P3A DC calculations problems

martinho

Member
2013-01-31 1:43 pm
Hello! I will construct P3A for my college project, but before constructing it i have to do some calculations so i started it with DC.

---> link to P3A amp 60-80W Power Amplifier

I am using Proteus ISIS on my college and i try to check all calculations with it.

THE BIGGEST problem is the calculation of Q5 base current, here is pic from Proteus:

Vbe multiplier #2.jpg

i dont have a clue how to calculate it. I understand how to get Vbe multipliers current, but i dont understand how to get that Q9 emmiter current (or Q5 base current).

and one minor problem is in differential amp - in Q1 and Q2 base currents.

Q1 base current is smaller then Q2`s because base resistance of Q1 is bigger (24200 ohm, while Q2 resistance is 22k). This is all clear but i still dont have an idea to calculate it.

Diffamp #1.jpg

Thank you very much
 

martinho

Member
2013-01-31 1:43 pm
It is impossible to solve this circuit without using Shockley equation and some iterative methods for solving nonlinear equations.

thanks for answering

you mean shockley equation for a current source because it has a diode? i know its stupid but i look at that diode as resistor in DC analysis (i measured current and voltage and got 1600 ohms resistance). I know thats impossible (diode=resistor) but it eased things little bit, i will go back on it later when i calculate other things.
 

jony

Member
2004-06-26 11:39 am
wroclaw
Well I was talking about this equation
Ic = Is*e^(Vbe/Vt) .
And you cannot replace a diode with a resistor. Because diode static resistance is not constant. It will change with the current. We have nonlinear device here.

But why you want to calculate all this currents with such a precision?
Normally we don't need to know the exact value of a current in the circuit.
It's almost impassible to know exact value. Because we don't know exact value of a Hfe, Vbe, Is... etc

In most of a time we do simplified DC analysis.
We start with Q3

IcQ3 = (Vled1 - VbeQ3)/R7
So if we assume Vled1 = 2V and Vbe = 0.65V we have

IcQ3 = (2V - 0.65V)/560R = 2.4mA

And

IcQ1 = IcQ2 = IcQ3/2 = 1.2mA.

Also we can easily find Q4 collector current

IcQ4 = (Vee - VbeQ6)/(R9 + R10) = (35V - 0.65V)/6.6K = 5.2mA

And Vbe multiplier voltage (between collector end emitter) is equal to
VceQ9 = VbeQ9*(1 + R16/VR1).
 

martinho

Member
2013-01-31 1:43 pm
Well I was talking about this equation
Ic = Is*e^(Vbe/Vt) .
And you cannot replace a diode with a resistor. Because diode static resistance is not constant. It will change with the current. We have nonlinear device here.

But why you want to calculate all this currents with such a precision?
Normally we don't need to know the exact value of a current in the circuit.
It's almost impassible to know exact value. Because we don't know exact value of a Hfe, Vbe, Is... etc

I wanna calculate precisely because it is a work for a college. I approximately calculated Hfe, Vbe, but no Is and Ut and so on.. So yeah, i will have to do it other way. Thanks for that!

In most of a time we do simplified DC analysis.
We start with Q3

IcQ3 = (Vled1 - VbeQ3)/R7
So if we assume Vled1 = 2V and Vbe = 0.65V we have

IcQ3 = (2V - 0.65V)/560R = 2.4mA

And

IcQ1 = IcQ2 = IcQ3/2 = 1.2mA.

Also we can easily find Q4 collector current

IcQ4 = (Vee - VbeQ6)/(R9 + R10) = (35V - 0.65V)/6.6K = 5.2mA

And Vbe multiplier voltage (between collector end emitter) is equal to
VceQ9 = VbeQ9*(1 + R16/VR1).

Thanks for taking your time, i really appreciate this