purpose of diode??

[new guy]

I always thought the purpose of a diode connected across the coil terminals of a relay was to protect some delicate electronic circuitry somewhere down the line from a voltage spike during the high voltage collapse of the coils magnetic field. But if there no delicate electronic circuitry(only other relys and mechanical switches and contactors) is there a need for relays with diodes or can relays witout diodes be substituted? Does the diode somehow assist in the collapse of the coil field thereby making it faster and more definite? I am talking about 48volt DC relays in a control circuit of an elevator about 1993 vintage.

[Dean Huster]

That is exactly the function of the diode across the relay coil or any other electromagnetic device for that matter, at least in DC circuits. However, just because there isn't any sensitive electronic drivers connected doesn't mean that the diode is pointless. Without the diode, the CEMF voltage is high enough to cause arcing of switches. Yes, tiny relays aren't much of a problem, but larger ones could be.

The "condenser" (capacitor) across the points in the old Kettering ignition system in automobiles served the same purpose and there were no sensitive electronics involved in 1947!

Dean

[MrAl]

Hi there,


I was going to say what Dean said, plus this...

You asked about the effect of the diode on the relay action.
Do the contacts open faster or not with the diode?
Well, the relay coil can be looked at as an inductor, and an
inductor stores energy. When this coil (inductor) has energy
in it, it holds the contacts closed because of the magnetic flux.
When the power is disconnected from the coil, the energy
decreases and as the energy decreases the magnetic flux decreases
until it gets to the point where it can no longer hold the relay
contacts closed. Also, as the field begins to collapse the voltage across
the coil changes polarity and if a diode is present it forward biases the
diode and the diode holds the voltage at about 0.7 volts, but if there
is no diode present than the voltage can go almost as high as it
wants, until something arcs or, and this is even worse, the enamel
wire insulation breaks down and shorts the coil out. This is yet
another reason for the diode, and that is the basic operation.

To find out what effect the diode has on the time it takes to
open the relay contacts, we can look at the coil as an inductor
that stores energy in volt-seconds, and look at the case with
a diode and the case without the diode.

Having a diode:
As the switch is opened, the voltage reverses and increases very
rapidly until it gets to around 0.7 volts, at which time the diode
conducts heavily and clamps the voltage to this level.
Now just for example, lets say when the switch was closed the
coil had 0.07 volt seconds stored in its core. Since the voltage
is 0.7 volts, it takes 0.1 seconds for the field to collapse, and the
contacts open.

No diode:
As the switch is opened, the voltage reverses and increases very
rapidly, only this time because there is no diode it can go to more
than 100 volts, but lets say that it somehow gets up to only 100
volts for now just so we have something to compare it to.
With the voltage at 100 volts and 0.07 volt seconds stored in the
coil it only takes 700us now.

Compare:
Diode: 100ms, or 100000us
No diode: 700us

So we see that the case without the diode opens the contacts faster.

In practice however, the inertia of the mass of the metal used to
create the armature dominates the response, so the speed also
depends on the physical relay construction. Some relays will open
faster than others without a diode, and some relays will open
nearly the same with a diode or without a diode.
It also helps sometimes to use a zener or multiple diodes because
the volt seconds dissipate faster as noted above in the case
that had no diode.
If for some reason you require speed, you can try using the zener
diode technique, possibly with a regular diode in series with it.

In any case, operating the relay without any voltage clamp (diode
or zener or other) could result in a short in the coil after many
operations, or burn up the switch due to arc over, as well as cause
RF interference.

[Externet]

Hello Al.
Can you please explain why do you say

Hi there,
..... the voltage across the coil changes polarity .....

.... As the switch is opened, the voltage reverses and .....

Miguel

[Robert Reed]

Exter
The magnetic feild around the coil is no longer supported when the coil circuit is opened and collapses inward acoss the coil. Since this collapsing feild is of opposite polarity to the operating feild, it causes a voltage in the opposite direction. Can be quite hi and ringing if there is no load or suppression devices on it.

[Externet]

I believe that is wrong.
A widely spread misconception.

