Electronic – Debouncing Circuit with Schottky Diode

The circuit is "strange" and, as far as I can see, far from optimum.

The diode does not make sense where it is located and serves no purpose. When the TIP42C (datasheet here) is turned on the inductor has ~= 12V applied. When the TIP42C turns off the diode side of the inductor will "ring" positively to a voltage above the 12V supply. The diode will be reverse biased as shown and does not affect the ringing/ flyback voltage. It does nothing and may as well not be there.

C1 is a filter or reservoir capacitor. It especially acts to reove noise from the supply when the motor is running and provides current peaks to the motor. (Both the previous actions are "different sides of the same coin").

The transistor is connected in an extremely unusual manner. The inductor is in its emitter - it is an "emitter follower. There is no obvius reason for doing this. The transistor is turned on by clamping its base to collector. This means that the emitter must be above ground by a Vbe drop. As the emitter approaches the base voltage the transistor starts toturn off (at about 0.6V to 1V above ground range - so the transistor "wastes" some of the supplu voltage.

A better arrangement would be to have the emitter connected to V+ and the motor/coil in the collector circuit.

Overall I conclude that either :

• The circuit is drawn incorrectly (happens :-) )

or

• The person who "designed" it has very little understanding of electronics.

Can you provide a link to an A3144 data sheet please? It is not fully clear how it works.

Input selector in one of my amps worked that way (until I blew the chip up and had to recreate the selector since I could not get another chip).

I do not particularly like programming microcontrollers (why else would I make a dual frequency counter with a lot discrete logic chips?), so I always look for a way to avoid using one.

I made this with real buttons in mind (so I added debouncing). If you actually want to control this with logic output from some detector or chip (that does not need debouncing) you can get rid of R1-R4 and C2-C5. If your control signal is active low ("0" when the button is pressed) then leave the inverters. If your control signal is active high ("1" when the button is pressed) then you do not need to use U2C-U2F and just connect the control signals to the anodes of D1-D4.

How this works: When "1" appears on the anode of, say, D1, it also goes to the input pin of U1. The flip-flop is triggered by the riding edge of the clock. To allow for the input pin voltages to settle to their final values, a circuit made of U2A, R5, C1 and U2B delay the riding edge by a few ms. Actually, you could probably get rid of R5 and C1 too, the propagation delay of U2A and u2B would probably be enough, but I have no way of testing it without building it.

R7 and C6 reset the flop-flop to "0" (all LEDs off) state when the power is turned on, otherwise at each turn on it would get a random state.