Jaguars vs Victors

[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 ?

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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 :-)