Author Topic: design simple amp  (Read 1034 times)

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Offline NASKTopic starter

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design simple amp
« on: July 16, 2018, 03:26:09 am »
i need to understand basic bjt amplifier concept therefor I design this simple amp but i didn't getting required out put using calculation values. could i know reason for it?  (i have attached the calculation & out puts)
furthermore, using Cin & Cout calculated value didn't come to at least near to the required output  value. but change it to Cin 10uF and Cout 15uF it pass the 1V peak but not come to 2V peak (design value) .Is it not possible to achieve Av=200 from Bc548
 

Offline oPossum

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Re: design simple amp
« Reply #1 on: July 16, 2018, 04:03:15 am »
2N2222 has much lower Hfe than BC548
 

Offline Zero999

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Re: design simple amp
« Reply #2 on: July 16, 2018, 08:09:56 am »
Use lower resistor values for the biasing potential divider, if possible, so the Hfe is less of a factor.
 

Online Kleinstein

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Re: design simple amp
« Reply #3 on: July 16, 2018, 08:16:08 am »
For AC the R5 is effectively parallel to R1 and thus the gain is reduced.
 
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Offline LvW

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Re: design simple amp
« Reply #4 on: July 16, 2018, 04:03:37 pm »
Without any external load - and assuming that all capacitors are large enough for coupling the 1kHz - the actual gain is Av=-6k/25=-240.
Because I do not see that the load is specified, the remaining task is to find the proper load resistor for reducing the gain to -200.
That is a simple task.
In case, the load is given the effective gain determining resistor is 6k||10k=3.75 kohms and the gain can never reach the value of -200.
(Av=-3.75k/25=-150) .
Comment: Modifying the base resistors cannot help at all because it is the transconductance (gm=1/re) that matters only (as far as gain is concerned).
« Last Edit: July 16, 2018, 05:40:57 pm by LvW »
 
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Offline NASKTopic starter

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Re: design simple amp
« Reply #5 on: July 17, 2018, 03:51:43 am »
Thank you all for support.

I did a mistake calculating Cin coupling cap & it needs to be correct as 1/zin = (1/R3) + (1/R4) + (1/β*re). Not the 1/zin = (1/R3)+ (1/R4)+ (1/β*(R2+re) now  Cin is 75nF.

Hi LvW
I made a mistake to choose external load thank you for reminding. therefor I use this equation Av=Zout/ (re+Ze) instead of Av=Zout/ (re); because if Zout < R1 voltage gain is reduce by R5 however if R5 >>R1 load make little effect on gain.

Therefor I take re+Ze value as 29.5Ω (true value is 30Ω) otherwise R5 become infinity.
Av=Zout/ (re+Ze)
Zout=5.9kΩ

R5=Zout*R1/(R1-Zout)
R5=354kΩ(proper load )

C1=1/2πf Zout = 27nF

 200mv out is coming @1mv input . when changing to 10mV input output is not symmetrical why is it? .

 


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