2.5A full FET bridge driver - who needs so much current? - Page 1
Forum section: electronics / beginners

author: fsky
author_rank: Contributor
timestamp: August 24, 2015, 06:36:46 pm
content: Hello, recently I was looking at Intersil HIP4081A full bridge driver:https://www.intersil.com/content/dam/Intersil/documents/hip4/hip4081a.pdfI am wondering, what does it actually mean 2.5A gate driver current limit? If we have a FET transistors' gates on all outputs in typical application circuit, why would anyone need so much current, since FET gates need so little of it? Is it designed to drive stuff directly from IC pins, without transistors?

author: mos6502
author_rank: Frequent Contributor
timestamp: August 24, 2015, 06:54:11 pm
content: A FET gate is basically a capacitor. So turn on the FET, you have to charge the capacitor. The faster you charge it, the faster the FET turns on. The faster you dischage it, the faster the FET turns off. The gate charge is given in the datasheet. To calculate how much current you need, just use the formula I (A) = Q (C) / s.To get fast sub-microsecond switching times on big FETs, you'll find you need a couple of amps of gate drive.

author: Ribster
author_rank: Frequent Contributor
timestamp: August 24, 2015, 06:57:31 pm
content: You can use a simple NPN and PNP transistor in a complementary circuit.Kind of like a type B amplifier. This is noisy on the power supply because of the short when they are on at the same time, but it's a cheap FET driver.

author: Zero999
author_rank: Super Contributor
timestamp: August 24, 2015, 07:02:26 pm
content: Quote from: Ribster on August 24, 2015, 06:57:31 pmAre you talking about configuring the BJTs as emitter followers? If so then there won't be any shoot through.

author: Ribster
author_rank: Frequent Contributor
timestamp: August 24, 2015, 07:21:35 pm
content: Quote from: Hero999 on August 24, 2015, 07:02:26 pmNope, kinda like this:

author: Chris C
author_rank: Frequent Contributor
timestamp: August 24, 2015, 07:30:04 pm
content: The gate does act as a capacitor, but that alone doesn't explain why such large gate drive currents are needed.A very simplified explanation: There's a second capacitor between the gate and the other terminals of the MOSFET. The faster you switch the MOSFET, the faster energy is transferred across this capacitor, back to the gate. The end result is that it appears to actively fight your attempts to switch it quickly. Once you get down to switching speeds of 1us is when it typically starts becoming noticeable. Below that, you soon get into territory where everydoublingof switching speed requires aboutfour times more current.That is why you see, what at first glance, appears to be insanely high current capabilities on fast MOSFET gate drivers. And is one major reason why switching power supplies fall short of 100% efficiency. Switching the MOSFET faster decreases resistive losses in it, and heating, caused as it passes through the linear (half-on) region. But at some point the exponentially rising gate current required to do accomplish that exceeds what you save.Of course, this exponential relationship works both ways. So you might be pleasantly surprised to find that DIY gate driver circuit, even though it'll probably top out at 500mA, will only be about 2-3 times as slow as the HIP4081A providing 5 times more current.

author: rx8pilot
author_rank: Super Contributor
timestamp: August 24, 2015, 07:31:12 pm
content: If you charge the FET gate very fast quickly, the FET spends very little time in the linear region of it's operation where huge amounts of heat is generated. In a bridge driver, the FET is generally being turned on/off quickly and continuously. High current drivers are good for a high efficiency design. The 2.5A is only delivered for a VERY brief moment to get the capacitance of the gate fully charged at which time the current flow effectively drops to zero.The down side is that when you turn it on/off really fast, parasitic inductance in the devices and the PCB (even tiny ones) can cause ringing that should be considered. Ringing can create conducted and radiated emissions that will interfere with the circuit and other circuits. It can also destroy the FETs if it is bad enough.

author: fsky
author_rank: Contributor
timestamp: August 24, 2015, 07:56:39 pm
content: Quote from: mos6502 on August 24, 2015, 06:54:11 pmThank you, it's clear for me now.I was stuck with ideal models of devices for too long now.Also, thanksChris Cfor your input, that's some very interesting behavior of FETs that I've never heard aboutI think that the topic might be closed now, thank you guys

author: max_torque
author_rank: Super Contributor
timestamp: August 24, 2015, 08:16:24 pm
content: It's also worth noting that how much current a FET driver can transiently handle is to some degree a result of the necessary low impedance of that driver, required to keep the FETs gate at a known voltage. ie, because the gate is capacitively and inductively coupled to the Drain and Source, high voltage transients on these pins also act to change the Gates charge. In order to prevent these transients from changing the state of the gate, those charges must be clamped, and this is what the low impedance of the driver achieves

author: Zero999
author_rank: Super Contributor
timestamp: August 24, 2015, 08:31:47 pm
content: Quote from: Ribster on August 24, 2015, 07:21:35 pmThat's what I thought. The BJTs are configured as emitter followers so there will be no shoot-through.

