Electric Clocks Sparks and Arcs

[Schlitzer]

Hello. I just completed figuring out the wiring circuits of my International Time Recording master clock, and it runs great!! I'm not sure that I have a problem, but I certainly don't like to see a spark each time the main relay fires (once each minute for 2 seconds) and winds the movement, advances the bell program unit, and pulses the slave clock circuits. The spark occurs on the break. The contacts of this relay are large carbon points about 3/8 inch in diameter. By their appearance, the arcing has been going on for a long time, perhaps from day 1. This is a large master clock made in 1927 and used in a school in Wisconsin. I've posted pics of the relay as well as other parts of the clock in my topic IBM (International Time Recording Co.) Master Clock Wiring, if you care to look at them.

The circuitry of the clock is 12 volts dc. What can I do to prevent the arcing? I was thinking of trying a 12v capacitor as used in auto ignition points. I'll try that unless someone can suggest a better way. Any ideas?

Thanks,
Lloyd Schlitzer NAWCC 24823, Retired but not tired!!

 

[Schlitzer]

Hello. I just completed figuring out the wiring circuits of my International Time Recording master clock, and it runs great!! I'm not sure that I have a problem, but I certainly don't like to see a spark each time the main relay fires (once each minute for 2 seconds) and winds the movement, advances the bell program unit, and pulses the slave clock circuits. The spark occurs on the break. The contacts of this relay are large carbon points about 3/8 inch in diameter. By their appearance, the arcing has been going on for a long time, perhaps from day 1. This is a large master clock made in 1927 and used in a school in Wisconsin. I've posted pics of the relay as well as other parts of the clock in my topic IBM (International Time Recording Co.) Master Clock Wiring, if you care to look at them.

The circuitry of the clock is 12 volts dc. What can I do to prevent the arcing? I was thinking of trying a 12v capacitor as used in auto ignition points. I'll try that unless someone can suggest a better way. Any ideas?

Thanks,
Lloyd Schlitzer NAWCC 24823, Retired but not tired!!

 

[Jeffrey R. Wood]

The traditional way is to use a resistor in parallel with the circuit, having an ohm value of about 10 times that of the circuit. Wattage is not too critical, but I would use a wirewound resistor not so large that it would look out-of-place. With 1,440 jolts a day, a small carbon resistor would theoretically be more likely to fail eventually. Of course there are more modern methods of spark suppression, but I would avoid a diode (for example) because then polarity must be rigidly observed.

 

[eskmill]

Wood is on target with his recommendation to use shunt resistors in parallel with each electromagnet to decrease the back emf when the circuit opens. Locate the resistor adjacent to the coil connections. One or two watt is ok.

About ten percent of the DC resistance of each electromagnet is what's been used successfully on other maker's clocks. Ten percent greater current consumption isn't going to hurt anything.

The relay, I believe is special with two sets of contacts.
http://home.earthlink.net/~lexmd/CarbonContactRelay.jpg
I believe the small adjustable contact is tungsten and would be adjusted to make after the carbon pair make and break before the carbon pair takes the arc.

IMO, IBM never did learn to accept the industry's methods of arc suppression although they tried every other scheme. Some were effective but overly complicated and like the carbon relay, required periodic attention.

 

[swolf]

Hi. contact me by Email and I will send you a semiconductor Varistor. These Varistors came in different voltage ratings and absorb the back emf voltage by essentially becoming a low resitance mometarly across the contacts when the rated voltage is exceeded thus absorbing the energy that create the arc.


Sherm

 

[Mike Phelan]

The old method using a resistor is inefficient, and a compromise between suppressing the arc effectively and wasting power,

It is a much better idea to put a diode across the coil (observe polarity!) to immediately kill the ringing of the coil that causes the arc. It will also be much smaller.

Not sure why Jeffrey is against this - there are only two possibilities for polarity, and you only fit it once!
The band on the diode needs to go the most positive end of the coil, and a 1N 4001 will do - probably a few cents.

