The parts i add are all standard
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
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
Make R4 and R5 the same value...and observe that
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
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
About R7... you already has the most important...the knowledge, the decision, about
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
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
.........................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................
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................- Status
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