Partial Jag failure?

[techhelpbb]

Originally Posted by Mike Betts

JW,

If you look at the spec sheet for the MX5-A40, you can pull 133A through the circuit breaker for about 3 seconds before it trips. This can be a long time for a resistor to gain heat depending on the thermal resistance of the resistor.

Add to this that the MX5-A40 is an auto-resetting CB meaning that the heating can be cumulative as it repeatedly trips and resets.

Likewise, the 120A can pull much more than 266A for quite a long time...

Regards,

Mike

Edit: I had to back a couple of years to get a spec sheet on the 120A CB but you can see that it can pull 266A for over 20 seconds...

This was why I considered putting a fast blow 120 Amp fuse in series with my design for measuring the current (just like most digital multimeters). I fully expected someone would eventually make the common mistake of putting the circuit itself across the battery (either with no other load to monitor or effectively shorting out the load) and burn out the resistor even at 30 Watts...even if a fan was blowing on it and it was on a heat sink.

So that every one can see why:

E = I x R
E = 140 Amps x 0.002 Ohms
E = 0.28 Volts

P = E x I
P = 0.28 Volts x 140 Amps
P = 39.2 Watts

Now...even at 140 Amps...assuming you don't ever turn off the load...that resistor in my circuit (not the Jaguar) has too small a Wattage. Never let us mind that the 4 Watt resistor in the Jaguar would be toast at this point.

So what would happen if we put the resistor itself across the battery and ignore the wire and current limiting device resistance?

E = I x R
I = E / R
I = 12 Volts / 0.002 Ohms
I = 6,000 Amps

P = E x I
P = 12 Volts x 6,000 Amps
P = 72,000 Watts

So right about that point even the 30 Watt resistor would be destroyed...quickly.
Never mind that the battery might do something unpleasant.

Some fast acting fuses like the CNL125 which are targeted at fork lift application:
http://www.littelfuse.com/products/F...L/CNL125..html

Would probably work...but then...it costs a little more than $20.

The resistor I ended up using is only about $4:
http://www.riedon.com/us/images/stor...f/FPR2T218.pdf

In the end...this is why I was concerned about finding what FIRST might call a 'COTS' resistor for this application. Obviously if someone used a great big piece of copper or even aluminum with a known electrical resistance...one could fabricate a resistor that could survive dissipating a pretty serious amount of heat. However, in doing so you create a situation where the end user (any team using it) might have to be able to measure it and it has a small resistance that makes a simple digital multimeter pretty ineffective as an Ohmeter. Of course one could try to use it as shunt to measure it...but then I fear they could hurt themselves if they aren't careful. This is probably why in the past a coil was used to measure the current (so I've been told)...but the downside with that is that small currents become harder to read and fast changes in current get removed.

I'm open to better solutions for this sort of trade off (perhaps a bunch of slightly larger resistors in parallel dividing the Wattage, even 2 of the 0.002 Ohm resistors I've linked above would increase the rated Wattage to basically 60 Watts)...but if someone wants to discuss them...let's not distract further from this topic.

[techhelpbb]

To get this right back on topic...I'll summarize (and with respect to the other mentors I'll include some of their input)...

The 0.001 Ohm, 4 Watt resistor (the part labeled R23 in the Jaguar schematic) that measures current in the Jaguar is fine for normal operating currents one expects to use with the Jaguar. Those currents are generally less than 50 Amps, with a very occasional spike over 50 Amps and not much higher than 100 Amps. That resistor itself crosses it's maximum Wattage rating just after 63 Amps in this circuit.

However, it's Wattage rating is too small for repeat overload situations especially with the CIM motors and it can be both damaged and accidentally desoldered when such situations occur (and it's not the only thing in that circuit that will start to take damage).

Such situations are possible precisely because the current limiting devices between the battery and the Jaguar react relatively slowly. So even if they do trip they probably won't trip before you start to do some damage (and that damage will add up if you keep doing that).

If that resistor were to no longer connect the top of the power MOSFET circuit to the power from the battery and current limiting devices...then the voltage measured across that circuit and reported back to you will drop to basically zero. Maybe you can just replace that resistor...or maybe not...it all depends on what other damage has also taken place.

