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# 24 Volt AC volt to 12 volt dc?

Hi guys

i have a 24volt ac motor and want to reduce it to 12v dc. I have worked out how to change it to dc but not how to reduce the voltage. Is this easy to do or do i need to purchase something from the electronics store.

any advice would be appreciated

sam301

The way to "reduce it to 12v dc" is to power it with 12V DC.

It is as simple as that (if it runs on DC as you say).

L

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You said you had "worked out how to" change AC into DC, which says to me you know how to build a rectifier circuit

of some kind.I am going to humbly suggest trying "half wave" rectification, that is what you get from using

just one diode, rather than "full wave" rectification which is what you get from that "bridge rectifier", made from four diodes.See:

http://en.wikipedia.org/wiki/Rectifier#Half-wave_rectification

The reason I am suggesting this is that a sine wave that is half wave rectified will, I think, have exactly half the RMS voltage of a sine wave that is full wave rectified.

Try it without any filter capacitors and see if that works. I mean just let the motor do the filtering.

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It doesn't work quite that way. A half-wave rect. will have the same voltage as a full-wave, just rougher.

A full wave rect. takes the negative peaks and folds them up to fill in the gaps. But the peak voltage is the same.

I couldn't get your link to work.

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Sorry about the link.

http://en.wikipedia.org/wiki/Rectifier#Half-wave_rectification

Yes indeed, a half wave rectified sine wave, and a full wave one, have the same peak voltage.We can see that from looking at the pictures of the waveforms.But I was wondering about

RMS voltage. I did not do the math before, but I just did it now for,y

_{full}(t) = sin(w*t), when 0<=w*t < pi= -1*sin(w*t), when pi<=w*t<2*pi

y

_{half}(t) = sin(w*t), when 0<=w*t < pi= 0, when pi<=w*t<2*pi

It turns out that RMS voltage of a half wave rectified sine wave is exactly (1/2)

^{(1/2)}~= 0.7071 times that of the full wave rectified sine wave.So it wasn't exactly (1/2).Rather,(y

_{full})_{RMS}= [ (1/4) + (1/4)]^{(1/2)}= (1/2)^{(1/2)}= 0.7071(y

_{half})_{RMS}= [ (1/4) + 0 ]^{(1/2)}= (1/4)^{(1/2)}= 0.5000(y

_{half})_{RMS}= 0.7071 * (y_{full})_{RMS}Note that both those waveforms have a peak value of 1.

If you want to be kind, you might say I was thinking of power instead. If I placed that voltage, yhalf(t), across a resistive load, R, the time-averaged dissipated power, y

_{RMS}^{2}/R, would be exactly (1/2) what it would be for yfull.And I think that makes sense intuitively, since the half wave rectified sine wave is fully

turned off(equal to zero) exactly half of the time.Select as Best AnswerUndo Best Answer

A very do-able voltage reduction technique that works especially well

for incandescent lighting.

There is a concern that under shaft loading, the half wave will deliver

high current spikes to the armature brushes where they short two

commutator bars as the rotor turns.

This will damage the brushes.

I have had such a situation where an SCR phase controlled 240 VAC

down to a 110 VAC motor.

The current spiking turned the brushes to an

Unwashablevery dirtyfine black powder spread all over everything in the room

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thsnkyou . i will play and no doubt with more answers. thanks again

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_{If it was a stepper you would know enough to answerer it yourself. }All brush AC motors can run on DC.

A

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