Formula for minimum drive current for mosfet
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From: Paul E. Schoen on 11 Apr 2008 01:54 "Jon Slaughter" <Jon_Slaughter(a)Hotmail.com> wrote in message news:3sALj.2560$%41.2295(a)nlpi064.nbdc.sbc.com... > > <bill.sloman(a)ieee.org> wrote in message > news:20e9d847-ea21-4aa1-bb24-4a2848d7a57a(a)q24g2000prf.googlegroups.com... >> On 11 apr, 01:33, "Jon Slaughter" <Jon_Slaugh...(a)Hotmail.com> wrote: >>> Whats the formula? >>> >>> I = 1/2*F*Q*V? >>> >>> Trying to figure out if I can drive >>> >>> http://www.fairchildsemi.com/pf/FD/FDD8424H.html >>> >>> with a uC directly? (I think it can supply up to 20mA or so) >>> >>> V = 12V if I use pullup and 5V if not. >> >> Check out the data sheet. Figure 7 shows the typical gate charge >> required to get the gate voltage up to a level where the part is >> turned on - something like 10nC. The worst case total gate charge >> listed earlier in the data sheet is 24uC. >> >> 20mA s going to take 1.2usec to deliver that 24uC of charge - this is >> slow switching by MOSFET standards, and you won't want to switch that >> slowly very often, because if you do there is a real risk that the >> switch will overheat. >> > > Why is slower going to going to cause it to heat up? Its less current so > less heat (same amount of charge). In fact it probably would be better > because its spread out over time. (like, say, charging a battery at 1A > for 1 year compared to 365A in one day. Same amount of charge but totally > diffrent results) > > My switching is at most 100khz(its for motor control so anything about > 20khz should be ok but I'm going for about 50khz). I figure I need about > 5 to 10 times this but really it shouldn't be that important(don't need > it to be exact). > > Really though, Can you explain to me why a slower switching speed will > cause it to heat up more? It contradict's everything I know about > transister switches and switching speed. The uC is not what will heat up, although at 100 kHz it could get a little warm. But the MOSFET will dissipate power during the time it is in its linear region during switching, and if you are controlling something like 12V and 6A, you can have as much as 18 watts during this time. At 100 kHz, you could have 1 uSec of high power dissipation on each transition, every 5 uSec, or 20% duty cycle, and about 3.6 watts of switching losses (as a rough estimate). I had problems driving an FQP24N08 at 100 kHz with the on-board 1 amp gate driver of a UC1843a, such that I was barely getting 75% efficiency with a 12 V input at 40 watts output. It also has about 25 nC maximum gate charge. I was also losing about 1.2 watts in conduction losses with 0.05 ohms RdsOn. So I used a 6 amp gate driver UC3710, with an HUF75645 MOSFET, and I got close to 90% efficiency. This MOSFET has up to 238 uC gate charge, but its RdsOn is only 0.016 ohms, so I had less conduction loss. The driver gets a bit warm, too, so this is probably not an optimal combination. Just charging and discharging the 3800 pF gate capacitance takes a fair amount of power, probably about half a watt. Paul
From: bill.sloman on 11 Apr 2008 02:40 On 11 apr, 05:56, "Jon Slaughter" <Jon_Slaugh...(a)Hotmail.com> wrote: > <bill.slo...(a)ieee.org> wrote in message > > news:20e9d847-ea21-4aa1-bb24-4a2848d7a57a(a)q24g2000prf.googlegroups.com... > > > > > > > On 11 apr, 01:33, "Jon Slaughter" <Jon_Slaugh...(a)Hotmail.com> wrote: > >> Whats the formula? > > >> I = 1/2*F*Q*V? > > >> Trying to figure out if I can drive > > >>http://www.fairchildsemi.com/pf/FD/FDD8424H.html > > >> with a uC directly? (I think it can supply up to 20mA or so) > > >> V = 12V if I use pullup and 5V if not. > > > Check out the data sheet. Figure 7 shows the typical gate charge > > required to get the gate voltage up to a level where the part is > > turned on - something like 10nC. The worst case total gate charge > > listed earlier in the data sheet is 24uC. > > > 20mA s going to take 1.2usec to deliver that 24uC of charge - this is > > slow switching by MOSFET standards, and you won't want to switch that > > slowly very often, because if you do there is a real risk that the > > switch will overheat. > > Why is slower going to going to cause it to heat up? Its