Partial Jag failure?
 

At the regional this weekend, I have a tan Jag that, after two days of use (and several over-current faults) refuses to output power anymore.
When I noticed this, I tested this with BDC-COMM, in voltage mode. Not only did it not move the motor, but its output voltage remained 0. The heartbeat was enabled. (Even when it's not, however, I should get a short pulse for 0.1 seconds)
The Jaguar said it was at 35 degrees Celsius.

I plan to check this again to make sure it wasn't some temporary thing, (and that I didn't make silly mistakes in testing). However, I'm curios if anyone else has experienced this.

Re: Partial Jag failure?
 

Have you considered inspecting the MOSFETs inside the unit?

You'll find them in a ring under the fan.

The screws to open the unit are in the bottom.

Re: Partial Jag failure?
 

I did not smell any burning, but it would not surprise me if this was the issue. I will check later.

What I find particularly strange is that it SAYS it's output is zero. I thought this was a calculated value, not a measured value.

Re: Partial Jag failure?
 

There's a voltage VM+ on the Jaguar schematic in the User Manual.

http://www.luminarymicro.com/jaguar

It's connected to a voltage divider to scale it down so it can be input into the microcontroller analog to digital converter.

The Jaguar also has an INA193 on the schematic. That is a current shunt monitor measuring across a 0.001 Ohm resistor on the high side of the H-Bridge. The INA193 is a unidirectional current shunt monitor, but it will survive the reversal of it's input voltage. They made a provision but didn't populate for a bidirectional current shunt monitor.

http://focus.ti.com/lit/an/slyt311/slyt311.pdf

In short, it can measure both voltage and current to the motor....but in this configuration it can't measure any currents going backwards either due to the body diodes of the MOSFETs or the collapse of the motor's field. This circuit is not 'regenerative' so it's not applicable (it might be...in say an electric vehicle application where the motors turn into generators to slow the vehicle down and use the generated power to charge the batteries).

Just keep in mind, however, that when I say it can measure that voltage...it sort of cheats at that. It assumes the MOSFETs are operational and approximately matched.

Re: Partial Jag failure?
 

To extend this...

If you look at page 21 of the manual you'll see that VM+ is actually not across the motor output...but between the current shunt on the high side and the high side MOSFETs.

If you note the top of page 22 under the heading 'Voltage Sensing' you'll note that they refer to VBUS with regards to this voltage.

I should like to point out that if you look at that current shunt named R23 in the BOM, you'll note it's: 0.001 Ohm, +/- 1% and only 4 Watt.

Now...the Jaguars claim they can withstand 100 Amp at motor startup.

If we take Ohm's law: E = I x R.

If I = 100 Amp and R = 0.001 Ohm.

The voltage (E) across that shunt will be: 0.1 Volt
(I should like to point out that in older analog current meter applications it was quite common to see current shunts designed to produce maximum voltage drops of 0.05 Volt or 0.1 Volt...this is likely not an accident.)

So the Wattage across that shunt will be: P = E x I
(I'm using DC analysis here...so I'm cheating a bit)

Therefore the Wattage would be: 10 Watts.

Now...why the mere 4 Watt resistor?

Well...the Jaguar only says 100 Amps at *startup* and once the motor starts moving it'll not need so much current to keep moving...or that's the theory.

So basically they hope you won't cook that resistor until it opens the circuit. If you did cook that resistor the only voltage that would appear on that analog to digital converter (for reading VM+) would be driven there from the converter's input bias current...and that might be very low since the low side of the voltage divider will pull to ground pretty hard.

In another topic up here I was talking about monitoring the circuit current with a shunt (resistor). Oddly I designed a whole circuit for this application...then went back and took a good look at the Jaguar schematic and noticed they used the same part (INA193). Only thing was...I designed it with a 0.002 Ohm resistor, +/-1% accurate, 30 Watt resistor. I assumed someone *would* draw upwards of 120 Amps through that circuit and often enough it would be a problem otherwise. I came to this decision because the CIM motors often used can pull over 120 Amps when stalled...and they'll survive that abuse for a good long time.

In your case...if you've beaten up the Jaguar often...you've been effectively testing the response time of your current limiting systems. So maybe you 'got lucky' and finally did some lasting damage. If you can check that resistor you might want to.

Re: Partial Jag failure?
 

To make what I wrote above a bit easier to read...

Current shunt....is a resistor often of a small value.

