1 Ohm's Law

　　　　　　E=I*R

　　This formula is read descent of potential is E when current I flow on R.
This is not current I flow on R when charged by potential E. The reason why
will be shown later.


2 Give any potential to any point of circuit.

　　We can set up any potential by deviding resister.



　If I want to get bias voltage of 2V, set R1:R2=10:2 and it will be all right.



3 Separate by vertical lines

　I will show you an example with this simple circuit.






Line 1 : +Vcc is devided by R1 and R2, and it makes a potential of point A.

　Potential of point B is 0.6V less than that of point A, because Vbe=0.6V in
bipolar transistor.

　If the potential of B is determined, we can set any value to the current at 2nd
stage by changing R.(Because potential of B is that of voltage regulator.)

4 Formula" I=E/R" is permitted only when E is a voltage regulator.

　　I will show you that current I flow on R when charged potential E is wrong.
In ordinary case power supply has an impedance.

　 　　When the impedance of power supply is r, +Vcc descends as current flow.
So the potential of C will be

　　　　E-I*r


　　Therefore formula "I=E/R" is permitted only when E is a voltage regulator or I is
small enough.

5 Line 2 is a class A stage

　　Now we can choose any value of R3, then what it should be?
The answer is in the fact this stage is class A. In case of 2SC1775, the upper limit of Ic at low
distortion is about less than 10mA. So it is adequet when Ic is set at 4mA. In that case Ic ranges
from 0 to 8 mA when signal is in.


　　


　　Imagine Ic's behavior
　　　　
　　Then choose R4 as power supply will be most efficient.



　　　 　
　　Potential of C is center point of Vcc and Ve.

6 Line 3 : Class B push-pull stage

What's about potential of D? Substract Vgs of N channnel FET from potential of C,
this is the potential of D. But this is thought to be potential C-0.6V, because of diode's
roll of voltage regulator.

　　

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