EMF supression diode - Electronic Circuits Projects Diagrams Free

billion_boi · 26th August 2009, 05:30 AM

Hello all,
From what i know, a motor(inductor) will generate a back emf when its current supply is cutoff, i believe the back EMF grows with time since V = L*(di/dt).
So if i were to place a diode in parallel with a motor (inductor), i know that ideally it should have a fast reverse recover time so that it can clamp down on the voltage quickly befor the EMF grows too large...

But what happens if the EMF grows larger than the .7 forward voltage drop of a PN diode...does the current occilitate until all the voltage is depleted?

dknguyen · 26th August 2009, 10:49 AM

THe inductor is trying to get rid of it's stored energy in any way it can- current is preferable, but voltage if needed.

The BEMF voltage spike becomes whatever is required so to force current to flow. In this case, that is when the diode turns on at 0.7V. WHen the diode does turn on, the voltage stops rising becase current starts flowing. It should be noted that the voltage spike does not "build up" to a point until current starts to flow. THe inductor must start releasing energy right away so the voltage spike isntantly jumps to the required voltage so current will flow since this is a diode which can only be on or off. With a parallel resistor any voltage means current flow so the voltage change gradually, but in a diode it snaps to the required voltage.

Basically the diode "clamps" the BEMF to 0.7V and the current flows to dissipate the energy as P = 0.7V*Current. So no, it does not oscillate.

billion_boi · 26th August 2009, 09:18 PM

dknguyen thanks for clearing that up. But if you or anyone else don't mind
i have just two smaller questions, based on this quote from wikipedia

Quote:

Schottky diodes are preferred in flyback diode applications, because they have the lowest forward drop (0.2V rather than 0.7V for low currents) and the fastest (but still "soft") reverse voltage recovery

Why does the forward voltage drop and reverse recovery time matter?
A .7V drop as opposed to .2v is still not enough to damage the transister.Also
why does recovery time matter since the the EMF cant grow larger than the forward drop of the diode anyways?

dknguyen · 26th August 2009, 10:53 PM

Reverse recovery time matters because it's like the time it takes for the diode to switch from blocking to conducting. Think about it, when the inductor is being powered normally, the diode must be blocking. But the instant power is cut and the inductor starts trying to dump energy, you want the diode to begin conducting as fast as possible. THe longer it takes to conduct, the higher the voltage spike will be and the more likely something is going to fry (remember, the voltage will instantly spike to a level to force current to flow, aka where something has got to give). Until the diode does turn on and allows low voltage, high current energy from the inductor to dump, a high voltage, low current energy is flowing through something else. Who knows what it might be? But it's probably not something that's supposed to handle it.
------
Voltage drop is important because it helps define what level the BEMF is clamped to. In your scenario with a diode in parallel with the inductor, once the diode conducts the BEMF Is clamped to the forward voltage drop. So there's not too much of a difference between the clamping voltage 0.7V of a regular diode and 0.3V schottky.

For an H-bridge though, it matters a bit more. Obviously you can't put a diode in parallel with a motor in an h-bridge because current can flow both ways. So you either have a short in one direction if you try and clamp the BEMF when cutting power in one direction only, or a short in both directions if you try and clamp the BEMF when cutting power for both directions. So instead diodes are placed in parallel with the switches/transistors. This provides the same flyback current path as before (from one the terminal of the motor to the other), but now it needs to take the path through two diodes instead of just one and the result is you don't have a shorts that would affect normal operation. There is a catch though...instead of clamping the BEMF to the forward voltage like your scenario, it now clamps it to Vsource+Vforward. It is only capable of clamping the BEMF to a voltage ABOVE the source/battery voltage. Do you see how a high forward voltage drop coudl be a problem in that case?

RCinFLA · 26th August 2009, 11:28 PM

Schottky diodes are faster and lower voltage drop but the down side is maximum reverse breakdown voltage is usually between 35-50 vdc.

For a motor or relay snubber the voltage drop is not a concern. The diode is to prevent a a high negative surge voltage that may destroy the power switching transistor. The speed of diode recovery (turn-off) is also not an issue in this application.

billion_boi · 27th August 2009, 04:03 AM

dknguyen, thanks you very much. Your second reply got the conept through to me.These quotes from your reply, really helped me understand..

