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Bypass capasitor distorts signal in CE amp.

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Increazon

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Hello. I trying to findout how to calculate Common Emmiter Amplifier.
When i simulating board, i get unexpected distorts of signal.
Ok. I built this scheme ->
NPN Common Emitter Amplifiers

Now look at pics.
This is with bypass emiter capasitor, range can be from 5 to ∞ µF, same result.


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This is oscilogram

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As you see - bad work of amplifier :mad:
Now lets remove capasitor


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And look at oscilograme

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Its just ideal! What we need.

More info. This Simulation - in Altium designer, also i check in Multisim and Proteus - same result!

  1. Why this happens?
  2. Why we need this capasitor?
  3. What right equation to calculate this capasitor?
 
hi,
I will ask you a question.:)

I see that your test frequency is about 5kHz, so what is the capacitive reactance of a 36uF capacitor at 5kHz.?

EDIT:
In case you are unable to find the formula its Xc = 1/[2 *pi* f *C ] = 1/ [ 2 * pi * 5000 * 0.000036] = ......
 
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And I will ask a second question:

What is the gain of the amplifier with the capacitor?

oh and the third question

What is the gain of the amplifier without the capacitor? and what is the difference in gain between the two set-ups, with and without capacitor?

Hint: you will have to reduce the input signal voltage to get a meaningfull answer here.

JimB
 
Without the bypass capacitor across the emitter resistor, the voltage gain of the circuit is 4. The input peak voltage of 0.25V is amplified to 1.0V.
With the bypass capacitor across the emitter resistor then the voltage gain is at least 120 so the 0.25V input is trying to make the output have a peak of at least 30V which is impossible so it is severely clipped.

Simply turn down the input signal level.
 
hi,
I will ask you a question.:)

I see that your test frequency is about 5kHz, so what is the capacitive reactance of a 36uF capacitor at 5kHz.?

EDIT:
In case you are unable to find the formula its Xc = 1/[2 *pi* f *C ] = 1/ [ 2 * pi * 5000 * 0.000036] = ......

Thanks for answer! My points of view:
  • The signal generator is set to 20 Hz, look at pics (or your arguments why?).
  • Ok. Xc = 0.88, but what does it give for us?
  • According to the recomendations, given on this page NPN Common Emitter Amplifier Bypass Design Ce >> 1/(2 * pi * f * Re). So it must be 1/(2 * 3.1415 * 20 * 220) = its 36 uF ! Look, if we have 5000 Hz as you sad, it will be 0.14uF. Ok. The next, equation sad Ce >> (its mean a much more, but how much more???), so?

I jusm simulated again: 20Hz, 0.14uF = very good, 5000 Hz, 36 uF = very bad. Where i fail?
 
Without the bypass capacitor across the emitter resistor, the voltage gain of the circuit is 4. The input peak voltage of 0.25V is amplified to 1.0V.
With the bypass capacitor across the emitter resistor then the voltage gain is at least 120 so the 0.25V input is trying to make the output have a peak of at least 30V which is impossible so it is severely clipped.

Simply turn down the input signal level.

Yeah, resistor in series with signal done good job. So we need bypass capasitor for increaze circuit gain? Like negative callback? Is there way to calculate "right" bypass capasitor for big signals?
 
Thanks for answer! My points of view:

  • The signal generator is set to 20 Hz, look at pics (or your arguments why?).
  • Ok. Xc = 0.88, but what does it give for us?
  • According to the recomendations, given on this page NPN Common Emitter Amplifier Bypass Design Ce >> 1/(2 * pi * f * Re). So it must be 1/(2 * 3.1415 * 20 * 220) = its 36 uF ! Look, if we have 5000 Hz as you sad, it will be 0.14uF. Ok. The next, equation sad Ce >> (its mean a much more, but how much more???), so?


I jusm simulated again: 20Hz, 0.14uF = very good, 5000 Hz, 36 uF = very bad. Where i fail?

hi,
Look at the waveform you have posted, I have marked the image.

A frequency of 20hZ would have a period of 50millisec.!!

