# Flyback diode question

1. Dec 6, 2011

### clecol

In a simple circuit consisting of nothing more than a voltage source and inductor (with flyback diode in parallel) and a normally closed switch...i understand that when the switch opens the magnetic field collapses and the inductor starts to build up voltage in opposition to the change in current, driving current through the flyback diode. however, does the current traveling through the diode flow through to the power supply or loop back through the inductor? I'm guessing tries to go through the power supply to ground. but doesn't this damage your power supply? this seems like such a common practice for protecting against your inductive switching device from back-emf (in this case a switch), but nobody ever seems to talk about where that current goes?

2. Dec 6, 2011

### Staff: Mentor

Welcome to the PF.

It goes into the decoupling capacitors on the power supply rail. If you don't have enough capacitance to absorb the current transient, then yes, it could overvoltage the power rail.

On a related note -- In the flyback transformer circuit on a CRT, there is an explicit flyback capacitor that helps to establish the boosted voltage, based on the peak current and how much charge that will deposit on that capacitance during the flyback part of the cycle.

3. Dec 6, 2011

### yungman

If you put the flyback diode across the inductor, current do not go through the supply, only from one end of the inductor through the diode back to the other end of the inductor!!!

4. Dec 6, 2011

### Staff: Mentor

The current is returned from the lowside of the inductor back to the power supply storage capacitance:

http://en.wikipedia.org/wiki/Flyback_diode

IF the diode is not there, the voltage at the lowside of the inductor snaps quite high, and can damage the lowside switching element (like an NPN transistor for example, if it is used for the switch). With the diode there, the inductor looks like a current source as the field collapses, returning the current to the power rail. At least that's how I've always viewed it...

5. Dec 6, 2011

### yungman

There is no schematic, but when I design the diode, it would be across the inductor. There is no reason to involve any other components.

If you have an inductor that connect to +V and you pull current by pulling low. YOu put the anode to the top of the inductor that connect to the +V. The cathode connect to the other side of the inductor that is being pull low. When you disconnect the low side pull, the bottom of the inductor will fly up, the diode will turn on and return the current to the top side of the inductor. No other component is involve. That's how everyone put the flyback diode for protection.

6. Dec 6, 2011

### Staff: Mentor

I agree that it can work in isolation. And even with my current source analogy, the head and tail of the current source are tied together through the diode, so the current could just circulate. And by continuity of current in a loop, I guess no current does flow into the power supply decoupling capacitors. Hmm.

My confusion probably comes from having worked with flyback and boost converters. There, the diode connection is to a separate capacitor, so the current does flow into that capacitor and charge it up.

Thanks for the clarification/correction, yungman!

7. Dec 7, 2011

8. Dec 7, 2011

### technician

I agree with yungman. The diode allows current to continue flowing through the coil and as a result the magnetic field does not collapse instantaneously and therefore a large emf is NOT induced. This current that flows is an induced current and does not come from the power supply (the switch is open)
These diodes are to be seen connected across relay coils in transistor circuits used to operate relays.