Jaguars vs Victors

[techhelpbb]

Originally Posted by Matt Krass

I was aware of the Jaguars synchronous behavior, but I thought it wouldn't be a problem because, if I'm looking at my sketch correctly...
Matt

Sorry, I'm not following. It might just be me, but I think I should take you up on your offer to provide that illustration of what you propose.

[Matt Krass]

Originally Posted by techhelpbb

Sorry, I'm not following. It might just be me, but I think I should take you up on your offer to provide that illustration of what you propose.

Excuse the crudity of this model, it's not to scale...

So, with the Jaguar in forward, and the Victor in reverse, in the on period for both, Node A will be at +12V, and Node C will be at 0V, flowing some current through dynamic load R1, and B would be at 0V and D at +12V, so current is flowing the opposite way through R2.

When the PWM goes off, let's assume the Victor stays on longer for the sake of argument, so C stays at 0V, D stays at +12V, and A also goes to 0V, and B goes to +12V... No current flows.

Ok, so what obvious thing did I miss?

[techhelpbb]

Originally Posted by Matt Krass

Excuse the crudity of this model, it's not to scale...

So, with the Jaguar in forward, and the Victor in reverse, in the on period for both, Node A will be at +12V, and Node C will be at 0V, flowing some current through dynamic load R1, and B would be at 0V and D at +12V, so current is flowing the opposite way through R2.

When the PWM goes off, let's assume the Victor stays on longer for the sake of argument, so C stays at 0V, D stays at +12V, and A also goes to 0V, and B goes to +12V... No current flows.

Ok, so what obvious thing did I miss?

Now I see what you intended. Thanks.

Well for one thing there's the current sensing resistor in the Jaguar which is not in your diagram.
If you look at Ether's link of the manual it's in the schematic on page 23 pretty much dead center.

[Matt Krass]

Originally Posted by techhelpbb

Now I see what you intended. Thanks.

Well for one thing there's the current sensing resistor in the Jaguar which is not in your diagram.

I left out things like that and the control electronics for brevity, I was more referring to any obvious reasons that my scheme wouldn't work that I overlooked.

I think in this scheme, the sychronous behaviour of the Jaguar is not a problem. I also suspect some inductive and capacitive components on each ESC might be wise to smooth out the output so the load is more stable, especially given the (wildly) different switching frequencies involved.

EDIT: I'm just curious, could you describe what you thought I meant?

Matt

[techhelpbb]

Okay, but the current sensing resistor in this case is limiting the current to the Victor as well. So you can't fully test the Victor's power range like that. Plus I don't think they have the same on-state resistance so the Jaguar might drop more voltage. Might not matter for what you plan.

I thought you planned on powering the same load with both speed controls, because you started off in the singular tense in the first proposition post. Clearly, however, you intend to use 2 loads and that's just fine.

The only thing is that the terrible things (as you put it before) involved here are not really inductive or motors?

I sort of ask because while I understand the heat concerns entirely, obviously there are quite a few other concerns in the comparison.

[Matt Krass]

Originally Posted by techhelpbb

Okay, but the current sensing resistor in this case is limiting the current to the Victor as well. So you can't fully test the Victor's power range like that. Plus I don't think they have the same on-state resistance so the Jaguar might drop more voltage. Might not matter for what you plan.

I thought you planned on powering the same load with both speed controls, because you started off in the singular tense in the first proposition post. Clearly, however, you intend to use 2 loads and that's just fine.

The only thing is that the terrible things (as you put it before) involved here are not really inductive or motors?

I sort of ask because while I understand the heat concerns entirely, obviously there are quite a few other concerns in the comparison.

I fully expect they don't have the same on-resistance, that is why I want to go for constant current loads, so we can measure the performance of the FETs, getting an empirical comparison of the 3 FET leg of the Victor vs the 2 FET leg of the Jag. If we measure the voltage drop and temperature across the FETs on each side we can see which heats up faster, which loses more power to heat, etc.

As long as the power source doesn't run out of volts and the load can compensate, the current sense resistor should be canceled out by the adaptive load.

Matt

[techhelpbb]

(Back to what I was communicating with Ether about...)

I wrote:

On what I just proposed I'm not sure but it would seem to me that dumping that energy into the low side of the bridge runs the risk to mess with the lower side reference 'ground'. Since the high side would have to be driven higher than the supply to saturate the MOSFET anyway, because the high side is N-Channel MOSFETs, wouldn't it less risky to dump that energy to the side that is already able to exceed the supply rail which might shift down anyway? I mean might not a shift in the lower reference cause the lower MOSFET to not be entirely saturated?

