Roff said:You must be switching at a fairly high frequency.
no its just dknguyen maths doesn't add upHero999 said:Roff, you've just said that. You must have pressed the reply button twice.
Styx said:no its just dknguyen maths doesn't add up
so... " You must be switching at a fairly high frequency."
252nC and switched in 50ns will require a gate-resistor of X-ohms ok BUT since you are switching a FET it only draws current during turn-ON and turn-OFF for ... 50ns
To calcuate the true wattage of the gate resistor needed you need to calculate the RMS current AND that is dependent on the switching freq, in this case 30Khz
How did you calc the gate charge for this FET? For a 1200Volt, 1000Amp IGBT's (and even then the gate-capacitance is that high cause there are 8 in parallel in the package) I get a gate-charge 5x that, so something not right there
EVEN switching such devices at 20KHz with a peak gate current of 15A, using +/-15V to switch the gate as well (and devices hitting Millar in 500ns and gate-current negligiable in 1us) I only need 2W worth of gate resistance
you seem to have gone astray in yr maths
please provide the FET you are using as well as the drive circuit (ie what voltage you are pulling the gate-source between)
I do these calculations weekly so I can give it a go for you
I have a nice excel spreadsheet that chucks out the numbers for me so give me some numbers
BUT to answer your question BJT's will be fine, put something like 6uF worth of capacitance near the BJT's to provide a source for the current-spike at turn-ON and turn-OFF. I use 44uF for my cct's to give you an idea
dknguyen said:Is there some confusion going about whether W means watts or ohms? I mean watts when I use W, and I use R or "ohms" for resistance. I've seen my power prof use W as ohms though and Styx seemed to just do that to. So when I say 3W and 4W, I mean the power dissipated due to the transition time is 3W at 50ns and the I^2R losses are 4W at full load.
Yer Zetex have some fantastic discrete components. That or use a good FET drive chip, I quite like the IR4426 (and its non-inv equiv) - although in this case it would be too low a powerdknguyen said:I know the power on the gate drive BJT is low, but I was having problems finding one that could actually had a large enough beta value at higher collector currents. THe signal ones I was looking at had betas that dropped to about 1 at 1A (but I found some that don't now from Zetex. It's like oversized BJTs in an undersized packaged for high current, low powwer apps)
dknguyen said:I read the gate charge off of the datasheet and used Q=IT where T is the transition time. I'm using 15V gate drive switching at 30kHz with a rise time of 50-100ns. Right now I have two 4.7uF decoupling caps for each driver. Each gate driver is being powered by a 15V@66mA floating power supply.
https://www.irf.com/product-info/datasheets/data/irf2804s-7p.pdf
I am using the Figure 2 drive circuit:
https://www.farnell.com/datasheets/62207.pdf
except there is a Q1 emitter resistor and Q2 emitter resistor that are used to control the gate resistance and shoot-through, but they may be zero. I haven't gone through gate resistance calculations yet.
The thing is this part has a rise time of 150ns, if you are trying todo it in 50ns you are more then likely going to setup a pearce oscilator so be warned. NOTE the operating area of the FET as well!!!dknguyen said:About that Figure 2 circuit:
I also finally figured out why the NPN is on the high-side and the PNP Is on the low-side- to keep the thing in quasi-saturation for faster switching. I remember some time ago I was telling a new member asking about his gate driver and he had them this way and I told him he needed to flip them around. THat was MOSFET half-bridge thinking and although it would have worked, the BJTs would have been in saturation so the efficiency would have been higher but the switching would have been slower, and the pulsed nature of the driver makes the efficiency not as important and would have reduced performance. Sorry!
na (I use "R" for resistance, not "W" as a std-keyboard lazy omega), the confusion (asumption on my behalf sorry) was unknown FET and for 3W worth of switching and using:dknguyen said:ALSO:
Is there some confusion going about whether W means watts or ohms? I mean watts when I use W, and I use R or "ohms" for resistance. I've seen my power prof use W as ohms though and Styx seemed to just do that to. So when I say 3W and 4W, I mean the power dissipated due to the transition time is 3W at 50ns and the I^2R losses are 4W at full load.
Nope, that's what he did for the homework. It took me a while to figure out that the load was 10R and not 10W (even though he typed it out as 10W) because the problem was impossible to solve otherwise.Nigel Goodwin said:No, W means watts - to be kind, perhaps your professor is using the lower case omega, which looks like a 'babys bum' as we were told at college!
See https://en.wikipedia.org/wiki/Omega
Styx said:mmm I tend to work with Q=CV and use both that figure (and the gate-drive voltage) as well as looking at the charge curve for the voltage & charge for the end of the millar plateux.
Styx said:Yer Zetex have some fantastic discrete components. That or use a good FET drive chip, I quite like the IR4426 (and its non-inv equiv) - although in this case it would be too low a power
THanks. I'll check that out. The continuous load is around 20-30A. I just assumed 50A for maximum. I also assumed max voltage for the circuit iself. In practicality the battery voltage is probably going to be half that so the losses are quite a bit less.Styx said:The thing is this part has a rise time of 150ns, if you are trying todo it in 50ns you are more then likely going to setup a pearce oscilator so be warned. NOTE the operating area of the FET as well!!!