 

[Schlitzer]

Hello again and thanks for your suggestions. I will try them next Monday when I get to Radio Shack and get some stuff. My transformers came in the mail yesterday and I need some rectifiers for them. I will post again as I persue this spark problem.

I am a little bit confused where exactly to put the diode or the resistor. The relay involves two seperate circuits. The circuit which operates the relay is one of them and when energized, closes the contacts and completes the second circuit. The second circuit is the one that winds the movement, advances the program unit, and pulses the slaves. These contacts spark when it breaks. So where does the diode go? It seems to me that I might need one on each of the components, that is, the program wind coil, the program unit coil, and each of the slave coils. Am I correct, or can I do the job at the location of the relay itself with a single diode or resistor? Thanks. Lloyd

 

[eskmill]

The old technique of using a parallel resistance to quench the back emf when the circuit to a ferro-magnetic circuit opens is, admittedly old and somewhat inefficient electrically. Too, it's not a hundred percent effective but it's simple.

On the other hand, the newest and cleanest scheme is to connect a diode in parallel with the electromagnetic element. It's efficient, simple and has little effect on the response of the electromagnet. But, it does introduce an opportunity for error on the part of the next care-taker of the old clock unless the alterations are very clearly documented because now, the clock's circuitry is polarity sensitive, a characteristic that was never intended when the clock was designed.

Horological conservators decry, "do no harm!"

I suppose that it's ok to add and alter the wiring of a clock or other device if you plan to take it along to your grave! But if you must make alterations, consider the next caretaker of the master clock. Fully document the wiring functional details and adjustment procedures for the future owner. Don't assume the next owner will be a guru. Your documentation will be appreciated.

Schlitzer is already having to undo the clever "handywork" modification of an ignorant previous owner. Why repeat the folly?

Again Schlitzer, I believe that careful adjustment of the tungsten and carbon contacts on the master relay of your clock will reduce the arcing to the extent originally designed.

An analysis of the double contact relay with the carbon contact making first and breaking last is effectively placing the resistive carbon contacts across and suppressing the back emf of the electromagnets in the circuit.

 

[Mike Phelan]

Lloyd
A diode would need to be across the relay coil that is causing the arcing - in this case, the one that is connected to the arcing contacts, not the one that is energised by it.

Les
I appreciate what you are saying, but making any change, for example putting a resistor across the coil, would be equally non-original.

Any future electric clock repairer who sees either of these additions and is worth his salt will know what a diode and what it does - if they do not, then really they ought not to be doing this, and are just as likely to cause worse damage than blowing a fuse because a diode or the supply is reversed because they have not spotted the diode- why would anyone want to change something that works? Yes, I know.......

That said, Lloyd's clock must have worked for a long while originally as it was. :biggrin:

 

[Schlitzer]

Hello again, and thanks again.

I believe the relay is working properly. That is, the carbon contacts are making first and breaking last. Les, From what you say, it seems that the arcing would be worse if the sequence with the tungsten contacts were reversed. That makes sense and I accept that as true. Thanks.

The arcing occurs when the current passing through the movement wind coil, and/or the program advance coil, and/or the slave clock relay coils is broken by the opening of the carbon contacts. It would seem that I might need a diode for each of these electromagnetic coils. If so, should each diode be in parallel with its wind coil or in series with it? I believe you indicate a parallel connection, but I don't see how that works. If it is hooked in series it should prevent any reverse flow of current. Isn't that correct?

As an alternative, how would it be if I used a diode in series with the carbon contact? Wouldn't that prevent any reverse flow?

Lloyd

 

[eskmill]

In my view the diodes, if you must, should be connected in parallel directly at the coil windings to be most effective. If connected downstream, there is a possibility of distributed capacitance and inductance that can bring unwanted circuit oscillations.

Some would advise a resistor in in series with the diode and others would suggest a resistor in parallel with the diode. This to avoid overtaxing the diode's ability.

The simplest diodes are to electric current as a check valve is to hydraulics. The applied current can only flow in one direction in a diode circuit within it's forward current specification. And don't overlook the inverse voltage specification of the diode either. The direct current circuits in a master clock aren't dead-smooth DC. There are voltage spikes equal to or greater than the source during switching. I think you could expect 200-300 volt spikes in both directions.