Hmmm, might not matter, guess it would depend on just how not ideal the lower reference 'ground' really is.

The high side current limit resistor is probably why they used the low side as the current going back up through the high side would heat that resistor. So that resistor makes the high side's path back to the battery positive less than ideal as long as it exists (even if the transistor remained fully saturated while it did that).

Trade off I guess.

[Ether]

Originally Posted by techhelpbb

The high side current limit resistor is probably why they used the low side as the current going back up through the high side would heat that resistor. So that resistor makes the high side's path back to the battery positive less than ideal as long as it exists (even if the transistor remained fully saturated while it did that).

Trade off I guess.

The current sense resistor would not be in the path of the inductive current if the high side were used for shunting

The inductive current would flow in a loop: through the motor, through one pair of high-side FETs, then through the other pair of high-side FETs back to the motor.

[dsirovica]

Guys,

if you use a Jag to drive a Victor you will get into a pickle. The two PWMs are free running and will produce all sorts of beating products. It will be very confusing to figure out whats going on. May be fun though!

What we have is really pretty simple. For the Jag we have decent documentation for the Vic we don't so we are guessing a bit for the Vics.

For the Jag, during the HIGH PWM cycle, we have:
Battery,
Current sensing Resistor,
Two Mosfets in parallel,
the Load (Motor),
Two more Mosfets in parallel.

During the LOW PWM cycle, we have:
the Motor that is shorted via two sets (in series) of two parallel Mosfets.

The Mosfets are driven hard-ON and hard-OFF at 15KHz by the driver chip.

All the other components are “in the noise” - that is they matter very little and can be ignored for our purposes here. It is fair to assume that these other components are engineered properly, and are not limiting the behavior we observe.

Based on the previous discussion in this thread it seems that the practical limitation we have is thermal overheat due to high amperage. The Jags have software that cuts out way before meltdown (perhaps too soon, sacrificing maximum power), while the Victors can burn in the field but as a result deliver more power than the Jags. This gives Victors an apparent power advantage, and it seems many teams prefer them for that. The sad thing is that the Jags actually could deliver more power than the Vics if allowed.

I've been playing with a simulator – it is really cool and answers many of the questions we posed here. I would advise all to play with this, see:
http://www.linear.com/designtools/software/
It is very noble for LT to provide this for free – its a great educational and practical tool.

LT doesn't have the exact MOSFETS we have, but you can get close, or design your own.
I entered a simplified model with just one Mosfet. I don't have a Motor model, so I assumed a Stalled-Motor equivalent of an Inductor and series Resistor. A good approximation for a CIM is L=200uH, R-80mOhms (please advise if you have better data on this). See:
http://i.imgur.com/1cNXB.png

You can see the Current through the Inductor is ~60A. In reality we were getting 55A see real picture of a PWM driven Jag and stalled CIM Search for the "Voltage vs. PercentVbus” thread for more motor pics:
http://imgur.com/0IjKf

To give more details on the simulation parameters if you want to play with it (it takes just minutes to do this):
V2 is 12VDC. R3 is the internal Battery resistance (ours measured about 20mOhms)
L1 is 200uH, and R2 is 80mOhms these numbers seem to match a stalled CIM based on some measurements we ran in the aforementioned thread. It would be great if someone has the SPICE models for FIRST motors!
I just used a Diode D1 for the flyback shunt. You can see this would dissipate 30W if the Jags didn't use the Mosfets for this! As the current through D1 is 60A during the LOW PWM.
V1 is the PWM:12V, Period=75us (for 15 Khz), I gave it a rise and fall time of 2.5us each. It runs for 200 cycles or 15ms.
R1 is an arbitrary value. You can see that the Mosfet's parasitic capacitance is spiked through this R and that is where the Capacitive loss is converted to heat – that stays in the Mosfet Driver chip – I will let you all run this to see the trace of the current through R1 – it is cool!
The Mosfet I picked is IRFH5004, this has an RDS of 2.2mOhms and I doubt this simulator takes heating into account so RDS probably stays at 2.2mOhms. In our reality we are heating these puppies driving RDS to about 150% of typical value.

The main things still missing are the forced-air cooling equations for the Jag&Vic (ie. What thermal C/W number should we use) and all the documentation that we don't have for the Vics.

