Capacitor to reduce DC motor start up current draw.

Hi folks,

I'm a noob when it comes to electronics but am trying to learn.

A while back I bought a cheap battery powered rotary tool / large engraver. It didn't run long on a chanrge and didn't have much power so I opened it up. It had three NiCad cells end-to-end as it's energy store, a simple switch and charged through a jack to which you'd connect the 5v DC power supply.

I decided to replace the NiCads with an 18650 Li-Ion cell and use the following module to charge the cell and prevent discharging;

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I've used those modules before with cells taken from old laptop battery packs and LEDs and they work well. However after I modded the case, soldered everything together and tried it (yeah I should have tested it outside the case...) I found it wouldn't switch on. :( However if I momentarilly shorted the DC out to the battery connection the motor would run just fine except if put under very heavy loads, in which case it'd stop. (I killed a couple of the modules messing around trying to fit a momentary contact switch to short across to start the motor.)

Some guy in a forum on that site suggested it was because of the high current draw on start-up of the motor and that I put a capacitor in parallel (?) with the motor it might fix it. However I was unable to engage him any further to get him to tell me more, what size and type of cap etc.

I'm not rich and don't have the funds to buy electronic components and try whenever possible to use salvaged parts. I keep old PCBs and tend to use them as my supply of parts whenever possible...

Can anyone tell me if doing this is likely to work and if so what type and value of capacitor to use please? I seem to recall seeing something similar somewhere, what appeared to be a tantalum capacitor wired across a DC motor. I can see quite a few of those on the various PCBs that I have. If someone can confirm that it might work them I'll get to unsoldering and try to work out what values the caps are that I can find.

Cheers, and TIA.

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Shaun. 

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~misfit~
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problem is your module isn't tough enough to start the motor. adding capacitors will likely make matters worse.

Adding a resistor in series with the motor might help, say 0.47 ohm 5W then put a switch parallel with the resistor to bypass it once the motor starts.

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Reply to
Jasen Betts

I can't help but point out that even the sales picture shows a board with an LED misaligned and almost shorting to a nearby resistor, which looking at the board traces, would probably damage the rearby IC. I'm not saying that the board is the problem, I just think it says something about the sellers of this stuff that they use a board like that as the demo.

That wouldn't work unless the capacitor's connection to the battery voltage bypassed the power switch, so that it was charging all the time. Even then it may trigger the current limiting circuitry, which may or may not cause a problem depending on design. If it uses foldback current limiting, then it would drop the voltage until the point that less that the maximum current is flowing. As the capacitor charged, the current flow at that voltage would be reduced, and as such the current limiting circuit would progressively raise the voltage until both charge circuit and capacitor were effectively at their full voltage. with a normal electrolytic capacitor, this would always take less than a second to occour.

If however, the current limiting actually cuts off the power completely when the maximum current is exceeded (for protection in the case of a short perhaps), you would need to limit the power going to the capacitor when it charges. This could be achieved by placing a current limiting resistor (47 ohms should allow a quick enough charge time with most large value electrolytics) in series with the capacitors connection to the battery voltage (which is still before the power switch), and a power diode (eg. 1N4004) in parallel with its Anode towards the capacitor. The diode would supply current from the capacitor when the power switch is turned on.

The capacitor value required would depend on the peak load exerted by the motor (Storage Oscilloscope measurement), but I'd recommend just trying the largest capacitance value electrolytic you have which is a suitable size for the application. You might need to try some in parallel if that isn't enough. You'd want the total capacitance well above 1,000uF in any case. Quite likely above 10,000uF.

Note that in both cases, the connection of the capacitor bypassing the power switch will cause a constant (though not relatively large) drain on the battery. The exact current would depend on the leakage of the capacitor, and therefore its capacitance, quality and age. The other suggestion of a switched resistor in series with the motor might be preferred if this is undesired. From the battery/module's point of view, that is simply a manual equivalent to the capacitor solution.

Indeed a circuit could be constructed to automate the switching of the motor series resistor if so desired. :)

Perhaps they were actually for reducing electrical noise from the motor, ceramic capacitors are sometimes used for that purpose.

Sorry for the long post. And I just realised that I can't really speak because I've got a cheap rotary tool I was given with a dead NiCad battery, and it's still sitting disassembled somewhere. :)

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Reply to
Computer Nerd Kev

Ok, thanks. Like I said I didn't understand what the guy meant and I'm quite new to electronics (and an 'old dog' - finding new tricks not as easy to learn as they used to be).

Thanks. There's very little room in the case. I've already crwoded it with the longer cell and charging module. The momentary swith I fitted was a thin thing, a recessed reset switch scavenged from a dead ADSL modem and hot-glued into the rotary tool case.

