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Startup Capacitor tiny 3-6 volt DC motor

kingdomoffurr

Feb 27, 2011
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Hello, I wonder if any of you could help me? I am a bit new to Electronics and I've got to make a motorised bubble blower for a degree project.

I've been given a DC motor of the above voltage that I want to run super slow.

The problem is the inertia current it requires requires it to be going almost full speed and I want it to run at a very slow rate.

I've thought about putting a capacity in series after doing some reading, just to get it up and running, but I've no idea what type I need, also will it work?

The motor is connected to a variable resistor, I assume the capacitor goes after the variable resistor?


Thanks

Ryan
 

(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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That's not going to work.

Have you thought of gearing the motor?
 

kingdomoffurr

Feb 27, 2011
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That's not going to work.

Have you thought of gearing the motor?

Its got worm gear, driving an axle at 90 degrees to it already as a kit So, I assume this means I am going to have to have it running full speed and put some kind of gear ratio on it?
 

Resqueline

Jul 31, 2009
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So it's already geared somewhat.
Did you mean that you have tested the motor with the variable resistor and that to get it started you have to turn the resistor down so low that the motor runs almost at full speed once it gets going?
What's your power supply voltage, and what are the resistance values?
Can you use more advanced electronics than a resistor?
 

kingdomoffurr

Feb 27, 2011
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So it's already geared somewhat.
Did you mean that you have tested the motor with the variable resistor and that to get it started you have to turn the resistor down so low that the motor runs almost at full speed once it gets going?
What's your power supply voltage, and what are the resistance values?
Can you use more advanced electronics than a resistor?

Ok, I'm using 6 v, a 1k pot with 6 150 ohm resistors in parallel, I know this might seem odd, but its all we had to hand.

This gives more control over the speed of the motor. I have to turn the 1k resistor to next to zero,to get the motor to shift, once it does I can crank up the resistance and it will run nice and slowly.

Does that make sense?
 

Resqueline

Jul 31, 2009
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Yes, that makes sense.
Are those six individual resistors connected in series before being connected in parallel with the pot or is it all connected in parallel?
It sounds like all you need is a jumpstart, and I think it can be achieved with a capacitor like you expected.
Below is how you connect it, in parallel with the resistors. I can't say (yet) how big it needs to be.
Some experiments may be needed. Try with 10, 100, or 1000 uF to begin with.
 

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kingdomoffurr

Feb 27, 2011
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Yes, that makes sense.
Are those six individual resistors connected in series before being connected in parallel with the pot or is it all connected in parallel?
It sounds like all you need is a jumpstart, and I think it can be achieved with a capacitor like you expected.
Below is how you connect it, in parallel with the resistors. I can't say (yet) how big it needs to be.
Some experiments may be needed. Try with 10, 100, or 1000 uF to begin with.

They are in parallel, we used it rather than a smaller resistor which would take less current.

Is that ok?

Thanks for your help
 

Resqueline

Jul 31, 2009
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Ok, that results in a resistance of 150Ω / 6 = 25Ω. This is such a small fraction of the 1kΩ pot resistance that it makes it hard to adjust.
It's ok but it would be easier using a 100 or 220 Ω pot. You'd then be able to use more of the track of the pot for useful adjustments.
To get a kick-start time period of about 0.1 second with that resistance you'll need to use a capacitor as big as 0.1s / 25Ω = 4000µF.
 

kingdomoffurr

Feb 27, 2011
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Ok, that results in a resistance of 150Ω / 6 = 25Ω. This is such a small fraction of the 1kΩ pot resistance that it makes it hard to adjust.
It's ok but it would be easier using a 100 or 220 Ω pot. You'd then be able to use more of the track of the pot for useful adjustments.
To get a kick-start time period of about 0.1 second with that resistance you'll need to use a capacitor as big as 0.1s / 25Ω = 4000µF.

So you are saying ditch the resistors. Go buy a 100 or 220 ohm
Pot? To be fair we just worked with what we had but I can see this is better.

I've no idea - but I'm slowly getting my head around all this - is a 4000 micro farad cap hard to find or expensive and big? Capacitors are a bit of a mystery to me at the mo.

Once again thanks for the help.
 

Resqueline

Jul 31, 2009
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No, I was thinking of just replacing one of the 150Ω resistors with a pot (ditching the 1kΩ). Replacing the whole resistor assembly you'd want a 25Ω wirewound pot.

Capacitors are like small batteries. Applying a current to them they'll charge and hence the voltage across them rises with time (it starts at zero).
A 4000µF will be very hard to find, but 1000, 2200, 3300 & 4700 are readily available.
If it's rated for only 6V it won't neccessarily be very big (maybe 10*30mm) and will only cost a dollar or so.
If you have a scrapped electronic device then chances are quite high that you'll find something in it you can use, for free.
 

kingdomoffurr

Feb 27, 2011
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No, I was thinking of just replacing one of the 150Ω resistors with a pot (ditching the 1kΩ). Replacing the whole resistor assembly you'd want a 25Ω wirewound pot.

Capacitors are like small batteries. Applying a current to them they'll charge and hence the voltage across them rises with time (it starts at zero).
A 4000µF will be very hard to find, but 1000, 2200, 3300 & 4700 are readily available.
If it's rated for only 6V it won't neccessarily be very big (maybe 10*30mm) and will only cost a dollar or so.
If you have a scrapped electronic device then chances are quite high that you'll find something in it you can use, for free.

IAh well often it seems with this electronics stuff you have to find the best fit :) as far as replacing one 150 ohm res with a pot I assume this would enable fine tuning of the fine tuning, so to speak?
 

Resqueline

Jul 31, 2009
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Now you're using only 10-20% of the 1kΩ track for effective adjustment. Ditch that and replace it with a 220Ω pot and you get to use 100% of the track for adjustment.
You get a finer and hence easier adjustment.
 

kingdomoffurr

Feb 27, 2011
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Now you're using only 10-20% of the 1kΩ track for effective adjustment. Ditch that and replace it with a 220Ω pot and you get to use 100% of the track for adjustment.
You get a finer and hence easier adjustment.

Right. Will swap out the 1 k for a 220 :)
 
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