And for the advanced student, here is the Fast Fourier Transform (FFT) of this PWM signal on V1 – it is mind boggling what LT Spice can do !!!
http://i.imgur.com/7Fjh2.png
You can see the main component at 15KHz, and the square wave's harmonics at 3X intervals.
This is much simpler and hugely more powerful than the Fortran punched card program I used for very primitive Fourier Analysis in my college days :-)

[techhelpbb]

Originally Posted by Ether

The current sense resistor would not be in the path of the inductive current if the high side were used for shunting

The inductive current would flow in a loop: through the motor, through one pair of high-side FETs, then through the other pair of high-side FETs back to the motor.

This is very true, but I would think only after the delay when the other half of the high side turns on.

( I hate to put this below a post like the one directly above, dsirovica that was a very good post.)

[Ether]

Originally Posted by techhelpbb

This is very true, but I would think only after the delay when the other half of the high side turns on.

Are you talking about the 2 microsecond shoot-through delay? During that time, the intrinsic diode of the non-conducting high-side FET pair would complete the current loop. The inductive current doesn't flow through the sense resistor, because there is no path to ground.

[techhelpbb]

Originally Posted by Ether

Are you talking about the 2 microsecond shoot-through delay? During that time, the intrinsic diode of the non-conducting high-side FET pair would complete the current loop. The inductive current doesn't flow through the sense resistor, because there is no path to ground.

Wouldn't the bias current of the INA193 reference ground to some extent?
Though it's certainly not going to see anything more than 20uA.

No you're right...while it might impart noise on the measurement it wouldn't produce any appreciable current flow in that resistor so it wouldn't heat it.

[dsirovica]

Techhelp & Ether, just to clarify and hopefully close on the “Dead Time” issue.

(BTW, you cannot use the high-side for the inductive current path as Q1's Intrinsic Diode is pointing the wrong way)

This is reasonable well explained in:
http://www.ti.com/lit/ug/spmu130c/spmu130c.pdf
On Page 18 under “Motor Output Stage” and “Switching Scheme”, and shown in Figure 4-4
on Page 19.

From there I quote:
“The 2μs DEAD_TIME period starts with Q1 turning OFF. Current continues to flow through the
load, with the path being completed by the Q2 intrinsic diode and Q4. The voltage drop across Q2
is equal to a forward-biased diode.”

In Figure 4-4 they show 3 red-current paths happening in this sequence:
1. “PWM_ON” during the PWM-HIGH cycle
2. “Dead_TIME” -the inner red arrow on the low-side. When Q1 shuts off the current HAS to continue flowing through the inductor(Motor) – (for a convincing and memorable learning experience -see below). This current is pushed out of the inductor(Motor) into the already ON Q4, then via the bottom wire (just happens to be connected to ground though that is not relevant here), through Q2, and sucked back into the inductor(Motor). Now Q2 is OFF, so the current passes through the “intrinsic diode” of Q2 (shown on the right of Q2's gate). The power dissipated in this diode is what this discussion was all about, and is P=I*V. V is the voltage drop across this diode, lets say this is 0.8V. I is say 60A, so the Power dissipated is 48W. But the duration of this is set 2us(microseconds). Therefore the energy per event is E=P*t, so E=48W*2*10^(-6), or 96uJ. Now there are two of these events per PWM cycle (the documentation doesn't state the duration of dead-time for the ON transition, but one assumes (danger!!!) it is also 2us, and there are 15,000 PWM cycles/second, so the total Power for this mechanism is P=E*2*15000/s=2.88W, or 1.5W per Mosfet – This IS rather significant given our earlier numbers in:
http://www.chiefdelphi.com/forums/sh...6&postcount=50
However – this 1.5W is produced in the non-driving low-side Mosfets. Now, these same Mosfets take the full inductor current in the next sequence:
3. “PWM_OFF” during the PWM-LOW cycle. This current is shown in the outer red arrow on the low-side. The Jag switches this Mosfet ON as it dissipates much less power than the intrinsic diode. So why not switch it on immediately after switching Q1 OFF. The reason is engineering safety margin of 2us to ensure both Mosfets are never ON at the same time. I am guessing that the designer chose 2us for good reasons, probably to allow for different PN of Mosfets etc. in the future so that the design need not be reexamined for each PN change.

