I made an infrared light board, and the instantaneous current can reach 5A when the light is turned on. This causes my power adapter 2A to fail to meet the requirements (the main control board cannot be started)
I want to add a current-limiting resistor for slow start, limit the current to less than 2A, and then turn on the lamp (the lamp time is 10ms) to provide current through a large number of capacitors.
Is it possible to do this?
Devine 게시 날짜 August 5, 2020
Your problem should be that the inrush current at the moment of starting is too large. Give you a circuit to try.
Adama 게시 날짜 August 5, 2020
This question should be changed to how to effectively suppress the huge current when the capacitor is charged?
A 12V power supply supplies power to a circuit, and the circuit has an input capacitance close to 120,000 uf (the result of multiple capacitors in parallel). When the power is on, the capacitor can be regarded as a short-circuit state with huge current.
Auburn 게시 날짜 August 5, 2020
The principle of the circuit diagram given above is the power-on delay. Q1 is a MOS switch chip, which is turned on by the front capacitor and the 4-pin delay switch.
