How this automatic powr on off circuit working?

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Herewith attached the automatic power on circuit of an car audio. It is used as auto power on when there is audio input. Audio amplifier starts working after some time of audio signal input.
When there is no signal for 30 seconds it becomes automatically off.

This circuit is not becoming off after 30 seconds. many times it is on for long hours in car even when audio signal is not present. Kindly explain me how to modify or remove this problem. So please guide about working of this circuit.
 

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Hi Jaya,

First, where you +5V marked, should be a ground. Second, i assume that circuit is powered by Vcc+= +5V and Vcc-=-5V bipolar power supply.

U2C is comparator. Its negative input is set around +1V by R59,55,56. So if signal on pin 10 is >1V, the output is at high, say +3.5V. Conversely, if input 10 is below 1V, the output goes close to the Vcc-, lets say -3.5V.

D11, R53 and C40 is a rectifier/peak detector. They rectify U2B output and present it to U2C input. So if you have approximately 0.7V ac RMS output from U2b, it will be rectified and trigger U2C to high level.

U2B is just a noninverting amp with DC gain =1 and AC gain approximately 700. Thus you need 1mV AC signal at the input of U2B to trigger the circuit on.

Now, how to trigger the circuit off? for this there is R57 and D12. when there is no signal, C40 slowly discharges through R53,R54 until voltage on C40 reaches 1V. that triggers U2C and its output becomes negative. thus D12 becomes conductive and discharges C40 completely through R57.

That's it in nutshell.

Now, I suspect you have some noise at the input (may be some hum). remember. you need only 1mV to turn the circuit on and after that it will stay on as long as the noise is there. So, this will be the first place I check.

Good luck.
 
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