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If I apply an current source into the the cathode of a diode (current flow into cathode), what can possibly happen to the diode.

For the cuicuit below, when the FET is OFF, how does the battery current flow. Will the current flow through the diode while it's in reverse region. enter image description here

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  • \$\begingroup\$ It does not flow through a perfect diode in reverse polarity. An inperfect diode has some very high reverse-polarity resistance and capacitance, so some negligibly low current will still flow. \$\endgroup\$ – sx107 Oct 13 '18 at 1:20
  • \$\begingroup\$ @sx107 Maybe It's not considered as perfect diode, there is a recovery time for diode called Trr. In this time, there is any change a current flow through the diode in reverse direction ? \$\endgroup\$ – Electronics newbie Oct 13 '18 at 1:31
  • \$\begingroup\$ If you apply a reverse current source to a diode, your diode stops being a diode and starts being a smoking piece of silicon. \$\endgroup\$ – Hearth Oct 13 '18 at 1:44
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If I apply an current source into the the cathode of a diode (current flow into cathode), what can possibly happen to the diode.

Which diode?

  1. The desaturation check diode
  2. The schotkey diode at the power supply
  3. The diode setting the the turn-off characteristic

I am going to assume you means #1 as the other's are not connected to anything and this appears to be part of either a phase leg or some pulled down load. In future please be concise...

Depends. Is this a theoretical case such that "ideal current sources" exist then the voltage associated with such an ideal current source could increase to infinite voltage to ensure the constant current would flow. This would cause a real diode to avalanche and be damage. But if we are contemplating ideal components then the diode could be ideal and thus could block infinite voltage and equally the FET could block infinite voltage.

If you are discussing a hypothetical load at the Drain of the FET then as long as the blocking voltage of the diode (and the FET) are higher than the forcing voltage behind whatever can source current... nothing will happen, at this part of the circuit. there will be some leakage current flowing, the value of which depends on the specific device.

For the cuicuit below, when the FET is OFF, how does the battery current flow. Will the current flow through the diode while it's in reverse region.

What battery? there is no battery.

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  • \$\begingroup\$ I'm not sure about the diode type, may be #1. For the circuit, the battery is connect to the Drain of the FET (also the Kathode of diode). This circuit is used to drive FET in motor, when the current to the FET exceeds the defined current limit, driver will turn OFF the FET. When the FET is OFF, I just don't know how the current will flow since the only path is through the diode in reverse direction. \$\endgroup\$ – Electronics newbie Oct 13 '18 at 1:27
  • \$\begingroup\$ the current won't flow. think of it like a light switch. When you turn off the switch does teh light stay on? The FET is like the switch. There is potential for current to flow just nowhere for it to flow. That Desaturation diode is there as part of the driver. it facilitate checking the FET is saturated when it is ON. if something goes wrong and the device starts to desaturate then Vds will rise and the driver will act. When the FET is off this diode is reverse bias and blocks current flow (bar leakage) \$\endgroup\$ – JonRB Oct 13 '18 at 1:34
  • \$\begingroup\$ When diode changes from forward bias to reverse bias there is small amount of time called "reverse recovery", during this time there is current flow through the diode in reverse direction before it completely stop flowing. I'm wondering that is ther current in this "reverse recovery" time can cause any damage to the IC or not (this current may be flow from the diode to 100ohm res to 5kohm res and flow into VCC2 pin or any other paths which I'm not sure). \$\endgroup\$ – Electronics newbie Oct 13 '18 at 1:46
  • \$\begingroup\$ The reverse recovery current is something different to what you were querying (some load). Yes the reverse recovery current, the charge stored in the PN has the capability to damage and this is something the designer has to design for. In this instance there is a 100R and a capacitor \$\endgroup\$ – JonRB Oct 13 '18 at 8:44
  • \$\begingroup\$ this recovery current is the internal characteristics of the diode or depending other external elements (such as reverse bias voltage), and is there any way to calculate this current (maybe I want to calculate the peak of the reverse current) ? \$\endgroup\$ – Electronics newbie Oct 13 '18 at 13:04

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