I need help understanding something on the data sheet of an IGBT
ElectricAye
Posts: 4,561
Hi all,
I hope to use an IGBT to drive a Peltier thermoelectric cooler using PWM and I'm trying to estimate how much power the IGBT will gobble up (and heat up) as it switches the current. The voltage and current being provided to the IGBT are 15 volts at 10 amps and I hope to switch it at roughly 2kHz. The data sheet does not list anything like an "internal resistance" but I noticed it has something called "switching loss". See data sheet attached, page 2. If I understand this correctly, with a switching loss of roughly 5.1 mJ per switching, I could expect no more than about 5.1 watts of loss at 1kHz. Therefore at 2kHz, I would see no more than about 10.2 watts. (I see these wattage numbers as being upper limits because the data sheet specifies the switching taking place at 600 volts at 40 amps).
Anyway, I don't know if this is the correct way to think about this since I really know diddly about transistors. But this is a critical question because the Peltier's have a resistance of only about 1.2 ohms, and when you are driving the Peltier at 15 volts and 10 amps, then if the IGBT has an effective resistance on the order of 1 or 2 ohms, then the IGBT itself starts to eat up the output of my power supply.
Also, while I'm at it, does anyone know why you would select one value of gate resistor vs. another when operating around 2kHz?
thanks for helping,
Mark
I hope to use an IGBT to drive a Peltier thermoelectric cooler using PWM and I'm trying to estimate how much power the IGBT will gobble up (and heat up) as it switches the current. The voltage and current being provided to the IGBT are 15 volts at 10 amps and I hope to switch it at roughly 2kHz. The data sheet does not list anything like an "internal resistance" but I noticed it has something called "switching loss". See data sheet attached, page 2. If I understand this correctly, with a switching loss of roughly 5.1 mJ per switching, I could expect no more than about 5.1 watts of loss at 1kHz. Therefore at 2kHz, I would see no more than about 10.2 watts. (I see these wattage numbers as being upper limits because the data sheet specifies the switching taking place at 600 volts at 40 amps).
Anyway, I don't know if this is the correct way to think about this since I really know diddly about transistors. But this is a critical question because the Peltier's have a resistance of only about 1.2 ohms, and when you are driving the Peltier at 15 volts and 10 amps, then if the IGBT has an effective resistance on the order of 1 or 2 ohms, then the IGBT itself starts to eat up the output of my power supply.
Also, while I'm at it, does anyone know why you would select one value of gate resistor vs. another when operating around 2kHz?
thanks for helping,
Mark
Comments
If you are only switching low voltage, I'd recommend using a MOSFET instead. Also, it might be a good idea to increase the switching frequency to at least 10kHz, AND use a filtering capacitor. I really don't like the idea of applying a full 15V across the Peltier. This probably won't cause any problems, but if you are aiming to regulate the current through the Peltier by PWM, a smoothing Capacitor should reduce any peak currents.
Philldapill,
would you mind elaborating on why you think this could be a problem?
thanks for answering,
Mark
Well, deciding on what sort of gate resistor to use turns out to be a trade-off between generating switching losses and generating EMI during switching. With low values of gate resistors, you tend to cut down on switching losses but at the cost of generating EMI. Higher gate resistors help cut down on EMI but cause more switching losses. You can read a little more about this in the attachment.
I hope that helps me.
Anyway, I think I had read your resistance value wrong. For some reason I saw "1.2 mOhms" rather than 1.2 ohms. Still, if the Peltier is rated for 10A max, you might want to reduce any excessive currents. I mean, 12.5A peak current isn't really that big of a deal.
And why would you use a capacitor anywhere in the Peltier circuit?· Just turn the thing on and off to control the amount of cooling you get.· It's a thermal device, probably with a heat sink, and its response is going to be very slow.· No need to make the drive thousands of times faster than any possible response.· I'd start out experimenting at about 1 Hz and go from there.
There's no need to smooth either the drive or the supply to the Peltier junction.· The heat-conducting parts of the system will do all the smoothing you need.
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· -- Carl, nn5i@arrl.net
Carl,
I'm certainly no expert on this, so I'm only going by what I've read on the internet. But apparently Peltier devices are vulnerable to mechanical fatigue, so the more often you thermally (and thus mechanically) cycle them, the faster they'll crack. When using PWM to control them, you have to get the frequency up to a level where the thermal "inertia" of the semiconductor material prevents the rapid bending stresses from fatiguing the semiconductor material. When researching whether or not to use PWM to control Peltiers, I came across some very heated debates about whether or not it was a good idea to use PWM, but it seems that the key is to use a PWM frequency of at least 2 kHz or maybe even higher. Apparently, driving them with PWM at low frequency is a quick road to thermal fatigue.
One problem I anticipate is that data sheets are from tests done with pure DC currents and so the maximum efficiencies, etc are drawn from those tests, but I'm guessing PWM will probably generate some other sweet spots because of the thermal behavior at a higher frequency.
Anyway, this is how I understand the phenomenon. I certainly invite others to give their opinion or share their experiences.
thanks for your interest and advice,
Mark
The switching loss will be negligible compared to the loss of pushing high peak currents through.
