Quick question. Need a more detailed explanation of how a suppression diode works in a DC circuit. Example would be horn circuit.
Pin 86 = B+ on control side
Pin 85 = Ground on control side
Pin 30 = B+ on load side
Pin 87 - Load
Questions:
a)Magnetic field collapses. Where does current go after switch is opened. Explain using conventional theory.
b)Why does it not just go to ground and if it did, where would it go from there.
c)What would happen if it heads back to power source?
d)What would happen if there was no diode?
e)Why can't the diode be placed in front of the coil instead of across it?
f)Why do some companies use a resistor?

I had this discussion with one of my students today. I want to see if I answered him correctly or whether you gents have a better viewpoint. They were all good questions. Thank you in advance

 

What were your answers?

 

Are you sure you are not the student?

 

Could you post the schematic so others may know what you are talking about.

 

Those are standard automotive relay numbering designations.
85 and 86 are the coil, 30 contact common, 87 N.O. if 87a present, this is N.C. contact.
I assume you want to know where and when to fit suppression diode?
An example connection would help.
Diodes are generally used in DC circuits for BEMF suppression, on AC the R/C snubber is used, as obviously a diode cannot be used in this instance.
For (e) for example, the diode cannot be used in series as it will not recirculate the reverse EMF.
Max.

 

First off, we are all students and I don't recall posting this under homework. I am the teacher and I have a student who asked some very good questions. I came on the forum to get some intelligent answers to what I thought were some well thought out questions. Because I teach DC mostly, I was caught a little off guard and had to put some serious thought behind my answers. My understanding is that the diode and the coil form a mini circuit and when the magnetic field collapses, around 200 volts is formed on the negative side(ground) of the coil. Upon collapse, this voltage is routed through the diode and back through the coil in a circular pattern until it dissipates(but why does it dissipate). When he asked why it didn't go through ground instead (path of least resistance), I was at a loss. I understand where the diode is located and its orientation so my answer regarding why the diode is not in series was twofold. First it would interfere with the coil getting full voltage for operation and secondly, the voltage would travel back through the ground side and would cause problems elsewhere.
For those that can answer these questions with a good answer, I thank you in advance.

 

There is something missing from the drawing, the switching method for turning the relay on, there will be no BEMF until the power is removed from the coil.
The diode is configured this way because the diode presents a high load to the reverse EMF generated by the collapsing field of the coil.
If not present, not only does the supply EMF oppose the coil BEMF but any high voltage generated will be either impressed on the switching device or cause arcing of any contact that may be operating the relay.
Max.

 

You are correct about the switch. There should be a SPST switch between the fuse and the coil. Sorry about that. Okay so when the switch opens CEMF comes from the coil and tries to head back towards source of power. Because switch is open, it cannot continue and therefore finds another path. But why does it not choose to go to ground. I get the first part and that is what I told the student. My problem in explaining it is, where does this two hundred or so volts of CEMF go. Does it just dissipate until it is gone? What actually happens to the CEMF.

 

With the diode in place effectively providing a short circuit to the EMF, it is quickly extinguished and not allowed to reach the value it would otherwise.
If no diode is present then there is no clear path to complete the circuit and the Potential build across the coil, there is no path for it to follow if one end of the coil is open circuit.
It then collapses and in doing so creates a potential in the other direction, in this manner the collapsing voltage will oscillate or 'ring', quickly decreasing to zero, in the mean time this very high voltage 'ring' can cause EMI and havoc on any semi conductor connected to it, also an arc can be struck across any contact that interrupts the coil voltage.
In the same way a capacitor stores energy even when disconnected from the supply, so does an inductance, albeit for a short period.
Max.

 

Large relays also need to extinguish the arc accross the contacts or the relay will wear out quickly. Especially when switching a high current motor.

I don't remember correctly if it was that way before I fixed one such relay board but after it, I had capacitors accross the contacts.

There was a worm in the circuit and no spares available so at first I took it apart completely and the rebuilt it without a schematic.

 

You are correct about the switch. There should be a SPST switch between the fuse and the coil. Sorry about that. Okay so when the switch opens CEMF comes from the coil and tries to head back towards source of power. Because switch is open, it cannot continue and therefore finds another path. But why does it not choose to go to ground. The ground, if there is one isn't part of the circuit. I get the first part and that is what I told the student. My problem in explaining it is, where does this two hundred or so volts of CEMF go. Does it just dissipate until it is gone? Yes in the R What actually happens to the CEMF.