This would suggest that the worst case power dissipation will be on the LOW-PWM cycle because then we have both the Mosfet ON conduction as well as this reverse diode power loss of 1.5W. Therefore a lower PWM duty cycle would yield a higher thermal dissipation. However, at low duty cycles we will not be able to attain high current as that is driven by the HIGH-PWM period. The following picture, with our simplified Mosfet driver, shows that at PWM duty cycle of 25% we only attain 30A on a stalled CIM – as compared to 60A for a 50% PWM duty cycle.
http://i.imgur.com/iO447.png

So lets try to find the worst case power dissipation – which will be our limiting factor for maximum power output.

From a prior post on this thread we concluded the following:
Jag FDP8441 RDStyp=2.1mOhm, Temp factor (150C)=1.55 => effective RDS=3.26mOhms
Power dissipated due to 15KHz switching = 0.5W
At current=60A with 99% duty cycle Pmosfet=3.4W
This is for each of the two high-side driving Mosfets.
So lets fix 60A into the motor, and a Junction Temp of sizzling 150C.

Now we know that the low-side Mosfet has an additional 1.5W due to the 2us Dead_Time. So we want a PWM duty cycle that uses the low-side more. Lets try 50%.

We then have the following power contributions to the low-side Mosfet:
P on_current= 50%*30A^2*3.26mOhms= 1.5W
P dead_time= 1.5W
P switching= 0.5W

Ptotal= 3.5W for each of the two low-side non-driving Mosfets
Ptotal=1.5W for each of the two low-side driving Mosfets (I am guessing these are full-ON all the time)
Ptotal= 2W for each of the two high-side driving Mosfets (this came down from 3.4W as we reduced the PWM cycle)
Ptotal= 0W for the non-driving high-side Mosfets

We might be able to dissipate a bit more power on the limiting Mosfets by going to a lower PWM cycle, but then we will start to see a lower max Amperage – so this late in the day – I am content to say the max power dissipation is 3.5W on the two non-driving low-side Mosfets – that is where I would stick a thermometer! Any offers for more than 3.5W ?

--------

So for the promised “memorable learning experience” above:

Lesson: an inductor always resists changes in the current.
Explanation from physics: A current in the coil builds a magnetic field inside the coil. Any changes in that magnetic field induce a current in the coil. Therefore once you “charge” a coil with magnetic energy, that energy will want to sustain the current in the coil. If the coil circuit is discontinued, then the voltage will rise until...

Take a 9V battery and any old transformer, such as one from a wall-wart. Connect the battery to the secondary windings – the low voltage side. Hold the two primary wires in one hand – make sure they touch your skin, but not each other. Then in a couple of seconds, disconnect the battery. DISCLAMER: never use anything bigger than a small 9V, or AA, AAA, battery. Never use any transformer bigger than what you can cup in one hand. Do not do this if you have any medical condition. Do not do this if you are in an environment where you can hurt yourself by falling or bumping into things due to a sudden jerk.

When I was a high-school freshman in a boarding school in England a couple of us played the following trick: during lunch we asked a dozen friends around the table to grab a fork in one hand and a knife in another and create a ring. We then proceeded to execute the experiment above with the dozen kids connected to the primary winding on a transformer. Knifes and forks went flying in the dining hall with screams to accompany the pandemonium. Lesson: be aware of unintended consequences :-)

[Ether]

Originally Posted by dsirovica

(BTW, you cannot use the high-side for the inductive current path as Q1's Intrinsic Diode is pointing the wrong way)

Not true.

See attachment.

To use high-side for inductive current path, switch as follows:

For positive (forward) motor current:

Q1	Q2	Q3	Q4

on			on	PWM ON

on		*		Dead Time *current flows thru Q3 diode

on		on		PWM OFF

For negative (reverse) motor current:

Q1	Q2	Q3	Q4

	on	on		PWM ON

*		on		Dead Time  *current flows thru Q1 diode

on		on		PWM OFF

This is reasonable well explained in:
http://www.ti.com/lit/ug/spmu130c/spmu130c.pdf
On Page 18 under “Motor Output Stage” and “Switching Scheme”, and shown in Figure 4-4 on Page 19.

Yes, we know. This was already mentioned in post#82
http://www.chiefdelphi.com/forums/sh...0&postcount=82

[dsirovica]

Quite right Ether,

The H-Bridge is a mirror image, so you can do with one half what you can with the other.

Thnaks for pointing that out.
Dean