In the circuit

(-)-----------uuuuuu--------------/ -------(+)
..............----|>|------

(ignore the dots, and the diode is connected parallel to the coil)

Electron current flows from (-) trough the coil, then to switch to (+), the diode does not conduct.

That is, the current in the coil flows from the left coil terminal, inside the coil, to the right coil terminal.

When the switch opens, the coil current flows from the left coil terminal, inside the coil, to the right coil terminal, enters the cathode, inside the diode, exits the diode anode, to the left coil terminal and continues the loop clockwise trough the coil and the diode until the magnetic field has fully collapsed.

The current direction in the coil never reverses. If the current never reverses, neither does the voltage.

If it reversed, the diode would never conduct.

Miguel.

[MrAl]

I believe that is wrong.
A widely spread misconception.

In the circuit

(-)-----------uuuuuu--------------/ -------(+)
..............----|>|------

(ignore the dots, and the diode is connected parallel to the coil)

Electron current flows from (-) trough the coil, then to switch to (+), the diode does not conduct.

That is, the current in the coil flows from the left coil terminal, inside the coil, to the right coil terminal.

When the switch opens, the coil current flows from the left coil terminal, inside the coil, to the right coil terminal, enters the cathode, inside the diode, exits the diode anode, to the left coil terminal and continues the loop clockwise trough the coil and the diode until the magnetic field has fully collapsed.

The current direction in the coil never reverses. If the current never reverses, neither does the voltage.

If it reversed, the diode would never conduct.

Miguel.

Hi Miguel,

Sorry to say, but that is not quite the way it works.
I agree 'almost' with Robert, except that the field itself does
not have to reverse to get the voltage to reverse. It is
the direction of the change in field (sometimes written as "dB/dt")
that causes the current flow one way or the other. It's when
the field *change* reverses (-dB/dt) that forces the current
to keep flowing in the same direction and the voltage reverses.
With the switch closed the field is induced in the coil caused by the
left-to-right flow of electrons, and once the switch opens the
field starts to decrease and that change in field induces a
current in the same direction until all of the energy is spent.
Another way of looking at this is that for a current flow I,
the change in field is dB/dt, and the opposite 'change' in field
-dB/dt would cause the same current I to flow.
We cant change the direction of the field instantaneously because
it represents a finite amount of energy that would need to be
dissipated and that either takes time or a huge amount of energy.
We dont have a huge amount of energy, but we do have some time
(like 100ms or less perhaps).

We must look at the bigger picture however...

The big difference is this:
In this kind of circuit, the coil has two basic 'modes' of
operation. It can either be acting as a passive element
(absorbing energy) or as an active element (sourcing energy).

With the switch closed, the coil is absorbing energy from the battery,
and so it acts as a passive element except that it can store energy,
and as the switch opens the coil almost instantly becomes an active
element, ie a current source, and so it actually sources current.
Because the current is in the same direction as it was before, the
only way the coil can source current in that direction is if it also
reverses the voltage across it. In other words, the left side
becomes positive and the right side becomes negative. This allows
electrons to flow out of the right side, through the diode, and back
into the left side (as per your schematic).
Another way to visualize this is to replace the inductor with a small
battery. Which way would you have to connect that battery to get
electron flow *out* of the right terminal (switch open)? Once you
connect the battery you will see that is the only possible configuration,
and since the coil acts as a small current source (for a short time)
the polarity of the coil must be the same as the additional battery.

As an example, lets say we use a 10v source to power a 10v relay
coil, switch closed. The voltage across the coil is as your schematic
shows, with the negative on the left, positive on the right, and
the current flows from left to right as you said. The current flow
is limited mostly by the dc resistance of the coil here, and say the
coil resistance is 100 ohms, so the current flow is 100ma.
Now we open the switch.
At the instant the switch opens, the inductor becomes a current source.
(you can call it a voltage source too for now, but current source is
a little more accurate). Ok, so we have 100ma flowing but now we have
this source in a circuit with almost infinite resistance. Lets say that
for some reason the resistance was not that high, but only 100k.
Using ohms law, 100k resistance with 100ma means there will be a
10,000 volt signal across the coil. What will the polarity be?
Since the current is still flowing left to right and now the coil is
*sourcing* (instead of sinking) current, the polarity must change,
so the voltage is now actually -10,000 volts. Of course with the
diode in the circuit and connected as you have shown, the diode will
conduct and this will greatly lower the external resistance so the
voltage will clamp to -0.7 volts or around there.

