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shaiqbashir

Class A amplifier Formulae help!!!

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HI guys!

well! i have got two question with me. Some forumlae i know and some i dont! please help me in solving it.

here is the 1st question:

There is a single stage class A amplifier which has

Vcc =20V
Vceq = 10V
Icq = 600 mA
RL = 16ohms
Ip=300mA.

We have to find:

1) DC Input Power
2) DC Power Loss in load resistor
3) AC power developed across load resistor
4) DC power delivered to the transistor
5) DC power wasted in transistor
6) OVerall efficiency

okz, now i know how to find (1) and (6). What should i do for remaining. Following im wrtitng what i think about them. Please tell me if im wrong or correct. If im wrong then what is the correct formula:

2) P = (Icq^2) RL
3) P=0.5 Icq x Vceq
4) dont know
5) dont know

Please tell me those formulae which i donot know and also make those confirmed which i have written above.

OKz here is another question to be considered:

Determine the following values for the class A amplifier when operated with maximum possible output signal:

1)Min transistor power rating
2) AC output power
3) Efficiency
4) Maximum Load Power

okz so here is what i think about these formulae:

1) this is the power at Quiscent point. that is
P = Icq x Vceq

where Icq is simply the Ic of the circuit and Vceq is the Vcc/2

2) that would be P = 0.5 Icq x Vceq

3) Efficiency = 0.5 Icq x Vceq/ Vcc x Icq

4) for max load power, it think i should use this formula :

P = (Vc^2)/RL

where i can find Vc = Vcc - Ic Rc

where RL is the capacitively coupled Load Resistor.

Please tell me that have i written the above formulae correctly or not. If not then what are the correct formulae.

I shall be thankful to u for this act of kindness.

take carez!

Good Bye!



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Hi Shaiq,
I simulated your 2nd circuit. Because it doesn't have negative feedback and because it has a resistor instead of a current source as its transistor's DC load, it has high distortion that changes its DC bias. This pic shows about 10% distortion on its output. The top part of the waveform is squashed.

post-1706-14279143187771_thumb.png

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Hi audioguru!

thanks for that precious simulation.

But very sorry to say that i couldnt understand what u r trying to say with this.

the circuit u have drawn is exactly correct. One mistake that i myself has made above is this that R1 = 4.7k not 5.7 k. so please make it correct in ur calculation.

Please explain me what u r trying to say with this simulation diagram. I shall be thankful to u for giving me these answers which i require. Because i desperately need them.

THanks a lot in advance!

take carez!

Good Bye!

SB--


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Hi Shaiq,
Transistor amplifiers have a very high amount of distortion without negative feedback. The 1st circuit I simulated had a voltage gain of about 120 and distortion of about 10%. The distortion is much higher when its levels are increased.

I removed the capacitor across the emitter resistor to give it plenty of negative feedback, so now its voltage gain is only 1.65. Even though its output level is much higher than before, its distortion is much less, maybe only 0.1%.

If a constant current source replaces R4 then its distortion will be much less and therefore its gain can be increased but it will still have low distortion.

I changed R1 from 5.7k like before to 4.7k like now and it made a small difference.

post-1706-1427914318796_thumb.png

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The transistor doesn't make much difference.
A transistor without negative feedback and a low impedance voltage input signal produces a very distorted output signal when the output level is high. Because the base voltage change does not produce a linear collector current change. It is exponential.

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Of course properly designed transistor amplifiers can have low distortion. They have negative feedback to reduce the distortion. The opamp with only 0.00008% distortion has transistors with plenty of negative feedback.

The circuit I simulated has a very low impedance voltage source as its input signal and the first circuit has a very high voltage gain. The base-emitter voltage change from the input signal does not produce a linear change in collector voltage, it is exponential.

If a series base resistor is added then the distortion is much less because then the transistor is driven with current instead of voltage. But the total voltage gain is reduced.

If an unbypassed emitter resistor is added to provide negative feedback as shown in my second simulation then the distortion is much less but the voltage gain is also much less.

Here is a single transistor with a voltage gain of only 3.7 without a load. It has plenty of negative feedback but its distortion is pretty bad when compared to designs with more transistors.

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