EDIT:
Added LTS image of frequency response
 

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Last edited:
And I will ask a second question:

What is the gain of the amplifier with the capacitor?

oh and the third question

What is the gain of the amplifier without the capacitor? and what is the difference in gain between the two set-ups, with and without capacitor?

Hint: you will have to reduce the input signal voltage to get a meaningfull answer here.

JimB

With Ce, gain = 26, without = 4.

Yes now i see problem. Thank you!.
Just one more. How can i calculate circuit gain (with and without Ce) to design amplifier for needed input signal. I need new equation!
:D
 
The voltage gain of a silicon transistor is the collector resistor value (in parallel with the load resistance) divided by the unbypassed emitter resistor value (in series with the internal emitter resistance of the transistor).
The internal emitter resistance is 0.026 divided by the emitter current in amps.

A transistor with its emitter resistor bypassed with a capacitor has a high voltage gain and high distortion.
An unbypassed emitter resistor adds negative feedback which reduces the voltage gain and reduces the distortion.
 
hi,
Look at the waveform you have posted, I have marked the image.

A frequency of 20hZ would have a period of 50millisec.!!

EDIT:
Added LTS image of frequency response

Yeah, i fail, because of Altium designer dont change circuit elements values, only label! I remake simulation in Multisim, everything working

Multisim.png

At this stage i cant analize in my ming LTS image, just know that if it lower -3 db (or 3)
is bad.

Now i will be searching equation, that show dependance of circuit gain vs bypass capasitor. Thanks.
 
The value of your emitter resistor's bypass capacitor is too low so the gain at 20Hz and distortion are low.
The capacitor value should be at least 10 times higher (try 470uF to 1000uF) for maximum voltage gain at 20Hz.
 
120 uF gives circuit gain 48. Output amplitude 12 V without distortion. What are methods to find "safe" capacistanse?
 
I have never heard of a "safe" capacitance. What is it?

If the emitter resistor bypass capacitor has a reactance that is one-half the transistor's internal emitter resistance then the gain is nearly as high as is possible. A transistor has high distortion when its gain is high because then it has no negative feedback.
 
Actually mistake was in signal amplitude. I using principles and equation for small signal. Small signal is 1 mV. I guessed that small signal is 0.25 V (its 250 times more!). Thats why i get all this.
At latest test i used bypass capasitor up to 1F! and still get no distortion, and voltage gain around 150. Thats because goal of bypass capasitor is make ground for AC signal. So it capasistance can be as high as possible. We calculating it for money and place saving. Because all you know, that capasitor with higher capasistance is more expensive ant size is greater
 
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You do not need a huge emitter bypass capacitor. The lowest frequency that you need full gain determines the value of the capacitance.
If you want full gain at 50Hz then the capacitance can be reasonable small and inexpensive.

A transistor with its emitter resistor bypassed with a capacitor has high gain and produces severe distortion as I show in my simulation. The distortion is about 40% when the input is 25mV peak and makes it difficut to measure the voltage gain.
 

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The voltage gain of a silicon transistor is the collector resistor value (in parallel with the load resistance) divided by the unbypassed emitter resistor value (in series with the internal emitter resistance of the transistor).
The internal emitter resistance is 0.026 divided by the emitter current in amps.

A transistor with its emitter resistor bypassed with a capacitor has a high voltage gain and high distortion.
An unbypassed emitter resistor adds negative feedback which reduces the voltage gain and reduces the distortion.

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When we using bypass capacitor, how the equation of gain looks likes? Of course if Ce is too big, Re is short circuited and deleting from denominator. But if i want regulate gain with Ce?
 
**broken link removed**

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When we using bypass capacitor, how the equation of gain looks likes? Of course if Ce is too big, Re is short circuited and deleting from denominator. But if i want regulate gain with Ce?
You can have an emitter resistance so that the transistor has some negative feedback and less gain and less distortion.
If you regulate gain with the value of Ce then low frequencies have plenty of negative feedback, low gain and low distortion and high frequencies have high gain and high distortion.
 
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