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I'm trying to make this long post a little easier to read, and this is the only way I know how. So I apologize if this looks too much like a school report.

What I'm trying to do:

As part of my final year college project, I am trying to build a driver circuit to drive my magnetorquer (Think of it as a solenoid within a metal coil) , I am using this magnetorquer : www.cubespace.co.za/cubetorquer . I intend to supply bi-directional current to it using a H-bridge motor driver, I am using this motor driver chip I will be controlling the input pins (InA and InB) using PWM signals from my arduino. As a test, I am just using a simple arduino program that varies the duty cycle of the pwm output in Pin 5, and sets Pin 6 to digital low. Shown below is the circuit connection of my circuit:

Magnetorquer Driver Circuit Note that the batteries would represent my DC Supply(I am using 5V to power my chip) , and the inductor represents my magnetorquer. Also, assume that the bottom left pin of the IC is pin 1.

What went wrong:

I have observed that as I am drawing an unusually low amount of current, compared to what I should get theoretically(V=IR , resistance of the magnetorquer is shown in the link above ). When the duty cycle of my PWM signal is set to 50%, the chip only draws 0.030A of current, in that case the resistor of my magnetorquer would be R=V/I = 2.5/0.03 = 83.33Ω! Which is almost thrice of what is stated in the link above!

I am using an oscilloscope to measure the output voltage across my magnetorquer, when I set my duty cycle to 50%, what I get on the oscilloscope is a square wave that has max and min voltages of 2.5V and -2.5V respectively. If Pin 5(OUTB) of my motor driver is always 0, I should expect to see my output voltage vary from 0-2.5V right?

Also, when I set the duty cycle to 100%, I should be expecting to see 5V on my oscilloscope, but what I see is a very small DC voltage ( about 100mV ) which is effectively 0.

What I've tried:

I tried connected the negative end of my magnetorquer to ground while keeping the positive end at Pin 8 (OUTA) of my motor driver chip, but nothing changed.

I have tried shorting Pin 5(OUTB) of my motor driver chip to ground, didn't work either.

I have also tried connecting a the second channel of my DC power supply ( Set to 2.5V ) to pins 6 and 7 (InB and InA ) of my motor driver chip, when I do this the current jumps up to about 0.150A and doesn't increase when I increase the supply voltage of the second channel of my DC power supply.

I'm currently really out of options, at this point I can really only assume that my motor driver chip is broken. Is there something that I am not doing right?

Thanks and sorry for the long post.

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When the duty cycle of my PWM signal is set to 50%, the chip only draws 0.030A of current, in that case the resistor of my magnetorquer would be R=V/I = 2.5/0.03 = 83.33Ω! Which is almost thrice of what is stated in the link above!

If you have a 50:50 duty cycle, the average voltage applied to the device is 2.5 volts and 2.5 volts divided by 30 ohms (resistance of the device) is a current of 83 mA BUT, because power-in roughly equals power-out, the current taken from the 5V supply will be half of this at about 40 mA. So you measure 30 mA - it's not too far off and might be explained by slightly incorrect assumptions about duty cycle, measurement errors and metering errors.

I get on the oscilloscope is a square wave that has max and min voltages of 2.5V and -2.5V respectively. If Pin 5(OUTB) of my motor driver is always 0, I should expect to see my output voltage vary from 0-2.5V right?

You've probably got your scope set to measure AC not DC.

Also, when I set the duty cycle to 100%, I should be expecting to see 5V on my oscilloscope, but what I see is a very small DC voltage ( about 100mV ) which is effectively 0.

Yup that's what happens when you have the scope measuring AC.

I have also tried connecting a the second channel of my DC power supply ( Set to 2.5V ) to pins 6 and 7 (InB and InA ) of my motor driver chip, when I do this the current jumps up to about 0.150A and doesn't increase when I increase the supply voltage of the second channel of my DC power supply.

I have no idea why you would do this.

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  • \$\begingroup\$ Hi Andy, could you elaborate on why the current taken in would be half if power in = power out? Is there anyway that I can increase this current without increasing the source voltage? Oh and I tried using the second DC supply channel as well, I was running out of ideas and getting desperate. \$\endgroup\$ – John Oct 21 '16 at 22:19
  • \$\begingroup\$ Average volts out at 50:50 duty cycle is 2.5 volts. Average volts in is constant at 5V. Therefore if average current taken on output is X, because there is nothing else to waste power (MOSFETs turn hard on or hard off) the supply current to the chip MUST be 50% of X current that is being fed to the coil thing. \$\endgroup\$ – Andy aka Oct 22 '16 at 0:33
  • \$\begingroup\$ Would that mean that the coil is still drawing 83mA worth of current? \$\endgroup\$ – John Oct 22 '16 at 2:54
  • \$\begingroup\$ Average volts on the coil is 2.5 volts, coil resistance is 30 ohms. Ohms law tells you the current is 2.5/30 amps but put a meter in if you doubt it. \$\endgroup\$ – Andy aka Oct 22 '16 at 9:02
  • \$\begingroup\$ Yeah I get that part. What I am confused about is why the chip draws half the current the is supplied to the coil. So if my duty cycle is 50% , the chip itself draws about 32mA, but in 83mA is supplied to the coil? So if my duty cycle was let's say 20% , I'd get 1V average voltage, which would mean that 33mA of current goes to my coil, then how much current would I expect my chip to draw? Sorry for asking so much, but I'm really quite confused. \$\endgroup\$ – John Oct 22 '16 at 9:09

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