10 watt output quasi design

Thread Starter

simpsonss

Joined Jul 8, 2008
173
i'm now trying an error on the bias part. The cd already played for 30 minutes still nothing happen yet.And i changed the R9 from 4k7 to 10K POT.Hope it works. will update soon if anything happen.haha...
 

Audioguru

Joined Dec 20, 2007
11,248
Changing R9 from 4.7k to 10k will not make any difference. When the pot is turned down as far as it goes then the slider connects to ground and the input of the amplifier gets no signal. But you said the amplifier still plays when the volume control is turned down so you have a wiring problem.

R10 adjusts the bias current for the output transistors and many almplifiers use this circuit. Q1 should be touching the heatsink of the output transistors for thermal regulation. When the heatsink gets hot then Q1 turns on a little more which reduces the current in the output transistors so they get cooler.

Without a signal then R10 should be adjusted for 12mVDC to 25mVDC across R7.
You think about how R10 works wrongly. When the resistance between the base and collector of Q1 is low then Q1 turns on a lot and reduces the current in the output transistors. Start the adjustment of R10 like this then slowly turn R10 until you measure 12mVDC to 25mVDC across R7.
 

Thread Starter

simpsonss

Joined Jul 8, 2008
173
hi,

Does this quasi works in this way? Q3 & Q5 amplify the positive signal while Q4 & Q6 amplify the negative signal.

thanks.
 

Audioguru

Joined Dec 20, 2007
11,248
Does this quasi works in this way? Q3 & Q5 amplify the positive signal while Q4 & Q6 amplify the negative signal.
They don't amplify since they have no voltage gain. They supply current with the polarities you said.
Q2 amplifies the voltage of the entire signal.
 

Thread Starter

simpsonss

Joined Jul 8, 2008
173
1. So i would get a higher current at Q5(e) compare with Q3(e).
2. R10 is for the bias tuning of Q3 and Q4?
3. Will i get a different polarity waveform at Q1(b) & Q4(b) since they are for different polarities waveform?

thanks.
 

Thread Starter

simpsonss

Joined Jul 8, 2008
173
hi guru,
u mentioned that if i tune the volume to lowest point but i still can hear the audio playing means i have wrong connection right?

what if, when i tune the R9, i can hear that the volume is getting lower when tune to low and getting louder when i tune higher. But when i tune to the lowest i still can hear a very very low audio playing through the speaker and when i tune just abit slightly higher not more than half of the POT the volume suddenly become very high and loud. Upon my situation, do u think that i'm still having the wrong connection?

thanks,
 

Audioguru

Joined Dec 20, 2007
11,248
1. So i would get a higher current at Q5(e) compare with Q3(e).
Correct. Q5 is the output and does most of the work. Q3 is the driver with a fairly low current.

2. R10 is for the bias tuning of Q3 and Q4?
Correct. R10 is adjusted with no signal for about 25mA to 50mA in the output transistors (measured across R7) to eliminate crossover distortion.

3. Will i get a different polarity waveform at Q1(b) & Q4(b) since they are for different polarities waveform?
You mean the drivers B3 (b) and Q4 (b).
No, because they both have the same input signal.

if i tune the volume to lowest point but i still can hear the audio playing means i have wrong connection right?
Correct.
But maybe the volume control is defective or your soldering is poor. Don't use a breadboard.

what if, when i tune the R9, i can hear that the volume is getting lower when tune to low and getting louder when i tune higher. But when i tune to the lowest i still can hear a very very low audio playing through the speaker and when i tune just abit slightly higher not more than half of the POT the volume suddenly become very high and loud. Upon my situation, do u think that i'm still having the wrong connection?
I think your pot has a broken track or you used a linear pot for R9 instead of a logarithmic audio taper volume control. A linear pot makes the volume very loud until it is turned to almost zero then the volume gets low. A logarithmic volume control reduces the volume when it is turned down a little then at halfway the volume is reduced a lot. At minimum the volume is zero.
 

Audioguru

Joined Dec 20, 2007
11,248
A volume control is supposed to be the type of pot that is logarithmic with an audio taper, because our hearing has a logarithmic response to loudness. I think you used a linear type of pot which causes the problems you have.
 

Thread Starter

simpsonss

Joined Jul 8, 2008
173
hi guru,

about the R9 part. it works like a voltage divider right? for example i'm using a 10K POT and after measure.i notice that when the resistance between input and the E-cap negative leg is ~10.23K ohm, the volume is low. which mean when the resistance between the E-cap negative leg and the GND is low ~3.3ohm, there will be no signal outputted. Am i right?
 

R!f@@

Joined Apr 2, 2009
9,918
simpsonss,
I got good news for you.
I got the permission from Rod, to post his schematics here and for you to refer.
So do you want to go my way and start with something that works for sure.
 

Audioguru

Joined Dec 20, 2007
11,248
about the R9 part. it works like a voltage divider right? for example i'm using a 10K POT and after measure.i notice that when the resistance between input and the E-cap negative leg is ~10.23K ohm, the volume is low. which mean when the resistance between the E-cap negative leg and the GND is low ~3.3ohm, there will be no signal outputted. Am i right?
The voltage divider of 10,233 to 3.3 is a reduction of 3100 times which is about -70dB. The output of the amplifier will be very low but maybe you can hear something with your ear to the speaker.
 

R!f@@

Joined Apr 2, 2009
9,918
OK.. Here's the circuit. I redrew it and if you are planning to build this I can suggest suitable replacement that are easily available.
This is a working circuit and will give you plenty of Power to a 8 ohm Speaker.
Since you wanted a quasi-complementary output stage amp to learn, I gave you this.
This amp needs a Power supply of around 60V at 4 amps per amplifier.

I did not write the values. If you want them let me know, I'll PM you the Part no.'s.

Remember this is a copyright material.
You are warned not to use this material for any other purpose other than to educate ur self.
Courtesy of Rod Elliot
 

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Thread Starter

simpsonss

Joined Jul 8, 2008
173
hi R!f@@,

Thanks for the help and afford, yeap i would like to build this and learn it. But i would also continue for the previous one. Can u please PM the part no's?

thank you.
 

Thread Starter

simpsonss

Joined Jul 8, 2008
173
The voltage divider of 10,233 to 3.3 is a reduction of 3100 times which is about -70dB. The output of the amplifier will be very low but maybe you can hear something with your ear to the speaker.
ok.so if i change the POT to 100K ohm, then the output will be -90db which will be more lower if compare with 10k ohm. So can i change the R9 to a more higher resistance value to solve the problem which there is still a very very low audio outputted when the volume is tune to the lowest?
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And regarding the log POT that i posted, is it the correct one that u mentioned? If yes i would plan to buy it.
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thank you.
 

Audioguru

Joined Dec 20, 2007
11,248
ok.so if i change the POT to 100K ohm, then the output will be -90db which will be more lower if compare with 10k ohm. So can i change the R9 to a more higher resistance value to solve the problem which there is still a very very low audio outputted when the volume is tune to the lowest?
The very simple input of the amplifier has a fairly low input impedance therefore will not work properly with a 100k volume control.
Every volume control will have a different "zero resistance". High quality volume controls are closer to zero.

And regarding the log POT that i posted, is it the correct one that u mentioned? If yes i would plan to buy it.
It will work well as a volume control because it is logarithmic. But it is made in China where many parts are of poor quality. It might have a high "zero resistance". They might be mixed up and when you order a logarithmic pot you might receive a linear pot.
 
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