Cheers.

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Shaun. 

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~misfit~

Actually I've noticed that with a lot of their products and so far nearly all of the things that I've bought have been better made than the pictures. I considered that maybe they get a prototype for the pic?

Yeah. That's a bit too complicated and fiddly for my liking. ;)

So about as big as the 18650 cell than? ;)

Ok. However the case is quite tight...

Heh, yep if I was up to the task and there was room.

Thanks.

LOL, sounds like me. It was well worth it re-powering with the Li-Ion cell. It's more powerful (and doesn't seem to overheat or anything) and of course, even with a salvaged cell last longer between charges and can sit in the drawer for a month between uses and still have charge.

I originally modded this with a simple Li-Ion charge module and it worked great. But then I got worried about over-discharging so changed to this thing. I'm starting to think that the easiest answer might be to run the motor from the battery terminals and connect an LED to the 'load' terminals. When the LED goes out then it's time to stop using the tool.

There is continuity between the load out + and battery + so I could switch positive to both the LED and the motor. Then have the LED - going to load and motor - going to the battery teminals of the module. When the battery drops below 2.5v the LED should go out. That should work don't you think?

I dunno, I'm winging it here. ;)

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Shaun. 

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~misfit~

Huh, you never can tell with this Chinese stuff.

You know most of that was describing the technicalities of how it worked? The actual circuit is just the capacitor with the existing power switch and control module:

Argh, I couldn't be bothered to do this last time... _._ CAP | | _____| |_________ + _________ | | -------- + | | | *__|_____| | | C | | |______|__________*\ ON | |---| E | | MOTOR | POWER SWITCH- \ | CONTROL | | L | | | *OFF | MODULE |---| L | | |------------------------------|_________| | | -------- - - |___|

The only added component is the capacitor.

CAP RESISTOR(47R) | | ______ -| |---|______|-- _._ | | | | | | | | | |\ | | | | | |-| >|------| + _________ | | -------- + |DIODE> |/ | *__|_____| | | C | | |______|__________*\ ON | |---| E | | MOTOR | POWER SWITCH> \ | CONTROL | | L | | | *OFF | MODULE |---| L | | |------------------------------|_________| | | -------- - - |___|

Quite likely (having just looked up the size of one of these mystical Lithium beasts). Possibly smaller if you can find 6V or 10V rated caps in the right range. Much smaller if you use a couple of Supercaps.

Oh boy, here we go again... LED ~120R | /| ______ ----|< |--|______|- _._ | | \| ________| + _________ | | -------- + | | *_______| | | C | | |______| *\ ON | |---| E | | MOTOR | POWER SWITCH- \ | CONTROL | | L | | | *OFF | MODULE |---| L | | |------------------------------|_________| | | -------- - - |___|

Hmm, if that's what you meant then we are in trouble...

I think this is really what you want: LED ~120R _This is a bridge, |\ | ______ / not a join ----| >|--|______|- / _._ | |/ | | / + _________ | | -------- + | *__|____| | | C | | |______|__________*\ ON | | |---| E | | MOTOR | POWER SWITCH- \ | | CONTROL | | L | | | *OFF | | MODULE |---| L | | |------------------------------|_________| | | -------- - / - |___| / That's the join (Resistor to -)

It's drawn a bit confusingly because I'm getting lazy(er), but the LED should be on until the voltage gets below ~2V, or the power switch is turned off.

Now looking at the notes from the a Lithium battery controller I recently designed (but also haven't got around to doing yet...), I had the cut-off voltage at 3V. Perhaps against best advise, I trust myself in my past reasoning and would suggest that 2V is too low a cut-off point. However I remember I got all my information from the website "Battery University" (I'm out of time now, so can't look it up myself to check).

In any case, a regular diode in series with the LED would make the cut-off point ~2.7V, and that might be acceptable. In practice it will grow dimmer towards the cut-off point instead of suddenly going out.

Sorry for the whining, don't feel you put me under any obligation to do all this.

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Reply to
Computer Nerd Kev

And before you redefine your understanding of capacitor theory, the capacitor in those diagrams should be connected with its negative to the battery negative, not to the motor positive. eg. _._ CAP | | _____| |_________ + _________ | | -------- + | | |+ *__|_____| | | C | | |______|__________*\ ON | |---| E | | MOTOR | | POWER- \ | CONTROL | | L | | | | SWITCH *OFF | MODULE |---| L | | |------------------------------|_________| | | -------- - - |___|

Sorry about that.

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Reply to
Computer Nerd Kev

Thanks for taking the time to help and sorry it's been so long since I visited here.

