Hi all
i have a 7amp power supply (for a project im doing) powering a wiper motor off a car. the constant current of the motor is about 1.5amp but obviously the start up current is rather more than 7 amps and it sometimes causes the power supply to momentarily shut down. Can i use a capacitor of a reasonable size to hold enough juice for start up and if so, what size?
cheers
steve
ps changing the power supply is not an easy option....
You would need a *very* large capacitor. C = I * dt/dv, where I is the starting current, dt is start-up time (guess it) and dv is the permissible voltage 'sag' before the PSU complains, which might be very small. I (max) will be determined by the DC resistance of the motor windings.
A small float-charged SLA battery might be a better bet, or mods to the PSU, perhaps.
You calculate the approx value from
CV=IT
or C= IT/V
C in farads , I in amps extra current over that supplied by the PSU, T the duration in seconds for the excess current and V the voltage dip that you can tolerate.
Say 10 amps, for 100mS and voltage dip of 5 volts.
so 0.2F or 200,000uF
Bob
Its not THAT bad. The model car racing boys use supercapacitors to get an instant startline boost.
Try a few thoushand 'muffs'
100ms is pretty long. The MOMENT the armature starts to move, the back EMF starts to be there.
"Step start" would be another option. Stick a resistor in series with the motor to limit the inrush current, and slap a relay across that. Power its coils from a small delay circuit so that the relay shorts out the resistor after say 50ms.
To hold all the exess startup current, you'd need a big one, but that's not needed. A cap that holds a yet unknown percentage of the start surge would be enough to aviod nuisance trips. Of course you then have the problem that the psu trips at power up from charging the cap, unless psu has soft start, or unless, as is more likely, the i and t are within its shutdown limits.
In plain english, put a few thousand mfds on it and see.
NT
To hold all the exess startup current, you'd need a big one, but that's not needed. A cap that holds a yet unknown percentage of the start surge would be enough to aviod nuisance trips. Of course you then have the problem that the psu trips at power up from charging the cap, unless psu has soft start, or unless, as is more likely, the i and t are within its shutdown limits.
In plain english, put a few thousand mfds on it and see.
I've got an old battery booster than don't work.... maybe thats got some big caps in it?
steve
If by that you mean an old car battery charger, it wont have any caps in.
NT
no im referring to those boosters you carry round to start your battery in the event of not having a charger.
steve
If by that you mean an old car battery charger, it wont have any caps in.
NT
In message , Mr Sandman writes
Does the motor have to maintain its torque when under load? If it's not that critical, a low-value series resistor might suffice.
For example, assuming a 12V supply, and that, when stationary, the motor resistance is 1 ohm, a 0.7 ohm series resistor would limit the switch-on current to 7A.
If the motor takes 1.5A when running, its effective resistance will be 8 ohms. This is relatively small compared with the 0.7 ohms, and when lightly loaded, and the voltage across the motor will be 11V (the resistor dropping 1V).
If you load the motor, it will slow down, and draw more current. If you keep on loading it, the resistor will drop more and more voltage, so the motor torque will begin to suffer. But, depending on your application, this might be acceptable.
If you think this simplistic approach might be adequate, you could measure the actual resistance of the motor, and calculate the correct value for the required resistor.
Does the motor have to maintain its torque when under load? If it's not that critical, a low-value series resistor might suffice.
For example, assuming a 12V supply, and that, when stationary, the motor resistance is 1 ohm, a 0.7 ohm series resistor would limit the switch-on current to 7A.
If the motor takes 1.5A when running, its effective resistance will be 8 ohms. This is relatively small compared with the 0.7 ohms, and when lightly loaded, and the voltage across the motor will be 11V (the resistor dropping 1V).
If you load the motor, it will slow down, and draw more current. If you keep on loading it, the resistor will drop more and more voltage, so the motor torque will begin to suffer. But, depending on your application, this might be acceptable.
If you think this simplistic approach might be adequate, you could measure the actual resistance of the motor, and calculate the correct value for the required resistor.
Thanks Ian,
it only stirs water at a very low speed (wiper motor out of a car on slow speed) if i put load on the motor by holding the output shaft the current increases but not in my application as i don't load the motor it only draws
1.5amp when running but on start up it is drawing more and tripping the power supply. The resistance of the motor is 1ohm (the faster speed is 1.5ohm...it'd be nice to have the option to use it too)What wattage will the resistor need to be to cope with the current going through it?
steve
In message , Mr Sandman writes
Using the 'I squared R' formula, 1 ohm at 1.5A is 2.25W. But you might be wise to allow for more current. The power rises rapidly (as the square of the current), so I'd go for 5W or more. Depending one what you have in you junk box, it might be more convenient to use several lower wattage resistors in parallel (eg at least five x 4.7 ohms or ten x 10 ohms). This enables you (if you need to) to alter the total value by adding or removing resistors.
Series, surely? A shunt would divert current from the motor, but would present an even greater load for the PSU.
That is correct.
The trouble is the thermistor is a bit wasteful as it has to dissipate power to maintain it's temperature.
I never saw one used to supply a 7A load, although they are probably available.
The main reservation I'd have is the amount of heat produced and from past experience the reliability due to the metal bonding not expanding and contracting in sympathy with the ceramic type medium of the body.
HN
reducing start current, thus reducing starting torque, is ok for stirring water.
NT
Now that is deep!
HN
In message , H. Neary
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