mosfet driver with seperate sink/source pins - Page 1
Forum section: electronics / beginners

author: Simon
author_rank: Unknown
timestamp: September 18, 2014, 08:57:59 pm
content: I'm looking at:http://uk.farnell.com/diodes-inc/zxgd3004e6ta/ic-gate-driver-mosfet-sot23-6/dp/1549141on the datasheet it has separate pins for source and sink and the application circuit shows a series resistor for each. Do i have to use these resistors or can I just join the two outputs together ? It claims no shoot through

author: SeanB
author_rank: Super Contributor
timestamp: September 19, 2014, 04:34:38 pm
content: If they say use the 2 resistors then at some combination of temperature, loading, supply voltage and rise and fall time both outputs can be turned on together and cook the chip. You can probably get by with the source resistor only if you are not worried about blowing the mosfet or driver from ringing on turn off, but 2 SMD resistors of 100R or so will not break the budget, and can also be used as jumpers or convenient test points.

author: Simon
author_rank: Unknown
timestamp: September 19, 2014, 04:57:54 pm
content: You mean I should use a resistor anyway ? in that case i might as well use the 100r as a pullup and pull the gate low with an N channel.

author: SeanB
author_rank: Super Contributor
timestamp: September 19, 2014, 05:04:23 pm
content: True, but 10k would work for DC or low frequency switching, and will not act as a heater. The advantage of the chip is it is convenient, and layout is easy with nice paths to and from it. It also allows turn on and off tailoring so you can reduce EMI from the switching at the expense of a little power being dissipated during transitions.

author: Simon
author_rank: Unknown
timestamp: September 19, 2014, 05:09:51 pm
content: I'm running around 1KHz with 3.3nF of gate capacitance switched with 10-12V. As it is I'm using paralleled power transistors to cope with loads from 20-70A so that I can "modularize" the design. To make life easier in laying around around large power traces and avoid long traces from driver to power mosfet I'll have a driver next to each mosfet

author: SeanB
author_rank: Super Contributor
timestamp: September 19, 2014, 05:42:11 pm
content: So adding the small package, 3 resistors ( a pull down at the input just in case the wiring is disconnected), a bypass capacitor of 100n on the power pins of the chip and then rinse and repeat for each power device will not add much extra in terms of board space. Makes driving easier, and you can control drive parameters with the resistors to have either slow turn on for DC loads or faster for PWM use.

author: Simon
author_rank: Unknown
timestamp: September 19, 2014, 06:00:22 pm
content: Well this will be a product produced on a large scale so no need to waste money on parts. It will always run PWM at 1KHz. I'm thinking around 15A per mosfet (10mR) so 2-5 blocks to go from 20 to 70A

author: David Hess
author_rank: Super Contributor
timestamp: September 27, 2014, 03:26:03 pm
content: The ZXGD3004E6 gate driver is apparently just an NPN/PNP emitter follower pair built using some of those magical ZETEX high current density transistors so if the inputs are tied together, it operates in class-B and no shoot through is possible. The application circuit shows two output resistors so that the rise and fall times can be adjusted separately but one resistor may be used if this is not a requirement.

author: Simon
author_rank: Unknown
timestamp: September 27, 2014, 05:47:15 pm
content: Right, I just wanted to be sure I'd not be bloswing it but I can't spare 10mA to drive it anyway so am just going to use a pull down mosfet and pullup resistor. I can run at a frequency as low as 100Hz if i want to i suppose I was over fussing.

author: David Hess
author_rank: Super Contributor
timestamp: September 27, 2014, 06:53:17 pm
content: Quote from: Simon on September 27, 2014, 05:47:15 pmIt only needs 10 milliamps to drive the output current to 1.9 amps typical which implies a current gain of 190 which is reasonable for their super transistor process. You can always add a single resistor on the input to limit the drive current and accept a lower output current. It will still allow switching of the MOSFET roughly 200 times faster than without the driver.I am a little surprised that such a device exists given that it just replaces a complementary pair of transistors with their bases and emitters tied together.

author: Simon
author_rank: Unknown
timestamp: September 27, 2014, 06:58:50 pm
content: Quote from: David Hess on September 27, 2014, 06:53:17 pmLikewise, I don't see the point I could just put them in discreetly and it would cost less.

author: SeanB
author_rank: Super Contributor
timestamp: September 27, 2014, 07:12:28 pm
content: Single component so easier to place, and you need one less feeder. Resistors could already be used on the board so another feeder saved. Makes sense if you are trying to stay in the limits of a PNP machine and are running short of feeder space. Plus the added bonus of controlled drive for the device.

author: Marco
author_rank: Super Contributor
timestamp: September 27, 2014, 07:23:32 pm
content: I doubt you will get the combination of peak current capabilities and relatively low transit times as easily in discretes.

