Help!!! Audio Amp
Help!!! Audio Amp
Can anyone please tell me how this circuit works please?!
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Re: Help!!! Audio Amp
It's an old post, but the slightly simplified answer is:
The input comes into J1 on the left, and C1 (probably about 0.1uF to 2.2uF - the higher the value the better the low frequency response) is a 'coupling capacitor' there to prevent any DC flowing to the op-amp.
P1 is a variable resistor to set the proportion of the input going to the op-amp, effectively setting the volume.
This signal goes into the non-inverting input on the op-amp. (The '+' and '-' labels correspond to 'non-inverting' and inverting', NOT the supply voltage which are the top and bottom vertical lines numbered 8 and 4)
The output of the op-amp flows via a potential divider network of R2, D1, R3, R4, to the bases of Q1 and Q2. C5 allows the full signal to flow to the base of Q2, while the potential divider network holds the transistors in a 'half-on' state. Therefore the signal applied to the base of each transistor will affect it's forward current (ie; how hard it is conducting). Q1 is an NPN transistor, while Q2 is a PNP transistor. NPN transistors conduct more as their base becomes more positive, and PNP transistors conduct more when their base becomes more negative.
This arrangement creates a 'push-pull' effect on the output.
Q3 and Q4 basically increase the power output. (When Q1 gets a positive input, it conducts more, bringing the base voltage of Q3 lower than its collector voltage, making Q3 conduct more also. This makes the output "harder" positive. The same positive input signal would be applied to Q2, which would make it conduct less, stopping the flow of current to Q4, turning it off. ...And vice versa)
The capacitor / resistor network shown at the bottom of that schematic is a negative feedback loop to the op-amp's inverting input. The more this network allows the signal to flow back to the inverting input, the less gain the op-amp has. The arrangement shown is for use as a tone adjuster. By adjusting P2, you can alter the level of higher frequencies being filtered back to the op-amp's inverting input.
That just about covers it I think....
Oh, apart from the power supply circuit below it!
This is just your typical dual-rail power supply circuit, with a power indicator LED on the positive rail. 230v AC in : +16v DC, 0v ground, -16v DC out.
The input comes into J1 on the left, and C1 (probably about 0.1uF to 2.2uF - the higher the value the better the low frequency response) is a 'coupling capacitor' there to prevent any DC flowing to the op-amp.
P1 is a variable resistor to set the proportion of the input going to the op-amp, effectively setting the volume.
This signal goes into the non-inverting input on the op-amp. (The '+' and '-' labels correspond to 'non-inverting' and inverting', NOT the supply voltage which are the top and bottom vertical lines numbered 8 and 4)
The output of the op-amp flows via a potential divider network of R2, D1, R3, R4, to the bases of Q1 and Q2. C5 allows the full signal to flow to the base of Q2, while the potential divider network holds the transistors in a 'half-on' state. Therefore the signal applied to the base of each transistor will affect it's forward current (ie; how hard it is conducting). Q1 is an NPN transistor, while Q2 is a PNP transistor. NPN transistors conduct more as their base becomes more positive, and PNP transistors conduct more when their base becomes more negative.
This arrangement creates a 'push-pull' effect on the output.
Q3 and Q4 basically increase the power output. (When Q1 gets a positive input, it conducts more, bringing the base voltage of Q3 lower than its collector voltage, making Q3 conduct more also. This makes the output "harder" positive. The same positive input signal would be applied to Q2, which would make it conduct less, stopping the flow of current to Q4, turning it off. ...And vice versa)
The capacitor / resistor network shown at the bottom of that schematic is a negative feedback loop to the op-amp's inverting input. The more this network allows the signal to flow back to the inverting input, the less gain the op-amp has. The arrangement shown is for use as a tone adjuster. By adjusting P2, you can alter the level of higher frequencies being filtered back to the op-amp's inverting input.
That just about covers it I think....
Oh, apart from the power supply circuit below it!
This is just your typical dual-rail power supply circuit, with a power indicator LED on the positive rail. 230v AC in : +16v DC, 0v ground, -16v DC out.