[techhelpbb]

So I was digging around in the older Jaguar schematics and discovered in the older gray (grey) model they used a 0.0005 Ohm, 2 Watt resistor (labeled R35) for measuring the current.

http://www.luminarymicro.com/products/rdk_bdc.html

The schematic and BOM in question is in the:
"Brushed DC Motor Module and RDK Design Package Rev B" dated 1/19/2010

The user manual in question is in the:
"Brushed DC Motor Control Module and RDK User's Manual" dated 1/19/2010

Additionally, instead of being on the high side (+ side) of the motor control circuit it's on the low side (- side) of the motor control circuit. This would have made it very hard to measure the reverse current that they don't measure anyway.

This altered resistor value, at half the Wattage still limits the safe operating current to 63 Amps or less continuous. However, it's lower resistance also limits the range of the voltage it will achieve across that resistor.

In that circuit they used an operational amplifier not designed for this purpose (not a big deal...except it might limit it's ability to survive overload).

That operational amplifier is configured as a non-inverting amplifier with a gain of 40. So any voltage that appears across that resistor will be multiplied by 40. So at maximum it'll report a voltage of about 1.26 Volts at 63 Amps. Still safe for the 3.3V microcontroller. However...this means that older Jaguars have a slightly reduced ability to measure current and still have the same possible failure situations.

Again...if that resistor burns out or becomes desoldered the circuit will still fail and you'll still read 0 Volts from it.
So my post above applies whether you use the gray (grey) or black Jaguars. Just modify it with the older Jaguars to disconnecting the low side (- side) of the motor driving circuit.

[Al Skierkiewicz]

Jim,
The spec sheets for the breakers we use can withstand 600% over current for several seconds prior to trip and can withstand something near 200% indefinitely. So for the 40 amp breakers, periodic stall currents are not enough to trip the breakers. As they are self resetting, the reset takes place almost immediately, and with sustained trip currents, the breakers will actually buzz. The drawback is that these devices are thermal by nature and so get hot under trip conditions. The main breaker is similar in design and function except it is not self resetting.

Tech,
The chip is called a shunt current monitor in that it was designed to be used in that application. The resistor in the Jag is merely in series with the output and as such is not really a shunt since it is the only path through which current flows. It is a convenient chip to use in this application since the gain is fixed, it is designed for single power supply and can swing to within 0.2 volts of the positive power rail. The output current of the battery is speced at around 600 amps for a few seconds and is primarily limited by the internal resistance of the battery which is .011 ohms. While it is easy to say currents in the Jag are normally around 50 amps, that is not correct. Many teams design mechanical systems that draw significantly higher currents. There has been at least one post in the past few days where the team started out draining their battery in less than two minutes. For the black Jags, the manufacturer states that the current monitor will fault the device based on current over specified time. I believe that the black Jag is more aggressive at current monitor due to the reduction of MOSFETs in the output stages. The Victors do not use a current monitor for a variety of reasons. I believe this is due to the fact that the sense resistor limits current delivered to the load, adds to the parts count, and is not needed with three 40 amp FETs in parallel in each leg of the output circuit. While the FETs have a higher series resistance than the Jag, in most of our applications, the added resistance is minimal as it amounts to about the resistance of two feet of #10 AWG wire. Please remember that the power rating on a resistor is based on it's temperature rise and sustained temperature over time. A 1/2 watt resistor can handle 10 watts for a few seconds while it can never handle 1 watt over a several days.

[techhelpbb]

Originally Posted by Al Skierkiewicz

Tech,
The chip is called a shunt current monitor in that it was designed to be used in that application. The resistor in the Jag is merely in series with the output and as such is not really a shunt since it is the only path through which current flows. It is a convenient chip to use in this application since the gain is fixed, it is designed for single power supply and can swing to within 0.2 volts of the positive power rail. The output current of the battery is speced at around 600 amps for a few seconds and is primarily limited by the internal resistance of the battery which is .011 ohms. While it is easy to say currents in the Jag are normally around 50 amps, that is not correct. Many teams design mechanical systems that draw significantly higher currents. There has been at least one post in the past few days where the team started out draining their battery in less than two minutes. For the black Jags, the manufacturer states that the current monitor will fault the device based on current over specified time. I believe that the black Jag is more aggressive at current monitor due to the reduction of MOSFETs in the output stages. The Victors do not use a current monitor for a variety of reasons. I believe this is due to the fact that the sense resistor limits current delivered to the load, adds to the parts count, and is not needed with three 40 amp FETs in parallel in each leg of the output circuit. While the FETs have a higher series resistance than the Jag, in most of our applications, the added resistance is minimal as it amounts to about the resistance of two feet of #10 AWG wire. Please remember that the power rating on a resistor is based on it's temperature rise and sustained temperature over time. A 1/2 watt resistor can handle 10 watts for a few seconds while it can never handle 1 watt over a several days.

