Mosfet gate current consumption? - Page 1
Forum section: electronics / beginners

author: mstfylcn333
author_rank: Newbie
timestamp: October 15, 2021, 08:36:00 am
content: Dear Friends,Altought I searched mosfet gate current consumption on Internet, ı couldn't find too much information about this topic.Is there any techniquies ,formula or application notes(how to calculate) about this topic.For example, how to calculate of any mosfet or IRFZ44 gate current consumptions at 20khz?Could you help me about this topic? :

author: CaptDon
author_rank: Super Contributor
timestamp: October 15, 2021, 03:09:10 pm
content: One estimate is to view the gate as a capacitor. The gate capacitance will be listed in thespec sheet and can be very large at 5000pf in some cases. Calculate the Xc of the capacitor(The capacitive reactance) and pretend that Xc is simply 'resistance'. Full saturation of powerFETs requires about 10 volts gate to source so calculate using ohms law the current requiredto induce 10 volts across the calculated capacitive reactance at the frequency you have chosen.In reality the peak current will be higher because you want the FET to turn on and off quickly.Most FET switching circuits will be driven with something resembling a square wave andcapacitors look like dead shorts to square waves. I am sure there is a much more accurateway to calculate, but my calculations have always gotten me into the ballpark and if I getmy answer and understand it will be a bit lower than what is required I just up the drivecurrent by a factor of 2 or so and usually things look very good on the oscilloscope. I havedone many switched mode supplies and FET class D switching amplifiers and all have runcooler than expected with no failures just doing my 'Poor Man' calculations. Many FET driverI.C.'s push and pull multiple amps in AND out of the FET gates they are driving and theFET driver circuit must have a very stiff supply to not sag under the high peak current demands.

author: Manul
author_rank: Super Contributor
timestamp: October 15, 2021, 03:27:00 pm
content: You mean DC leakage or dynamic consumption? DC leakage should be inside datasheet, it is very small and negligible for almost any application. The dynamic consumption can be calculated from total gate charge at some specified conditions and switching frequency.If you follow datasheet values given for IRFZ44 at Vgs = 10V, Vds = 30V, Id = 51A you see they give total gate charge 67nC. This is charge which needs to be transfered into gate 20,000 times per second (at 20kHz). So do the math of average current.But the problem is that total gate charge depends on many factors when mosfet is used in real circuits. You can try to calculate, but it will be more or less ballpark figure. Do you really need this figure? I would say better to measure experimentally if you need any kind of precision.

author: Siwastaja
author_rank: Super Contributor
timestamp: October 15, 2021, 03:53:17 pm
content: Total gate charge Qgtot is directly available.One Coulomb is equal to A*s.Rest is left as homework. Tip: if you have trouble with it, don't worry, it's really too simple. The solution should fit on one line.This being said, it will be an approximation with some assumptions. The Qgtot is measured at some specific Vgs, and you may be using different Vgs.If any of this really matters, you can make things better using a modern MOSFET. IRFZ44 has massive gate charge compared to its power handling. Modern parts have better ratios.

author: Picuino
author_rank: Super Contributor
timestamp: October 15, 2021, 04:12:39 pm
content: https://www.vishay.com/docs/73217/an608a.pdfThere are two capacitors in a mosfet and you must charge both.The first capacitor (Cgs from gate to source) is charged first, before switching.The second capacitor (Cgd from gate to drain) is charged during switching, with as much voltage as there is from drain to source. This is also called a miller capacitor.It is usually the one that needs the most current and charge, and depends on switched voltage by the mosfet.

author: Picuino
author_rank: Super Contributor
timestamp: October 15, 2021, 04:26:28 pm
content: The peak current drawn by the gate does not depend on the frequency, but on the switching speed.In practice Cgd is voltage dependent and the gate of the transistor must be charged with an almost constant amount.I_gate_peak = Q_switch / t_switchFor example:Q_switch = 10nQ (IRF530 Id=14A)t_switch = 100nsI_gate_peak = 10nQ / 100ns = 0.1 Ampere (during switching)https://www.vishay.com/docs/91019/irf530.pdf

author: mstfylcn333
author_rank: Newbie
timestamp: October 16, 2021, 11:28:19 am
content: Quote from: Manul on October 15, 2021, 03:27:00 pmThank you Dear "Captdon" for your advice.I will search your meaning "Calculate the Xc of the capacitor(The capacitive reactance) and pretend that Xc is simply 'resistance'. Full saturation of powerFETs requires about 10 volts gate to source so calculate using ohms law the current requiredto induce 10 volts across the calculated capacitive reactance at the frequency you have chosen."By the way What was the max frequency you have driven mosfet without facing any problem?

author: mstfylcn333
author_rank: Newbie
timestamp: October 16, 2021, 11:29:27 am
content: Quote from: Siwastaja on October 15, 2021, 03:53:17 pmI am talking about dynamic comcuption my Dear friend.So I am concern that about peak current of mosfet gate charge. I mean, now I am designing BMS(battery managment system) for li ion so I am using 8 pieces of mosfets to charge batteries.As we know to drive those mosfets we need mosfet driver and voltage regulators. If I know peak current (not avarage current) of mosfet gate charge , I pick voltage regulator and mosfet according to peak curent of mosfet gate charge .

author: mstfylcn333
author_rank: Newbie
timestamp: October 16, 2021, 11:36:22 am
content: Quote from: Picuino on October 15, 2021, 04:26:28 pmThank you Dear "Picuino" for gave me example but I would like ask t_switch that means duty on of periode.And is 0.1 ampere peak current of gate or avarege current?Becouse I saw on some app notes, avarage current of mosfet gate charge as you did.Best Regards.

