How to match Mosfet Gate Driver with Mosfet - Page 1
Forum section: electronics / beginners

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: August 28, 2018, 06:36:20 pm
content: I currently have a mosfet gate driver (half bridge) IR2111.Max output current is 200mA, min rise time is 80nS, max is 130nSI want to drive a IRF540 mosfet with this and turn it on as fast as I reasonably can, the faster the better. The Qg (total gate charge) of this mosfet is 71nC.By using I = Qg/ t, and using t as 80nS we get a current of 0.9A which is more than four times than that of the gate driver. I could of course choose a higher current rated gate driver however currently I only have this in my kitty.How do I ensure that I do drive the mosfet as fast as I can and yet stay within the limits of the gate drivers max output current?Is it as simple as adding a series gate resistor at the IRF540 gate?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 28, 2018, 06:38:45 pm
content: Are you sure you actually want to turn it on "as fast as possible"?Many things, probably far more extreme that you are expecting, are possible. Particularly given enough budget. A woefully underconstrained problem like "as fast as possible" leads to very interesting, but not ultimately very helpful, proposals.If you're worried about the driver itself, it is fine to connect to a minimal resistance load. Read the datasheet, it's measured with -- I forget what for IR2111 exactly, but 1nF or 10nF load, no resistance, is very common.BTW, the equivalent gate capacitance is Qg / Vgs(on), or about 7nF here.Tim

author: jmelson
author_rank: Super Contributor
timestamp: August 28, 2018, 06:48:44 pm
content: Quote from: ZeroResistance on August 28, 2018, 06:36:20 pmIf you want fast, you need a stronger gate driver. I think IR has up to about 6 A capability. 200 mA is really weak, and will lead to long turn-on and turn-off times, causing extra heating in the transistors. I use the IR2113 (6A) and the IR2181 (I think) at 2 A.Jon

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: August 28, 2018, 06:49:40 pm
content: Quote from: T3sl4co1l on August 28, 2018, 06:38:45 pmOK so maybe I worded it wrongly, however I checked the datasheet and saw that the rise time of the IRF540N is around 35nS @ 50V /16A with a gate resistor of 5 ohm.BTW I do f ind a CL = 1000pF mentioned in the IR2111 datasheet.

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: August 28, 2018, 06:54:08 pm
content: Quote from: jmelson on August 28, 2018, 06:48:44 pmI will move to a faster driver as soon as I can but for my immediate tests I need to get this driver working. My frequency of turning on is not too high here maybe just a momentary pulse so I'm not so concerned about the heating.I just don't want to damage the gate driver, so how do I ensure that the current is within the 200mA limit and how much turn on time should I expect in the mosfet.

author: Marco
author_rank: Super Contributor
timestamp: August 28, 2018, 07:15:55 pm
content: Quote from: ZeroResistance on August 28, 2018, 06:36:20 pmIt's a short circuit current, the resistance is internal.

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: August 28, 2018, 07:21:42 pm
content: Quote from: Marco on August 28, 2018, 07:15:55 pmOk! So what you are trying to say is that there is an internal resistor inside the IR2111 that will limit the current to 200mA? and I don't need to add an exteral resistor?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 28, 2018, 07:53:04 pm
content: Yes, or something roughly equivalent to that.IR2111 is a CMOS device, so it has MOSFETs driving the output, and those MOSFETs will have an Rds(on) figure.The I_O and V_OH/V_OL parameters proscribe what that output resistance is. (Note that it's I_O at specified V_O, and V_OH/L at specified I_O. Don't mix I_O and V_OH/L together because they aren't measured under the same condition!)Bipolar drivers have more of a constant-current characteristic, but the same idea applies.Tim

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: August 28, 2018, 07:57:38 pm
content: Quote from: T3sl4co1l on August 28, 2018, 07:53:04 pmOk! So its getting a bit clearer. However I also rechecked again and the Io is specified at a PW of <= 10uS. What if I have longer pulse widths'?Ok I got it, duh! that short circuit current will only last till the gate capacitor is charged, So it doesn't matter how long the pulse width is.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 28, 2018, 08:14:34 pm
content: Right.I've seen a "quasi-static" measurement from Fairchild, where they load it with a big stinkin' cap (0.1uF or thereabouts) and use dV/dt to measure output current. It's not a steady state (static) measurement, so... "quasi".There should be no particular reason to suspect that pulse width, or slow rate-of-rise, has any effect. It would be a rather poor driver if it did, at least. So, granting that assumption -- these methods should be as good as using a dumb resistor.(Note thatfastrate-of-rise definitely does reduce output current, simply in that, the internal circuitry itself only moves so fast. Say for a capacitor load, the rise time versus capacitance relationship. As you decrease capacitance below a certain range, rise time doesn't continue to drop, it flattens out. You might expect 10nF to take about twice as long as 5nF, but it would be unreasonable to expect 0.1nF to be 1/100th of the time!)Tim

author: jmelson
author_rank: Super Contributor
timestamp: August 28, 2018, 09:23:03 pm
content: Quote from: ZeroResistance on August 28, 2018, 06:49:40 pmGiving a rise time without specifying the gate driver is just meaningless! I seriously doubt you will get any 35 ns risetime with a 200 mA gate driver. I was getting 400+ ns risetimes with a similar triple half-bridge driver all in one package. I then changed to using 3 separate driver chips with a 2A spec, and cut the risetimes to 100 ns or so.Jon

