Audio Amplifier Design

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Electrical09

Joined Dec 5, 2009
1
I am working on a problem to design an audio amplifier. It must have a voltage gain Av = 952, an input impedance of Z = 50Ω, and a load impedance of RL = 50Ω. To complete this design I am able to use resistors of any value, capacitors of any value, 9 volt batteries, and npn transistors with β =100 . It is assumed that the temperature is 300K.

In my calculations so far I have been unable to get a voltage gain anywhere near 900. I am having trouble combining the transistors together which I believe is the way to go about solving it.

Any help would really be appreciated.

Thanks
 
I am working on the exact same problem, let me guess Jazi's class for Industrial Electronics?

What do u think the load impedance is? The load resistance or RL || RC ?

I need help with this too, just can't figure out the resistance values to get the right Voltage gain.
 

TrevorP

Joined Dec 8, 2006
55
You could do something like two or three biased and cascaded NPNs with by-pass capacitors to get really high gain. Then use an emitter-follower to drive the low 50ohm resistance, it will reduce your gain a bit but not as bad as without it.
 
just confirming, if we use a darlington pair, only the voltage drop doubles and the DC gain (beta) increases to beta squared?

does it affect anything else?
 

ftsolutions

Joined Nov 21, 2009
48
Just be aware of the total power and current you are passing through the pair configuration - the power dissipation of either transistor must not be violated. It may be easier to just use (2) similar common emitter stages each with a gain of ~ 31 or so, followed with an emitter follower as suggested.
 
we can only use npn transistors.

If I use a gain of say 30 (sqrt of 900 Voltage Gain) for each, how do i go about calculating values for the resistors? we aren't given anything for the values of resitors

Av = (RL' * beta)/rpi

How do i get Icq if i dont have any resistor values? is it best if i just use trial and error?
 
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so i did a few calculations. we are allowed only 9 volt batteries.

Vcc = 9V

Vb = 4.5V also implies R1 = R2

Rl' = RL || RC where RL = 50 according to question (is load impedance RL?)

i'm just really confused, I can't even get the PSpice simulation to work with a gain of 100 for which already have R values.

what am i doing wrong?
 

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Audioguru

Joined Dec 20, 2007
11,248
Didn't your teacher tell you that the voltage gain is about Rc/Re?
So your gain is only about 5.5 without the 50 ohms load. If the external load is 50 ohms then the gain is almost nothing.

You don't want the base of the transistor to be biased at half the supply voltage, you want the collector at half the supply voltage so it can have the maximum amount of voltage swing.
 
So the load impedance is RL || RC ? since RL can't be 50 ohms?

the question kinda confuses me.

according to my professor Av = ((RL||RC) * beta) / Rpi, also in one of the example he solved RC = RE and voltage gain is -106.
 
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JoeJester

Joined Apr 26, 2005
4,390
Jony,

He has corrected it in his posting. His diagram didn't have the junction of R1/R2 connected to the base of the transistor, so the transistor only had the input signal for bias.
 
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