motor current sense question

[Mike]

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

 

[Mike]

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

 

[Mochuelo]

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),

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?

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.

What do you measure, and how?

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

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

 

[Mike]

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?

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.

What do you measure, and how?

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

 

[Mochuelo]

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.

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.

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

And what are the numbers for each of them?

 

[Mike]

Ok. You could bypass the high-side FET, and connect that side of the
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.

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

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

 

[Spehro Pefhany]

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

Why do you think it should be the same?


Best regards,
Spehro Pefhany

 

[Mochuelo]

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)

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
---------------------------
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,

 

[Mike]

Why do you think it should be the same?

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.

 

[Mochuelo]

Maybe this picture helps:
http://personal.cucafera.org/ng/ng_20060118A01.png

Best,

 

[Mike]

Maybe this picture helps:
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

 

[Mochuelo]

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.

If I understood you correctly, yes.

What is the distinction between dis/continuous conduction mode?

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,

 

[Mike]

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

 

[Mochuelo]

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.

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

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

Best,

 

[Mike]

It must be the same, unless there are more components than the ones
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.

 

[Mochuelo]

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.

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,

 

[Mike]

If you sketch a schematic of your entire circuit, with two arrows
http://web.njit.edu/~mep4187/circuit.bmp

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)

 

[Mochuelo]

http://web.njit.edu/~mep4187/circuit.bmp

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)

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.

Is any of the FETs dead? How about the diodes?

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,

 

[Mike]

If the I_source you drew does not include the current that goes to the
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

 

[Mike]

Sorry, forgot to paste the link:

http://web.njit.edu/~mep4187/circuit2.bmp