AB class poweramp help understanding ?????

[prorms]

I have already buit the poweramp but wondering could this style of ab class poweramp be better .I know some part of the poweramp are stander no mater what the voltage on both rails are as long as thier both have the same value. other parts need to be caculated to exact value .

[prorms]

the part i add are all standerd values this parts cant realy be changed but the transistor can changed or crossed reffence with ease the other parts is were iam unsure of the math to get the right value

R4 and R5 have set at about 1.5mA each
R7 has to be about 3mA
math is A ? HERE
R9 AND R2 must be the same value 10k to 47k
how to know which is the right value is unknow to me


I realy very interseted know math to find the rest of this parts
that dont have value

[analog_sa]

Sorry for bringing this up but you had already started another thread about this exciting amp. Why not continue in the old thread?

At least you had figured out an alternative to zipping bitmaps

[prorms]

this thread will let me know how caculate the math for the other parts resistor if i every wanted to make a louder amp

[Dzsoni]

55Meg BMP Zipped, oh my god...
Why not convert that to an about 10kB GIF?!

[destroyer X]

the voltage you will have over R4 extremes will be the same voltage you will need from Q4 base to the positive supply line.

Q4 is using a emitter resistance...this is used, normally, because the mirror side you have with 1.2 volts..... so.... Q4 base to positive line will measure 1.2 volts....good idea to split this in two parts...0.6V to be from Q4 base to emitter and 0.6 volts to be over R10.

So.... you will need to produce 1.2 volts over R4 and R5... and the current you choice was 1.5 miliamps to each one.... 0.0015A to each one.

What will be the resistance that will have 1.2 volts when 1.5 miliamps will be crossing it..... this is the use of ohms law.

Resistance (in ohms) is equal the voltage (in volts) divided by the current (in amperes).

So.... the resistance will be the result of 1.2 divided by 0.0015A... the resistance value.... for R4 and R5 is 1000 ohms

[destroyer X]

the current that will cross Q3 junction.... the colector to emitter junction.... it is 3 miliamps as you concluded.

Observe that exist 2 diodes into the base.... this means 1.2 volts from base to the negative power rail...because each diode will keep around 0.6 volts.

Once again you will divide the voltage.... half of that (0.6 V) will be to the base to emitter junction...and 0.6V will be over R7.

So... again ohms law; the resistance you want will be 0.6 volts divided by 0.003A..... the result is 200 ohms.

regards,

Carlos

[destroyer X]

to calculate the amplifier DC points.

Go ahead... and good luck.

I gave you my contribution..forum has thousands of guys....they can help you too.

regards,

Carlos

[CBS240]

.........................Q6 is the constant current source for the single ended VAS, Q4. VR11 will be the same as VR7 at 0.6V for the same reason. This means R10 and R11 are the same value, 0.6V/Iq. BTW, adding a base resistor to Q6 would keep the voltage constant on R7 in case Q6 becomes saturated so it dosen't affect LTP current via Q3. R9/R6 obviously sets Av.(voltage gain) R2 & R6 should be same value to help keep the LTP balanced. The collector to base caps on Q4 & Q6 is miller compensation. This sets the frequency poles to prevent occilation. Q5 is the Vbe multiplier and adjusts output bias with temperature to prevent thermal run-away. R12, 13, & 14 bias Q5. C5 provides an AC current path around Q5 to sink the current needed to turn off Q9 & 10, and Q11 & 12. Transistors require current to turn off as well as turn on................

[prorms]

thxs