less current so > less heat (same amount of charge). In fact it probably would be better > because its spread out over time. (like, say, charging a battery at 1A for 1 > year compared to 365A in one day. Same amount of charge but totally diffrent > results). The maximum heat dissipation during switching occurs when the voltage across the switch is half way between the rails. The longer the time the drain takes to get from one rail to the other, the longer this dissipation keeps on happening. Spice can work it out for you if you model your load tolerably accurately > My switching is at most 100kHz(its for motor control so anything about 20khz > should be ok but I'm going for about 50khz). I figure I need about 5 to 10 > times this but really it shouldn't be that important (don't need it to be > exact). > > Really though, Can you explain to me why a slower switching speed will cause > it to heat up more? It contradict's everything I know about transister > switches and switching speed. It ain't what you don't know that screws you up, but what you think you know that ain't so. You need to develop a better understanding of what is going on while the current through the switch moves from off to full on, and the voltage across the switch moves from the full rail voltage to practically nothing. In the middle of this process the instantaneous power dissipation in the swtich gets pretty high - you can work out exactly how high - and since the process takes a finite time - 10 to 20nsec if you know what you are doing, a microsecond or so if you cheapskate on the driver - each switching opertion dumps a predictable amount of energy into the switch junction. Work it out. -- Bill Sloman, Nijmegen
From: Paul Hovnanian P.E. on 11 Apr 2008 21:09 bill.sloman(a)ieee.org wrote: > > On 11 apr, 05:56, "Jon Slaughter" <Jon_Slaugh...(a)Hotmail.com> wrote: > > <bill.slo...(a)ieee.org> wrote in message > > > > news:20e9d847-ea21-4aa1-bb24-4a2848d7a57a(a)q24g2000prf.googlegroups.com... > > > > > > > > > > > > > On 11 apr, 01:33, "Jon Slaughter" <Jon_Slaugh...(a)Hotmail.com> wrote: > > >> Whats the formula? > > > > >> I = 1/2*F*Q*V? > > > > >> Trying to figure out if I can drive > > > > >>http://www.fairchildsemi.com/pf/FD/FDD8424H.html > > > > >> with a uC directly? (I think it can supply up to 20mA or so) > > > > >> V = 12V if I use pullup and 5V if not. > > > > > Check out the data sheet. Figure 7 shows the typical gate charge > > > required to get the gate voltage up to a level where the part is > > > turned on - something like 10nC. The worst case total gate charge > > > listed earlier in the data sheet is 24uC. > > > > > 20mA s going to take 1.2usec to deliver that 24uC of charge - this is > > > slow switching by MOSFET standards, and you won't want to switch that > > > slowly very often, because if you do there is a real risk that the > > > switch will overheat. > > > > Why is slower going to going to cause it to heat up? Its less current so > > less heat (same amount of charge). In fact it probably would be better > > because its spread out over time. (like, say, charging a battery at 1A for 1 > > year compared to 365A in one day. Same amount of charge but totally diffrent > > results). > > The maximum heat dissipation during switching occurs when the voltage > across the switch is half way between the rails. The longer the time > the drain takes to get from one rail to the other, the longer this > dissipation keeps on happening. > > Spice can work it out for you if you model your load tolerably > accurately > > > My switching is at most 100kHz(its for motor control so anything about 20khz > > should be ok but I'm going for about 50khz). I figure I need about 5 to 10 > > times this but really it shouldn't be that important (don't need it to be > > exact). > > > > Really though, Can you explain to me why a slower switching speed will cause > > it to heat up more? It contradict's everything I know about transister > > switches and switching speed. > > It ain't what you don't know that screws you up, but what you think > you know that ain't so. > > You need to develop a better understanding of what is going on while > the current through the switch moves from off to full on, and the > voltage across the switch moves