I'll draw a picture of the coil when it is sourcing current as a battery
so the polarity becomes self-evident:

Code: Select all

        + |    -
    ------| |------
    |     |       |
    |             |
    ------|>|------
           

but more accurately is this:


        +     -
    ------>>-------
    |             |
    |             |
    ------|>|------

where ">>" represents a current source, and the voltage
across it goes to whatever it has to be, which here happens to
be as the + and - indicates.

Darn i wish we had a way to post schematics.

[dyarker]

The current never changes direction, exactly right.

But the coil changes from being a load to a source while the field collapses. Therefore the voltage must reverse.

Try this, a resistor connected to a battery. The electrons flow from battery negative through the resistor to battery positive. Current flow from negative to positive just like you expect. But wait, the electrons don't just disappear from the universe, they continue their journey through the battery from positive to negative. That's right, in the voltage source electrons (current) travels from positive to negative.

Any help?

[Externet]

Regards, Dale.

...Therefore the voltage must reverse....

That makes no sense to me.
The coil makes electron flow behave as if they had 'momentum'
The polarity in the coil does not reverse, it always flows from left to right on the schematic.
Yes, it becomes a current souce interrupting the supply, but of the same polarity.

At what node you see the voltage reversal ? Can you explain in idiot-proof mode ?
Miguel

[Robert Reed]

Mr Al
Good presentation! Can you give a breif disertation on the coil with no suppression devices or parrelell loads?

Dyarker
Same holds true with transformers. Always been a bit confusing .

[Chris Smith]

Its called dampening the "Chatter".

Diodes are not perfect one way devices, but they do catch most of the AC oscillations faster than the build up, clamping the AC faster than the reactance of the wire over time.

If you look at any whip, the wave form is AC, then it stops, then it reverses back at you as a AC wave form.

All electrons have mass and momentum, if they didn’t the choke wouldn’t work like a flywheel, and the cap would just react with out loss.

Even electrons change coarse because electrons have mass. It cant come to a perfect rest with out a means.

Only the photon is seen as having no mass and just pure energy, which accounts for it annihilation or converting of the photon [energy] into another form.

[MrAl]

I believe that is wrong.
A widely spread misconception.

In the circuit

(-)-----------uuuuuu--------------/ -------(+)
..............----|>|------

(ignore the dots, and the diode is connected parallel to the coil)

Electron current flows from (-) trough the coil, then to switch to (+), the diode does not conduct.

That is, the current in the coil flows from the left coil terminal, inside the coil, to the right coil terminal.

When the switch opens, the coil current flows from the left coil terminal, inside the coil, to the right coil terminal, enters the cathode, inside the diode, exits the diode anode, to the left coil terminal and continues the loop clockwise trough the coil and the diode until the magnetic field has fully collapsed.

The current direction in the coil never reverses. If the current never reverses, neither does the voltage.

If it reversed, the diode would never conduct.

Miguel.

Hi Miguel,

Sorry to say, but that is not quite the way it works.
I agree 'almost' with Robert, except that the field itself does
not have to reverse to get the voltage to reverse. It is
the direction of the change in field (sometimes written as "dB/dt")
that causes the current flow one way or the other. It's when
the field *change* reverses (-dB/dt) that forces the current
to keep flowing in the same direction and the voltage reverses.
With the switch closed the field is induced in the coil caused by the
left-to-right flow of electrons, and once the switch opens the
field starts to decrease and that change in field induces a
current in the same direction until all of the energy is spent.
Another way of looking at this is that for a current flow I,
the change in field is dB/dt, and the opposite 'change' in field
-dB/dt would cause the same current I to flow.
We cant change the direction of the field instantaneously because
it represents a finite amount of energy that would need to be
dissipated and that either takes time or a huge amount of energy.
We dont have a huge amount of energy, but we do have some time
(like 100ms or less perhaps).

