Need some help on BJT amp design problem

Discussion in 'Homework Help' started by cwilli85, Sep 10, 2008.

1. cwilli85 Thread Starter New Member

Sep 10, 2008
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I'm having to design a BJT single stage amp for my electronics II class and can't get it even started.
The supply is suppose to be 14V
and the signal is 200mVp-p with rs=1.5k
The ouput is to be 6 Vp-p across a 47k ohm load
The lower cutoff freq has to be less than 35Hz while the upper cutoff freq must be between 50kHz and 200kHz
I have to use a 2N2222 trasistor: hfe=75

Anyone have any suggestions on how to go about getting this done?

Last edited: Sep 10, 2008
2. mik3 Senior Member

Feb 4, 2008
4,846
63
First, think what bias you want to apply to the transistor and then calculate your Q-point to be in the middle of the supply voltage as to have the maximum excursion of the amplified signal. Then calculate the input capacitor reactance to be ten times less than the thevenin equivalent resistance on the base of the transistor at 35 Hz. This is to ensure that signals with frequencies above 35 Hz are amplified with almost no losses. If you will use a common emitter amplifier then calculate the bypass emitter capacitor a proper value as not too lose gain at frequencies as low as 35Hz. Also, calculate the output capacitor with the same thinking as the input capacitor. For the upper cutoff frequency consider how the gain of the transistor changes with frequency.

3. cwilli85 Thread Starter New Member

Sep 10, 2008
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I will be using common emitter biasing

4. JoeJester AAC Fanatic!

Apr 26, 2005
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Sketch out your schematic and post it. You can label the componets without their values.

Then everyone would be on the same page as you work your way to the solution.

5. cwilli85 Thread Starter New Member

Sep 10, 2008
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The circuit will be set up as above

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Last edited: Sep 14, 2008

Sep 10, 2008
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7. Audioguru New Member

Dec 20, 2007
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896
Your circuit over at PhotoBucket (it should have been posted here) will work fine if you calculate the value of its parts correctly. Use simple arithmatic.
Use Ohm's Law and the formula for a coupling capacitor, 1/(2 pi RC).

8. cwilli85 Thread Starter New Member

Sep 10, 2008
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0
I can't get the freq right

9. cwilli85 Thread Starter New Member

Sep 10, 2008
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None of the values i plug in work out right

10. Audioguru New Member

Dec 20, 2007
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896
The calculation for a coupling capacitor's value is simple arithmatic. It is 1/(2 pi RC) which is 0.16 times RC. Then the response is down 3dB (half power or 0.707 voltage).