The supply of circuit becomes from a AC line in the J1. This voltage can be from a separate transformer 2X12V (the prices of materials that I give it is for 2X12V AC), from existing coil 12V in their M/T of power amplifier or if it cannot become somebody from the two, then from the coils of mainly supply final amplifier, adapting always the prices of resistances R1/2 and R3, proportionally the price of voltage that is supplied the amplifier, according to the law of Ohm and the fall of voltage that we want to achieve (R=V/i). The voltage that it should we have in point A, before the IC2, should is bigger than + 15V 200mA, the IC2 supplies all the relay and led.
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The remainder circuit is supplied by the R3/D9. When we supply the amplifier with voltage of network (220V AC), charge the C6 via the R4, the price in the entry of IC1a is (H) exit (L) Q1- RL1, is in cutting off. In line with being first the M/T of power supply, intervenes the RX, which ensures smooth connection the M/T in the network, avoiding the burn of fuses, specifically if the force power supply, is big. After 1sec after charge the C6, his negative pole goes to 0V, the entry of IC1A becomes 0V (L), conduct Q1 closes the RL1, short the resistance RX and all the voltage of network is applied in the M/T. Simultaneously turns on LD 1. Via the R5 charge slow the C7 (~5sec), when charge the situation in the pin5 of IC1b become (H), (the other are already (H) from the R23), exit is (L) and the exit of IC1C (H), the Q2 drive the RL2, giving the output of amplifiers in loudspeaker. Simultaneously via the R13 charge the C8 (~2 sec). Hardly charge the C8, conduct the Q3 and close the contacts of RL3, at the same time with those of RL2. The circuit is in complete operation. If we interrupt the line of network all the supply?s fall very fast, with result all relay is cut off, very rapidly cut off, him loudspeakers. If are presented some continuous voltage in entries J2/1 and J2/4, the two circuits of detection DC, then the Q5 or Q6 conduct and lead the entry of IC1b to pin 5 to 0V (L), with result the exit is become (H), the exit...
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