} = +9V ## for ##U_{e}## smaller than the minimal Base-Emitter-Voltage which is 0,5V .
I think you missunderstand. The base emitter ON voltage could be as high as 0.77 according to the data sheet. So transistor could be off with Ue =O.5V.

This parameter isn't the value you must feed it. It's the threshold at which it turns ON. That could lie anywhere between 0.5 and 0.77 depending on production tolerance. So you need at least 0.77 to be sure it's on and less than 0.5 to be sure it's off.
 
  • #7
LvW
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It is the voltage gain of the stage that is requested - and the gain is a small-signal parameter.
Hence, the BJT must first be DC biased in order to allow an operation point within the quasi-linear region of the transfer characteristic.
Otherwise, no voltage gain can be defined..
The most simple solution would be to have an input voltage that contains a suitable DC portion (a rather unrealistic case).
 
  • #8
CWatters
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Ok you could be correct. Lets ask the OP...

Are you just expecting the transistor to switch its output from 0-9V or amplify a small input signal?
 
  • #9
Ok you could be correct. Lets ask the OP...

Are you just expecting the transistor to switch its output from 0-9V or amplify a small input signal?
I expect it to amplify a small imput signal
 
  • #10
I think you missunderstand. The base emitter ON voltage could be as high as 0.77 according to the data sheet. So transistor could be off with Ue =O.5V.

This parameter isn't the value you must feed it. It's the threshold at which it turns ON. That could lie anywhere between 0.5 and 0.77 depending on production tolerance. So you need at least 0.77 to be sure it's on and less than 0.5 to be sure it's off.
So it may be that the voltage i fed the transistor with was to small, hence the small gain?
 
  • #11
You say with "Ue > 0.5" the collector drops to 6.5V but what are you actually making Ue?. The base voltage might have to be >0.7 to stsrt turning the transistor on. What happens if Ue is say >2V?
Unfortunately i will not be abel to reproduce the experiment untill wednsday, but what you are saying makes sense to me
 
  • #12
CWatters
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I expect it to amplify a small imput signal
Ok ignore what I said in #5. You do need to bias the transistor into the correct region as others have said. There are several tutorials on the web and youtube. For example..

 
  • #13
rude man
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if you apply a small signal to your circuit the output will be practically zero since even in the positive halves of the input signal the base current will be negligible.

As someone else said, this probleem has no sensible answer.
 
  • #14
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Ok ignore what I said in #5. You do need to bias the transistor into the correct region as others have said.
I must admit that I do not understand the above quoted comment. The questioner has asked for the "voltage gain of a common-emitter amplifier".
CWatters, can you please explain WHY the transistor does not need a dc biasing "into the correct region as otheres have said"?
 
  • #15
CWatters
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Read the bit you quoted again. It says "You do need to bias...."
 
  • #16
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Sorry - I do not know why I have misinterpreted your answer.. Perhaps because - as a non native english speaker - it is to me quite unusual to say "you do need" instaed of "you need". Please excuse me.
 
  • #17
CWatters
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No problem.
 
  • #18
rude man
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You definitely need a bias network including feedback in oirder for the ciruit to work.

If someone asked you to design a common-emitter amplifier you simply would not design the circuit shown in post 1. That circuit is not an amplifier. it is a practical joke.
 

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