I was thinking of something along the lines of:

LED ~120R _This is a bridge, |\ | ______ / not a join ----| >|--|______|- / _._ | |/ | | / + _________ | | -------- + | *__|____| | | C | | |______|__________*\ ON | | |---| E | | MOTOR | POWER SWITCH- \ | | CONTROL | | L | | | *OFF | | MODULE |---| L | | | ----|_________| | | | -------- ------------------------------------------ |___|

Where the motor positive and LED positive are switched but motor negative goes straight to the battery bypassing the protection circuit (which seems to operate on the negative - the battery positive and load out positive are common).

That way only the LED is on the protection circuit (so motor shouldn't trip overcurrent protection) and would turn off when the voltage drops below the predefined low for the 18650 (18mm diameter, 650 long, the most ubiquitous Li-Ion cell there is - it's in flashlights, laptop batteries, powertool batteries and even powers Prius and Tesla cars).

Cheers,

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Shaun. 

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~misfit~

D'oh! Of course 18650s aren't 650mm long they are 65mm long. Sorry about that. :-/

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Shaun. 

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~misfit~

A lot of 18650s on the internet have a capacity only a fraction of that claimed. It is probably better to pay the extra for known brands (and hope that they are actually from that manufacturer).

Reply to
keithr

OK, that's alright if you actually don't mind pulling over 3A from the Lithium cell. For the breif moment at start-up it probably wouldn't be too much of a problem (but you might want to check the technical specs, because I'm no battery expert). The issue might be if you put too much load on the tool and cause it to stall, then the full stall current will flow and may damage the cell.

You've also left out the second diode in series with the LED, that I recommended to get the low voltage warning to trigger closer to 3V.

I know there's a page about the correct discharge cut-off voltages for Lithium-Ion bateries at this website:

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But the bookmark is on another computer and I can't find it in the index.

OK, just tried looking on other computer, written notes, forehead. Can't find that bookmark anywhere, perhaps it doesn't exist, I'm sure it does though. Argh, try another website.

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Reply to
Computer Nerd Kev

Thanks, I know. The only ones I've bought (other than ones supplied with flashlights etc.) have been Panasonic NCR18650s from a reputable NZ source. Otherwise I've obtained quite a few by opening laptop battery packs and indivudually testing each cell. I've got a couple of dozen that way (Sanyo, Sony and Panasonic) that hold in excess of 1.5 A/hr out of bad battery packs where it's turned out only one cell (or row*) was dead, the rest of the cells being still quite usable.

[*] It seems that most laptop batteries have electronics in them that use electricity, even on the shelf. Usually a PU for controlling charge / discharge parameters and a ROM which records data such as date of manufacture, design capacity, cell manufacturer and power cycles. It's been my experience that most packs run their cells in a mix of parallel and series (i.e. a six cell pack would be 2P3S). All the of the 'dead' packs that I've opened have run their internal circuitry from a single parallel row of cells. Often if a pack is left flat for a time then that row will drop below the voltage threshold that the PU will recharge so the pack is essentially 'dead'. However the other parallel rows are often salvageable if trickle charged and regain most of their capacity.

(Needless to say I have a charging / discharging set up that records how much energy a cell takes to reach 4.2v and how long it takes to discharge to

2.6v using various drains - usually LEDs.)

Now I just need to find a use for a few dozen ~1.5 A/hr 18650s. I have a couple of flashlights but they're not using all of the cells which, once tested and capacity recorded, are currently just sitting on a shelf.

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Shaun 

"Humans will have advanced a long, long way when religious belief has a cozy  
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Reply to
~misfit~

The cell can handle it. Also unlikely to have full stall current for more than a second if that as it's hand-held. Even 'low draw' 18650s can handle

5A continuous draw (although it will shorten their lives somewhat) and the higher ones as much as 20A.

I wouldn't mind a slightly higher trigger voltage to extend cell life but it's a trade-off with storage capacity. As it is I've got so many usable salvaged cells that they're likely to die of old age (even stored at ~40% charged in the fridge) before I find uses for them all. ;)

Cheers, and thanks again for the input.

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Shaun. 

"Humans will have advanced a long, long way when religious belief has a cozy  
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~misfit~

One of the things it said on the page was that 95% of the energy is exhausted from the cell when it reaches 3V. I actually did finally find the page yesterday (they'd changed it on me so I didn't realise when I found it), but of course I bookmarked it on another computer and now I don't have time to find it again... It's in the index under some fairly obvious name.

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Reply to
Computer Nerd Kev

Yes, you're right - it was the other end that I was thinking of (~4.2v). I'd be quite happy to stop draining my cells at 3v. Trouble is almost all control modules I can find pull them down to around 2.6v.

Cheers,

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Shaun. 

"Humans will have advanced a long, long way when religious belief has a cozy  
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~misfit~

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