The power supply
VR1, C29 and C28 supply the regulated DC voltage for the circuit. VR1 regulates the input, which should be around 12VDC input to J5, to a level of 10VDC for use in the unit. Yes the internal supply voltage indicates it is 9VDC and here is the reason why. You will notice that diode D1 is connected between VR1 and the power switch. This diode is to prevent voltage being applied to the output of VR1 from battery BAT1. Diodes are neat things because they will only conduct current in one direction such as in this case when a positive voltage is applied to the anode of the diode with respect to the cathode. (OK… the cathode is the side of diode to which the arrow points in the schematic or the end with a band or ‘mark’ on it in the actual device. The anode is the other side. The cathode likes to be negative with respect to the anode.) Anyway, you never get something for nothing and in the case of the diode when it is conducting there is about .6 volts lost across it so the voltage supplied to the circuitry is around 9.4 volts but we’ll just call it 9VDC.
You will notice that the power jack, J5, has a switch contact which disconnects the battery from the circuit when external power is being used. This prevents the battery from attempting to charge when you are using an external power supply. It is not a good idea to attempt to charge a battery, in fact it can cause a battery to explode, if the battery is not designed to be charged. NEVER attempt to charge a standard alkaline or carbon/zinc battery.
SHA2 • 7