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1st August 2007, 03:28 PM  #1 
diyAudio Member

AB class poweramp help understanding ?????
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 .

2nd August 2007, 12:02 AM  #2 
diyAudio Member

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 
3rd August 2007, 01:53 PM  #4 
diyAudio Member

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

3rd August 2007, 02:14 PM  #5 
diyAudio Member
Join Date: Jun 2007

55Meg BMP Zipped, oh my god...
Why not convert that to an about 10kB GIF?! 
3rd August 2007, 09:46 PM  #6 
diyAudio Member
Join Date: Feb 2004
Location: Recife  Brasil Northeast

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
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Power supply; ripple, filter, noise, stability and the use of capacitance multiplier... Portuguese and English; https://www.youtube.com/watch?v=sSm0ku1eIgg 
3rd August 2007, 10:01 PM  #7 
diyAudio Member
Join Date: Feb 2004
Location: Recife  Brasil Northeast

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
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Power supply; ripple, filter, noise, stability and the use of capacitance multiplier... Portuguese and English; https://www.youtube.com/watch?v=sSm0ku1eIgg 
3rd August 2007, 10:04 PM  #8 
diyAudio Member
Join Date: Feb 2004
Location: Recife  Brasil Northeast

Well.... this is a very long thing..other folks will continue, as you want to learn
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
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Power supply; ripple, filter, noise, stability and the use of capacitance multiplier... Portuguese and English; https://www.youtube.com/watch?v=sSm0ku1eIgg 
4th August 2007, 12:39 AM  #9 
diyAudio Member
Join Date: Aug 2005
Location: Ktown

.........................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 runaway. 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................
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All the trouble I've ever been in started out as fun...... 
4th August 2007, 12:47 PM  #10 
diyAudio Member

thxs

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