Wiring Diagram

In operation, with no input, the DC voltage at pins 1,2 and 3 of the op-amp will be about 4 volts, and the voltage on the (+) inputs to the 3 comparators (pins 5,10,12) will be about a half volt less due to the 1N914 diode drop. The voltage on the (-) comparator inputs will be around 5.1 and 6.5 which is set by the 560 and 750 ohm resistors.

When an audio signal is present, the 10uF capacitor connected to the diode will charge toward the peak audio level at the op-amp output at pin 1. As the volume increases, the DC voltage on the capacitor and also (+) comparator inputs will increase and the lamp will turn on when the (+) input goes above the (-) input. As the volume decreases, the capacitor discharges through the parallel 100K resistor and the lamps go out. You can change the response time with a larger or smaller capacitor.

Circuit operation is as follows. The 555 timer is set up as a multi-vibrator 12hz. The output on pin 3 drives the 4027 flip-flop. This flip-flop divides the input frequency by two and delivers complementary voltage outputs to pin 15 and 14. The outputs are connected to LED1 and LED2 through the current limiting resistor R3. The LED's are Arranged so Pls That the polarity across the circuit is one way only one LED will from light and Pls the polarity reverses the other LED light earnest, therefore Pls no transistor is connected to the tester the LED's will from alternately flash. Also The 4027 outputs are connected to resistors R4 and R5 with the junction of these two resistors connected to the base of the transistor being tested. With a good transistor connected to the tester, the transistor will of turn on and Produce a short across the LED pair. If a good NPN transistor is connected then LED1 will from flash by Itself and if a good PNP transistor is connected then LED2 will from flash by Itself. If the transistor is open both LED's will from flash and if the transistor is shorted then neither LED will from flash.

I've drawn the circuit with a single pole relay. But you can use a multi-pole relay if it suits your application. Only one half of the Cmos 4013 is used. So you could construct two independent toggle switches with a single IC. The circuit will work at anything from 5 to 15-volts. All you need do is select a relay with a coil voltage that suits your supply.

The LED provides a visual indication that the relay is energized. In effect - it tells you whether the switch is on or off. It's not necessary to the operation of the circuit. If you wish you may leave out R3 and the LED.

The collector of the NPN feeds the base of a larger PNP power transistor which supplies most of the current to the load. The output voltage will be about 0.7 volts below the voltage of the wiper of the 1K pot so the output can be adjusted from 0 to the full supply voltage minus 0.7 volts. Using two transistors provides a current gain of around 1000 or more so that only a couple milliamps of current is drawn from the voltage divider to supply a couple amps of current at the output.

Note that this circuit is much less efficient than the 555 timer dimmer circuit using a variabe duty cycle switching approach. A fairly large heat sink is required to prevent the PNP power transistor from overheating. The advantage of the circuit is simplicity, and also that it doesn't generate any RF interference as a switching regulator does. The circuit can be used as a voltage regulator if the input voltage remains constant.