Quote:

THe inductor is trying to get rid of it's stored energy in any way it can- current is preferable, but voltage if needed.

Quote:

Until the diode does turn on and allows low voltage, high current energy from the inductor to dump, a high voltage, low current energy is flowing through something else.

Thanks again! Im not to worried about designin an Hbridge as im just starting out. my first project will be powering a motor with a simple switch.

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26th August 2009, 05:30 AM	#1
billion_boi	EMF supression diode Hello all, From what i know, a motor(inductor) will generate a back emf when its current supply is cutoff, i believe the back EMF grows with time since V = L*(di/dt). So if i were to place a diode in parallel with a motor (inductor), i know that ideally it should have a fast reverse recover time so that it can clamp down on the voltage quickly befor the EMF grows too large... But what happens if the EMF grows larger than the .7 forward voltage drop of a PN diode...does the current occilitate until all the voltage is depleted?

26th August 2009, 10:49 AM	#2
dknguyen	THe inductor is trying to get rid of it's stored energy in any way it can- current is preferable, but voltage if needed. The BEMF voltage spike becomes whatever is required so to force current to flow. In this case, that is when the diode turns on at 0.7V. WHen the diode does turn on, the voltage stops rising becase current starts flowing. It should be noted that the voltage spike does not "build up" to a point until current starts to flow. THe inductor must start releasing energy right away so the voltage spike isntantly jumps to the required voltage so current will flow since this is a diode which can only be on or off. With a parallel resistor any voltage means current flow so the voltage change gradually, but in a diode it snaps to the required voltage. Basically the diode "clamps" the BEMF to 0.7V and the current flows to dissipate the energy as P = 0.7VCurrent. So no, it does not oscillate. __________________ Tanaka Sensei (avatar) says: Please spell it "ridiculous" correctly! Not "rediculous". ^^ Last edited by dknguyen; 26th August 2009 at 11:01 AM.*

26th August 2009, 10:53 PM	#4
dknguyen	Reverse recovery time matters because it's like the time it takes for the diode to switch from blocking to conducting. Think about it, when the inductor is being powered normally, the diode must be blocking. But the instant power is cut and the inductor starts trying to dump energy, you want the diode to begin conducting as fast as possible. THe longer it takes to conduct, the higher the voltage spike will be and the more likely something is going to fry (remember, the voltage will instantly spike to a level to force current to flow, aka where something has got to give). Until the diode does turn on and allows low voltage, high current energy from the inductor to dump, a high voltage, low current energy is flowing through something else. Who knows what it might be? But it's probably not something that's supposed to handle it. ------ Voltage drop is important because it helps define what level the BEMF is clamped to. In your scenario with a diode in parallel with the inductor, once the diode conducts the BEMF Is clamped to the forward voltage drop. So there's not too much of a difference between the clamping voltage 0.7V of a regular diode and 0.3V schottky. For an H-bridge though, it matters a bit more. Obviously you can't put a diode in parallel with a motor in an h-bridge because current can flow both ways. So you either have a short in one direction if you try and clamp the BEMF when cutting power in one direction only, or a short in both directions if you try and clamp the BEMF when cutting power for both directions. So instead diodes are placed in parallel with the switches/transistors. This provides the same flyback current path as before (from one the terminal of the motor to the other), but now it needs to take the path through two diodes instead of just one and the result is you don't have a shorts that would affect normal operation. There is a catch though...instead of clamping the BEMF to the forward voltage like your scenario, it now clamps it to Vsource+Vforward. It is only capable of clamping the BEMF to a voltage ABOVE the source/battery voltage. Do you see how a high forward voltage drop coudl be a problem in that case? __________________ Tanaka Sensei (avatar) says: Please spell it "ridiculous" correctly! Not "rediculous". ^^ Last edited by dknguyen; 26th August 2009 at 11:11 PM.

26th August 2009, 11:28 PM	#5
RCinFLA	Schottky diodes are faster and lower voltage drop but the down side is maximum reverse breakdown voltage is usually between 35-50 vdc. For a motor or relay snubber the voltage drop is not a concern. The diode is to prevent a a high negative surge voltage that may destroy the power switching transistor. The speed of diode recovery (turn-off) is also not an issue in this application. Last edited by RCinFLA; 26th August 2009 at 11:35 PM.