Soft start DC power-on circuit

Hoping someone can help me verify my logic/math here. Honestly... I may be a tad over my head.

I am working on a soft-start circuit for some self-powered speakers that I am building.

The ciurcuit will function 2-fold:

1) It has a 3 or 4 second delay before turning on a relay which connects the speaker wires to the amp output - preventing any power-on pop to the speakers

2) It will also activate a small MOSFET at the same time the relay is turned on that is part of a soft-start setup that limits inrush current given the large capacitor employed.

I got #1 working perfectly. So now, I am "piggy backing" off #1 in order to employ #2.

So this project uses a TPA3255 amp board in combination with a large 20,000uF filter capacitor on the power rail. Needless to say, upon applying power - the inrush current is pretty severe and will weld the contacts of nearly any switch I use. Given the max input voltage of 40V and the speaker configuration I have chosen - the max wattage that this amp board will ever draw during normal operation is around 200W (absolute max values here). So the total DC load will likely never exceed 5 amp.

It may be worth noting that this speaker is meant to be powered by either DeWalt batteries or a laptop-style power supply - so switching the mains/AC is not exactly an option.

What I am planning to do is simple:

1) When the switch is turned on, the full 36-40 volts is sent THROUGH a 10 ohm resistor and to the 20,000uF capacitor (to charge it up).
2) Once my [already existing] power-on-delay circuit kicks on (to energize the speaker relay).. I can have that same circuit ALSO energize the gate of a IRFZ46N that basically bypasses the 10 ohm pre-charge resistor.
3) The "bypass MOSFET" will bypass the 10 ohm resistor direclty from the DC input jack to the capacitor (so operational current will NOT flow through the switch).

Question 1:
As mentioned - the input voltage will typically be between 36-40 volts. If I am calculating things correctly - this would limit the inrush current to around 4 amps (max) to charge up the capacitor, correct? Assuming worse-case scenario: the discharged 20,000uF capacitor is "seen" as a dead short for an instant and current is only limited by the 10 ohm resistor. Ohms law says this would be max of 4 amps in an instant and quickly drops. I ran a time-constant calc from an online calculator and seems the entire charge is done in less than a quarter of a second (.2 seconds).

Question 2:
Even this 4 amp figure is a theoretical max and only for a few milliseconds, after which it very quickly drops as the cap charges. Meaning the switch does not have to be aything crazy - correct? Would a switch rated for a few [DC] amps be enough here?

Question 3:
I am thinking of breadboarding this wtih the IRFZ46N simply because I currently have a handful sitting on the bench right now. Is there any issue with using this particular device for the application at hand? I have several other various MOSFETS in my parts bins if needed - but the spec sheet seems to look OK for what I am doing (again, making a BIG assumption that I understood it all correctly).

Question 4:
If there is a simpler solution here, care to let me know? I saw thermistors and have the chance to buy a 10-pack of 14amp (10 ohm) for pretty cheap. But I ALSO read many calling them "not the best solution in most cases". Also some concern about how long it takes them to "reset" (or cool down?).

-Dean
 
Last edited:
If there is a simpler solution...

Use a power supply with a built-in current limit. It would run at foldback current until the current drops to normal values. No need for precharging resistor. That's probably the simplest way.

IRFZ46N ... Is there any issue with using this particular device for the application at hand?
IRFZ46N is a rugged device and would do the job neatly.
 
Thermistors will ultimately stop thermisting when stressed over time and not always reproducible. Not a robust solution.

You might think about a current source that limits inrush current. A current mirror with a power MOSFET on the high current side may be good. Current through a capacitor makes a linear voltage ramp, so you will get a softer start than any RC circuit that will have a high start 'angle' (as you note 4 amps). And you can just lower the current mirror to make a slower startup time. It's like the current limited power supply Newvirus recommended.