author: Ribster
author_rank: Frequent Contributor
timestamp: August 24, 2015, 09:21:32 pm
content: Quote from: Hero999 on August 24, 2015, 08:31:47 pmAm i wrong that the NPN and PNP will be on at the same time ?At the transition from high to low and low to high, doesn't the commutation of the output state makes the NPN and PNP conduct at the same time?

author: nuno
author_rank: Frequent Contributor
timestamp: August 24, 2015, 09:23:00 pm
content: You'll find drivers for even more than those 2.5A.Some (big) FETs have like nFs of gate capacitance, but in fact it is a little more complex than that; the capacitance will change while you charge it, so we usually talk about charge (in Coulombs, or nano Coulombs nC - that's the useful information you'll find in the datasheets).And then you may have 2 or more FETs in parallel, driving your current drive requirements even further up.

author: Zero999
author_rank: Super Contributor
timestamp: August 24, 2015, 09:56:57 pm
content: Quote from: Ribster on August 24, 2015, 09:21:32 pmIt's not possible for both the NPN and PNP transistor to be switched on simultaneously.This is the reason why there will be crossover distortion, if the circuit is used as a linear amplifier

author: eneuro
author_rank: Super Contributor
timestamp: August 24, 2015, 11:26:04 pm
content: Quote from: Hero999 on August 24, 2015, 09:56:57 pmI've used BC337/BC327 in this 3 phase 5V MPU (ATTiny85) based driver, where three MPU pins are used to controll (on/off) low power 3 phase loadHowever, fast 1N4148 diodes are added and additional resistors to be able switch desired phase on/off and protect transistors from inductive spikes, etc.I've used also successfully two of those phases to drive simple isolated DC-DC low power full-bridge SMPS, so I hope there is no shoot-through since circuit simulators in simple cases ar efine and I've got similar results in other sim too

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 24, 2015, 11:52:54 pm
content: 2.5A isn't nearly as much as it sounds -- it's obviously not 18V * 2.5A = 45W continuous output power, or even 1/4 of that (which would be realistic for a circuit at those full ratings into an AC load). It is that much capacity, peak, but this doesn't get you as far as you might imagine when considering it has to be done in a hundred nanoseconds or less.As mentioned, the gate has a capacitive characteristic, so it takes peak current to make its voltage move. It also has some internal resistance, which is due to the gate being constructed from a thin layer of aluminum or polysilicon, as well as other internal losses.The important datasheet figure is Qgs(tot), the total gate charge required to go from off to on (the voltages are specified; since E = Q*V, it takes more energy to reach a higher voltage, so the voltage is also important information for this parameter).The internal losses of both gate and driver (which has an equivalent Rds(on) or output resistance specified in the datasheet) add in series, and this limits how fast the gate can be swung.Typically, during turn-on or turn-off, a transistor will experience Iload * Vsupply for approximately the duration of switching. If it's a power transistor on the mains side of an SMPS, that might be 20A and 320V, so the peak power is extreme! To achieve useful efficiency, the duration of this power dissipation must be kept to a very small fraction of the cycle, never more than 1/20th and typically under 1/100th. Or for a typical 100kHz switching frequency (Tcycle = 10us), under 100ns.Combining all these facts:We can estimate the switching speed as:t_r ~= 2 * (Rdriver + Rgate) * Qg(tot) / (Vgs(on) - Vgs(off))A typical example might have Rdriver = 4.5 ohms (typical of parts like the HIP4081*), Rgate = 5 ohms, Qg = 100nC, and Vgs(on) - Vgs(off) = 10V (usually 10 and 0V, respectively). This gives an equivalent capacitance Qg/Vgs = 10nF, a time constant of 95ns, and a full switching time about 190ns.(*Note, by the way, that 2.5A peak is only possible when the output is shorted to the opposite supply, which is only true for the most brief of instants during switching -- much of the time is spent at half that voltage, so the average current during switching must be lower, which is why it's often valuable to use even *larger* drivers!)And that's without using a series gate resistor, which is often desirable to avoid parasitic oscillations. (The AC small signal equivalent of the gate terminal, actually looks like a small negative resistance, giving rise to oscillations in the 20-100MHz range, if the gate circuit has too much inductance and too little resistance.)At 100kHz, 190ns is 1.9% of a cycle, so if we're dissipating 1000W during switching (typical of a 100V, 10A application), the average switching loss is 19W -- better have a heatsink! But not only that; this adds with conduction losses, so the total will be higher.Conduction losses are due to Rds(on), which might be 75mohm for a device in this class. P = I^2 * R = 7.5W, but this is only drawn during half the waveform, so the dissipation per transistor is half this, or 3.75W.So you can see, the switching losses can be quite substantial, and there is very good reason to keep it fast!But speed comes at a price, and you must deal with the consequences of that risetime. All inductances and capacitances in the circuit get involved, whether you thought they existed or not! That includes stray wiring inductance as well as component parasitics. A simplified (first order) analysis of this can be made as well, but goes well beyond scope here.Tim

