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I am building a buck converter with Vin = 12V to 36V and a duty cycle D=25% to 100%. I am using an Arduino to drive an N-channel MOSFET to perform low-side switching as here. In my case the buck has a resistive load. I'd like to switch at say 32.25Khz (i.e. setting the prescaler to 2). I selected the IRLZ44 MOSFET which has the following characteristics:

  • Rds(on) = 28 mOhm @ Vgs = 5 V
  • Total gate charge = 66 nC

Do I need to a MOSFET driver? I think the answer is no and here is my reasoning: Since the PWM output should remain less than 40 mA this means I want to charge the gate in no faster than 66 nC / 40 mA = 1.65 us.

Now if my PWM is 32.25kHz, the total cycle time is 40 us. In my application the smallest duty cycle = 25%, in which case the on time is 10 us, of which I'd spend 1.65 us charging.

Sounds reasonable to me... but this is my first time using a MOSFET and I see lots of articles insisting that a driver is necessary for high frequency switching. Do I need a driver? If so, can you recommend one? If no, do I need a resistor to limit the Arduino PWM pin to its rated 40mA?

saucerful
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    Hi! Not quite clear to me what you are trying to do... Are you building a buck converter (= DC-DC converter) or a simple switch for driving a resistive load? The two are not the same. About your actual question: to keep the losses in the MOSFET low, you want to turn it on or off *much faster* than the on-time. Whether or not 1.6 us are enough depends on how much switching loss you can tolerate. What's your current? – polwel Apr 18 '22 at 21:45
  • Thanks @polwel, maybe you can help me estimate the switching loss? Yes it is a buck converter (actually I am trying to do MPPT on a simple nichrome resistive heating element). The switched input will range from about 16 to 36 V and 4-9 A. – saucerful Apr 18 '22 at 22:05
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    Please _edit your question_ with a schematic of your proposed circuit. – TimWescott Apr 18 '22 at 22:08
  • Thanks @TimWescott I don't have a complete schematic yet, but I am going off of the one in this question: https://electronics.stackexchange.com/questions/330471/low-side-n-mosfet-buck-converter but with inductance and capacitance values from plugging in my specs to a calculator – saucerful Apr 18 '22 at 22:48
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    Let's try this again. Welcome to Stackexchange! Stackexchange likes nice, well-formed questions that are *complete*, and that *don't require looking through comments* to figure out the whole question. Please **edit your question** with that information. By preference, copy & paste the schematic, but even just _editing your question_ with just the link will make it complete. – TimWescott Apr 19 '22 at 02:10
  • Thanks, I added the link. I didn't drop in the schematic because the voltages, resistances, etc. are not what I will have. In the future I will get better at drawing schematics and include. – saucerful Apr 19 '22 at 13:54

1 Answers1

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If you are trying to PWM a resistor, the considerations might be a little different than a SMPS.

If you have a 36 V supply and 9 A load, then you have a 4 ohm load. Even without switching, this wil dissipate I^2*R = 2.7 W in your FET which will heat it up significantly.

In the transition, during worst case, you will dissipate V/2I/2 = 184.5 = 81 W in the FET. If you spent 2 us there every 30 kHz (~ 30 us) (rememberthere is turn-on and -off), you would add additional power of 2u/30u*81W = 5.4 W. This won't work well (these calculations are approximations, but you are not in a safe area).

You will need to keep FET power much less than a few W. Therefore you will need turn-on and -off times of < 100 ns, and so gate currents (66 nC in 66 ns == 1 A) of about 1 A. Therefore you will need a gate driver.

Note that if you are driging a long wire (more than a few cm), you may also have inductive transients if you turn off too fast. A 1 m wire will have abouy 1 uH of inductance, and turning off 9 A in 100 ns will generate a voltage spike of V=L.di/dt = 1u*9/0.1u = 90 V -- note this is not very well controlled and in fact hte FET may switch off faster than this. You will need a clamp circuit (that can clamp 9 A).

jp314
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    Thanks! @polwel's comment tipped me off to switching losses (I had only looked at Rds(on) loss before) and I estimated similarly unacceptable losses as yours. I started searching for gate driver ICs on Newark and came across MIC4420ZT which is rated for 6A with a 48ns output delay and conveniently is a through-hole package. Do you think that would work? – saucerful Apr 19 '22 at 00:56
  • Regarding inductance and clamping, the wire to load is probably less than 10cm but this is actually a buck circuit so (in addition to the resistive load) there is a ~100uH inductor. But there is also a capacitor and diode. Does that affect the need for protection? – saucerful Apr 19 '22 at 00:59
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    As long as you have good decoupling on the power supply inputs, you're OK. The gate driver seems suitable. 32 kHz for a DCDC is very low and will lead to a large inductor being needed. If your load is resistive, perhaps you don't need a full DCDC ? – jp314 Apr 19 '22 at 02:40
  • regarding wire inductance, are you referring to the wire to the wire from the supply to the buck or the buck to the load? In any case I plan to put capacitors across both – saucerful Apr 19 '22 at 14:20