With no R it would ring forever.
The "ground" side could actually be considered the source of the counter (reverse) emf.

 

There is no such thing as a perfect coil.

windings have cross capacitance even if it is very low. This is what I believe where the electric power goes and of course, it is dissipated immediately + it takes a high voltage to have any effect.

Otherwise, it would increase until an actual arc forms.

 

Some good info about suppression diodes and contact switching in the attached pdfs. More on te.com

 

I'm going to try to make this short, partially as a convenience to you.

If there is no diode, the energy in the coil (collapsing magnetic field) arcs across the quickly opening switch, dissipating the energy as heat, ionized air, and melting of the contact surfaces.

If the diode is in place, the collapsing magnetic field keeps the current going the same way it was going before you moved the switch and it goes through the diode, back into the coil, round and round, and is slowly changed into heat by the resistance of the coil.

By, "slowly", I mean milliseconds instead of microseconds.

 

#12, I have to say, for my student, that will be a perfectly acceptable answer. I could always explain about arcing across the switch and the circuit going round and round, but I forgot about the transformation to heat. Well done and I thank you. As for the rest of the answers, I thank all of you as I learn more and more about these things from the electronic side or perspective. It is quite different from teaching DC theory to College students.

 

We all try to answer the best way we know how. Eventually, (almost always), one of us hits the right combination of words to get the explanation to make sense. I did not address all of the questions. I formed my answer in terms of where the energy goes because that is a BIG aspect of my day job. If you want to point out some of the other aspects that need more examination, feel free to continue the conversation from here.

 

(Dang it, new computer, and I can't find the pics from the last 12 times I explained this).

The figure on the left shows the relay ON. The coil symbol is a shorthand for both the resistive part (which limits the current I) and an inductive part (which provides the magnetic field to move the contactor. That part is fairly simple and easy to understand.

Now turn the relay off by interrupting the current from B+ to ground. The inductive part of the coil will dominate here, and an inductor has the voltage/current relationship of:

V = L * (rate of change of current)

where V is voltage, L is the coil inductance, and (rate of change of current) is (rate of change of current)

If you know calculus you should be familiar with the actual expression, but this version is fine for this. When you open the switch the current change is extremely large, and the current is decreasing, so the "(rate of change of current)" term is a very large negative number, which means the voltage is also a large negative number.

I've drawn the polarity of the coil voltage on the figures so you can see how this reversal of voltage happens.

There's little to limit this voltage. I've seen 350 volts on an oscilloscope. But with the diode across the coil the voltage gets limited... to the diode voltage itself.

It's interesting to look back at the voltage/current relationship for the discharge:

V = 0.7 = L * (rate of change of current)

Since the voltage is a constant, the rate of change is a constant, meaning the current decays in a linear manor.

Some relay companies are concerned about how fast their device turns off, and it cannot turn off until the current decays. To speed this up they put a zener diode in series with the diode to discharge thru a larger voltage, giving a larger rate of change. I've built such devices for two major relay manufacturers and both use a 36 volt zener with a 1,000 volt diode.

One other thing to note is when the inductor kicks back and drives the diode, the inductor voltage (and thus the diode voltage) is in a direction to add to the B+ voltage.

 

Good answers and I am getting a better sense of how it dissipates but what happens when the switch is located on the B+ side and not the ground side. Again, why does it not head for the ground post of the battery?

 

When the switch opens the circuit from the positive supply, the collapsing field can not get current from the positive supply because you just opened a switch that has a lot more resistance than the diode and the coil. When the switch opens the circuit from the negative supply, the collapsing field can not get current from the negative supply because you just opened a switch that has a lot more resistance than the diode and the coil. Then, post #14 happens.

 

Perhaps I am not saying this correctly. If the switch is on the positive side of the circuit (conventional) and the switch is opened, there is too much resistance across the circuit for it to travel across that gap. I get that. If the switch is closed, current normally travels through the relay coil to ground. I get that. But when the switch is opened and the diode comes into play, rather than taking the diode route, it would seem to be an easier path to just take CEMF to ground. Why does it choose the diode and not ground? Perhaps I'm just not getting it.