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This is a follow up from Driving IRF540N with darlington

I came up with a different circuit that uses jfet op amps and smaller fets instead of gate pull up resistor:

enter image description here

The basic idea is that the opamps create a normal and an inverted 12V pwm signal. They open and close the 2N7000 fets. Those fets charge or discharge the gate of the power mosfet. Because the smaller fets are never opened at the same time, almost no current is needed to keep the power mosfet opened or closed. Much less power is discipated. And because of their very low open resistance, the power mosfet's stays less time in half opened state, allowing higher pwm frequency and increasing efficiency.

The jfet opamps are fast enough to be used at higher frequencies (slew 13V/us). I'm not sure how fast can they fully open or cose the smaller fets. Max output current of the opamps is limited but I do not see it in the datasheet. (At least the gate of the 2N7000 has far less capacitance than the IRF540N.)

Would this work better than the previous design? Would it work at all?

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  • \$\begingroup\$ I suspect that the lower 2N7000 will be closed after the gate of the irf is discharged. Maybe if I use -12V on the opamps and on the drain of the lower 2N7000? \$\endgroup\$ – nagylzs Nov 19 '16 at 17:06
  • \$\begingroup\$ most likely your 2N7000's will have shoot-thru failures with 1A for x us. How can you expect to design anything without a load spec? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Nov 19 '16 at 17:29
  • \$\begingroup\$ It was not a design only an idea. Sorry \$\endgroup\$ – nagylzs Nov 19 '16 at 18:43
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    \$\begingroup\$ I cannot stress enough, the critical importance of writing down all input output specs for; V, I , Z load before doing any design!! Then add timing and functions. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Nov 19 '16 at 18:59
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Or save yourself all that messing around and just use one of these: -

enter image description here

They cost 44 pence from mouser here and they are in a SOT23 package, Vcc can be as high as 40 volts,

Using a TL074 to drive a MOSFET is going to be problematic because with its negative supply pin at 0V the output won't swing down below about 2V and you are likely to be never able to switch the lower MOSFET off. No, I'd just go for an integrated MOSFET driver and save time, money and space.

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  • \$\begingroup\$ yes out swing is a problem 1.5V offset typ for >10k in this case >1M and Vgs is 2.5typ but 0.8 worst case \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Nov 19 '16 at 17:42
  • \$\begingroup\$ Pretty tasty, who says we shouldn't recommend speciifc components! \$\endgroup\$ – Neil_UK Nov 19 '16 at 18:05
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Look at the shoot-through charge in datasheets for MOSFET gate drivers.

Some ICs have break-before-make implemented on the silicon.

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