Driving differently-sized mosfets at different frequencies - Page 1
Forum section: electronics / beginners

author: derGoldstein
author_rank: Regular Contributor
timestamp: September 16, 2017, 01:32:46 am
content: I've been reading about driving mosfets and the different methods to do it. Not all of it I understand, so I'm trying to clarify a few things so that I get a better idea of how to go about it (in other words, there's no one question, I'm trying to see if I understood what I've been trying to learn... There's no other way for me to compose these questions without having to post multiple questions on the same topic).Mosfets have a gate capacitance which is analogous to a small capacitor that has to be either filled or depleted in order to turn the mosfet on or off. Assuming that a middle-point is undesirable when PWM-ing a mosfet (it should either be completely on or completely off), the capacitor has to be filled or drained as fast as possible (the driving waveform has to be as close to square as possible). A partially-filled capacitor would lead to the mosfet being in a high-resistance state, "burning" energy unnecessarily.At very low switching speeds, something like an IO pin from an MCU would be enough. A slight step up would be passing the IO signal through an inverter/buffer which has slightly more driving power.At all but the lowest frequencies you need to dedicate some external components to charge and drain the gate. In the simplest form those would be an NPN and PNP BJTs. At higher switching speeds you'd use a dedicated mosfet driver.Assuming that driving voltage isn't a problem (either you're driving a low-side n-channel or a high-side p-channel, in both cases there's no need to bootstrap), is current the only factor to driving a mosfet at high frequencies?A slow transition time on the part of the driver would mean a less sharp waveform, which would lead to more half-on time, leading to more energy lost as heat. Is it possible for a driver (in whatever form it may take) to have a high current capability, and yet not be able to produce a "square-enough" waveform?It seems like the larger the mosfet the higher its gate capacitance (is this always the case?). It's easier to drive smaller mosfets than larger ones because you have to move less current into and out of its gate. The faster you switch the mosfet on and off, the more times (per-time-interval) you have to fill and drain its gate.So there are 2 factors when determining how much power is required to switch a mosfet: the gate capacitance and the frequency. Is there a practical way to calculate this? If I have the specs for a particular mosfet, and I know at what frequency I want to drive it, is it possible to determine how much current I'm going to need in order to effectively drive it?This seems like a matter of calculating the amount of current required to fill and drain the gate capacitance at the necessary rate, but almost every document on the subject insists that this is an oversimplification. In what way?Is driving high-power mosfets at high frequencies always going to mean a lot of energy "lost" to filling and draining the gate? Even if the mosfet is driving very little current, as long as it's being driven at a high frequency (and it's a large mosfet), the circuit will constantly be "burning" energy to switch the mosfet on and off?I apologize in advance for the meandering post... Any feedback would be welcome.

author: Circlotron
author_rank: Super Contributor
timestamp: September 16, 2017, 04:06:46 am
content: Quote from: evb149 on September 16, 2017, 01:57:26 amYep.Drive the gate x volts above ground to switch them on and the same x volts below ground to switch them off. AFAICT negative gate voltage is no problem for a MOSFET. Resonance converts a positive voltage on the gate-source junction capacitance to a negative voltage and vice versa theoretically in a lossless manner.

author: David Hess
author_rank: Super Contributor
timestamp: September 16, 2017, 01:41:22 pm
content: Quote from: derGoldstein on September 16, 2017, 01:32:46 amIt may be calculated if the input capacitance and reverse transfer capacitance are specified.QuoteIt is an oversimplification because the voltage swing on the drain which is often high makes the reverse transfer capacitance look larger than it really is. So the total capacitance that needs to be charged also depends on the drain voltage.QuoteThere are some tricky designs which recycle part of the energy stored in the gate capacitance to lower the total power required.

author: derGoldstein
author_rank: Regular Contributor
timestamp: September 17, 2017, 04:48:40 pm
content: Quote from: evb149 on September 16, 2017, 01:57:26 amI never paid attention to that rating, but now I see that it's typically over an ohm. Not all datasheets seem to show this value, though (especially for older mosfets). So I should really model the mosfet as a component with a resistor leading to a capacitor leading to a gate (all 3 are the same, but I mean just as a way to visualize it)...Quote from: evb149 on September 16, 2017, 01:57:26 amThis is the "Qg" characteristic rated in nC (nanocoulombs)? Do I also have to consider the Gate-Source Charge(Qgs) and Gate-Drain Charge(Qgd), or is the Qg the important one I should be taking into account?The old IRF540 has terrible RdsOn (0.077ohm, while newer mosfets drop below 2mOhms), but relatively low Qg (72nC). Is this why it's lasted as a (relatively) common component for so many years, or is that just inertia because it was cited in so many books? (and probably manufactured in massive amounts)Quote from: evb149 on September 16, 2017, 01:57:26 amI never considered using a logic-level mosfet for this reason, but it makes sense. The total amount of energy required to switch it completely on is less than a typical 10 Vgs.Quote from: David Hess on September 16, 2017, 01:41:22 pmI was wondering if it was possible to do something else with that gate capacitance apart from shorting it to ground. Maybe store it in another capacitor? Or move the charge to another mosfet's gate? This is beyond what I'd be able to implement, but it's an interesting problem...

author: Cerebus
author_rank: Super Contributor
timestamp: September 17, 2017, 05:40:14 pm
content: I'll pick up on just one point that I haven't seen others explicitly mention.Quote from: derGoldstein on September 16, 2017, 01:32:46 amYes that is possible. If you think of it in terms of the output impedance of the driver versus frequency it's clear that a driver that has potentially massive current drive capability at low frequency may not necessarily be able to bring it to bear quickly enough at high frequency.Imagine you're using a high power audio amplifier as your driver. This is something that can deliver several amps at low frequency (a 100W amplifier would deliver over 3 amps into an 8 ohm load at 1kHz). What it can't do is deliver that kind of current at 1 MHz as its output impedance climbs with frequency.The easiest way to represent this and do calculations is to look at the complex output impedance of your driver and the complex input impedance of your load (in this case a MOSFET gate). If you're not comfortable with complex representation then you can still make a decent fist of it by examining your parts as simple equivalent circuits and calculating the reactance of parts at a few spot frequencies of interest.