When properly connected in parallel across an electromagnet, the diode acts like it wasn't even there during voltage application to activate the electromagnet but microseconds after the current is cut-off, the diode becomes a dead-short-circuit to any current that flows backward from the self-inductance of the electromagnet. It turns any back emf into heat.

Too, don't overlook the varistor that Sherm suggested. Some have described the varistor as a whole wad of avalance diodes connected helter-skelter in a two lead package. They are widely used to protect sensitive electronics from voltage spikes imposed on household mains. They are not polarity sensitive until the applied voltage reaches a threshold limit. The down-side of the varistor in this application is that they tend to dead-short when their limit is exceeded but are deceiving as their visual appearance doesn't change when overloaded to death to become a piece of wire.

Is this horology?

 

[Schlitzer]

Hi Les,
Okay, I understand the parallel connection now. I had the idea that the diode should be connected to pass current in the direction of the circuit current, but just the opposite must be the case.
Certainly this is horology. I'm reminded of the song Grandfather's clock. You know. It stopped short never to run again when the old man died! Such is the case here. In fact, the large oak case of this clock is almost big enough for my casket. My wife could simply stuff me in, lock the door, and plant me in Mother Nature's garden.

 

[Mike Phelan]

Originally posted by Schlitzer:
Hello again, and thanks again.

I believe the relay is working properly. That is, the carbon contacts are making first and breaking last. Les, From what you say, it seems that the arcing would be worse if the sequence with the tungsten contacts were reversed. That makes sense and I accept that as true. Thanks.

Well done, Lloyd - you do not need to worry about the diodes and stuff now!

If it is hooked in series it should prevent any reverse flow of current. Isn't that correct?

Lloyd

Although you need not worry about this now, and at the risk of making my ears burn but for completeness' sake, the diode is not really to prevent reverse flow, but to stop the coil "ringing".
Anyones' eyes that are glazing over, look away now. IMHO it is relevant to horology, though.

When a current that flows through a coil, it becomes magnetised. Turning off the current suddenly makes the magnetic field collapse - this acts as a generator, giving voltage across the coil, just like a dynamo. You could say that energy is transferred from the magnetic field to the coil winding. The faster the current is switched off, the higher the voltage - nothing to do with the supply.
As this is an unstable arrangement, the voltage generates a magnetic field again, and so on and so forth, until losses eventually stop all this after a few cycles or so.
This first voltage spike is of opposite polarity from the supply, so the diode shorts it out and stops any further oscillation.
This is just a brief explanation without going into E=L dt/dt and stuff!
Your clock will now be OK. :biggrin:

 

[K Reindel]

I agree with Les but for a different reason. Adding semiconductor components to these clocks takes finesse and care. It can be done successfully but requires experimentation and analysis. What seems simple at first, isn't.

The good news is that a diode correctly placed across the coil will not see any more than the supply voltage (eg, 3V, or 12V, or 24V etc.)across it. And, wiring to the diode isn't really critical (but it is polarity sensitive).

The bad news is when you add a diode across a coil, you're also potentially increasing the time it takes to de-energize the magnet, slowing it down. It may not affect you, but it must be considered. For magnets and solenoids in SW clocks, it could add 5-20 milliseconds to the turn off time of the magnet.

The reason for the arc is this. When the switch instantaneously opens, the energizing current will want to keep flowing (the coil is inductive), and the voltage across the contacts will go to whatever value needed to keep this current flowing, which is theoretically infinity but practically under 1000V. This voltage will arc across contacts as they are opening.

The old self-winders used the resistor about 10x the value of the magnet's dc resistance to limit the voltage developed across the contacts while dissipating the energy in the coil QUICKLY. A 2 watt carbon resistor will work reliably and indefinitely, but a wirewound would be more authentic. A noninductively wound wirewound would be best but now we're getting fancy.

Also note:

"Ringing" occurs if there is enough winding capacitance in the coil to create a tuned LC circuit. Adding capacitance across contacts will create a tuned LC response, which if not done right, will cause additional problems.