author: rx8pilot
author_rank: Super Contributor
timestamp: August 25, 2015, 02:37:36 am
content: Quote from: T3sl4co1l on August 24, 2015, 11:52:54 pmI learned that lesson the hard way after laying out a few  SMPS circuits. I carefully calculated everything except the parasitic inductance and capacitance of the PCB itself. It was a mess that I could not figure out. I found some resources that focused on PCB layout that were very helpful. With the exact same schematic and components - a super tight layout turned out to deliver pretty good performance. The effort primarily surrounded the issues of fast FET switching.

author: bingo600
author_rank: Super Contributor
timestamp: August 25, 2015, 07:50:07 am
content: Quote from: eneuro on August 24, 2015, 11:26:04 pmCould we have a schematic of your driver please ?/Bingo

author: nuno
author_rank: Frequent Contributor
timestamp: August 25, 2015, 08:57:14 am
content: Quote from: eneuro on August 24, 2015, 11:26:04 pmFET/IGBT driver design is a very interesting topic on its own. That transistor topology only gives you 5V - Vbe(~4.3V) output when "high", since you're driving it with a microcontroller (assuming its running at 5V); may be perfectly fine in your application, but 4.3V gate drive is probably not the typical case when you need to use such circuit to up the drive current.Quote from: T3sl4co1l on August 24, 2015, 11:52:54 pmActually I think is double that value, 38W, you forgot about the "fall" transition.bingo600, here's another example of discrete FET drivers with low latency you can look at if you like the subject (it's one driver for a NMOS and another for a PMOS, both repeated):https://github.com/vnevoa/DiffTrike/blob/MarkIV_RasPi_NJAY/Electronics/PowerBridge/hw/bridge-fet-drivers-schematics.pdfThey're far from perfect, but they do drive wheels

author: Chris C
author_rank: Frequent Contributor
timestamp: August 25, 2015, 09:38:14 am
content: I just wanted to say that I love this quote.Quote from: T3sl4co1l on August 24, 2015, 11:52:54 pm

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 25, 2015, 12:52:22 pm
content: Quote from: nuno on August 25, 2015, 08:57:14 amI cheated.It's actually more like a triangle, where voltage is rising and then current is falling (and vice versa at turn-on), so that the area is half that of a rectangle having the same width and height. But it happens twice per cycle, so it still comes out right. (The same occurs when you look at gate drive power, where you charge and discharge the equivalent capacitance Ceq = Qg/Vgs, which holds energy e = 0.5 CV^2, half of which is stored, half dissipated in the gate resistances; but it doubles per cycle so you can use e directly as gate switching energy.)38W of course is the figure for two devices, which is more important on a system level.Tim

author: nuno
author_rank: Frequent Contributor
timestamp: August 25, 2015, 01:25:06 pm
content: Ahh! Yes you're right. I use that formula (Pd(sw) = V x I x f x (tr + tf) * 0.5) too.You may not believe, but many cheap chinese electric bicycle BLDC motor controllers do asynchronous rectification instead of synchronous, even though they have the all hw in there.

