Does Anyone Know How To Use a Variable Voltage Controller.

I Drink It Up!

Hi.

I bought one of these and it is supposed to allow me to adjust the volts and amps being drawn.

However all I can adjust at the moment is volts, except when I have something attached to the controller, and then it allows me to adjust Amps.

I have a small fan, rated 12v 0.5A and the controller is powering it grand, but it is only drawing 0.05A and not 0.5A.

If I jack the voltage up, it begins to draw more Amperage, maybe up to 0.1, but then the motor in the fan begins to burn out.

This controller of mine is supposed to be 20v 5A rated.

What role does the load play in the setting of Amperage?

Thanks.

dahamster

I=V/R

the supply can give upto 5A. The motor will simply draw what current it needs.

I Drink It Up!

dahamster wrote: »

I=V/R

the supply can give upto 5A. The motor will simply draw what current it needs.

Ah grand thanks. So I need a bigger application, say, a large capacitor or the like before I begin to see it drawing the full whack of 5A?

Thanks again.;)

DublinDilbert

I Drink It Up! wrote: »

What role does the load play in the setting of Amperage?

Two factors will effect the current that flows in the load:-
1) The resistance (impedance) of the load
2) The applied voltage

Voltage = Current x Resistance
=> Current = Voltage/Resistance
Increase the voltage, the current will rise.
Increase the resistance, the current will fall.

I'm not sure about your fan problem, can you draw a circuit diagram?
Typically you set the voltage correctly, and don't worry about the current, once it doesn't exceed the max for the converter.

DublinDilbert

Just an aside.... The equations i give above are purely for a resistive load, a motor is different.

The current drawn by a motor will also be proportional to the load on the motor. IE the current into the motor is proportional to the torque its producing.

If you increase the load on your motor you'll see the current increasing too.

I Drink It Up!

DublinDilbert wrote: »

Just an aside.... The equations i give above are purely for a resistive load, a motor is different.

The current drawn by a motor will also be proportional to the load on the motor. IE the current into the motor is proportional to the torque its producing.

If you increase the load on your motor you'll see the current increasing too.

Hi and thanks for responding.....my only problem with the fan was that it is is 0.5A rated but is only drawing 0.05A at 12v {it is also 12v rated}

Pushing the voltage to 15v, the current rises to 0.1A but at this stage the fan begins to burn out.

Maybe it is just a mis-print on the sticker.....anyways, at 12v, the fan refuses to draw more than 0.06A without me manually forcing it to by increasing the voltage beyond the 12v rated.

pljudge321

I Drink It Up! wrote: »

Hi and thanks for responding.....my only problem with the fan was that it is is 0.5A rated but is only drawing 0.05A at 12v {it is also 12v rated}

Pushing the voltage to 15v, the current rises to 0.1A but at this stage the fan begins to burn out.

Maybe it is just a mis-print on the sticker.....anyways, at 12v, the fan refuses to draw more than 0.06A without me manually forcing it to by increasing the voltage beyond the 12v rated.

Does the fan have a power rating, either in watts or volt-amperes (not applicable if its a dc voltage supply).

Power is the product of voltage times current, P = VI. In a simple circuit like you have the power supplied by the source should equal the power consumed by the load, in this case your fan.

If your fan is rated for say .6 watts and your are powering it with a 12V source then the max current that the fan will be able to take before bad things start to happen is I = P/V = .6/12 = .05A

Bear in mind my knowledge is mostly theoretical and someone more knowledgeable might come along and tell me I'm doing something stupid here.

I Drink It Up!

pljudge321 wrote: »

Does the fan have a power rating, either in watts or volt-amperes (not applicable if its a dc voltage supply).

Power is the product of voltage times current, P = VI. In a simple circuit like you have the power supplied by the source should equal the power consumed by the load, in this case your fan.

If your fan is rated for say .6 watts and your are powering it with a 12V source then the max current that the fan will be able to take before bad things start to happen is I = P/V = .6/12 = .05A

Bear in mind my knowledge is mostly theoretical and someone more knowledgeable might come along and tell me I'm doing something stupid here.

No what you said makes perfect sense. The fan is obviously only supposed to pull 0.5A as it burns out beyond that much. I was also looking around the house for something that would pull more Amperage than that but no joy. I will buy an audio capacitor and hook it up to my mini-welder {18v rated} and see what current that drags out of the power supply. I will post results here, cheers.

DublinDilbert

I Drink It Up! wrote: »

Hi and thanks for responding.....my only problem with the fan was that it is is 0.5A rated but is only drawing 0.05A at 12v {it is also 12v rated}

Maybe it is just a mis-print on the sticker.....anyways, at 12v, the fan refuses to draw more than 0.06A without me manually forcing it to by increasing the voltage beyond the 12v rated.

Your Fan is not rated at 0.5A

0.5A is the max current that will be drawn by the fan, you should see this current flowing when the fan starts, but it will quickly drop off to something like 0.05A.

Why are you obsessed with the 0.5A figure? Is the fan not doing its job correctly?

I Drink It Up!

DublinDilbert wrote: »

Your Fan is not rated at 0.5A

0.5A is the max current that will be drawn by the fan, you should see this current flowing when the fan starts, but it will quickly drop off to something like 0.05A.

Why are you obsessed with the 0.5A figure? Is the fan not doing its job correctly?

I was not "obsessed" as such, it's just that there is a great big old sticker on the back which says "12v 0.5A".

Thats all.:cool: