Details of driving MOSFETs and the required current. - Page 1
Forum section: electronics / beginners

author: David_
author_rank: Contributor
timestamp: February 09, 2018, 06:46:46 pm
content: Hello.I have been trying to figure out what sorts of MOSFETs I can drive with a microcontroller pin, or specific MOSFET driver IC's. But there are a few question marks that makes me really confused, the application is a 4-switch non-inverting buck-boost converter controlled by a digital feedback-loop implemented on a microcontroller(so a driver of some kind will be required to enable me to use one specific N-channel MOSFET for all 4 switches), I am currently looking atLTC1157 - 3.3V Dual Micropower High-Side/Low-Side MOSFET Driver.The hole circuit will be used with one of two different supplies, ether a single cell Li-Ion battery, or two series cell Li-Ion batteries. so ether 1 or 2 18650/26650 batteries.I could put the question like this, if I know that my driver can supply "this" much current then what sort of values for total gate charge may I consider for the MOSFETs to be used?I have read about this for quite some time now and I know that:IG= QG/t(transition)where:QGis the total gate charge.t(transition)= the desired transition time.IG= the gate current required to turn the MOSFET on in time period t(transition).This specific information was from a rather simplistic Microchip document, but what I do not understand is what t(transition)is. In my specific case the switching frequency is 100kHz, but 10µS is 1 period and I would think that t(transition)would have to be a lot smaller than that. I would have thought that the t(transition)should be as small as possible, so a better question might be "how long may t(transition)be without negatively impacting the circuit?".I would think that the t(transition)in a 1MHz switcher would need to be smaller than that of a 100kHz switcher, but really they could as well be the same for all I know, I think I know that this is a question about losses in the driver and MOSFET but I guess I am looking for a much simpler approach to estimate a reasonable t(transition)time.If I would understand better what t(transition)is I could determine what sort of t(transition)I am concerned with and then calculate the maximum gate charge by:IG/t(transition)= QGWhile looking in the LTC1157 datasheet it looks like the gate current is around 100µA... which seems like a very tiny current, but another thing that confuses me is the difference in the current being supplied to a driver IC and it's peak current supplied to the gate, I at least think that it is correct that the supply current may be very small, a few mA's while the peak output current may be very much higher, since capacitors are used to store the low current that is supplied all the time and then discharged quickly while reaching much higher currents.Regards

author: Audioguru
author_rank: Super Contributor
timestamp: February 09, 2018, 06:49:27 pm
content: The gate of a Mosfet does not conduct DC. But it has a high capacitance that takes a fairly high current to charge or discharge quickly.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 09, 2018, 07:35:13 pm
content: You're on the right track.t_t can't be much longer than, say, t_clk / 10 or 20, else so much time is spent transitioning that you're better off with a class C oscillator as such!Typical times are under 1us, with under 50ns being typical for higher frequency and higher efficiency circuits.Note that the output (drain) transition time is shorter than the total gate transition time, which helps a lot. Drain voltage is only transitioning during the Miller plateau (where Vgs moves very little), which is usually about half the total gate rise/fall time.Note that assumes inductive switching, which is very typical. For a 4-switch converter, typically in any given cycle, half the transistors will be handling the transition, while the other half will be sitting around passively. This is because a synchronous rectifier doesn't have to deal with Miller effect -- which is not to say it's a slam dunk to drive, because the same phenomenon still arises, it's just behind a threshold. You need a low impedance gate drive, to firmly hold it off, otherwise when the main switching transistor turns on, some turn-on gets capacitively coupled to the sync rect gate and switching loss can go up massively.Typical MCU output pins are similar to 74HC or LVC logic pins, having an equivalent output resistance around 30 ohms.For t_t = 50ns, and using a ratio of 1 time constant, at 5V, this gives Qg max ~ 8.3nC (at 5V).Oh, that introduces some other things --1. Resistance. CMOS output pins are largely resistive, so it's not just that you're applying a current and stuff happens; it's more like a switch with resistance. So you get an RC time constant, rather than a constant slope and t = Q / I.2. Time constants. Voltage swings 95% of the way after 3 time constants. But we don't need that much. Two time constants is 87% of the way, which will more or less include the Miller plateau in either rising or falling direction, and is a good figure for what you'll typically see at the gate waveform. Again, about half of that, or one time constant, is the edge speed you'll see on the drain/output, and is where most of the switching loss is dissipated. I wouldn't design based on one time constant, but you might consider this a worst-case scenario (so, 8.3nC really is the most you should ever consider, for this transition time).3. Voltage. We need this to convert gate charge to effective gate capacitance. Q = V*C is the fundamental equation of the capacitor (or (dQ/dt ==) I = C * dV/dt is its more commonly seen derivative -- same fundamental thing). The incremental capacitance changes with voltage (namely, it's small at low gate voltages and high drain voltages, massive during the Miller plateau due to feedback effect, then higher at high gate voltages and low drain voltages), but as long as we don't try to look too closely at the waveform (because of these bumps), the overall effect will be consistent with using this effective capacitance as the C in a regular linear RC time constant.Some other tweaks:- You have to use logic level FETs, obviously!- There's no fundamental difference between logic and regular level types. It's just Vgs shifted down to be more convenient, but the gain is the same. Vgs(on) = 5V is about equivalent to a regular one at 7 or 8V. Which is to say, not very on -- you get what's in the datasheet (Rds(on), and usually Qg(tot) as well, guaranteed at 4.5V or so), but it's not exactly at peak performance. Keep this in mind for selection.- The lower Vgs(th) means the Miller plateau falls at a much lower voltage, like 1 or 2V instead of 3 or 4V, where you have that much less current to pull through it. Logic-level FETs have poor turn-off characteristics for this reason.- If you are switching low voltages, try to push towards 30V or lower Vds(max) parts, because they have more gain, and also less Rds(on) tempco. For a marginal case -- say a 12V power supply -- you might opt for an H-bridge of 20V parts, versus push-pull with 40V parts. Down at battery voltages, some quite excellent parts are available (there's a MOSFET with something like 300uohm Rds(on), Vds(max) = 8V and Vgs(on) = 1.8V, IIRC).HTH,Tim

author: danadak
author_rank: Super Contributor
timestamp: February 09, 2018, 11:53:58 pm
content: Some ref material -http://www.ti.com/lit/ml/slua618/slua618.pdfhttp://www.ixys.com/Documents/AppNotes/IXAN0010.pdfhttp://www.ti.com/lit/an/slua105/slua105.pdfhttps://www.infineon.com/dgdl/an-978.pdf?fileId=5546d462533600a40153559f7cf21200https://www.fairchildsemi.com/application-notes/AN/AN-6076.pdfhttps://gate-driver.power.com/sites/default/files/product_document/application_note/AN-1001_IGBT_and_MOSFET_Drivers_Correctly_Calculated.pdfGo to archive. org, search on MOSFET, tons of papers, databooks, apps info.Regards, Dana.

author: David Hess
author_rank: Super Contributor
timestamp: February 10, 2018, 03:41:26 pm
content: The LTC1157 is a low power driver intended for low and high side load switches which essentially operate at DC. Something like theLTC1693with its 1.5 amp output current would be more suitable.At high source currents, other factors come into play which will limit switching speed. The gate structure itself has series resistance which limits charge and discharge current into the gate. At high source currents, source inductance from the layout and packaging converts the source dI/dT into a voltage opposing the gate drive. (1) So increasing the gate drive current beyond some point loses effectiveness.Excessive switching speeds also conflict with reverse recovery time of any switching diodes if used and contribute to EMI.(1) There is something to be said for returning the gate drive current directly to the source pin to minimize the loop area minimizing inductance to allow faster switching. MOSFET gate drivers may be located close to their respective MOSFETs for this very reason and this is very easy to do with discrete gate drivers. If integration is not required to save space, then they make small high current transistors now which are very suitable for driving large gate capacitances.

author: David_
author_rank: Contributor
timestamp: February 16, 2018, 07:14:59 pm
content: Thank you for your answers, I haven't really processed it all jet though but I will read it all a couple of more times.I am actually really interested in discrete gate driver circuits, but so far I have had a difficult time with the high-side switch gate drive requirements. All the bootstrap circuits seems to rely on the switching of the low-side switch in order to replenish the bootstrap capacitor, which might not be a dead-end but it introduces enough complexity into the situation that I get stopped in my track, I guess the other solution would be a dedicated charge-pump to double to voltage to be switched. But given that this is for a 4-switch buck-boost converter, wouldn't that require a even higher voltage for the boost high-side switch in cases when the boost function is used?

author: David_
author_rank: Contributor
timestamp: February 26, 2018, 11:24:58 am
content: I think I have determined that my view on the boost section was faulty, I am much more familiar with buck converters but it would appear that the boost section doesn't make the situation any different.So I have chosen to go with some integrated gate driver however there is one issue that is somewhat frustrating and I am not sure of my options.I can find many suitable half-bridge drivers but there high-side switch relies upon a boot capacitor which requires the low-side switch to activate in order to replenish the charge on the boot capacitor. But in this application the buck-sections high-side switch will sometimes need to be on 100% of the time... Is there any standard or well known solution to that problem?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 26, 2018, 11:49:51 am
content: You can charge pump them.Add an extra oscillator (could be a CMOS 555), and from its output, connect a series capacitor (and current limiting resistor) to a series diode pair on the bootstrap supply (A1 = SW, K1A2 = capacitor, K2 = +VBS).Maybe you have such a signal already in circuit (any alternating signal will do), in which case only a buffer might be needed.One caveat is switching dV/dt back-feeding the oscillator or buffer; you might want schottky diodes on its output to handle this. The current limiting resistor helps here.Or you can use the switching nodes: an R+C from each, to an individual series diode pair, and duplicate this for each bootstrap supply. This can deliver excessive voltage of course (full switching node peak-peak voltage), so add a zener to shunt excess. This is less efficient, of course, and a burst mode controller may be desirable for light loads.Or use a gate drive circuit that is tolerant of wide +VBS swings, and just make a static "over-the-top" rail to power the driver (which at this point is more of a level shifter / amplifier than a bootstrap, i.e., instead of making 10V of drive, it's making 10+Vsw of drive; same thing to the transistor though!). Run both drivers from this, no bootstrapping needed anymore.You can also get gate drivers with integral charge pump, I think Intersil does a few? Kinda slightly boutique, YMMV.Tim