author: eneuro
author_rank: Super Contributor
timestamp: August 25, 2015, 03:17:02 pm
content: Quote from: nuno on August 25, 2015, 08:57:14 amYep, you are right, it is closer to 4V than 5V, but it doesn't matter in this simple SMPS, while even using such poor inductive coupling in quick made transformer for tests at ratio 1:2.5 using ferrite beard (pri: 8T sec: 20T) I was able easy get in this galvanic isolated DC-DC 11.37Vmax after full wave rectifier using ATTiny MPU shown above to drive this toroid primary using additional AC coupling with 100nF ceramic caps at programmed 400kHz switching frequencyQuote from: bingo600 on August 25, 2015, 07:50:07 amI wouldn't like reveal detailed schematics for the moment, while after one of my projects is finisched PCBs made in China will be available on my website and maybe I'll publish in local electronics magazine snapshot of this work, so the requirement is that it can not be published before official publication in magazine, so I wouldn't like to talk to much about it, for the momentHowever, nothing fancy in the case of two phase there and more tricky part is driving those pins when we need three phase, since we need leave one of the MPU control pins floating, so knowledge of assembler can be helpfull if we wanted on not too fast MPU (1MHz-8MHz) get nice symetric timing of all phases-at 400kHz speeds things like number of RJMP cycles matters (compilators can generate optimized code which will not get you perfect timings) and I've made compensation for things like this in implemented MPU delays to get exact timing of all phases at given switching frequency

author: Chris C
author_rank: Frequent Contributor
timestamp: August 25, 2015, 07:19:32 pm
content: Quote from: nuno on August 25, 2015, 08:57:14 amNice drivers. I use something almost the same. I haven't yet figured out how to make the schematic large enough to read the part labels, but see that first resistor, through which the input signal passes to the pull-down transistor's base? I add a Schottky diode in parallel with that. This accomplishes two things:1) When connected to an MCU or other device which can sink current, and it switches to logic low, the Schottky conducts and bypasses the resistor. Allowing the full available sink current to drain the base charge, and improving transistor turn-off time. It does so even though there's an anti-saturation diode between that transistor's base and collector, they work great together.2) When it switches to logic high, the Schottky is reverse biased and does not conduct. Instead, it acts as a capacitor, in the neighborhood of 10pF. This is a significant portion of the the transistor's base capacitance. And so, by again bypassing the resistor, it improves turn-on time some too.Both of which make the the driver more efficient, and reduce losses in the MOSFET. Not bad for one extra component.Additional improvements? Add a little more capacitance, 10-20pF, in parallel with the Schottky, so that the total exceeds that of the transistor's base capacitance. And replace the resistor that feeds the pull-up transistor's base with a constant current source. But these provide smaller improvements, and I usually don't bother. I always use the Schottky, though.

author: poorchava
author_rank: Super Contributor
timestamp: August 25, 2015, 07:42:54 pm
content: For those who might now know: the key term is 'Miller capacitance' or 'Miller effect'. In datasheets it's usually referred to as Cgd. In great simplification: as your fet starts to turn on or off, the voltage at the drain changes rapidly and that high frequency signal propagates through Cgd and counteracts the change of gate potential.This is why you can find drivers capable of 20A+ peak currents (eg. Microchip makes those). I have once designed a driver which had to provide over 40A of peak drive current (and yes, the EMI from this was extreme). Actually the gate drive consumed about 60 W of power by itself

author: eneuro
author_rank: Super Contributor
timestamp: August 25, 2015, 08:05:02 pm
content: Quote from: Chris C on August 25, 2015, 07:19:32 pmUnfortunatelly, I can't see any of them capable to provide negative Vgs, which should help overcome many problems eg. in half-bridge in synchronous buck converterIsn't negative Vgs a must in more demanding applications?Probably, i would use rather isolated DC-DC to provide positive and negative +/-15Vgs ON/OFF with optoisolated a few Amps gate driver configured to work with eg. +/- 12Vgs.I've used cheap TLP351F succesfull in such configuration, since its VCC is above 24V as I rememberUpdate: Yep, TLp351F has:Code:Power supply voltage: 10 to 30 Vso I've choosen +/12V to operate within safe 24Vcc and +/-12Vgs

author: nuno
author_rank: Frequent Contributor
timestamp: August 25, 2015, 08:34:11 pm
content: Quote from: Chris C on August 25, 2015, 07:19:32 pmThank you. It's the "Raw" button on the right on top of the image, but here's a direct link:https://github.com/vnevoa/DiffTrike/raw/MarkIV_RasPi_NJAY/Electronics/PowerBridge/hw/bridge-fet-drivers-schematics.pdfI just noticed the schottky diodes's symbols somehow ended up the same as PN ones, I'll have to change it when time permits. The github project under ".../PowerBridge/..." has a 25V 25A DC motor controller project with all design files for schematics (KiCAD but there are PDFs of all schematics), pcbs and fw for whoever wants to look.There are a few "problems" with these drivers, but I also wanted to keep component count down, as much as desperately possible. For example, the low driver doesn't keep the FET off when the input is floating (in my application this isn't a problem because the high side ones are guaranteed off) and the top output transistor is always dropping 12V (and it needs a constant, although tiny, load, guaranteed by the FET's gate-source resistors).Thanks for the suggestions below.Quote from: Chris CI currently don't have any of those drivers with me, but I was pretty happy with their switching latency, basically "zero" (switching time in my app is decent too). I have some scope shots, including comparison with no anti-sat diode at all and with 1N4148 as anti-sat, but not here with me. When I get my hands on the drivers again I'll try that 2nd schottky to see what's the difference - as it is, the driver's output (loaded with the FET) starts reacting "instantly", but from your description I guess it goes up/down faster (it is actually irrelevant in my application, because the phase's di/dt is being limited by parasitic inductance at the FET's source (which is bad and good) - hey, I'm still learning!).Quote from: Chris CI see. The input's 0->1 transition, being fast, goes through the "schottky's capacitance" bypassing the resistor.

