Help required in some amplifiere problem

[shaiqbashir]

OKz here is the 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!

 

[berkeman]

Hi again. As I said in your other thread about class A amps, we need more info and hopefully a schematic. Can you post something?

 

[shaiqbashir]

Hi audioguru!


we do have a schematic, due to some problems, i cannot attach a schematic, but help u visualize that.


It is a simple common emitter configuration class A amplifier.


With R1 = 5.7 k , R2= 1K

R1 and R2 forms a voltage divider bias.

Vcc = 24V

Rc= 330 ohms
RE = 100 ohms

RL= 330 ohms

beta = 150

there is a capacitor bypassing RE to ground. the value is not given
there is a capacitor coupling RL to the collector, the value is not given.
I hope this is much info, please visualize this circuit. and remember it is for the second question.
I shall be looking forward for ur precious help.

Take carez!

Good Bye!

SB--

 

[Averagesupernova]

The quiescent power consumed by the whole circuit is the power supply voltage multiplied by the current drawn from the power supply. What you have written in your first post is the power dissipated by the transistor. The collector and emitter resistor will dissipate significant power also.
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I also find that the collector voltage will be about 9.5 volts. So over half the power supply voltage is already lost across the collector resistor. So the voltage between the collector and emitter CANNOT be one half of Vcc.
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The power in the load is ONLY AC. You cannot use the quiescent voltage on the collector to determine this. When there is no AC (signal) on the collector, no power is dissipated in the load.
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There may be other mistakes, I skimmed it kinda fast and don't have the time right now.