EDIT: Ha! You beat my response, Mark. You're right about the thermal fatigue... I didn't even think of that. I suppose ideally, you want the LEAST amount of thermal fatigue as possible - even more reason for a capacitor in parallel(acts like a low pass filter).
Post Edited (Philldapill) : 8/24/2009 1:01:42 AM GMT
Phill,
Intellectually what you're telling me doesn't make sense (yet), but intuitively I'm sure you're right - I just can't explain it to myself just yet. Intellectually speaking, I see this: the maximum efficiency of my particular Peltier is supposed to happen at about 15 volts at 10 amps (which was tested at DC). When the PWM duty cycle is ON, I see the Peltier as reaching that point of max efficiency, so any capacitor would just dampen the magical spirits of the Peliter. Intuitively, it does seem brutal to jackhammer the sucker on every cycle. Maybe the only way to know for sure is to see what parts of it blow up first via experimentation.
thanks for your inputs,
Mark
You can analyze the Peltier as a device with at least two components. The first being the 100% efficient "heat mover". Basically, for every bit of charge that moves across the junction, it will pull some heat from one side to the other. Again, we'll consider this part to be 100% efficient and will not generate any EXTRA heat, other than what was moved from one side to the other.
The second part to the device is a series resistance. This series resistance is the reason for the inefficiency of the device, and the reason that the device is warm after you turn it off - it generates heat. Think of it as just a resistor in series that will eat up power.
Now, if the first element of the device is 100% efficient, then the amount of heat transfer(power) from one side to the other is strictly a function of how much charge was moved - not how FAST the charge was moved. On the other hand, the power that the resistance uses is a function of how FAST the charge is moved, i.e. current. If you move 10A through the 100% Peltier part of the device, in 1 second, then you will move exactly the same amount of charge, and thus heat, as if you moved 1A for 10 seconds.
The resistance is different. Let's say for example, your resistance is 1 ohm. If you move 10A through it for 1 second(10 coulombs total), you dissipate 100W for 1 second = 100 Joules. If you move 1A through for 10 seconds, you only generate 1W for 10 seconds or 10 Joules. I hope that clearly shows that pushing LESS current over a longer period of time, causes less power dissipation.
Now, as for your example, if you were to push 12.5A for 0.8 seconds, that would be the same amount of charge as pushing 10A for 1 second. Let's just assume your internal resistance is a nice even 1 ohm for simplicity. Doing the math, the power dissipation for your 12.5A vs. 10A is: 12.5A^2*(1 ohm)*0.8seconds = 125W. If you were to feed a steady 10A through, that would only be 10A^2*1second = 100W.
You should achieve the same amount of heat transfer using 12.5A 80% duty cycle vs. 10A 100% duty cycle. However, you will be generating roughly 25% MORE heat if you slam it with 12.5A 80% duty square waves.
Make sense? It sure was long enough... [noparse]:)[/noparse]
Honestly, I'm going to have to chew on this a while. As I understand Peltiers, they have a max efficiency at, say, 15 volts at 10 amps. You can reduce the self-heating of the device by lowering the current through it, but then I think you also greatly reduce its ability to move heat (because you're now operating off the sweet spot), so you wouldn't get the device to do what you want. Surely, the more heat you need to move, the lower the efficiencies go on these devices. They aren't very efficient at all, maybe 5 to 10 % overall. I confess I'm totally mystified how they move heat at all - something about electrons being allowed to expand as the packing of electrons decreases down the voltage gradient, wow - so I'm having a hard time creating a mental model of how to think about them.
I will definitely ponder your explanation and see if I can clear my head over this thing.
thanks again,
[noparse][[/noparse]Edit: My particular Peliter is a CP1-12710 that I bought off ebay. I have attached its data sheet in case that helps anyone who might be following this discussion. I confess that the order of the Delta T degrees across the bottom of the one graph makes no sense to me and the manufacturer has never responded to my request for clarification. I don't know how a Delta T axis can go from -33... -13... 7... 27... 0. Beats the heck out of me.]
Post Edited (ElectricAye) : 8/24/2009 2:13:24 AM GMT
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· -- Carl, nn5i@arrl.net
Post Edited (Carl Hayes) : 8/25/2009 1:25:49 AM GMT
Thanks, Carl,
I'm not being sarcastic when I say that your opinion provides me with some much-needed comfort. I have read through a lot of debates about this PWM thing for Peltiers, and I still don't know what to really think about it. I guess I'll find out the hard way once I get my IGBT to work hopefully this week. I think the tricky thing about a Peltier, and perhaps the thing that makes it hardest to intuitively grasp, is that it pumps heat across such a small space. And when it's not actively pumping heat, it instantly starts CONDUCTING it. And because it's so thin, it doesn't have far to conduct the heat and turn the heavenly cool side into a raging inferno. I suspect I'm going to encounter some interesting non-linearities when I actually get the PWM up and running. I'll have this thing instrumented to the hilt, so maybe I'll be able to answer some hardcore questions about efficiencies and so forth and clear up this conundrum (once and for all)?
many thanks,
Mark