You are absolutely correct. How long these currents are drawn is a vital factor with regard to the resistor (hence why I noted that using this power calculation is cheating to some degree). One of the problems being that outside the case of Jaguar you can't see what's going on with that resistor. There is to my knowledge no measurement of it's temperature and you can't make contact while it's assembled or even measure that temperature with infrared. You can't really trust the digital measurement of it either, because once it's too far you might not have time from the CAN bus to back off...unless you back off early. Now I'm not clear on exactly when the Jaguars will current limit and I haven't looked yet in the software. They state openly they'll let you drag 100 Amps through them at motor start...but that implies that there's a loose limit there somewhere...probably based on time (as you say they note the limit as being a function of how long it's over the limit). The Jaguar microcontroller itself is obviously quite fast and I'd think that when it does finally decide things are out of hand it would react pretty fast. In short it's much more practical for the Jaguar software to stop this situation than the software in the cRIO.

The real problem using time to determine the longevity of this resistor is that you might not be accounting for heat that is already baking the resistor from past overloads or ambient air temperature. I don't think they have any mechanism to force you to back off until things cool off. Even if they did...it would probably depend on the MOSFETs working correctly and if they turn into shorts (or fully saturate with no relief) and complete the circuit...the situation will not be possible to salvage.

Things is, there's also the issue with the Victor that they don't have a current mode. With the Jaguars they implemented that mode and therefore needed some way to measure current as well (hard to know which decision came first...did they start off wanting a current set point or current limit...and then need the resistor...or did they put the resistor in and then decide they could monitor it for more than over current).

Personally I find the empty slots in the ring that surrounds the Jaguar's MOSFETs under the fan to be sort of a dead giveaway that they thought they might add more MOSFET like the Victor.

[kamocat]

Getting back to the question of how the Jaguar failed:

Al's explanation of the desoldered current shunt resistor would make sense, except that the Jaguar says it's outputting 0v.

I re-checked the diagram: the Jaguar measures its supply voltage, but not its output voltage. The output voltage is calculated from duty cycle. (Indeed, in the first revisions of the firmware, the output voltage was only GIVEN in duty cycle, on a range of -1 to 1.)

Apparently nobody has had this exact issue before.

I will check that the resistor is still connected, but I will also try reflashing the Jaguar, in case the firmware got corrupted.

More info on the over-current: the Jaguar goes into over-current fault at 60A after 2 seconds. I was running an RS-775. (There were some binding issues with the jack screw on our arm)

[kamocat]

Well, it worked fine today. I even held it in a 41 Amp stall for 5 seconds.
There's only two causes I can think of.

One is that perhaps it had tripped the breaker and needed "voltage enable" to be sent again. However, I'm pretty sure I would have seen movement out of it before that happened.

The other is that today I'm outside at 15 degrees Celcius. Friday I was inside at 30 degrees. The Jaguar only said its internal temperature was 37 degrees on Friday. I thought it would operate up to 50.

(Note: I haven't opened it up yet.)

[Al Skierkiewicz]

Marshall,
This kind of action leads me to believe there is a foreign body floating around inside. A look inside is warranted.

[kamocat]

I just checked inside.
No debris.
No signs of heat.
The resistance from 0v to vLow measures 0.3 ohms on my multimeter (the same as when I touch the leads together). I measured this between the 0v 6-32 screw and the lead of a low-side MOSFET.

[techhelpbb]

Originally Posted by kamocat

I just checked inside.
No debris.
No signs of heat.
The resistance from 0v to vLow measures 0.3 ohms on my multimeter (the same as when I touch the leads together). I measured this between the 0v 6-32 screw and the lead of a low-side MOSFET.

It would be unlikely a simple digital multimeter could measure that small a resistance reliably with the currents it uses when measuring resistance. Though checking for continuity would confirm that there is connectivity and we know from the fact that it does work there must be a connection currently.

