I have two diodes one has an voltage drop of 0.2v while other has a drop of 0.55v in the same circuit. While I use the 0.55v drop diode for freewheeling (35 Amps diode
) and the other diode remains in the same circuit. Is it possible for the freewheeling current to flow through diode with lower voltage drop (in5408)
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\$\begingroup\$ maaaaybe for a very brief moment, during the reverse recovery time of the 5408, if the switch-off, as seen from the coil, was fast enough. \$\endgroup\$– Pete WDec 23, 2020 at 16:22
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\$\begingroup\$ Thanks Pete, what if I use a diode with voltage drop more than the 35A diode? Say 0.6V \$\endgroup\$– GowthamanDec 24, 2020 at 4:02
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\$\begingroup\$ Not sure what you mean. Different issue from the reverse recovery. If mean you want the coil to de-energize faster, a larger voltage drop will indeed make faster dI/dt. The way to do it is to add an additional element in series with the antiparallel diode, often a zener. However! the peak power in this element will be proportionally higher with whatever voltage drop you set it up with -- remember, the discharge starts with all 30A going through the antiparallel path. So size it accordingly. Also check the thermal limit lines, as faster discharge means more thermal loading on the 35A too. \$\endgroup\$– Pete WDec 24, 2020 at 14:54
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\$\begingroup\$ Also 0.6V is diode drop at ~1mA. More at 30A. \$\endgroup\$– Pete WDec 24, 2020 at 14:59
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No it wont. During freewheeling the inductor will act as a current sink to -24V. This means the top of the inductor will be a little lower than -24V. The IN5408 should be reverse biased.