Why do you divide by 6? THe switching losses formula I use is:Styx said:na (I use "R" for resistance, not "W" as a std-keyboard lazy omega), the confusion (asumption on my behalf sorry) was unknown FET and for 3W worth of switching and using:
W = 1/6*V*I*Ts*Fs lead to some more real voltage and some more real amperage and well kinda didn't think that was going to happen
Seeing the FET in question however
P = (1/2)(V*I)f(ton+toff), the average of the on/off state power averaged out over time and frequency.
I had originally gone with the Direct Back EMF sensing scheme, but the power losses (since the sampling speed is related to the PWM frequency which is related to motor speed) was increasing too much. Lately I've been trying to change that scheme to the regular virtual ground formed by 3 resistors, except I am making a sensing circuit that floats at that neutral point rather than regular ground so I don't lose sensitivity and deal with the common-mode noise of the changing netural point a bit better. Then I can use whatever switching speed I want while grabbing BEMF data whenever I want. Requires more floating supplies though, but whatever.
IT would have been dead easy if there was a DC transformer to differentiall re-reference the neutral-to-phase BEMF reading for the electronics so no attenuation, filtering would be needed, or worrying about voltage ratings or accuracies of differential amplifiers. But that'd be too good to be true.
dknguyen said:Nope, that's what he did for the homework. It took me a while to figure out that the load was 10R and not 10W (even though he typed it out as 10W) because the problem was impossible to solve otherwise.
Thats why you don't use the stated capacitance BUT use the gate-charge curves for your operating voltagedknguyen said:I read some stuff about how the capacitance can be misleading so I try and just stick to the using Q rather than C. Something about C changing or whatever. I think I also read that if capacitance is given and Q wasn't then it's best to take C, calculate Q from it and then use that to recalculate C at your operating voltage.
This is from the 2nd page and is ONLY valid for: Id = 160A, Vds=32V, Vgs=10VQg Total Gate Charge ––– 170 260
Sorry equation wrong (looked at the wrong cell in my excel sheet). I usedknguyen said:Why do you divide by 6? THe switching losses formula I use is:
P = (1/2)(V*I)f(ton+toff), the average of the on/off state power averaged out over time and frequency.
mmm getting interesting, capturing the backEMF can be tricky and must be sync with your PWM and modulation depth (so you sample during periods of no switching). Still there is only two instances when you would go to the effort of trying to measure the BackEMFdknguyen said:I had originally gone with the Direct Back EMF sensing scheme, but the power losses (since the sampling speed is related to the PWM frequency which is related to motor speed) was increasing too much. Lately I've been trying to change that scheme to the regular virtual ground formed by 3 resistors, except I am making a sensing circuit that floats at that neutral point rather than regular ground so I don't lose sensitivity and deal with the common-mode noise of the changing netural point a bit better. Then I can use whatever switching speed I want while grabbing BEMF data whenever I want. Requires more floating supplies though, but whatever.
IT would have been dead easy if there was a DC transformer to differentiall re-reference the neutral-to-phase BEMF reading for the electronics so no attenuation, filtering would be needed, or worrying about voltage ratings or accuracies of differential amplifiers. But that'd be too good to be true.
THe gate driver power dissipation is something like that isn't it? WHat I have been talking about is the switching losses due to the MOSFET passing through the intermediate on/off region in a finite amount of time.Roff said:I don't understand why gate driver power dissipation is not simply F*Q*V when the charge is given, or F*C*V^2, where only total equivalent capacitance is given. F is the switching frequency, Q is the gate charge, and V is the p-p swing of the gate voltage. If there is a discrete series resistance between the driver and the gate, the dissipation would be distributed proportionally between that resistance and the internal resistance of the driver.
Can someone comment on that?
Styx said:mmm getting interesting, capturing the backEMF can be tricky and must be sync with your PWM and modulation depth (so you sample during periods of no switching). Still there is only two instances when you would go to the effort of trying to measure the BackEMF
1) space-vector modulation of a BLAC machine
2) a means to sense speed
#1 is bad, there are better ways to derive it indirectly
#2 alot of hassle, might as well strap an encoder disk and count pulses
That's what I was originally talking about, but styx went off on a tangent about gate resistor dissipation:dknguyen said:WHat I have been talking about is the switching losses due to the MOSFET passing through the intermediate on/off region in a finite amount of time.
Then he posted an equation,To calcuate the true wattage of the gate resistor needed you need to calculate the RMS current AND that is dependent on the switching freq, in this case 30Khz
but as far as I can tell, he never said what it applied to. Maybe I missed that part. I thought the equation was for gate resistor dissipation, since that was the topic of his previous post.W = 1/6*V*I*Ts*Fs
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