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Can anyone give me a brief run-down of some methods I could use for this?

enter image description here

I want to drive a MOSFET or BJT with a 12v 555 circuit or oscillator. The MOSFET will drive a high voltage circuit that can have both large positives and large negatives occuring on both sides of the transistor. As a result, a MOSFET will require a floating voltage divider to ensure the gate isn't destroyed. But that won't ensure that my control circuitry is protected.

I'm pretty sure that I really need to completely electrically isolate the control circuitry if possible. Can anyone throw out some common techniques so I have an idea of ways to go about it? I need this to operate between 50kHz and 800kHz.

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  • \$\begingroup\$ I'd use a solid state relay. There is a good chance that a MOSFET won't survive due to the reverse body diode inside the MOSFET-- but I can't say for certain given the lack of information in the question. \$\endgroup\$ – user3624 May 30 '13 at 3:50
  • \$\begingroup\$ David, I think I read this somewhere before but forgot about it. You're saying there are limitations in how much reverse current the body diode can conduct. Is that right? \$\endgroup\$ – JamesHoux May 30 '13 at 4:06
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    \$\begingroup\$ The reverse body diode is a crappy diode. Slow, inefficient, large Vf, etc. If the properties of the diode is even remotely important then put a real diode in parallel with it. \$\endgroup\$ – user3624 May 30 '13 at 4:09
  • \$\begingroup\$ David, I'm looking at the datasheet for my mosfet and it shows the same reverse body diode current as the regular forward drain current. st.com/st-web-ui/static/active/en/resource/technical/document/… I'll definitely consider the real diode in parallel. I suppose if I use a real diode with a lower voltage bias then it should kick in first and carry most of the current. \$\endgroup\$ – JamesHoux May 30 '13 at 4:11
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    \$\begingroup\$ Look closely. Forward on voltage is a terrible 1.5v, and the reverse recovery time is 85 ns. Depending on your app, the reverse recovery current might also be an issue for you. While the source-drain-current is 180A, if you actually had that much current flowing then you would be dissipating 270 watts! \$\endgroup\$ – user3624 May 30 '13 at 4:27
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That's exactly the application optocouplers are made for.

Optocouplers consist of a LED on one side and a photodiode or a phototransistor on the other side. Both are put into one package but are kept well insulated from each other.

The electrical signal is transformed into light by the LED and back into an electrical signal by the photodiode/phototransistor.

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