Please advise on the current limiting and stabilizing problem between the power supply and the load capacitor.

乐君羊

I bought a 24V/15A switching power supply. My load is a capacitor. I just charge it with a large current for a short time (about 10-14A for 3ms). I need the charging current to be as large as possible. However, if the charging current reaches 15A, will the over-current protection inside the switching power supply be generated, thereby cutting off the power by itself? So I want to add a current limiting device between the switching power supply and the capacitor. backcolor]chipor design a current limitingcircuit, which is required to withstand 24V voltage and keep the current stable at about 14A. Does anyone have any good recommendations or guidance? Thank you everyone. I saw these current limiting circuits when I was looking up information. Let's take a look. Are these feasible?

maychang

"If the charging current reaches 15A, will the switching power supply generate overcurrent protection and cut off the power automatically?" The overcurrent protection action of each manufacturer and model is not exactly the same. Some models will shut down and can only be restored manually; some models will try to restart after a period of time after shutting down; some models will reduce the output voltage to limit the current.

chunyang

Whether the power supply has overcurrent protection depends on the power supply manual. Generally, power supplies with output protection will not cut off the power when the limit is reached, but will achieve protection through current limiting. If this is the case, there is no need to add an external circuit, but you must check the manual to confirm. If the power supply only has a fuse, such as a 15A fuse, then you need to add a current limiting circuit yourself. It is more convenient to use a transistor to build a current limiting circuit, but it will cause a voltage drop, and the input voltage needs to be increased to compensate. The simplest way is to add an auxiliary energy storage capacitor and then connect a self-recovery fuse in series, but this method is not accurate enough and requires the power supply to allow a certain overload.

gmchen

Get two capacitors, charge one of them first, you can take a little longer until it is fully charged. Then use this capacitor to charge the other one, the instantaneous current can be very large, to meet your requirement of the bigger the better, as long as the transistor acting as a switch can withstand the large current.

PowerAnts

Switching power supplies basically have their own current limiting overcurrent protection, such as the self-excited flyback represented by RCC and the other-excited flyback represented by UC384X. They all use sampling resistors in series at the emitter or source of the switch tube to perform cycle-by-cycle current sampling. The peak current of any switching cycle is limited. Since the transformer turns ratio is fixed, the peak current of any cycle is limited, which also limits the output current. The half-bridge represented by TL494 uses a transformer for proportional drive. If the output current is overcurrent, the drive current will also exceed the standard. Detecting the current of the drive transformer can achieve periodic current limiting or average current overcurrent protection.

乐君羊

maychang posted on 2018-6-10 13:52 "If the charging current reaches 15A, will the switching power supply generate over-current protection and cut off the power by itself?" The over-current protection action...

Thank you!

乐君羊

gmchen posted on 2018-6-10 16:19 Get two capacitors, charge one of them first, it can take a little longer until it is fully charged. Then use this capacitor to charge the other one, the instantaneous current can be very large, ...

Thank you,

luck_gfb

The capacitor is free of charge before power is applied. Since the voltage across the capacitor cannot change suddenly, it is equivalent to a short circuit, that is, 24V/15A=1.6 ohms. A 1.6 ohm 5W or 10W resistor can be connected in series between the positive end of the capacitor and the positive end of the power supply.