I want to design buck converter for 385VDC to 48VDC for that I calculated the values of component and simulated it with proteus but I am not getting the desired output. What is wrong? PWM has a 20v amplitude as given in IRF740 data sheet. And 20kHz frequency and 12% duty cycle.enter image description here

  • \$\begingroup\$ For an N channel device, the gate must rise above Vout by a large enough voltage to fully enhance the device. A boost circuit is required to get what you want. \$\endgroup\$ – Peter Smith Mar 24 '19 at 13:30
  • \$\begingroup\$ I did not get it. Means I need to boost gate pwm pulses amplitude? I am getting 385vdc already from a boost converter. \$\endgroup\$ – Deep Mar 24 '19 at 13:44
  • \$\begingroup\$ You have your N-channel MOSFET configured as a "source follower". This means that the source terminal cannot rise higher than \$V_{gate} - V_{threshold}\$. Threshold voltage for the IRF740 is in the range of 2 to 4 V. If you want to turn it on fully, bringing the source terminal to +385 V, you'll need to drive the gate with a signal that rises to +390 V. This can be done with a "charge pump". \$\endgroup\$ – Dave Tweed Mar 24 '19 at 13:49
  • \$\begingroup\$ Can you please suggest changes in circuit. \$\endgroup\$ – Deep Mar 24 '19 at 13:53
  • \$\begingroup\$ The gate needs to be driven to Vout + the fully on Vgs at a minimum, so about 53V in this case. Note that this will have significant heating effects on the FET. \$\endgroup\$ – Peter Smith Mar 24 '19 at 14:05

To turn on Q1 you need to drive the gate about 10 volts positive with respect to the source. Given that you expect the source to rise to around 385 volts, this means that gate has to rise to 395 volts to adequately and efficiently turn-on the MOSFET.

I’d strongly consider using a proper boot strapped driver circuit to do the job. Additionally, you shouldn’t drive it more than +15 to +18 volts above the source because the absolute max limit is stated as 20 volts in the IRF740 data sheet.

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  • \$\begingroup\$ The source appears to be connected to the load (as with an ordinary non synchronous buck) so I would think the gate needs to rise to Vout + perhaps 10V or am I missing something? \$\endgroup\$ – Peter Smith Mar 24 '19 at 14:49
  • \$\begingroup\$ @PeterSmith I believe that is what I said.... \$\endgroup\$ – Andy aka Mar 24 '19 at 14:52
  • \$\begingroup\$ Yes load is connected to source. \$\endgroup\$ – Deep Mar 24 '19 at 15:00
  • \$\begingroup\$ @Andyaka; the OP wanted a Vout of 48V \$\endgroup\$ – Peter Smith Mar 24 '19 at 15:20
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    \$\begingroup\$ @PeterSmith yes I know, but with a buck regulator, to efficiently drive it you need the MOSFET to turn on hard and rely on the inductor and capacitor (and importantly the control loop) for averaging the PWM waveform (peaking at 385 volt) to be a DC level of 48 volts. That's how switching regulators work. Linear ones on the other hand have to control the MOSFET linearly and hence the output at the source would need to be 48 volts. The question title refers to buck and the circuit is a buck regulator. \$\endgroup\$ – Andy aka Mar 24 '19 at 19:19

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