Why would you use a clamping diode when using a relay?

I think its got something to do with emf and the clamping diode protects the transistors and other components from being damaged.

Could anyone please give me a fuller explanation of the question above.

 

When the relay is energized, a magnetic field is generated around the coil. When the relay is de-energized, the magnetic field collapses and generates a voltage in the coil. This voltage can be hundreds of volts depending on the inductance, stray capacitance and shunt resistance. The diode limits the voltage to .5 volts and the diode must be able to handle the forward current.

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Russlk is correct, but I would like to add that when the field collapses the polarity of the voltage is opposite, that is why the diode is installed the reverse polarity of the energizing voltage.

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One effect of NOT using a clamping diode is stray spikes in the rest of the circuitry causing erratic behavoir...

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gerty

 

Russlk and k7elp60 are both correct. The only point I would add is this.

When the circuit is opened, the inductance of the relay coil wants to keep the same current flowing (like Newton's first law of motion - a moving body wants to keep moving). So it generates an EMF in order to achieve this aim. So if the shunt resistance is large and the stray capacitance is small, then the EMF is very high and can damage the transistor, ie. the transistor breaks down and the current flows through it.

The shunt diode therefore provides a circuit for the current and so the EMF is limited to around 0.5 to 0.8 Volt as Russlk said.

Len

 

How does it get the name "clamping diode?" I've always wondered about that one. I guess it has something to do with the diode "clamping down" on the back-EMF and creating a short(er) circuit for the surge?

 

Well more importantly the ONE effect of not having a clamping diode (or free-wheel diode as it is named in power-electronic circles) is that the switch blows on over-voltage - then other things start to go wrong!!!

Since you have just HARD open-circuited an inductor (equally you must never hard short-circuit a charged capacitor)

via the inductor equation:

V = L di/dt

you can see that you must not change the di/dt too fast or you will get a large amount of volts - It is this back-EMF (so to speak) the magnetic field of the inductor tries to build up to maintian the current flow. It will carry on building up the voltage until that current flows, be it due to the switch re-switching (and then it is dead due to extreamly increased voltage thus switching losses through the roof) or because or reverse-breakdown of the switch again killing the switch.


Even with a freewheeling diode you will still get some overshoot due to the finite time it takes for the diode to start conducting. But the point is that overshoot is under control/clamped. you can see an overshoot of abt 10v for a 100v link for well match diode-switch.

 

The term "damping" comes from damped oscillations. If you stimulate a RLC circuit into oscillation, the amplitude decreases with time due to the energy lost as heat in the resistor.

There are mechanical equivalents. For example - suspend a spring from a beam with a weight attached at the other end. If you pull the weight down and let it go it oscillates up and down. But since energy is lost in the spring and due to air resistance, the system gradually looses energy so the amplitude of the oscillation gradually decreases. It is therefore damped.

Len

 

lol he was telking about CLAMPING not DAMPING

 

Len[/quote]

lol he was telking about CLAMPING not DAMPING[/quote]

Thanks. I must have read the cl as a d.

A clamping diode is where a diode is used to prevent a signal from exceeding (or alternatively going below) a certain voltage, i.e., it is "clamped" to that voltage.

However, in this case it is not really a clamping diode. It is providing a path for the coil currnet when the transistor is turned off thus protecting the transistor from damage.

 

Aren't these two conditions contradictory? If the inductance of the relay is trying to keep the same current flowing when the circuit is opened then why would it reverse polarity? Wouldn't reversing the polarity mean the inductance is reversing the current flow through the coil rather than trying to maintain the current flow that existed before the circuit was opened?

 

Er no. The current doesnt change polarity (it cant).

But if you open-circuit an inductor that has current flowing you no longer has a driving voltage. The voltage across the inductor changes polarity - it has change from a load to a source to ensure that current continues to flow

 

[/quote]
Aren't these two conditions contradictory? If the inductance of the relay is trying to keep the same current flowing when the circuit is opened then why would it reverse polarity? Wouldn't reversing the polarity mean the inductance is reversing the current flow through the coil rather than trying to maintain the current flow that existed before the circuit was opened?[/quote]

No they are not contradictory. The EMF generated in the coil when the transistor is turned off is in the direction necessary to keep the current flowing in the same direction. So the lower end of the coil becomes positive in order to keep the same direction of current flow. The coil becomes the energy source. The energy stored in the magnetic field has to be dissipated. Whereas, when the transistor was on, the upper end was positive.

It may be easier to visualise if there was a resistor across the coil rather than a diode. In order to maintain the same current direction through the coil, the current through the resistor must flow from the lower end (of the resistor) to the upper end. (I'm using conventional current flow)

Len

 

Its not very intuitive (reversing the polarity but maintaining the same current direction), but I go it now. Thanks for clearing that up.

 

While the transistor is on, the current comes from the power supply. When the transistor is turned off, the energy source is the magnetic field around the coil.

Think of a battery that is supplying current. The current flows (conventional flow) from the positive terminal through the load to the negative terminal and inside the battery from the negative to the positive terminal.

Len