Again: Avoid indiscriminately adding capacitors and diodes. The safest, simplest bet is the resistor.

We now return to our regularly scheduled horology.

Kindest regards
Ken

http://www.kensclockclinic.com/SWC.htm

 

[Ralph]

They have a good dissertation on Arc Suppression in the document found here:

Relay Contact Life

Ralph

 

[K Reindel]

Great article. It points out a problem with using a capacitor across the coil to suppress the arc--Instead of a high voltage arc the contacts see a high current arc. The solution to the high current arc is to add a resistor in series with the capacitor. Once again, the resistor acts to dissipate the stored energy.

Either way, the solution is to add a shunt resistor to dissipate energy stored in the inductor quickly. Our forefathers had the right idea. I've always been impressed with how much they accomplished with minimum resources.

Kind regards,
Ken

http://www.kensclockclinic.com/SWC.htm

 

[Schlitzer]

I want to thank everyone again who participated in resolving my issue with my ITR master clock. I have solved my spark arcing problem and this post is to cap it off so to speak and tell you what I did.

First, I installed a IN4005 diode in parallel with the movement winding coil, and another with the program unit coil. This resolved the arcing, but produced a rather annoying hum whenever the coils were energized. I don't know what caused the hum, but I suspect it might be some incompatibility with the power supply. I'm using a 12v 13watt dc unit. Does this make sense? Then, I removed the diodes and replaced each with a .5 watt 270 ohm resistor. I had previously measured approximately 30 ohms resistance across each of the coils, so 270 ohms seemed about right. This worked fine. The arcing is gone and so is the hum.

Now I am in the process of installing a slave and a bell. Hopefully, all will go smoothly, but Murphy's law rules most of the time.

View attachment 880

View attachment 881

 

[harold bain]

LLoyd, have you figured out how to program the disc programmer yet? It is quite easy once you figure out the relationship between the small wheel and the large wheel.
Is your slave clock 12 volts DC?
Harold

 

[Schlitzer]

Originally posted by harold bain:
LLoyd, have you figured out how to program the disc programmer yet? It is quite easy once you figure out the relationship between the small wheel and the large wheel.
Is your slave clock 12 volts DC?
Harold

Hi Harold. My slave clock is 24 volt, so I will use a relay. I bought a small transformer from Radio Shack and put a bridge rectifier on it. By the way, I bought a large 19 inch slave on ebay a few days ago. It's #180008154500 if you care to look at the listing. I got a nice place for it right beside the master clock so I'm looking forward to its receipt.
Thanks,
Lloyd

 

[harold bain]

Excellent wall clock Lloyd. The movements in these early clocks will last forever, I think. I have never seen one wear out. You will need 2 twelve volt relays to run the correction properly, one on the A line and one on the B line (switching the 24 volt power supply). Or, you could do without the correction, and twist the A and B lines together. That is how I run my slaves off a Stromberg master. Only have to impulse them during time change season.
Harold

 

[K Reindel]

Lloyd,

Congratulations on solving your problem and on a most beautiful clock.

You stated that you are using a transformer with a bridge rectifier. If you did not have about 100uF of capacitance across the rectifier output, you're feeding rectified AC (AC +DC) to the magnet coil. It will work, but you'll probably hear a buzz. The addition of the diode probably changes the harmonic content somewhat which makes the buzz more audible.

Ain't electronics wonderful?

Ken
http://www.kensclockclinic.com/SWC.htm

 

[harold bain]

LLoyd, just had a closer look at your ebay purchase. The movement is not original. It looks to me like a hybrid, part IBM, part Simplex. The correction contacts and cam are Simplex, the movement later (1950's) IBM. Strange that there are no wires off the movement to hook up to the system. Still, will be a nice looking slave clock.
Harold

 

[H Daniels]

Quote from very old post confuses me. Is the most positive end of the coil the side which connects to the positive battery terminal? Doesn't band on diode go to negative side?