author: nuno
author_rank: Frequent Contributor
timestamp: August 25, 2015, 08:37:57 pm
content: Quote from: eneuro on August 25, 2015, 08:05:02 pmNope, they do not provide negative turn off voltage. It wasn't necessary in my application, and as far as I've seen, that is used, typically, only with IGBTs (because they are more stubborn to turn off I guess, with that "current tail" or "trail" or whatever).Quote from: poorchava on August 25, 2015, 07:42:54 pmJust to add another Miller term: "Miller plateau"Quote from: poorchavaNow that seems like a lot of fun!

author: Chris C
author_rank: Frequent Contributor
timestamp: August 25, 2015, 09:16:35 pm
content: Quote from: nuno on August 25, 2015, 08:34:11 pmI only recently found out a Schottky was an option for the anti-sat diode. I'd always assumed it would limit the base voltage too much, and therefore limit current through the transistor. But I have not yet had a chance to actually try it and see. It sounds like you've performed this experiment. Don't go through too much trouble finding scope traces, but since you went with the BAT54, I assume it did in fact work out better than the 1N4148?Also, now that I can see the schematic (thanks for the help on that), and realize the anti-sat is actually a Schottky, then the Schottky I suggested adding is not the optimal choice. The base voltage will never be high enough that my Schottky will significantly assist with turn-off, other than acting as a bypass capacitor. Better to just use a capacitor instead! If you get a chance to try it, I'd recommend starting with a 22pF.

author: nuno
author_rank: Frequent Contributor
timestamp: August 25, 2015, 09:22:28 pm
content: Quote from: Chris C on August 25, 2015, 09:16:35 pmMuch better, no comparison possible as far as I remember. No visible latency at all with the schottky (no time spent removing junction charge). I have the shots easy to reach, it's just that I'm not at home, I'll post them later.Quote from: Chris COk, thanks. I'll sure try it. I have seen the cap usage to assist removing the bjt from saturation but I think when I saw it it was something like "you can use the caporyou can use the diode", never gave it a thought about using both methods.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 25, 2015, 10:51:03 pm
content: Quote from: poorchava on August 25, 2015, 07:42:54 pmThis is absolutely true, and a large part of the operation of MOSFETs!But be careful with the quantities. Measuring MOSFETs by capacitance is highly ineffective...The charge equivalent capacitance is typically more than quadruple the value of Cgss!The reason for this is dual: not only do you have to charge Cgd (Miller effect), but all terminal capacitances (Cgs, Cds and Cgd) vary considerably with drain and gate voltages. Take careful note of what voltages Cgss is measured at, then refer to the plot of capacitance versus drain voltage. Ask yourself... is that *really* useful?So, while you can use capacitance, do make sure it's the charge equivalent: Ceq = Qg(tot) / Vgs(on), as in my example.Tim

author: nuno
author_rank: Frequent Contributor
timestamp: August 25, 2015, 11:41:16 pm
content: 3 shots, without any anti-saturation diodes nor other prevention techniques (note the time scale, different from the next 2 shots), with a 1N4148 and with a BAT54, on the PMOS driver (this one is non-inverting). I can't remember exactly but some of the interference is probably measurement artifacts, is not physically possible to measure these 2 signals with short GND clips. Yellow is driver input, can't remember anymore what the blue trace is, the hump at the end looks like the Vcc + Vfw(intrinsic-diode) which should be visible at Vd, but dropping only 9V suggests it's Vg. Anyway, for the comparison at hand I think Vg or Vd is irrelevant, it represents the driver's output.

author: Chris C
author_rank: Frequent Contributor
timestamp: August 26, 2015, 12:13:53 am
content: Oh yeah, that BAT54 is performing very nicely. Really about as good as I've seen with a standard switching diode for anti-sat, and either the Schottky or capacitor bypassing the input resistor. Thanks!