# motor current sense question

• posted

When driving a DC brush motor with a MOSFET h-bridge ...

Applying a PWM signal to the low-side FET (the high-side FET is always on), with a duty cycle such that the RMS voltage at the motor terminals is say .. 2V-rms.

I am driving a high power resistor / inductor in series to simulate a motor. I know my DC resistance is ~0.25 ohm. Calculating my Imotor-rms, I get 4 amps, which is consistent with what I measure.

Why is the rms current I measure/calculate through the 'motor', not the same as my power supply current?

Thanks, Mike

• posted

Correction on the calcuated motor current ... should be 16 amps.

• posted

Sorry, it is a FULL-BRIDGE. There is a MOSFET driver controlling the bridge. It is set up such that the high side FET is always on, the low side on (on the other side) is the one which is PWM'd. The other half of the bridge is not in use. Consider half of the bridge.

I am measuring current through the load with a current probe. Also at the source.

• posted

If it was like you are saying, you would be creating a short circuit every time that the low-side FET is on. Maybe you are actually not talking about a half-bridge.

I'm not sure that I understood your circuit. One side of the motor is connected to the node that is common to both FETs, and what about the other side of the motor? Where is it connected?

What do you measure, and how?

Isn't this the first time that you mention the power supply current?

• posted

Ok. You could bypass the high-side FET, and connect that side of the motor directly to the power rail.

Maybe you already have it, but remember that there should be a path for the inductor current to continue existing right after the FET turns off. The current through an inductor has a high "inertia." This task is usually done by one diode in anti-parallel with the high-side FET that you are not using.

And what are the numbers for each of them?

• posted

This is not possible because I need to maintain the full-bridge

All of the MOSFETs (IRFS4310) have reverse biased diodes built into them. In addition to this I have also incorporated transient voltage suppressors across both the low and high side FETs to further divert back emf voltage transients from reverse biasing the MOSFET.

OK, here are the numbers:

V-motor(rms) ~= 9.36 [volts] (measured with dmm) I-motor(rms) ~= 36 [amps] (measured with scope current probe) I-motor(peak-peak) ~= 12.5 [amps} (measured with scope current probe) R-motor ~= 0.27 [ohms] (measured with dmm)

V-source(rms) ~= 48.0 [volts] (measured with dmm) I-source(rms) ~= 9.4 [amps] (measured with scope current probe) I-source(peak-peak) ~= 4.0 [amps] (measured with scope current probe)

And just to make sure we're on the same page:

I have a full-bridge MOSFET, but I am only driving the motor in one direction or using half of the bridge. The high-side FET (top left of bridge) is always ON. The low-side FET is being PWM'd such that the the V-motor(rms) is as mentioned above. A MOSFET driver is being used.

Thanks for your time, any suggestions / insight is appreciated.

Mike

• posted

I just re-did some tests, to confirm the numbers I'm getting. I assumed that the two should be the same. It's obvious now that I'm missing something fundamental. Honestly, I'm just flat out confused.

• posted

Mochuelo, the picture definetly helped.

So you're saying that because after the inductor is charged , and its current can not immediately go to zero when the end of the pwm ON-time is reached (I-source goes to 0), the inductor maintains current (decaying of course) until the next period when it begins to charge again (and I-source > 0). So my inductor is always higher because the inductor can't change its current fast enough. But the I-source > 0, only when charging the inductor at which time their currents are the same.

What is the distinction between dis/continuous conduction mode?

Thanks

• posted

Why do you think it should be the same?

Best regards, Spehro Pefhany

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• posted

One more question ...

I got the jist of what you are saying, but there is something that does not correlate to my measured results.

If I look at the peak currents of my source and inductor, they do not match. What I'm saying is that while the inductor is charging the source current is not the same as I understood in your explanation.

The source current is in fact increasing with increasing inductor current, but the peak levels attained to not explain why the inductor current is higher?

Am I missing something?

Thanks

• posted

I_motor(rms) > I_source(rms), and that is because the percentage of the time that the motor is traversed by a current is always larger than the percentage of the time that the source is traversed by a current and because whenever there is source current, it is equal to the motor current.

Step by step:

---------------------------

1) When the FET is on: I_inductor > 0 I_source = I_inductor > 0 I_diode = 0

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2) When the FET is off, we have two cases: 2a) Continuous current mode. I_inductor does not have time to go down to zero. I_inductor > 0 I_source = 0 I_diode = I_inductor > 0 2b) Discontinuous current mode. We will further divide this time period (FET off) into two sub-periods: 2b1 and 2b2. 2b1) (the inductor is being discharged through the diode, but this current does not flow through the source) I_inductor > 0 I_source = 0 I_diode = I_inductor > 0 2b2) (the inductor is and stays completely discharged) I_inductor = 0 I_source = 0 I_diode = I_inductor = 0

---------------------------

In other words, whenever I_source is nonzero, there is also I_inductor, and both currents are identical. But there are also times (2a or 2b1) during which I_source is zero and I_inductor is nonzero, and these are the times that make I_source(rms) < I_inductor(rms), as you measured.

