Using a diode to protect a relay

Armin_Tamzarian

I understand the reason for putting a diode across the -/+ contacts of a relay to protect it.
From what I know when the voltage is removed a back EMF is generated which turns into current which tries to flow back across the coil.
Please correct me if I'm wrong about any of this.

I'm confused about how exactly a diode achieves this.
My knowledge of them is quite low.
I understand that diodes allow current flow in one direction only.
I also understand that they have a forward voltage drop property.

Looking at the way the diode is connected across the relay I don't see how cany current could flow through it from the positive side to the negative side.
From my limited knowledge of diodes it just looks like the diode would prevent current flow in that direction.

Also, why doesn't current flow back through the coil the same as if the diode wasn't there?

Any help, explanations would be gratefully received.
-Thanks

DublinDilbert

The diode is normally reversed biased when the relay coil is energised (on) and only conducts for a very short period after the relay coil has been turned off. It is used to dissipate the energy stored in the coil.

To put in another way, the coil is an inductor. An inductor is a device which will resist the current in it changing. When the relay coil is energised, the coil will have a steady DC current flowing in it. When you turn the coil off, the inductance will want to try keep the same current flowing in it. So at the instant of turn off the current in the coil does not change (or go to zero), what happens is the voltage across the coil reverses (from a volt drop to a voltage rise) and will try drive the same current that was flowing in the inductor prior to switch off. This current then free-wheels through the reverse biased diode. As the current is free wheeling it starts to decay, as power is dissipated in the resistive elements of the circuit.


Typical the reverse biased diode on the coil is not present to protect the coil as such, but usually to protect the device driving the coil...

Armin_Tamzarian

DublinDilbert wrote: »

The diode is normally reversed biased when the relay coil is energised (on).

From my understanding the cathode is on the + side and the anode is connected to the - side of the relay, correct?
I don't really follow what you mean by the above sentence.

DublinDilbert wrote: »

This current then free-wheels through the reverse biased diode. As the current is free wheeling it starts to decay, as power is dissipated in the resistive elements of the circuit.

What do you mean by 'free wheel'?
Why does the current flow through the diode rather than back through the coil?
Presumably there are two parallel paths for the current to flow through.
Through the coil and through the diode.
AFAIK with parallel paths that contain resistors, IxR is the same on all paths.
How does current get split across parallel paths when a diode is involved.

Also, would the reverse current coming back off the relay not have to flow through the diodes cathode to the anode, i.e. the way that the diode is supposed to prevent it from flowing.

Thanks by the way, for taking the time to reply to me.


Typical the reverse biased diode on the coil is not present to protect the coil as such, but usually to protect the device driving the coil...[/quote]

DublinDilbert

Armin_Tamzarian wrote: »

From my understanding the cathode is on the + side and the anode is connected to the - side of the relay, correct?
I don't really follow what you mean by the above sentence.

I've corrected your sentence to what i think you meant to say. I've attached a diagram, you need to look at this at 3 points in time:-

1. Switch Closed,diode reverse biased, current flowing in relay coil
2. Switch Opens, coil polarity reverses, causing diode to become forward biased and conduct
3. Switch Open, current in coil free wheels to zero.

Armin_Tamzarian wrote: »

What do you mean by 'free wheel'?

Its a term used to describe this type of diode, the current "free wheels" in it when the coil is turned off. As the current free wheels it will decay to zero.

Armin_Tamzarian wrote: »

Why does the current flow through the diode rather than back through the coil?

I'm not sure what your asking.

The current will want to flow in a loop, there's 2 different loops, depending on the state of the switch:-

• When the switch is closed the loop in which current flows contains the battery + relay coil.
• When the switch opens the loop which the current flows contains is the coil and the diode.

Armin_Tamzarian

DublinDilbert wrote: »

I've corrected your sentence to what i think you meant to say.

Yes, you're correct, I'd meant to type anode, I've edited my post, cheers.

I'm getting close to understanding now, thanks to you.
I was overlooking the fact that the +12v is removed from the circuit at the point when the back EMF is created.

Ok, so if you can bare with me.
The current caused by the back EMF runs into the diode's anode.
As the current can pass through the diode it should then run back into the coil and so on in a loop.
Is it the diodes voltage drop property that is achieving the 'free wheeling' effect?

Why does the current not just run straight back to the negative terminal of the battery?

DublinDilbert

Armin_Tamzarian wrote: »

Yes, you're correct, I'd meant to type anode, I've edited my post, cheers.

I'm getting close to understanding now, thanks to you.
I was overlooking the fact that the +12v is removed from the circuit at the point when the back EMF is created.

Ok, so if you can bare with me.
The current caused by the back EMF runs into the diode's anode.
As the current can pass through the diode it should then run back into the coil and so on in a loop.
Is it the diodes voltage drop property that is achieving the 'free wheeling' effect?

Well the volt drop property of the diode, will clamp the coil voltage @ 0.6V during free-wheeling....

Think of the diode like a one-way valve. It allows current flow in one direction. During free-wheeling, it is like a valve opening allowing the current to pass, in one direction only.

Armin_Tamzarian wrote: »

Why does the current not just run straight back to the negative terminal of the battery?

Good Question!!!! Current always wants to flow in a loop, as the switch is open there is no closed loop containing the battery any more.

Aside 1:- not wanting to confuse things, but if the free-wheeling diode is not present in the circuit i've presented above, the energy from the coil will go back to the battery. As the switch opens an arc will form in the switch. The arc will continue till the current in the coil falls to zero.

Aside 2:-If the coil is large this Arcing can damage the switch. This is the reason they fit a capacitor across the points in a traditional car ignition circuit. Otherwise the tips of the points will burn out from arcing.