We must look at the bigger picture however...

The big difference is this:
In this kind of circuit, the coil has two basic 'modes' of
operation. It can either be acting as a passive element
(absorbing energy) or as an active element (sourcing energy).

With the switch closed, the coil is absorbing energy from the battery,
and so it acts as a passive element except that it can store energy,
and as the switch opens the coil almost instantly becomes an active
element, ie a current source, and so it actually sources current.
Because the current is in the same direction as it was before, the
only way the coil can source current in that direction is if it also
reverses the voltage across it. In other words, the left side
becomes positive and the right side becomes negative. This allows
electrons to flow out of the right side, through the diode, and back
into the left side (as per your schematic).
Another way to visualize this is to replace the inductor with a small
battery. Which way would you have to connect that battery to get
electron flow *out* of the right terminal (switch open)? Once you
connect the battery you will see that is the only possible configuration,
and since the coil acts as a small current source (for a short time)
the polarity of the coil must be the same as the additional battery.

As an example, lets say we use a 10v source to power a 10v relay
coil, switch closed. The voltage across the coil is as your schematic
shows, with the negative on the left, positive on the right, and
the current flows from left to right as you said. The current flow
is limited mostly by the dc resistance of the coil here, and say the
coil resistance is 100 ohms, so the current flow is 100ma.
Now we open the switch.
At the instant the switch opens, the inductor becomes a current source.
(you can call it a voltage source too for now, but current source is
a little more accurate). Ok, so we have 100ma flowing but now we have
this source in a circuit with almost infinite resistance. Lets say that
for some reason the resistance was not that high, but only 100k.
Using ohms law, 100k resistance with 100ma means there will be a
10,000 volt signal across the coil. What will the polarity be?
Since the current is still flowing left to right and now the coil is
*sourcing* (instead of sinking) current, the polarity must change,
so the voltage is now actually -10,000 volts. Of course with the
diode in the circuit and connected as you have shown, the diode will
conduct and this will greatly lower the external resistance so the
voltage will clamp to -0.7 volts or around there.

I'll draw a picture of the coil when it is sourcing current as a battery
so the polarity becomes self-evident:

Code: Select all

        + |    -
    ------| |------
    |     |       |
    |             |
    ------|>|------
           

but more accurately is this:


        +     -
    ------>>-------
    |             |
    |             |
    ------|>|------

where ">>" represents a current source, and the voltage
across it goes to whatever it has to be, which here happens to
be as the + and - indicates.

Darn i wish we had a way to post schematics.

For a full explanation, check above Miguel.
There is another point to be made here, and that is that
just because the electrons are moving in the same direction
does not mean the voltage can not reverse. The voltage
acts according to how the current is flowing and whether or
not the device is sourcing or sinking. Since the current
is flowing through some sort of resistance and that resistance
makes a voltage drop because of E=I*R, the voltage and its
polarity follows that Law (Ohms Law).

Mr Al
Good presentation! Can you give a breif disertation on the coil with no suppression devices or parrelell loads?

Robert, Thank you, and did you mean a coil without any load, when the switch
is opened?

[Chris Smith]

Don’t forget that when you load a coil with DC, and the current is flowing in one direction, then when you release it,... the coil is permitted to make AC with out a clamping or steering direction.

[It always makes ac, its just under control]

The common Alternator produces AC with each phase line that is intersected, first its built up then chopped, so does the standard coil when you first compress it, then release it with no direction to head to.

The same goes for feeding in DC pulsating into a transformer, AC comes out the other end because there are no rectifiers.

Bottom line is in any coil that has a charge that is then released, AC will be the wave form coming out,..... and then under a control using diodes.

Ac still comes out,... but its minimize and controlled because the diode is not perfect.

[Robert Reed]

Mr AlL
Yes, upon opening the switch.

[dyarker]

"At what node you see the voltage reversal ? Can you explain in idiot-proof mode ?
Miguel"

Okay, after work I'll put up some commented drawings.

In the meantime re-read Mr Al slowly and carefully.

Cheers,