PWM Arduino Uno Speed Control for 180V 2.2Hp Treadmill PMDC Motor with MOSFET Issues

Ok, so I have a 180Vdc 2.2Hp Treadmill PMDC Motor. I am trying to design a speed controller for it. Attached is a schematic that I am testing currently. I have an arduino uno generating a PWM wave@16khz I can vary the duty cycle using a Pot from 0 to 100% it switches the 2n2222 transistor which in turn switches the MOS@12V (The Max VGS is 20V) The problem is that across the motor I am getting only a max of 56Vdc @ 100% Duty Cycle the rest is being dumped across the light bulb. Am I not biasing the MOS correctly ? Should I be getting around 200Vdc at 100% duty cycle. Help would be Appreciated.

The light bulb is dividing the voltage and needs to be moved. BUt, before I tell you where to move it, I need to know:

What is the intended function of the light?

Also, your use of resistors in the gate circuit may be slowing down the switch rate. This is passive drive and 10K is probably not enough to overcome the gate capacitance and the drain reverse capacitance (see the following document on MOSFET drive techniques: http://www.ti.com/lit/ml/slup169/slup169.pdf)

Well without the light bulb the circuit trips the circuit breakers(10A) the bulb is so that whatever the extra voltage there is it drops across it (just like a resistor really). Should I increase the resistor on gate to 1Mohm.

Ideally The voltage drop across the bulb should be around a few volts.

I am not expert in motor control, but a couple of things from basic principles:

Motor: 180Vdc 2.2Hp
2.2Hp = 746W x 2.2 = 1641 Watts

1641W/180V = 9.1 Amps (current required by motor)

As the current in the circuit rises, so does the voltage across the bulb, limiting the voltage available to the motor.

Even if the motor is shorted, the maximum current that can flow in the motor circuit is limited by the bulb:
200W/220V = 0.9 Amps

Does this help ?

(Please check working to see if its me that's made a mistake.)

Hi Frank,

Your calculations are correct, but I have used this same 200W bulb in series with the Whole Treadmill, with the motor plugged into the Treadmill Controller. The motor was running at full torque and speed.

The current through the bulb was 10A but the voltage drop across the bulb was 5Volts AC.

Power Dissipated across bulb = 5*10 = 50 Watts (No, the bulb does not light up in this case)

200Watts is what the bulb is maximum rated for.

and the Voltage across treadmill was 234.5VAc (My Mains reading was 240VAC)

The bulb in this case is to prevent short circuits in the circuit by allowing the voltage to drop across it.

Well, it's either the light bulb that is dropping the voltage down to 56V or the MOSFET turning on too slowly, or both (and I would wager that it's more the light bulb -- especially since you're driving the MOSFET at 16KHz--a low enough frequency that the slow turn-on time may not be as much of a factor).

Does the MOSFET get hot?

And, increasing R1 to 1MΩ will make it worse -- i.e. wrong direction. You need to charge the _parasitic_ capacitors in the MOSFET. T=RC, thus, greater resistance = greater time. You need it to turn on in _less_ time. Also, I'm assuming that the intent for R1 and R2 is to create a voltage divider to prevent the gate voltage from going above 12 volts. But using two 10K resistors will divide the voltage down to 7.5V. I would make that more like 10V (unless you are using a "logic level" MOSFET, in which case 7.5V is plenty).

So how about:

# choose 1K ¼W for R2 (chosen to avoid resistors higher than ¼W also, highest, typical, capacitance involved is around 500pf, T=3RC [time for 95% charge], T = 3*1K*500pf = 1.5µS (plus around 20% more for charging the gate capacitance) -- still kind of slow, but probably fast enough for this.
# 10V across R1 leaves 5V across R2 (15V - 10V = 5V)
# 5V / 1K = 5ma (the current in the series circuit formed by R1 and R2)
# R2 = 10V / 5ma = 2K

BUT, the real issue is the fact that you are trying to drive a 180 Volt motor on a 240VAC line (240V * 1.414 = 339V - diode drops ≈ 335VDC). Putting a light bulb in series isn't going to drop the voltage consistently enough to make this work. You are dealing with "stall current", i.e. the current it is drawn by the motor as it starts up--has to over come static friction, inertia, etc. Also, IF the light bulb trick will, by some remote absurdity, then you need a 120VAC lamp, not a 220VAC lamp, and it needs to be more like 1600W [9A*180V] -- i.e. you want to drop half the power.

Also, 1 horsepower = 746 watts. P=IE, thus 2.2HP*746W = I*180. Whip a little algebra on that and you get a little more than 9 Amps!! BUT, that's probably the peak HP, and when it's up an running, as long as the mechanical load is not excessive, the running current may be less, but I must assume that it will run at 9Amps, until I have further data. If you determine otherwise, then extrapolate.