author: Zero999
author_rank: Super Contributor
timestamp: October 16, 2021, 12:34:11 pm
content: Quote from: mstfylcn333 on October 16, 2021, 11:28:19 amSpacing fixed. Please allow the forum to wrap the text and avoid using unnecessary new lines. It means the reader can decide the line width and stops it from being mangled, when displayed on a small screen, or when the window is resized. If you find short lines easier to read, resize the window.The gate capacitance, listed on the data sheet, only applies when the MOSFET is off. As the MOSFET turns on and goes into the active region, the gate capacitance increases significantly, due to the Miller effect. When the gate voltage approaches the gate threshold, the MOSFET starts to turn on, causing the drain voltage to fall. The parasitic capacitance between the gate and drain, provides a negative feedback path, which tries to turn the MOSFET off. This means that when the MOSFET is in the active region, its gate capacitance is much higher, than the figure listed on the data sheet. You need the gate charge figure to estimate how much energy is used to turn the MOSFET on.E = QV2/2Where: E is energy in Joules and Q, charge, in Columbs.Power is the amount of energy delivered to a load, per second. To get power, multiply by the switching frequency.P = FQV2/2

author: Manul
author_rank: Super Contributor
timestamp: October 16, 2021, 02:27:48 pm
content: Quote from: mstfylcn333 on October 16, 2021, 11:29:27 amThere is some misunderstanding here. If you ask about peak current, then:1. You can control it with gate resistor (this also controls rise/fall times and ultimately switching speed). There are pitfalls - too high current may cause ringing, too low current may cause other problems (higher power dissipation, high dv/dt susceptibility). Sometimes you want asymetrical drive.2. Generally you don't need to worry about peak current while picking voltage regulator (mosfet driver supply I assume?), you just add some capacitance, most importantly - small mlcc capacitor which has low impedance at high frequencies to supply these short, high surges of current. Average current is usually quite small.Quote from: mstfylcn333 on October 16, 2021, 11:28:19 amFrom your question, it seems to me, that you have some misconceptions. There are many problems which you may face driving mosfets (ringing, dv/dt effects, etc), it is not that there is just some frequency at which problems starts. Of course, there will be some frequency limit based on rise/fall times and switching speed of particular mosfet, but it is not a proper way to look at it (usually). Most important thing to keep in mind is just that for any given switching speed, increase in frequency will increase power dissipation inside mosfet, because mosfet is spending more time in linear region.

author: Siwastaja
author_rank: Super Contributor
timestamp: October 16, 2021, 03:02:09 pm
content: Peak current is calculated using sum of all resistances in the path, typically gate driver equivalent resistance, explicit gate resistance, and MOSFET parasitic (internal) gate resistance.Adjusting this resistance value does NOT affect the current consumption (dynamic current, dynamic power, whatever you call it); same amount of charge is put in and out every cycle, just at different speed, but same number of times per second.Each switching cycle means Qgtot is consumed once from the supply (when turning MOSFET on, this is consumed from the source; when it's turned off, it's wasted by "shorting" the gate). Seeing Coulomb is one A*s, all you need to do is to divide by the period in s to get A.Finally, often gate current consumption is not interesting because it tends to be small compared to switching loss. But sometimes it does matter.

author: Picuino
author_rank: Super Contributor
timestamp: October 16, 2021, 03:05:46 pm
content: Quote from: mstfylcn333 on October 16, 2021, 11:36:22 amThis formula is used as a thumb rule to calculate the peak current that enters the gate during the transition from on to off or vice versa (only during the switch that has a duration of 100ns).It is not an average current (0.1 amps is too much for an average current of a mosfet gate)

author: Siwastaja
author_rank: Super Contributor
timestamp: October 16, 2021, 03:13:27 pm
content: I don't understand the usage of that formula to estimate peak current because peak current can be calculated directly from the resistances fairly accurately, but switching speed is unknown. You can do the opposite though, estimate the switching speed from the peak current using that formula. It won't be very accurate but good for first-order napkin approximation. For good results, simulate in Spice and build prototypes.

author: Picuino
author_rank: Super Contributor
timestamp: October 16, 2021, 03:13:54 pm
content: At 20kHz it has a 50us period, maybe 20us on and 30us off, but the switch-on or switch-off time should be much lower, 100 ns or less to avoid switching losses.Here is when the peak current occurs, during the switch-off and during switch-on of the mosfet.

author: Siwastaja
author_rank: Super Contributor
timestamp: October 16, 2021, 03:15:15 pm
content: Yes, and switching time lends itself to some rule-of-thumb simplifications, like "spend 0.5% of the time switching". Good for very rough napkin estimations.

author: Picuino
author_rank: Super Contributor
timestamp: October 16, 2021, 03:18:28 pm
content: Quote from: Siwastaja on October 16, 2021, 03:13:27 pmFirst I estimate the peak current needed for designed t_on or t_off switching times.Second, I estimate the resistance that meets this current.In the example abobe, if I_gate=0.1Amp, V_gate=20Volts, V_gp=6V,R = (20 - 6)/ 0.1 = 140 Ohms

author: Siwastaja
author_rank: Super Contributor
timestamp: October 16, 2021, 03:42:26 pm
content: Yes, that makes sense, it's a good initial estimate for Rg. Then adjust Rg by prototyping. Because EMI is very very hard to "calculate" or even simulate beforehand. And if EMI is no concern, Rg would be close to 0.Choosing "appropriate" switching times is a similar problem; how to do it? Answer is prototyping and EMC measurements.

author: mstfylcn333
author_rank: Newbie
timestamp: October 19, 2021, 08:29:09 am
content: I would like to thank everyone for your all advice suggestion and advice my dear friends.Best Regards