Also...if the resistor got very hot...it would heat the PCB and itself. That combined mass might take a while to cool off even with the fan.

Could it be that you melted the paste on that component and when it finally cooled off it closed again? One could also ask this question about the points to which the body of the resistor meet the areas designed to be soldered down.

In other applications I've seen surface mount resistors recover from overload before but usually it alters the resistance which you might not be able to measure with that meter. If it has altered that resistance by some appreciable amount it'll alter your current measurements. Do these current measurements seem proper for the mechanism in question?

Would there be some value in using a fixed, high power resistor as a load on the Jaguar and taking comparison measurements with another Jaguar? How about just trying to measure the current with another meter with a fixed resistive load?

[Al Skierkiewicz]

Marshall,
I am still going with my foreign body analysis. It is certainly possible for it to have been blown out by the fan.

[kamocat]

That works as a partial explanation.

Any idea why it would refuse to output power? (That is, it reported its calculated output voltage as zero)
I'm guessing it's one of the fault conditions. I know of four:

• Current fault
• Temperature fault
• Bus Voltage fault
• Gate Driver fault (I don't know how it detects this)

I can rule out the Current and Bus Voltage faults right now. I know it tells me of those. The battery was fine, and the Current Fault is temporary.

Does the BDC-COMM display a Temperature fault or a Gate Driver fault in red, as it does the Current fault?

[Al Skierkiewicz]

Marshall,
I don't remember you mentioning a fault condition. It is possible that a particular fault will shutdown the voltage mode and the software may be written to report 0 volts under that condition. It is possible for the any of the sense circuits to have had a fragment causing a short to supply. Say for instance, a short at the input of the current monitor could have artificially raised the voltage input to beyond current rating. In the Tan Jag there is no feedback from the output terminals so the zero volt condition must be derived mathematically from internals.

[techhelpbb]

In response to your question in a previous post above:
"Does the BDC-COMM display a Temperature fault or a Gate Driver fault in red, as it does the Current fault?"

In BDC-COMM itself...we find this in: bdc-comm.cxx

//
// Update the status items on the GUI.
//
g_pStatusVout->value(g_sBoardStatus.fVout);
g_pStatusVbus->value(g_sBoardStatus.fVbus);
g_pStatusCurrent->value(g_sBoardStatus.fCurrent);
g_pStatusTemperature->value(g_sBoardStatus.fTemperature);
g_pStatusPosition->value(g_sBoardStatus.fPosition);
g_pStatusSpeed->value(g_sBoardStatus.fSpeed);
g_pStatusLimit->value(g_sBoardStatus.pcLimit);
g_pStatusPower->value(g_sBoardStatus.lPower);

if(g_sBoardStatus.lFault)
{
switch(g_sBoardStatus.lFault)
{
case 1:
{
strcpy(g_pcFaultTxt, "CUR FAULT");
break;
}

case 2:
{
strcpy(g_pcFaultTxt, "TEMP FAULT");
break;
}

case 4:
{
strcpy(g_pcFaultTxt, "VBUS FAULT");
break;
}

case 8:
{
strcpy(g_pcFaultTxt, "GATE FAULT");
break;
}
}
g_pStatusFault->label(g_pcFaultTxt);
g_pStatusFault->show();
}
else
{
g_pStatusFault->hide();
}

So BDC-COMM can trap that gate fault...(I deleted a post above...because apparently they have some unrelated gadget that can't trap the gate fault condition...didn't want to confuse anyone).

This is where they draw and hide the control for the fault indicator in FLTK in the file: gui.fl

Fl_Box g_pStatusFault {
label FAULT
xywh {325 62 100 25} box ENGRAVED_BOX color 1 labelcolor 7 hide
}

So it does not appear that BDC-COMM considers the type of fault in setting this control's color.
Have you seen it do otherwise?

[EricVanWyk]

Marshall - does it respond to PWM normally? Are there any signs of physical damage?

Originally Posted by Al Skierkiewicz

...

Al -

Thank you for the very clear and informative post.

The difference in total on-resistance between the Jaguar and the Victor is inconsequentially small. I think the Jag wins the numbers game, but not by enough to make an actual difference.

The reason that the Jaguar uses fewer FETs is that it uses slightly more modern FETs that can handle more current per device.