from the full rail voltage to > practically nothing. > > In the middle of this process the instantaneous power dissipation in > the swtich gets pretty high - you can work out exactly how high - and > since the process takes a finite time - 10 to 20nsec if you know what > you are doing, a microsecond or so if you cheapskate on the driver - > each switching opertion dumps a predictable amount of energy into the > switch junction. > > Work it out. Yep. Its the basis of Class D amplifier operation: http://en.wikipedia.org/wiki/Class_d_amplifier#Advantages The dissipation problem is the same for bipolar and MOSFET transistors (even though the underlying reasons for slow switching differ). > -- > Bill Sloman, Nijmegen -- Paul Hovnanian mailto:Paul(a)Hovnanian.com ------------------------------------------------------------------ When cryptography is outlawed, bayl bhgynjf jvyy unir cevinpl. -- Etaoin Shrdlu
From: gearhead on 12 Apr 2008 16:06 On Apr 10, 4:44 pm, Fred Bloggs <nos...(a)nospam.com> wrote: > bill.slo...(a)ieee.org wrote: > > On 11 apr, 01:33, "Jon Slaughter" <Jon_Slaugh...(a)Hotmail.com> wrote: > > >>Whats the formula? > > >>I = 1/2*F*Q*V? > > >>Trying to figure out if I can drive > > >>http://www.fairchildsemi.com/pf/FD/FDD8424H.html > > >>with a uC directly? (I think it can supply up to 20mA or so) > > >>V = 12V if I use pullup and 5V if not. > > > Check out the data sheet. Figure 7 shows the typical gate charge > > required to get the gate voltage up to a level where the part is > > turned on - something like 10nC. The worst case total gate charge > > listed earlier in the data sheet is 24uC. > > > 20mA s going to take 1.2usec to deliver that 24uC of charge - this is > > slow switching by MOSFET standards, and you won't want to switch that > > slowly very often, because if you do there is a real risk that the > > switch will overheat. > > > -- > > Bill Sloman, Nijmegen > > That thing looks like a cross conduction hazard and half at that > switching speed, both FETs come on at less than 2V from their source > rails, wonder if he's tying the gates together, definitely would want to > switch as fast as possible then...- Hide quoted text - > > - Show quoted text - Jon, the problem is the amount of time it takes the mosfet(s) to transition from on to off and back from off to on again. During each switching action of a mosfet it acts like a resistor for the duration. Heat! So you want the duration of each switching event as short as possible, regardless of whether these events occur at 100Hz or 100kHz. But that's only part of the problem. Fred brings up a good point about cross-conduction. Now, when your drive has the gates pulled all the way to the rail (either one), that's no problem. But it IS a problem during the switching transition, because both mosfets are partly turned on providing a path -- not through the load, but directly across the power rails. You don't want slow switching here. So a weak drive is bad. For motor drive, a kilohertz is probably way plenty, and this lower frequency is better so that you don't put your mosfets in the hot seat so often.
From: Jon Slaughter on 12 Apr 2008 16:31
Ok guys, it seems we are talking about two different "Frequencies" here. One is the speed at which a transition occurs and the other is the number of times those transitions occur per second. The first really isn't a frequency since a frequency deals with something that is periodic(or at least that tends to be the way people think about frequencies). I do realize that the faster the transition the less power dissipation. That is pretty basic. And this is independent of frequency(up to a point). But even having this then the frequency, F, the # of transitions / second, will increase the power dissipation because we are simply transitioning more times per second. So what confused me was when you guys called the frequency of the transition and I thought it was the frequency of the # of transitions per second. (or maybe I added that to it or just confused the two). So we are both right ;) Its true that if you increase the transition speed that the power dissipation goes down, but if you increase the # of transitions/per second then the power dissipation goes up. Hopefully its obvious now... |