First, I'm a carpenter who knows little about electronics. I'm trying to build a miniature table saw from this motor. I tried powering it with the charger adapter and it will spin the motor until I put a load on it. I figure the adapter is not putting out enough amps. (Motor pulls 2.4 amps free spinning). I got a power supply with a rated output of 19.4 v dc at 11.5 amps. As soon as I hook it up the motor spins very briefly then something trips in the power supply. After a minute or so it comes back online. Do I need to add capacitor(s) in the line to take up the initial amperage draw? if so, one on the positive, one on both legs and/or one across the motor terminals? What kind of farad rating would I need? Any suggestions are appreciated. This is a low budget project. Thanks in advance.
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\$\begingroup\$ you need to identify the problem before you start thinking of throwing parts at the problem ... please provide a datasheet for the power supply \$\endgroup\$– jsotolaApr 2, 2022 at 23:59
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\$\begingroup\$ Good luck getting it to start from 11A. I retro-fitted my drill with LFP batteries after its NiCds died, and it trips a 30A BMS every time, unless I squeeze the speed control trigger gently for half a second at startup to soft start it. \$\endgroup\$– user16324Apr 3, 2022 at 12:56
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1 Answer
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This is a common problem. Brushed DC motors draw high current, like 6 to 10 times rated current, at start and during acceleration. Electronic power supplies are usually not capable of more than 1.5 times rated current for even a very short time.
Possible Solutions:
- Insert one or more resistors in series and short them out at timed intervals as the motor accelerated to full speed.
- Use a power supply, like a battery or an electronic supply with a really high current rating, that can supply high current for a short time.
- Insert an NTC resistor. The resistance drops as its temperature increases.
- Use a power supply that electronically limits the current quickly enough to prevent damage or shut down.
- Use an electronic motor speed controller -- option 4 plus adjustability.
- A capacitor might work, but the capacitance needs to be high enough to supply the acceleration energy but not too high for the power supply to charge.
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\$\begingroup\$ Might be able to just put an 18 V lead acid battery in parallel with the power supply. It can supply the extra amps during startup. The power supply voltage seems low enough to be an OK float voltage for the lead acid battery. \$\endgroup\$ Apr 3, 2022 at 0:30
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\$\begingroup\$ A capacitor that’s big enough to help here would likely be outside a low-budget solution. A 2 ohm resistor would limit the current to a little under 10A and that may well be enough to get the motor up to speed. Power rating for the resistor would be be 200W worst case but you could probably manage with something rather smaller, say 50W as it would only be in circuit briefly. You’d want the resistor in series with the motor and then a switch to short across the resistor once the motor is up to speed. \$\endgroup\$– FrogApr 3, 2022 at 3:05
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1\$\begingroup\$ Re, "AKA foldback," I thought that foldback was different from current limiting. My understanding is that a foldback power supply effectively will shut down—and stay shut down until the load is removed—if the threshold current is exceeded. (See electronics.stackexchange.com/a/14965/69405) \$\endgroup\$ Apr 3, 2022 at 13:48
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1\$\begingroup\$ @Soloman Slow: You are probably correct. I removed that from my answer. Thank you. \$\endgroup\$– user80875Apr 3, 2022 at 14:59
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\$\begingroup\$ I was hoping for a simple solution. Maybe my idea is just not practical. The fact that the charger wall wart for the saw can spin the motor no problem, but not under a load makes me think the power supply I'm trying to use is the issue, especially after Charles Cowies explanation of dc motor current draw at startup. A lot of viable solutions but not all are practical for a craft saw to cut little pieces of wood. last ditch question: could I use a rotary dimmer switch for lighting to "sneak up" on the motor? Thanks for all the input. \$\endgroup\$– drewApr 3, 2022 at 16:13