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: September 01, 2018, 06:45:48 pm
content: So, let's say I use a typical logic level Mosfet, and I don't use a driver and directly drive it from a pin of a microcontroller(Arduino) say with 5V logic and the pin can source a max current of 25mA. How much time can I realistically expect for the mosfet to turn on in?Second point, if the mosfet is connected to a high drain voltage lets say above 80V, will switching drain inductive loads, need any extra circuitry to protect the logic level gate drive?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: September 01, 2018, 06:55:48 pm
content: Switching an inductive load without protection, will require much more robust drive circuitry, to account for when the transistor fails in a three-way short, dumping some fraction of Vds into the gate terminal.You'll most likely just put a TVS diode from ground to drain, to deal with that.Tim

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: September 01, 2018, 07:35:00 pm
content: Quote from: T3sl4co1l on September 01, 2018, 06:55:48 pmOkay! Thanks for your valuable inputs.Having said that are there mosfet packages available with built in driver? preferably say a Mosfet Half bridge with high side + low side driver.Or even a mosfet full bridge with integrated gate drivers.I have seen lots of motor drivers with mosfet h bridges however they seem to have extra pins for current sensing etc. I am just looking for a plain old H bridge with integrated drivers.

author: blundar
author_rank: Contributor
timestamp: November 12, 2018, 05:10:40 am
content: There are a plethora of such parts on the market. It's going to boil down to how much current you need to switch.Some of my favorite half bridges:ST VNH5019AHInfineon BTN8962A BTN8982ASome of my favorite full(ish) bridges:ST VNH7013AHInfineon IFX007All of these parts are designed to drive a decent amount of current with an inductive load more or less direct from a MCU. They feature drivers, protection circuitry, integrated FETs and sometimes current sensing and/or diagnostics. They're designed to be an "easy" button for designers compared to using a gate driver, discrete FETs and thinking.

author: Rx7man
author_rank: Regular Contributor
timestamp: November 12, 2018, 06:30:51 pm
content: Quote from: blundar on November 12, 2018, 05:10:40 amI just found this device in a board I was reverse engineering... seems like a handy unit, optocoupler with push/pull output mosfets and hysteresishttps://www.mouser.ca/ProductDetail/782-VO3120

author: T3sl4co1l
author_rank: Super Contributor
timestamp: November 12, 2018, 10:55:02 pm
content: Quote from: blueskull on November 12, 2018, 07:06:13 pmAnd what's more than that, note that Ciss is a small-signal, zero-bias condition that's basically meaningless for switching -- the average value over the switching event is usually about four times larger! Fortunately, the datasheet also provides Qg(tot), which gives Cequiv = Qg/Vgs(on).The exact situation that you're trying to avoid (VHF oscillation) is a bit different from the (~baseband) gate switching situation. Mainly, it's important to avoid modest value capacitances on the gate trace, which leads to a tuned-gate oscillator as the MOSFET goes through the linear range (during the Miller plateau). An IRF540 might oscillate at 50 to 100MHz; a modern SuperJunction type can oscillate at 400MHz, maybe more.You don't usually have a capacitance nearby, so it's not usually an issue, but long trace lengths (= transmission line stubs) can do it, or attached components like zener diodes (for protection).One easy way to address this, is putting a small ferrite bead in series with the gate lead. During gate transition, the peak current causes the bead to saturate, so it has little effect on switching (basically it introduces a slight (~1ns?) delay), but it presents just enough lossy impedance to stop oscillation.Tim

author: T3sl4co1l
author_rank: Super Contributor
timestamp: November 13, 2018, 05:39:32 am
content: If there is no resistance, then the energy stored in the inductance corresponds to the charge deposited in C_eq.If there is considerable resistance, then the (small signal) C_iss (at whatever Vds state is being considered) does dominate.C_iss is only given as the smaller value (natch); you have to find it plotted in a graph to figure out the low-Vds condition. Graphs are always typical data unless otherwise noted, so be careful with that, too.C_eq being larger, will give a conservative estimate, suitable for design purposes, but not for more precise predictions.If you are looking for accuracy, then indeed, simulation, or better yet real testing, is needed.Tim