Best,

• posted

Maybe this picture helps:

Best,

• posted

If I understood you correctly, yes.

This distinction applies to each inductor and transformer present in any switched-mode circuit that includes such elements. A magnetic storage element works in continuous conduction mode (CCM) if its magnetic flux (which is proportional to the electric current) is always nonzero in steady state. Otherwise, it works in discontinuous conduction mode (DCM). If your circuit includes only one magnetic element, then you say that the circuit works in CCM or DCM.

Many factors determine whether your circuit works in CCM or DCM: frequency, inductance, switch on and off times, voltage levels seen by the magnetic element, etc.

For instance, look at the bottom half of the figure I posted. If you reduce the FET off time (time during which I_source = 0), keeping the on time constant, you will reach a point where I_inductor will never be zero, in steady state. At that point, you will have toggled from DCM to CCM.

It is important to know that there are two conduction modes, and to know where your circuit is, because the equations that state the relationships between input and output average magnitudes are different in each case. The mode with the simplest equations is CCM.

Best,

• posted

There are capacitors across the supply (2x 200uF), that's it.

The peak levels are no where near each other. The current probe is set on DC coupling.

• posted

It must be the same, unless there are more components than the ones you mentioned. Do you have a capacitor somewhere? The I_source I have been talking about is the one that you would measure right at the inputs to the full bridge.

The peak levels must be identical. Can your current probe detect a dc current?

Best,

• posted

If you sketch a schematic of your entire circuit, with two arrows showing where you are measuring I_source and I_inductor, and post it here (or send me an email), I'll probably be able to tell you what's wrong.

Best,

• posted

Here's some new numbers, I re-did the test to make sure everything is accurate.

I-motor(rms) ~= 2.98 [amps] (measured with scope current probe) I-motor(rms) ~= 2.99 [amps] (measured with dmm)

R-motor ~= 0.27 [ohms] (measured with dmm) L-motor ~= 0.3 [mH]

V-source(rms) ~= 48.0 [volts] (measured with dmm) I-source(rms) ~= 0.279 [amps] (measured with scope current probe)

• posted

If the I_source you drew does not include the current that goes to the capacitors, yes, I_source must be identical to I_motor whenever I_source is nonzero. Most important, you should be convinced of this, regardless of what I'm saying.

While the PWM FET is on, the diode in anti-parallel with the high-side FET that is off is reverse-biased so there can't be current through it, and therefore I_source = I_motor. They must be identical, because there is no other possible path for I_source. When the PWM FET turns off, I_motor cannot grow further, because the voltage that the motor sees is suddenly reversed (it goes from +48 V down to -0.7 V, approximately), and the voltage across an inductor is proportional to the slope of its current. So the peak levels of I_source and I_motor must be equal.

Are you sure the other two FETs are off? How do you keep them off? If they are enhancement mode MOSFETs, the best way is to connect gate to source.

Are you using two current probes, or just one? If you are using two, put them to measure the same current, and make sure you see the same waveform. Also, make sure that not only the probe(s), but also the scope are dc coupled.

Can you sketch the waveforms that you are seeing in your scope, for I_source and I_motor? Please draw both on the same {i(t),t} axes.

Best,

• posted

The corrected sketch is at the following link:

The h-bridge and the capacitors share the same ground path.

How does this affect me? Is it because the capacitors are a source of current when the end of the PWM ON time is reached. Instead of the source current going to zero, it decays slower because of the capacitors? I still don't see how this accounts for the mis-match in my peak currents.

I certainly have a better understanding now of what's going on. I thank you for this, I'm much further along that I was a few days ago.

I agree with what you are saying, but something is just not right. My peak levels of current are NO WHERE near eachother?

The HIGH side (top right) FET is actually always ON (if) the LOW side (low right) is OFF. The mosfet driver I am using does this automatically. There are 5 inputs to the mosfet driver, A-LOW, A-HIGH, B-LOW, B-HIGH, and a DISABLE. I only have control of three of the inputs from my microprocessor. I did this to reduce the number of opto's I needed. So what I do is, pull the A-HIGH, and B-HIGH inputs UP (hardwired always active), and just worry about controlling the LOW-SIDEs from my microprocessor. This works because the mosfet driver automatically commands of the HIGH side FET when the same side LOW FET is commanded on. There is an adjustable delay built in for this, I have it maxed out to approx. 0.5us. The mosfet driver also has a bootstrap circuit, and of course pumps up the HIGH side gate voltages so that the Vgs is adequate to turn the fet ON/OFF.

As far as I can tell, NO. I had plently of experience burning FETs.

I have a single current probe. What I can do is capture the wave forms from the scope screen (save them as .bmp or whatever), one for the source, one for the motor. The current probe is a tek 'smart' probe, it automatically adjusts the scope settings. I double check to confirm this.

I'm not in my lab now. I will get the wave forms first thing tomorrow. Thanks for all your continuing help.

Mike

• posted

Sorry, forgot to paste the link: