start at the output and work back to the input transistors.
Assume that the output bias is XmA. (2mA is a good starting guess but I suggest you look at 1mA, 2mA and 4mA of output bias to see what happens to input Vbe).
The output Vre=[4.7*X]mV
The base of the output devices sits at output [Vbe + 4.7X]mV away from output voltage.
The voltage @ the bases of the output pair exactly equals the Vbe of the input pair.
Input Vbe=output Vbe+4.7X
Now look up the datasheet and find what current is required in the input stage to require that input Vbe to exist.
If it does not equal the input bias current then select another output bias current and try again.
But, go back and read PMA again. Set 12mA for the output bias and do the calcs for the devices & values PMA has selected.
This is simple arithmetic, no EE qualifications needed here.
Assume that the output bias is XmA. (2mA is a good starting guess but I suggest you look at 1mA, 2mA and 4mA of output bias to see what happens to input Vbe).
The output Vre=[4.7*X]mV
The base of the output devices sits at output [Vbe + 4.7X]mV away from output voltage.
The voltage @ the bases of the output pair exactly equals the Vbe of the input pair.
Input Vbe=output Vbe+4.7X
Now look up the datasheet and find what current is required in the input stage to require that input Vbe to exist.
If it does not equal the input bias current then select another output bias current and try again.
But, go back and read PMA again. Set 12mA for the output bias and do the calcs for the devices & values PMA has selected.
This is simple arithmetic, no EE qualifications needed here.
Hello again,
I wanted to let you know I found my answer. I just forgot to really take into account the Vbe serie internal resistor in my calculus because I thought it is so little it could be quasi neglected. I was wrong! I would have liked someone here to point it to me but whatever... I could "do-it-myself" after all and that's the real point...
Thanx
I wanted to let you know I found my answer. I just forgot to really take into account the Vbe serie internal resistor in my calculus because I thought it is so little it could be quasi neglected. I was wrong! I would have liked someone here to point it to me but whatever... I could "do-it-myself" after all and that's the real point...
Thanx
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