Two-Stages BJT CE Amplifier.

Thread Starter

Hamnah Malik

Joined May 13, 2024
5
I am trying to design a two stage pre amplifier and facing clipping in the output can u help me solve this. I need at least 25m Vpp as input . i would really appreciate any help
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ronsimpson

Joined Oct 7, 2019
3,223
How much gain are you trying for?
Input 24mV and the output can't go past 4V. I think you need a gain of 13 for each stage.
Right now, you have a gain of 200 in each stage so 200 X 200 = 40000. Just guessing.
I added two resistor.
I did the math in my head so I might be wrong. Please check my work.
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Thread Starter

Hamnah Malik

Joined May 13, 2024
5
thank u it helped a lot, and my problem is solved
How much gain are you trying for?
Input 24mV and the output can't go past 4V. I think you need a gain of 13 for each stage.
Right now, you have a gain of 200 in each stage so 200 X 200 = 40000. Just guessing.
I added two resistor.
I did the math in my head so I might be wrong. Please check my work.
View attachment 322219
thank u my problem got solved
 

MrChips

Joined Oct 2, 2009
31,091
Remove C2 and C4.
The gain of the first stage is approximately R6/R3.
Similarly, the gain of the second stage is R9/R4.
 

ronsimpson

Joined Oct 7, 2019
3,223
You know the gain is related to Collector Resistor and Emitter Resistor. Approximately the same current flows through each resistor.

You have two different gain formulas to work on. DC and AC
At DC all the capacitors are open circuit.
The Base voltage is set by a voltage divider R7, R1.
The Emitter voltage is 0.65 volts less. Now we know the voltage on the emitter resistor. (and the current)
This same current is in the collector. (I know there is a very small difference, don't think about it for now)
In you case the voltage across the Emitter Resistor, times 3.3, will be across the Collector Resistor. Gain of 3.3.

Now lets work on the AC gain. At 1khz all the capacitors have a very low impedance. Lets say o ohms.
In your circuit you have a Collector resistor of 3.3k and an Emitter resistor of 0 ohms. Remember C2=short.
(Actually the Emitter of the transistor has about 30 ohms hiding inside but don't think about that now.) This is why you have very high gain.
I added a resistor to C2 to set the AC gain to about 13. Now we have an Emitter resistor of 1k//260. Which is about 240 ohms. (doing the math in my head) ("//" is parallel)

Hope this helps. Note I did not use any formulas.
 

LvW

Joined Jun 13, 2013
1,778
The gain will be dependent on the transistor's beta which could be 100-300.
For my opinion, the gain depends NOT on the transistors beta but on the transconductance gm=Ic/Ut.
The beta-value has an influence on the input resistance only.
Two gain stages with different beta values will have exactly the same gain (for identical DC operational points).
Background: The BJT is a transconductance devive (Ic controlled by Vbe).
 

Audioguru again

Joined Oct 21, 2019
6,785
Both transistors have the same parts. (Since they have no negative feedback their distortion is high.)
The first transistor has a measured output of 2.65Vp-p and an input of 29.9mVp-p, then its voltage gain is 2.65/0.0299= 88.63 times.
Then both transistors together produce a total voltage gain of 88.63 x 88.63= 7855 times and the output is trying to be 235Vp-p!
 

dovo

Joined Dec 12, 2019
72
Hamnah, here is your amplifier with two added components to establish the gain of each stage. A 48 mV P-P input results in a clean looking 5.2 V P-P output swing. The 300 ohm resistors set the voltage gain of each stage at roughly 10. I adjusted the capacitor values to roll each stage off below 20 Hz.

The distortion products are shown for a 48 mV P-P, 1 kHz sinewave input. The 2nd harmonic is 48 dB below the fundamental and the 3rd harmonic is down 39 dB. Each harmonic can be summed and the THD calculated. The video shows how to calculate THD from the FFT plot.


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