The stall current will definitely cause the motor to draw way more than 9Amp (unless, of course, as I said, the 2.2HP rating is the peak, in which case the stall current may be 9A)-- BUt, it probably can't draw that current because of the light bulb in series. What does your circuit breaker trip at?

Also, that 470uF capacitor isn't going to do squat to filter out the ripple at those currents.

C ≈ IT/V where I is the current (whatever the stall current is -- let's say 25amps), T is roughly the time gap between peeks (≈8ms), and V is the ripple voltage -- lets call it 12 Volts

C ≈ 25A*8ms/12 = 16,000uF at around 250WV.

You need some kind of regulator or step down transformer -- one that can handle the highest current involved. That'll be one hefty transformer. Let's say a 240 to 120VAC voltage converter, because those will be more readily available than a 240 to 127V (180/1.414=127VAC) transformer, so if we use the estimated stall current (might be more), then the transformer needs to handle: 25A*165V ≈ 4KW.

Again, redo the math, if you determine that the stall current is more like 9A and the running current is less.

OK, a lot happened while I was writing my last reply.

So, the thing works with the bulb in series. Then that suggests that the motor is drawing less current than I thought. So, try focusing on the MOSFET driving circuit.

Also, that 470uF capacitor is probably inadequate. I would expect there to be quite a bit of ripple and that could be diminishing the motors mechanical power output, especially at fractional power.

Also, consider this. The motor may not be able to supply enough torque to run your treadmill at fractional power. Or, perhaps a linear reduction curve is not the answer, you may need something more logarithmic or you may need some sort of feedback loop (i.e. a tachometer that sends a signal back to your Arduino that is used to adjust the PWM duty cycle). But, like Frank, I have little experience in motor control.

Thinking back to college, incandescent lamp characteristics . . .

From Wikipedia:
"The cold resistance of tungsten-filament lamps is about 1/15 the hot-filament resistance when the lamp is operating. For example, a 100-watt, 120-volt lamp has a resistance of 144 ohms when lit, but the cold resistance is much lower (about 9.5 ohms)."

This may explain some of what's happening.

The Mosfet is mounted to a standard sized heat-sink but it does not get Hot.

I have already done the math on the transformer, its actually quite big and sourcing one is quite expensive.

The MOS isn't a logic Level one I tryout your suggestion.

I didn't have time to measure the current when it tripped, because as soon as I turned it on every single circuit breaker tripped (though the 10A fuse didn't blow ~ Murphy's Law). I had one 10 A breaker on the live line that tripped and the 16A breaker in my mains box that tripped. I mean come on the stall current can't be greater than 16A thats more than whats the motor rated for.

I still don't know why it tripped - Stall Current ?.

I realized that early on that the capacitor is doing squat, I just didn't bother to remove it because the motor actually ran quite smooth on unrectified directly from the bridge (Its odd my bridge was giving a total of 240Vdc rectified voltage OPEN although it should have given ~330 Vdc)

The other thing I noticed is that Treadmil Motor Controllers don't have transformers, they directly rectify 220VAC and what I believe have a half H-bridge circuit switching the motor.

You might ask why not use the treadmill controller to control the motor, I do have one but I dont have a console presently(the board where you select your speed level etc) so I cant replicate the signal it sends to the controller to get the motor working.

P.S. I am not going to make a pitching machine out of it if I get the motor working.

I plan to add feedback in the future, but First I am trying to get the motor to run.

Yeah I know the capacitors does nothing I didnt bother to remove it.

Yeah but at lower resistance the bulb should allow for a greater of inrush current, which is actually required isn't it when the motor starts up(inrush current).

Edit: I am going to make a pitching machine, its not going back into a treadmill.

Yes Umair, a motor takes a much larger current when it is starting up, but the bulb is acting as a variable resistor (with a slight time delay !), its resistance increasing massively as it heats up, making your calculations much more complicated !

You say, "Its odd my bridge was giving a total of 240Vdc rectified voltage OPEN although it should have given ~330 Vdc"

Your meter on the open bridge should show about the same figure for the rectified DC voltage as for the AC mains supply, because it is a pulsed DC with the same RMS value as the AC voltage, _until_ you put a capacitor on it that charges up to the peak value of the waveform.

A solution may be to use the full rectified 220 VDC, but limit the duty cycle of your PWM control circuit to keep the power down to an acceptable level. Or perhaps just a basic triac AC power control circuit before the bridge, with a means of limiting the maximum power it allows through.