Dan

The old method using a resistor is inefficient, and a compromise between suppressing the arc effectively and wasting power,

It is a much better idea to put a diode across the coil (observe polarity!) to immediately kill the ringing of the coil that causes the arc. It will also be much smaller.

Not sure why Jeffrey is against this - there are only two possibilities for polarity, and you only fit it once!
The band on the diode needs to go the most positive end of the coil, and a 1N 4001 will do - probably a few cents.

The old method using a resistor is inefficient, and a compromise between suppressing the arc effectively and wasting power,

It is a much better idea to put a diode across the coil (observe polarity!) to immediately kill the ringing of the coil that causes the arc. It will also be much smaller.

Not sure why Jeffrey is against this - there are only two possibilities for polarity, and you only fit it once!
The band on the diode needs to go the most positive end of the coil, and a 1N 4001 will do - probably a few cents.

 

[Toughtool]

This post is 14 years old, but to answer you question:"

Is the most positive end of the coil the side which connects to the positive battery terminal?

Yes. The cathode (the diode's band) should be connected to the positive side of the coil.You only want conduction to short the counter EMF the coil produces when the magnetic field collapses, after the master relay contact is opened.

IBM actually did a lot of things to reduce "sparking", which is inherent to switching inductive loads, including switching the AC side of the 12 and 24 volt rectifiers supplying power to the secondaries and other coils. They also switch the contacts of secondaries before the current is applied to their coils so these contacts were not damaged by sparking.

Remember back in 1927, silicon diodes did not exist. Rectifiers were Copper Oxide and later Selenium. Silicon diodes and transistors were invented by Dr. Shockley of Bell Labs, back in the mid 1940's. He and two others were awarded a Nobel prize for their work on semiconductors in 1956.

 

[H Daniels]

Thanks for reply, so if I understand correctly the diode is orientated to deal with the collapsing current when the points have just opened not when points are closed and current flowing from +ve terminal to -ve of battery. Hence cathode band on diode is closest to positive terminal of battery.
Perhaps I should stick with a resistor!
Thanks again
Dan

 

[Toughtool]

I would use a diode. No current used until the relay drops. Then the coil's induced energy (counter EMF) is just shorted across the coil. In a coil, current lags voltage by 90 degrees, because the coil is resisting a current change by inducing a voltage in the opposite direction to keep the current from changing. A resistor just slows this process down and converts some of the energy into heat. Arcing (IBM service instructions call it sparking) happens when ANY contact closes or opens under load. The greater the current, the more severe the arcing. A resistor reduces the current which reduces the arcing, but it is still there. I am working on a triac version of a master relay so that the relay contacts can be eliminated altogether.

Here is a circuit where I added the diode to protect the DC to DC converter. So far it is working where it failed without the diode.

 

[mxfrank]

You can use a bid directional TVS (transient voltage suppressor). These are specialized diodes that "clamp" at a threshold voltage. If you're switching 24V, you can use a 30V TVS, which would shunts surge current before voltage rises high enough to spark. The advantage is that you don't have to be concerned about polarity, and it's a robust device made specifically for this purpose. You can buy them wherever you buy diodes:

https://www.arrow.com/en/products/smcj30caq-13-f/diodes-incorporated

 

[Toughtool]

You can use a bid directional TVS

Is it available in something other than surface mount?

Perhaps I should stick with a resistor!

I woke up this morning thinking about the resistor. Not really knowing how it is connected to the circuit or the value of the resistor. If it is across the coil, then the resistance would be in parallel with the DC coil wire resistance. Two resistors in parallel would reduce the resistance and according to law, lowering the resistance would increase the current, therefore increase arcing. Am I wrong here? I=E/R and of course R=R1*R2 /R1+R2

 

[praezis]

You can use a bid directional TVS (transient voltage suppressor).

Much better than the "diode" method and slightly better (but more expensive and harder to find) than the "resistor" method!

Resistor: mounted parallel to the coil, value 10x coil resistance.

Frank

 

[mxfrank]

Is it available in something other than surface mount?

Sorry, my error. Try this one:

https://www.arrow.com/en/products/p6ke39ca/eic-semiconductor

Is it available in something other than surface mount?


