# : Voltage Regulator 1.5 amps using PWM

Discussion in 'Electronic Projects Design/Ideas/Reviews' started by dinofx, Jan 2, 2006.

1. ### dinofxNew Member

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I am trying to control the voltage going into an 18W brushless water pump via a PWM control signal provided on my PC's motherboard, and a 12V power source. Here is what I've research so far:

SOLUTION: Use a MOSFET to quickly turn the motor on and off
PROBLEM: Tachometer stops reporting RPMs correctly. I'm not even sure if a brushless motor is capable of running off of a pulsed power, since my understanding is that it has some controller circuitry.

SOLUTION: Use a MOSFET to charge a capacitor in brief spurts, thus providing a continuous voltage to the motor
PROBLEM: I don't see how this would work. First, let's say the motor consumes 1 amp, and the signal is on 25% of the time. What is to prevent the transistor from sending 4 amps of current to the capacitor, allowing it to provide a constant 1amp to the motor? Let's pretend the capacitor were infinite, it would simple charge to 12 volts and then run the motor at the full 12 volts.

Does anyone know how to do this correctly? I have been thinking about using a voltage regulator, a VFET, or adding a low resistor into to the charging of the capacitor to limit how fast it can charge.

EDIT: I guess what I don't really understand yet is how to model the resistance of a capacitor. Obviously a capacitor charged to 12 volts has infinite resistance in the presence of 12 volts being sent to it. I don't know its "instantaneous" resistance at any other value. For example, let's say I'm trying to run the motor at approximately 8 volts. When the MOSFET switches on, what is the rate at which it would begin to charge a .1 uFarad capacitor?

2. ### Nigel GoodwinSuper ModeratorMost Helpful Member

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As you suggested, DC brushless motors don't work well with PWM, because of the electronics built in the motor - it would probably be best to control it's speed using the internal electronics, this is how brushless motors in VCR's work!.

3. ### StyxActive Member

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Err DC brushless work extreamly well with PWM control. What wont is if there is electronics front end.

Just to correct this

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5. ### dinofxNew Member

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Here is what the pump sort of looks like inside. Direct pulsing of the motor is not an option, like I said before, since it makes the tachometer output less than reliable. Let's get back to a way of smoothing out pulsing into a quasi-steady voltage source.

6. ### Nigel GoodwinSuper ModeratorMost Helpful Member

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Try a capacitor across the motor, this will act as a simple low pass filter, adding a series inductor would help as well - but for directly feeding a motor values are likely to be high. Usually you would use small values feeding the input of an opamp buffer, and get a smooth DC out of the opamp (which you could then buffer with a driver transistor to feed the motor).

But for a start, try different value electrolytics across the motor.

7. ### dinofxNew Member

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OK, that's a new term to me. Is that the same thing as a VFET? In other words, a power transistor, which instead of turning on and off, tracks the control's voltage?

I've been searching patents and I found a diagram of what is claimed to be a "simple PWM controller". It's very similar to what you are saying. Basically, a capacitor across the load, and an inductor (or rectifier?), but the inductor is in series before the capacitor. Finally, it has a low-side transistor which I assume is turned on when the high side is off. So in other words, you create A/C. I guess the inductor needs to be able to draw current from ground when the PWM signal is switched to low.

US Patent

You need a TIFF viewer activex plug-in to view images.

Here is a description of the circuit from the patent:

"Electrical power for an integrated circuit (IC) is typically supplied by one or more direct current (battery) power sources, such as a pulse width modulation (PWM)-based, DC-DC converter. As diagrammatically illustrated in FIG. 1, this type of converter contains a PWM signal generator 1 that supplies a synchronous PWM signal to a switching circuit driver 2. Such a PWM-based converter architecture is ideally intended to deliver constant energy to an output node regardless of the input voltage. To this end, the switching circuit driver 2 controls the on-time and off-time of a pair of electronic power switching devices 3 and 4 (typically external NFETs) connected between power supply rails Vin and ground (GND). A common or PHASE node 5 between the two FETs is coupled through an inductor 6 to a load reservoir capacitor 7, with the connection 8 between inductor 6 and capacitor 7 serving as an output node from which a desired (regulated) DC output voltage is applied to load 9"

8. ### Nigel GoodwinSuper ModeratorMost Helpful Member

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It could be either bipolar or a FET, it depends on how you design it.

Yes, the inductor needs to be before the capacitor, it's a standard low-pass filter.

9. ### dinofxNew Member

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Ok, so here's an inductor in series. When the MOSFET is turned off, where does the current go? Do I just stick a diode in to allow the current to loop back around?

Will this circuit create too much RFI for a PC? I'm pretty sure a benign frequency has been chosen for the PWM, but switching 1 amp can create a lot of noise, can't it?

10. ### dinofxNew Member

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Is it possible to use a voltage regulator, and have the PWM signal used to vary the control signal to the voltage regulator? It would still be some kind of transistor+capacitor combination, but it would be at very low current and not produce as much RFI. As a bonus, I could wire the voltage regulator such that I could guarantee some minimum voltage at 0% pulse width.

Anyone? What is the maximum output voltage I could get from a voltage regulator with 12V going in? Will it damage the voltage regulator if I short the input and outputs together, so that Vout = Vin?

11. ### dinofxNew Member

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Here's my idea:

Use a voltage regulator and PWM to modify the input to the reg.

When the signal is off, the VReg's input is determined by the two 10K and 30K resistors. This would make the output 1.75V times 4. When the signal is completely on, the VReg's input is determined by 3 resisters, two 10Ks in parallel, i.e. 5K, and the 30K. Which means the output voltage would be 1.75V times 7, or basically the maximum output of the VReg.

A 50% duty cycle would charge the capacitor half the time, hopefully resulting in some voltage between 7 and 12.

Can anyone help me in choosing components such as a low drop regulator, and capacitor sizes. THANKS

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