# Can someone explain the purpose of this transistor?

I have a schematic for a THT design. All diodes are rated for 1A (1N4004). The resistors are 1/4W and the transistor is a BD679 NPN darlington transistor (I am aware that I used a wrong symbol).

This thing is called a Capacitor Discharge Unit or CDU for short. What it does, is to buffer energy in the capacitors and when either one of the 2 momentary switches (labeled with 'krom' and 'recht') is pressed. The energy in the capacitors is released through an inductive load.

Sometimes these inductive loads can be as high as 9A. One of the goals of the circuit is to relieve the PSU. I am not sure what the maxmium current draw of this circuit draw is. I believe it was under 500mA but I am not entirely sure.

A 2nd goal is that it prevents coils from overheating as the capacitors discharge quite fast.

I want to make an SMD design of this circuit but I do not fully grasp grasp how that transistor is working. It is rated as 4A but I doubt if that much current is ever passing through it. Considering that this circuit does not draw more than 500mA at any given time

So can somebody tell me, what the purpose of the transistor is? (I believe it should increase the charge time)

I would just use a circuit like this one. But it appears that the normal CDU performs quite well.

And does it have to be 4A darlington transistor? I would prefer to use a mosfet in sot23 casing.

At start up, the capacitors are empty. So, every positive cycle of the input sine the D1 conducts and turns the transistor on, current flows through collector to emitter to charge the caps to a voltage of one diode drop less than the input's peak. So, And does it have to be 4A darlington transistor? It might, because the peak currents at start up will be high, even though they last within a few cycle.
When you close one of the switches (krom or recht) the capacitors will discharge through the coils and the discharge current will flow through D? (one to the up). Shorting one of the switches also makes the base voltage of the Q1 zero, making VBE negative (because emitter is at a positive voltage thanks to the capacitors) and turning Q1 off. This guarantees a safe discharge of the capacitors.