Armin_Tamzarian

Ok, I measured the diode value with a multi-meter and got a value of 0.666.
I incorrectly assumed this meant is dropped the voltage by 66.6%.
So it actuall drops the voltage to .666V D.C?

Thanks for your patience, I'm getting there...

Once the +12V has been removed does the current caused by the back EMF look around and around, throught the diode and back through the coil until it is completely dissipated?

Presumably, energy is used every time the current passes through the diode and this is how the energy is dissipated?

Would tThis system would still work if the connection to the negative
terminal of the battery was removed as well as the connection to the positive terminal of the battery?

Michael Collins

Armin_Tamzarian wrote: »

Ok, I measured the diode value with a multi-meter and got a value of 0.666.
I incorrectly assumed this meant is dropped the voltage by 66.6%.
So it actuall drops the voltage to .666V D.C?

Yep, that is the forward voltage drop of your diode. It is usually around that value for most diodes.

Armin_Tamzarian wrote: »

Once the +12V has been removed does the current caused by the back EMF flow around and around, through the diode and back through the coil until it is completely dissipated?

Yes. And note that it is going in the same direction through the coil as the original current was, because the current through an inductor (i.e. coil) likes to remain constant.

Armin_Tamzarian wrote: »

Presumably, energy is used every time the current passes through the diode and this is how the energy is dissipated?

Exactly.

Armin_Tamzarian wrote: »

Would tThis system would still work if the connection to the negative
terminal of the battery was removed as well as the connection to the positive terminal of the battery?

Yep, it doesn't matter at all. Once that outer loop is broken, it doesn't matter how, the protection system will work - as decribed very well by DublinDilbert...

DublinDilbert

Armin_Tamzarian wrote: »

Ok, I measured the diode value with a multi-meter and got a value of 0.666.
I incorrectly assumed this meant is dropped the voltage by 66.6%.
So it actuall drops the voltage to .666V D.C?

Yep a typical diode will have a volt drop of approx 0.6V.

Armin_Tamzarian wrote: »

Once the +12V has been removed does the current caused by the back EMF look around and around, throught the diode and back through the coil until it is completely dissipated?

Yep, thats exactly what happens.

Armin_Tamzarian wrote: »

Presumably, energy is used every time the current passes through the diode and this is how the energy is dissipated?

The energy in the coil (in joules) will be 1/2 I^2 L, where I = current in coil, L = inductance. The energy will be dissipated in the resistance of the coil (at a rate of I^2 x R) and in the diode (at rate 0.6 x I).

Armin_Tamzarian wrote: »

Would tThis system would still work if the connection to the negative
terminal of the battery was removed as well as the connection to the positive terminal of the battery?

Exactly the same thing would happen. Remember current flows in loops, so if you break the positive or the negative (or both) sides of the battery, the same would happen...

Armin_Tamzarian

Me understand now, that's unpossible.
Thanks guys, just 2 final questions (I hope).

Michael Collins wrote: »

Yes. And note that it is going in the same direction through the coil as the original current was, because the current through an inductor (i.e. coil) likes to remain constant.

This I don't get.
My understanding was that the problematic current was caused by a back EMF which send the current in the opposite direction to that which the current was originally flowing in when the relay was closed?


Also, the only diode I had to replace a .6 diode, the only one I could find was a .4 diode, would this work ok?

Michael Collins

Armin_Tamzarian wrote: »

This I don't get. My understanding was that the problematic current was caused by a back EMF which send the current in the opposite direction to that which the current was originally flowing in when the relay was closed?

No. The current goes in the same direction through the relay coil as before. It's not the current that's problematic, it's the huge voltage that's generated when you interrupt the original current that's the danger, as DublinDilbert said, this can damage the device that's driving the relay coil, i.e. a transistor.

Also, the only diode I had to replace a .6 diode, the only one I could find was a .4 diode, would this work ok?

Yep, should be fine.

Armin_Tamzarian

1. Switch Closed,diode reverse biased, current flowing in relay coil
2. Switch Opens, coil polarity reverses, causing diode to become forward biased and conduct
3. Switch Open, current in coil free wheels to zero. .

How this make sense as DublinDilbert has stated that the polarity reverses
and also, how could current flow through the diode unless it had changed direction?

DublinDilbert

Armin_Tamzarian wrote: »

[/LIST]How this make sense as DublinDilbert has stated that the polarity reverses
and also, how could current flow through the diode unless it had changed direction?

I was talking about the polarity of the voltage on the coil, not the current in it. As Michael says above the current in the coil always flows in the same direction.

When the switch is closed there is a volt drop on the coil.
When the switch opens, the voltage on the coil changed to a volt rise, which then forward biases the diode....

Michael Collins

Armin_Tamzarian wrote: »

[/LIST]How this make sense as DublinDilbert has stated that the polarity reverses
and also, how could current flow through the diode unless it had changed direction?

All good questions! The admittedly suprising fact is the polarity of the emf on the coil can reverse, without the current through the coil itself reversing.

OK, see DublinDilbert's post above with the attached diagram. It shows the diode connected to either end of the relay coil. Now imagine the switch is opened i.e. battery is disconnected, so we can take away the outer loop completely - it has no effect anymore. The polarity of the voltage on the coil reverses, this is the back EMF acting to keep the current constant. How does this keep the current constant? Well, we see a positive voltage on the anode of the diode, and a negative voltage on the cathode. The current will flow through the diode from positive (the lower terminal) to negative (the upper terminal). Now where does the current go? Since the outer circuit is gone, it can only go one way - down through the relay coil, as before!

This is quite a complicated idea to grasp actually, there are lots of subtle points here. But hopefully this makes sense?

Armin_Tamzarian

Eureka, now it makes perfect sense.
Cheers lads, excellently explained...