author: Simon
author_rank: Unknown
timestamp: September 27, 2014, 08:18:56 pm
content: Quote from: SeanB on September 27, 2014, 07:12:28 pmWell both the pulldown transistor and the pull up resistor were already part present on the board anyway so it's just more components for the same bom lines

author: tautech
author_rank: Super Contributor
timestamp: September 27, 2014, 09:16:39 pm
content: Have a look at IRF5851PbF dual mosfet.Although it's only a 20 V device, it's current characteristics are similar.Sure not a dedicated driver but I have used them for some time as a driver from logic levels for up to 100 A IGBT's and all manner of n-mosfets.It has a implemented track layout that I prefer, the only catch that GND and VDD must be routed to pins 2 & 5 which is a little tight on a SS board, but very doable.Pins 1 & 3 become input and 4 & 6 gate drive output.Advantages: no quiescent current, low input threshold, only gate drive resistor required.http://www.irf.com/product-info/datasheets/data/irf5851pbf.pdf

author: David Hess
author_rank: Super Contributor
timestamp: September 28, 2014, 07:50:16 pm
content: Quote from: Marco on September 27, 2014, 07:23:32 pmThe characteristics come from using Zetex's process for Super E-Line transistors or whatever they are calling them these days and a quick check of their available surface mount transistors shows lots of discrete options. On and Fairchild and probably others have similar transistors now as well.Quote from: tautech on September 27, 2014, 09:16:39 pmWhen switching a capacitive load with a class-B common drain/collector amplifier, there is no quiescent current anyway. Their input threshold voltage is more than twice as high as a bipolar transistor. The same output resistors are needed to control drive strength; there is actually a disadvantage here because the bipolar transistors could be operated without the output resistors and with the input resistor used for beta current limiting which is not as simple to implement with MOSFETs.QuoteRunning a complementary MOSFET transistor pair in common source (or a bipolar pair in common emitter) will induce huge shoot-through currents during switching. The IRF5851PbF datasheet forbids this kind of operation unless Vgs is held below 2.5 volts or Vds is held below 1 volt to prevent excessive drain current. CMOS logic gets away with this because of the high impedances involved and low capacitive loading. This could be solved by adding . . . two source degeneration resistors on the output to limit the current during switching.Further, Vgs has an absolute maximum rating of +/- 12 volts so the output voltage range will be significantly restricted.If they are configured in common drain, then that eliminates the shoot through and Vgs problem but the gate threshold voltages are higher than Vbe of a bipolar transistor (twice as high or more) and will create a larger dead zone in class-B operation limiting the output voltage range. It might be fun to fix this by biasing them closer to class-AB operation but that add complexity.The IRF5851PbF dual MOSFET will certainly work as a replacement in some cases but not all. I might use it in common drain with a level shifter driving it but in that case, it provides no advantage over a good bipolar pair and it will almost certainly be more expensive for a given power level. It looks like the IRF5851PbF is discontinued anyway.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 28, 2014, 08:25:42 pm
content: Ah, yes...hmm...If you use IRF5851 as common source (i.e., just a regular unbuffered CMOS inverter), consider what happens for gate voltages between the threshold voltages. Suppose VDD = 10V. At Vg < 0.6V or Vg > 9.55V, no current flows (Id < 0.25mA). For gate voltages inbetween, one transistor will be fully saturated (~0.1 ohms) and the other stays in the linear range.Referring to Fig.1 and Fig.16, as Vgs becomes biased on, the transistor turns on more and more, drawingamperesfairly quickly. Curiously, the highest curve shown is 2.5V, drawing just under 10A in both cases; neither transistor is even rated for any higher peak current, for any duration -- yet they will quite clearly be capable of 20, maybe 30A peak current at 5V!What's wrong with that? Besides the ludicrous demands on the supply bypass (you will need large ceramic capacitors and aluminum polymers to hold up the supply during such a massive transient!), exceeding peak rated current likely means spooky failures, like metal migration: the average temperature might not be exceeded, but with such extreme current density, the device will nonetheless fail after some number of cycles, maybe one, maybe a billion. But it is sure to fail.Typical application as a gate driver replacing one of those "bipolar pair" devices might have ~20mA source drive, to expect 2A peak output. The total gate charge is around 10nC for 4.5V gate swing, but we're looking at a 10V application, so it'll be more, maybe 15nC total. A 20mA driver will then take 0.75us to commutate the gates, which means about 0.75us of >20A shoot-through. (Which will dissipate about 10V * 20A = 200W between the two transistors, or *0.75us = 150uJ per transition, for a maximum frequency of only 6.4kHz at rated power dissipation. Nevermind, maybe it will fail thermally first!) The output will swing a little bit faster than 0.75us, because most of the gate charge occurs during the Miller step, but it will still be much slower than with the bipolar device.The fundamental problem with such a choice is that it's more thanten times larger than necessary. Try a 1-2 ohm MOSFET instead; you'll still have to deal with 5A peak shoot-through, but it can be switched fast enough that it's not as big of a deal. This is comparable to the output drivers in proper monolithic CMOS driver chips. Even the largest (15A+) driver chips boast only of 0.2-0.5 ohm outputs; using as little as 0.09 ohms for a couple-amps application is ludicrous, as its performance shows.Moral of the story: always try to pick parts with are suitably matched to the application. Excessive size leads to wasted money and poor dynamics, and rarely yields improved efficiency!Tim