Partial Jag failure?

[techhelpbb]

Originally Posted by kamocat

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.

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.

[techhelpbb]

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.

[techhelpbb]

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.