They don't amplify since they have no voltage gain. They supply current with the polarities you said.Does this quasi works in this way? Q3 & Q5 amplify the positive signal while Q4 & Q6 amplify the negative signal.
Correct. Q5 is the output and does most of the work. Q3 is the driver with a fairly low current.1. So i would get a higher current at Q5(e) compare with Q3(e).
Correct. R10 is adjusted with no signal for about 25mA to 50mA in the output transistors (measured across R7) to eliminate crossover distortion.2. R10 is for the bias tuning of Q3 and Q4?
You mean the drivers B3 (b) and Q4 (b).3. Will i get a different polarity waveform at Q1(b) & Q4(b) since they are for different polarities waveform?
Correct.if i tune the volume to lowest point but i still can hear the audio playing means i have wrong connection right?
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.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?
hi guru,I think your pot has a broken track or you used a linear pot for R9 instead of a logarithmic audio taper volume control.
Cant get what u mean. U mean the link i post or??That is a real volume control. It is too bad it is Chinese and they don't have any.
yes of course.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.
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.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?
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 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.
The very simple input of the amplifier has a fairly low input impedance therefore will not work properly with a 100k volume control.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?
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.And regarding the log POT that i posted, is it the correct one that u mentioned? If yes i would plan to buy it.
by Aaron Carman
by Duane Benson
by Jeff Child