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DlStreamnet
 
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Car electronic gawds

Just retrofitted some powerfolding wing mirrors (lol what) and found out i'm possibly using the wrong relay, and the correct one is FAR too complex (as in its integrated into a general electronic module containing hundreds of connections for things like boot release, wipers, car audio etc [fords easy way of wiring]).

Anyway, basically the drivers side mirror is drawing too much power. How can I reduce the power easily?
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Old 01-12-2010, 03:35 AM DlStreamnet is offline  
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grjr
 
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how much power is too much power? Electric motors will draw more current as mechanical load is increased... why do you need to reduce the power draw?
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Old 01-12-2010, 05:40 AM grjr is offline  
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DlStreamnet
 
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Because they are folding out too quick, therefore the timer is out of sync. The relay is solid so I can't change the timer, therefore I need to reduce power, so it folds out slower, so it doesn't overfold.
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Old 01-12-2010, 06:25 AM DlStreamnet is offline  
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Typhoon43
 
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Sounds like you need an inline resistor. Paging gee to the thread.
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Old 01-12-2010, 07:32 AM Typhoon43 is offline  
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gee
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typhoon's right, going with a resistor is probably the best way to do things.

As for how to pick a resistor and size it... first grab a DMM, set it to the 10A setting and use it to measure what the current draw of the motor is. From this you can figure out what the motor's effective resistance is; R=E/I, E is 12 and I is the current in amps that you measure. If the mirror draws an amp when it moves, R will be 12V/1A = 12 ohms.

In this case, start with a resistor around 10 ohms; this will put the motor voltage a bit above 6 volts. It will also reduce the current into the motor, which will work out to around 12V/(12+10 ohms) = ~0.55 amps.

And from this amperage, you can figure out the power rating required in your series resistor, which is I^2 * R... or 0.55^2 * 10 or around 3 watts.

Practically, you'll want to get a handful of different value resistors and try various combinations of them until you get the motor going at the speed you want. Alternatively you could borrow a bench power supply and use that to power the mirror, figure out a voltage that works well and calculate a resistance that works out to the correct voltage.
Old 01-12-2010, 09:21 AM gee is offline  
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Typhoon43
 
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^ what a stud

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Old 01-12-2010, 09:23 AM Typhoon43 is offline  
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gee
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^ what a stud

I really should get myself a spandex suit, and make a list of bad electrical puns.

GEE TO THE RESCUE!.... *ahem*, looks like we've got a POTENTIAL problem here!
Old 01-12-2010, 09:28 AM gee is offline  
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Typhoon43
 
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I heard you get all "amped up" when you get to help.

badump clang!
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Old 01-12-2010, 10:14 AM Typhoon43 is offline  
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Stereodude
 
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Resistance is Futile!
Old 01-12-2010, 10:59 AM Stereodude is offline  
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DvBoard
 
Quote:
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typhoon's right, going with a resistor is probably the best way to do things.

As for how to pick a resistor and size it... first grab a DMM, set it to the 10A setting and use it to measure what the current draw of the motor is. From this you can figure out what the motor's effective resistance is; R=E/I, E is 12 and I is the current in amps that you measure. If the mirror draws an amp when it moves, R will be 12V/1A = 12 ohms.

In this case, start with a resistor around 10 ohms; this will put the motor voltage a bit above 6 volts. It will also reduce the current into the motor, which will work out to around 12V/(12+10 ohms) = ~0.55 amps.

And from this amperage, you can figure out the power rating required in your series resistor, which is I^2 * R... or 0.55^2 * 10 or around 3 watts.

Practically, you'll want to get a handful of different value resistors and try various combinations of them until you get the motor going at the speed you want. Alternatively you could borrow a bench power supply and use that to power the mirror, figure out a voltage that works well and calculate a resistance that works out to the correct voltage.
Agreed. You need a lower voltage to the motor so it will turn slower. The above will fix this.
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Old 01-12-2010, 11:02 AM DvBoard is offline  
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gee
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Quote:
Originally Posted by typhoon43 View Post
I heard you get all "amped up" when you get to help.

badump clang!
Yeah, rectifying a negative situation can definitely charge me up.

Anyway, enough of that phase... *ahem* so, watts up?
Old 01-12-2010, 11:51 AM gee is offline  
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Gee pretty much covered it.

I'd personally put a fairly large wattage value resistor in there. The I = V / R thing is only valid while the motor is spinning, and generating a back-EMF. When the motor's shaft is locked, it will draw a lot more current.

In a nutshell: that means that if the mirror ever gets stuck (tries to fold while frozen in winter, for instance), it will be drawing a lot more current, and that resistor will heat up -fast-.

Be sure to measure the current to the motor when the shaft is locked. You will get a radically different number. Size the resistor for that value.
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Old 01-16-2010, 04:41 PM kstokes is offline  
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