The resistor is put in series with the load (in this case the circuit controlling the motor), and because of it's small value it develops a small but proportional voltage across it. That voltage is proportional to the current in the circuit because of Ohm's law.

A 'current shunt monitor' or 'current sense amplifier' or 'current sense monitor' is an analog circuit designed to monitor the small voltage that develops across a current shunt resistor and usually to amplify it to some extent. For example the INA193 will multiply that voltage by a fixed 20x.

So if the 0.001 Ohm resistor in the Jaguar had a voltage of 0.1 Volts across it (the maximum) the INA193 that measure it would have a voltage of 2 Volts across it. That's okay the microcontroller operates on 3.3 Volts.

Also for those that don't know...when a motor field collapses the current that flows will be reversed from the direction of flow of the current that charged that field. So basically, as the Jaguars are built now they can only measure the current they put into the motor's fields. Not the current that flows back from the fields collapsing. If they put the 'bidirectional' current shunt monitor into that circuit they could measure that reverse current if they were able to do so fast enough...I suspect this was considered to be not essential.

Re: Partial Jag failure?
 

Dude,
This is not always the failure mode on the Tan or Black Jag. Your calculations on the current sense resistor are mostly correct and it is likely the source of the failure if nothing else went sour at the same time. I have noted that many tan Jag failures occurred when the current sense became so hot, it unsoldered itself from the board and in the movement of the robot actually moved off it's circuit pads. This opens the current flow inside the Jag. Just to be accurate, CIMs draw 133 amps in stall which would dissipate about 17.6 watts in the current sense resistor. Over an extended amount of time, this would cause severe heating of the resistor. Normally, CIMs are run at 40 amps or less so normal dissipation is about 1.6 watts.

The current shunts in analog meters (and many digital meters with current measuring) is to prevent damage to the meter (either causing the pointer to wrap itself around the pins or damaging analog to digital convertors). The more sensitive an analog meter, the less current it would need to go full scale. I have an old Midland analog meter that has a full scale current of 60 microamps. In order to measure current the meter needed a current shunt across the terminals to prevent damage. Say you wanted to measure 10 amps full scale, then the shunt had to be sized such that 9.99994 amps flowed in the shunt. This shunt is not the same as the series current sense resistor used in the Jags. My Fluke uses a #10 wire as a shunt on the high current (10 amp) range.

Re: Partial Jag failure?
 

Al:

Where does circuit protection factor into this? Individual Jags are protected by breakers at 40A max. How can 133A continue "over an extended amount of time" without something tripping?

Similarly, the main breaker on the bot is rated at 120A. Two or 4 jags in this situation should equal 266 or more amps. Why no tripping there?

JW

Re: Partial Jag failure?
 

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...

Re: Partial Jag failure?
 

I will not disagree with you about your experience in the matter with the analog meters (you are not incorrect about it, I am merely trying to use the terms in this way because I had hoped they would be most clear in this manner). However, different strokes for different folks.

Texas Instruments, the designer of the part INA193, specifically calls it a 'current shunt monitor':

http://focus.ti.com/lit/ds/symlink/ina193.pdf

I specifically provided an official application note above in which those parts are used in a replacement function over a current sensing application...and in the last post of the series...openly noted that it could also be called a 'current sense monitor' or 'current sense amplifier'. In point of fact, other manufacturers do...indeed...use the alternative naming.

For example:

Here national semiconductor refers to the same basic circuit by the alternative name...
http://www.national.com/analog/ampli...ent_sense_amps

Here WikiPedia furthers calling this a current shunt:

http://en.wikipedia.org/wiki/Shunt_%28electrical%29
"An ammeter shunt allows the measurement of current values too large to be directly measured by a particular ammeter. In this case the shunt, a manganin resistor of accurately known resistance, is placed in series with the load so that all of the current to be measured will flow through it. The voltage drop across the shunt is proportional to the current flowing through it and since its resistance is known, a millivoltmeter connected across the shunt can be scaled to directly display the current value."

Wonderful world of engineering...same idea, different words...and we are all using the same language. Let's try it in Korean or Japanese.

This is merely a matter of difference of terminology. This is why I posted that 3rd post.

Further, you are correct...of course...it's not always the MOSFETs that go.
However, it never hurts to ask because the MOSFETs have even more reasons they can go.

Re: Partial Jag failure?
 

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.

Re: Partial Jag failure?
 

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.

Re: Partial Jag failure?
 

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.

Re: Partial Jag failure?
 

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.

Re: Partial Jag failure?
 

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.