Transistor power question? - Page 1
Forum section: electronics / beginners

author: rickselectricalprojects
author_rank: Regular Contributor
timestamp: May 03, 2015, 11:51:27 am
content: Hi everyone!I was looking for some igbts on rs components and i see alot of these to-220 igbts that have very high ratings. As an example:http://au.rs-online.com/web/p/igbt-transistors/8303247/That igbt can apparently switch 600v at 60a and it says the max power dissipation is 160w. I know that 600v at 60a is a lot more that 160w ao does it mean that this igbt can handle a 160w load at 600v or a 160w load at 60aThanks

author: Rerouter
author_rank: Super Contributor
timestamp: May 03, 2015, 11:53:26 am
content: It means it can handle a 600V differential across its pins, a 60A current flow through them, and when switching or due to running in some form of linear region, can dissipate 160W of power into a suitable heatsink, or so i would assume,

author: SeanB
author_rank: Super Contributor
timestamp: May 03, 2015, 12:00:21 pm
content: Means as a switch it can handle either 60A in the on condition, or 600V in the off condition. When switching between them, it can dissipate __UP_TO__ 160W of power, the actual power being dependant on the heatsink used. When on it will dissipate power, from the voltage drop across the device ( typically a volt or so) multiplied by the current. So when off it dissipates no power, but when on it will dissipate up to around 100W at full current, depending on the part.The load however can be anything, provided the supply voltage is 600v or under and the load current is 60A or under, so the maximum power of the load is 36000W, or 36kW of resistive load. If it is inductive, or capacitive, then other derating factors apply.

author: TimFox
author_rank: Super Contributor
timestamp: May 03, 2015, 01:14:22 pm
content: In general, the "power" rating in the headline of the specifications for a power semiconductor is the maximum power dissipation with the case or tab maintained a "room temperature", such as 25 C. This is sometimes referred to as "welded to the hull of a battleship in the North Atlantic". In practice, the heatsink will warm up with power flowing through it to the ambient air. Sometimes, the calculations are easier if you change the power rating to a thermal resistance from junction to case: (Tmax - Troom)/Pd in K/W. You then add the heatsink's thermal resistance and any extra resistance due to the washer or grease between case and heatsink, use the maximum air temperature you expect, and calculate the power required to heat the junction up to Tmax. Depending on the device, Tmax can be between 100 and 200 C, and is a separate specification in the data sheet.

author: Seekonk
author_rank: Super Contributor
timestamp: May 03, 2015, 02:55:26 pm
content: Keep in mind that Vce of 1.32V @ 32A. You never drop below that. That voltage never gets lower. You are always generating heat even at "saturation." Driving a motor at 500V this is an accepted tradeoff. At low voltage high current, IGBT are avoided like the plague.

author: Mechatrommer
author_rank: Super Contributor
timestamp: May 03, 2015, 03:00:25 pm
content: Quote from: Rerouter on May 03, 2015, 11:53:26 ami assume the power rating is including non switching saturation mode. SOA graph will show maximum amperage capability in very low Vdrop region, that is saturation mode i assume.

author: Laura
author_rank: Contributor
timestamp: May 04, 2015, 08:18:21 am
content: Quote from: rickselectricalprojects on May 03, 2015, 11:51:27 amBack in the early 1980's as a summer intern, I was working on a circuit driving a brushless DC motor. It was a delta (or maybe Y) wired motor and hence had three pairs of transistors driving it.Every now and then, the controlling circuit wasn't able to turn the transistors on hard enough and they would run in their linear region. At that instant, they could be conducting 10A with a 100V drop across the transistor, i.e. 1000W of heat! The little TO220 packages however didn't get hot. That was because the linear region excursion only lasted say 1ms once every second.BTW, this circuit taught me not to trust digital storage scopes!

author: T3sl4co1l
author_rank: Super Contributor
timestamp: May 04, 2015, 09:07:05 pm
content: IGBTs typically only withstand full SOA (i.e., rated amps * rated volts -- a dead short condition) for microseconds at a time.It takes time for the power to heat up the silicon die, and for that heat to diffuse through the die, metal backing plate, and heatsink, so it makes sense that the power rating should vary with how long the pulse is.The DC rating is 160W range, but you're not free from physics just yet: IGBTs are very poor at DC power handling, because of instabilities in their design. They work pretty well under saturated conditions (Vce < 5V or so), but at higher voltages (in the "desat" (non-saturated) condition), it tends to be that natural variations in current flow, across the device, cause some spots to heat up more. Which magnifies the problem, because the device has a negative voltage coefficient. Pretty quickly (100us to 10ms+ time scale, depending on applied power), one spot gets really really hot, burns through, and it's all over, magic smoke.The same physics befall BJTs (where the phenomenon is called "second breakdown"), and most modern-process MOSFETs. Historically, MOSFETs had been called "free from second breakdown", but I'm pretty sure this was expressly because MOSFETs were so terrible on density that they simply couldn't be made to dissipate enough power to induce the effect. Modern devices are nearly as dense as BJTs, so are frequently just as sensitive. IGBTs have higher density than both, and are even more sensitive, so they're particularly poor at dissipating power at higher voltages.As for the capacity of a device, it's very complicated. Yes you can sink 60A through that transistor, assuming you heatsink it as well as the manufacturer did when they were testing it (which is often preposterous, but that's another discussion). But that's DC, which isn't very useful for a device that's supposed to be transisting (presumably you want to turn it off and on at some point, too!). But when you turn it off, it doesn't just slam off, it goes gradually, and as it goes, it slides through all those areas of massive power dissipation. Even if it's only switching in 100 nanoseconds, the power is so great that you must take this into account when designing a power switching circuit.Tim