| Electronics > Projects, Designs, and Technical Stuff |
| Is this MOSFET driver circuit ok? |
| << < (3/6) > >> |
| magic:
--- Quote from: DBecker on January 13, 2020, 01:42:33 am ---The clamp will allow the voltage to rise to 40V-70V at turn-off, and then turn on the MOSFET slightly to limit the voltage spike. --- End quote --- Any particular advantage over using, say, a 55V MOSFET and letting it avalanche? --- Quote from: Chriss on January 13, 2020, 06:55:24 am ---I didn't understand correctly what you mean by the clamping circuit vs the fly back diode. --- End quote --- It's about the need to apply reverse voltage to an inductive load to stop current flow through it. This voltage is helpfully produced by the load itself when the FET turns off (the - connection to the load starts to rise above 12V) and the more you allow it to rise above 12V, the faster will the load turn off. The 1N4001 diode permits only about 0.7V. |
| Chriss:
Ok. I understand now the clamping... If I just replace the D1 with a fast recovery one, like someone wrote before, does that would be enough to make the proper clamping circuit? |
| Zero999:
--- Quote from: floobydust on January 12, 2020, 11:44:17 pm ---Best to look at a logic-level N-ch mosfet that is OK with 5V drive. IRF1010 needs well over 5V drive. --- End quote --- I agree it's not ideal, but it's rated for 60A and the original poster is only using it for 2A, which it can happily do, even with only 4.5V of gate drive. https://www.infineon.com/dgdl/irf1010npbf.pdf?fileId=5546d462533600a4015355da754e188b --- Quote from: Chriss on January 13, 2020, 01:21:55 pm ---Ok. I understand now the clamping... If I just replace the D1 with a fast recovery one, like someone wrote before, does that would be enough to make the proper clamping circuit? --- End quote --- If you want to turn the solenoid off as quickly as possible, then any freewheeling diode will slow it down. When the transistor switches off, the current keeps flowing through the inductor and diode. The voltage drop across the diode will be just over a volt or so, at 2A, so the current will decay very slowly. A zener diode in reverse parallel with the MOSFET will dissipate the energy more quickly than a free-wheeling diode. You need to choose a zener diode with a slightly lower breakdown voltage, than the MOSFET. |
| Chriss:
Ok. What you mean by " A zener diode in reverse parallel with the MOSFET" the zen should be connected between S&D of the mosfet? Assume not. Or? Should I leav the D1 and add the zener in paralell with the mosfet S&D? Thanks. |
| Zero999:
The higher the voltage across the inductor, the quicker the current will decay. An ordinary diode has a voltage drop of a volt, at high currents, compared to its rating, so the current will decay slowly. When the MOSFET turns off, the voltage on the negative side of the inductor is just a bit higher than the supply voltage. A zener diode clamps the voltage across the MOSFET to around 55V in this case, so the current falls more rapidly. The voltage across the inductor will be the zener voltage (the 1N5369B is a 51V, but the voltage is higher, at higher currents) minus the supply voltage. |
| Navigation |
| Message Index |
| Next page |
| Previous page |