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I would like to ask what is causing the following phenomenon. I've spotted this on my real buck converter at a low duty cycle first, then tried in my SPICE simulation as well which confirmed this.

Let's have a following simple buck converter: Buck converter simulation schematic

I've set V2 to 50kHz and 1% duty cycle. The FET on time should be 200ns at this point. But the transient analysis shows this:

Transient analysis

Voltage pulse at V2 really takes 200 ns but voltage Q1-D roughly 600ns! Interesting is that the voltage Q1-D rises almost instantly at turn-on but takes a long delay at turn-off.

Datasheet for given FET says:

  • tdon 22ns
  • tr 99ns
  • tdoff 110ns
  • tf 92ns

These are for slightly different test conditions but that should not be the matter here. At least not the reason for such a long turn-off time. For these times I would expect the Q1-D pulse to last ca 300ns instead of 600ns.

Is there some additional restriction about switching MOSFETs on for a very short time?

Note: This simulation has been done in Simplis Elements 8.1. In real application I use a different MOSFET even with shorter datasheet times but the off-delay is even longer.

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  • \$\begingroup\$ You confused me for a moment. Your node labelled Q1-D is actually Q1's source (IRFP264 is NMOS). \$\endgroup\$
    – Tom Carpenter
    Commented Apr 24, 2018 at 23:29
  • \$\begingroup\$ Sorry,my mistake,should be Q1-S. \$\endgroup\$
    – tk_
    Commented Apr 24, 2018 at 23:39
  • \$\begingroup\$ If the pulse generator has a 50 to 75 ohm output, the fall time should be quick. If it is 600 ohm, that is much too high. You may need a PNP to drive the base low at a fast rate. Remember MOSFET's need substantial drive and sink current. Is R9 actually 1.0 ohm? \$\endgroup\$
    – user105652
    Commented Apr 24, 2018 at 23:47
  • \$\begingroup\$ You cannot set internal impedance for V1/V2. They are ideal voltage sources with Ri=0. R9 is 1 Ohm,when I set 10 Ohm virtually nothing changes. I would put the simulation file (*.wxsch) as attachment here but it seems this forum does not support this,only hyperlinks or images.It runs in Elements demo which is free. \$\endgroup\$
    – tk_
    Commented Apr 24, 2018 at 23:54

1 Answer 1

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As a check, remove C1, D1, and L1, and connect Q1.S directly to R2 so that Q1 drives a purely resistive load (R2). Now run your simulation and probe Q1's gate current.

You'll notice when voltage source V2 shuts off (transitions from +15V to 0V), the accumulated charge on Q1's gate discharges as current through R9, and Q1 remains ON until the gate charge level exits the Miller plateau, at which point the transistor goes into cutoff.

And of course, increasing R9's value slows down the discharge of Q1's gate, which causes Q1 to stay on longer after V2 transitions from +15V to 0V.

So what's the solution? To expedite the discharge of Q1's gate and thereby turn off Q1 sooner, try this. Replace voltage source V2 with a piece-wise linear voltage source (VPWL) and configure its time versus voltage values as follows:

time    voltage
 0s        0V
 50us      0V
 +5ns      15V
 50.2us    15V
 +10ns     -15V
 +40ns     -15V
 +5ns      0V

At t=50us, V2's voltage transitions from 0V to +15V, which turns on Q1. At t=50.2us, V2's voltage transitions from +15V to -15V, and V2's voltage remains at -15V for 40 ns, after which time V2's voltage rises from -15V back to zero volts. This -15V 40ns pulse expedites the discharge of Q1's gate (reduces the Miller plateau transition time), and Q1 shuts off much sooner.

HINT: Research MOSFET & IGBT gate driver ICs.

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  • \$\begingroup\$ I've checked this, thank you. I see now,never inspected the gate current before. When I replaced the V2 the pulse had ca 550ns instead of >600ns, when I removed C1,D1 and L1 the pulse has now about 400ns. Does that mean that even the inductor L1 prolongs the gate discharging? Btw. I use isolated mosfet driver ADuM4136 (driver must provide a DESAT function) and have no option for negative voltage in my real gate driving circuit. So no other options for shorting this unless going with Rg to zero and/or slowing down the base frequency? \$\endgroup\$
    – tk_
    Commented Apr 25, 2018 at 8:09
  • \$\begingroup\$ check the difference also when you decrease the value of R9 (which is already very small in your case) to see the discharging times. Also instread of 0V, setting the switch off voltage to -15V will increase discharging current and hence your switch will turn off faster ! \$\endgroup\$
    – HerrderElektronik
    Commented Apr 25, 2018 at 13:01
  • \$\begingroup\$ Actually, according to simulation the best option seems to be to set +15V for entire on time and -15V for entire off time for V2 and use R9=10 Ohm. This increases a switch on delay and decreases switch off delay substantially. I need the pulse at Q1.S to be as much similar to the pulse at Q1.G as possible, some shift does not matter. Now I need to figure out which DC/DC with dual voltage output to use, but it seems I'll use MEJ1D0515SC. The ADuM4136 I've used so far actually supports bipolar operation. \$\endgroup\$
    – tk_
    Commented Apr 25, 2018 at 13:23
  • \$\begingroup\$ The negative voltage pulse from my example is not a strict requirement. The important take-away for reducing the MOSFET's shutdown delay is this: to minimize the MOSFET's turn ON and turn OFF delays, you need an external circuit that rapidly drives charge onto, and rapidly removes charge from, the MOSFET's gate--e.g., a MOSFET or IGBT gate driver IC--which you are already using. And yes, inductor L1 will affect MOSFET Q1's turn ON/OFF times. \$\endgroup\$
    – Jim Fischer
    Commented Apr 25, 2018 at 14:53

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