I woke up this morning thinking about the resistor. Not really knowing how it is connected to the circuit or the value of the resistor. If it is across the coil, then the resistance would be in parallel with the DC coil wire resistance. Two resistors in parallel would reduce the resistance and according to law, lowering the resistance would increase the current, therefore increase arcing. Am I wrong here? I=E/R and of course R=R1*R2 /R1+R2

Whatever device you use is usually connected across the coil, but it could also be connected across the switch. If it's across the switch, some consideration is needed to avoid or reduce leakage when the switch is open. Voltage is what's critical to initiating an arc, current will impact the total energy discharged. The idea is to shunt voltage before it rises above the point where an arc can be struck. The reverse voltage spike created when the coil is disconnected is the problem, as it can rise well above 28v. The suppression circuit just creates a shunt which prevents voltage rising to the point that it can jump the gap.

The problem is more complicated than calculating ohm's law values, as you have an RLC circuit that will create a complex waveform before it gradually damps out. So in addition to V, I and R, release time is important. The various devices you can use differ in the cutoff voltage and the time it takes for voltage to decay to zero. Here are a few suppression methods compared:

https://jaxxeninc.com/pdfs/pcb-relays/tech_info.pdf

(A TVS functions like a combination of a diode and a zener)

 

[Toughtool]

"The total elimination of arcing is never possible, although arcing becomes insignificant if the current carried by a switch is low..." ( Switch principles:Arc suppression )​

I agree with this statement. The elimination of arcing is never possible. Just because you don't see it doesn't mean there is no arc and absolutely no damage to contacts when they open a circuit under load.​

This page (link) supports my theory of using a diode: Inductors - Inductive Kickback

and of course there is this: From the Q and A

 

[praezis]

It was already discussed here many times, and this is the essential part in your link:

The only disadvantage of this protection circuit is that it lengthens the decay of current through the inductor, since the rate of change of inductor current is proportional to the voltage across it.

For applications where the current must decay quickly (high-speed impact printers, high-speed relays, etc.), it may be better to put a resistor across the inductor,...

One of several reasons, why a diode protection is not the best solution in clocks.

Frank

 

[Toughtool]

"The only disadvantage of this protection circuit is that it lengthens the decay of current through the inductor, since the rate of change of inductor current is proportional to the voltage across it. "

I don't know where you got the above quote that supports your argument but to me a diode that is forwarded biased by the counter EMF of a collapsing magnetic field is a dead short to voltage and current when connected across the coil.
So I guess we will just disagree. Joe

 

[praezis]

Voltage goes down to 0.7V, current will stay and slowly decay until stored energy in the coil is zero. Just basics.

You are free to agree or disagree to an advice

Frank

 

[Toughtool]

Voltage goes down to 0.7V,

Yes, and 0.7 volts is not going to be a problem for the contacts that just opened.

current will stay and slowly decay until stored energy in the coil is zero

Slowly is a relative term here and so what. The contacts are open, electromagnetic energy is converted and absorbed, and the arc is suppressed. Not to mention that all this happens near the speed of light.

 

[praezis]

Amazing! I am out and wish good luck

Frank

 

[Wayne A]

Was reading one of the links above that referred to very high speed devices like line printers where diodes were not the best choice. Brought back a memory about commercial line printers made by DEC in the 80's. Those machines printed very fast and printed on every pass of the carriage both left and right and I've never seen anything go through reams of paper so quickly. Printer was in a 4'x4'x4' acoustic insulated box, open the hood and you needed hearing protection!

I've used diodes on high speed mercury relays that were pulsed in the 20 pulse per second range without issues. Funny thing about these relays is every few years one would stop working, just pull it out the socket give it a shake and it was good for a few more years. Mercury must have separated into little balls and shaking brings it back together.

Installed a diode on a kienzle automatic winder a while back and the clock already had a drain resistor which I just left in there. Only thought to check it with a scope after the diode was in place. Screen shot attached where I had used a jumper wire to pulse it. Looks good to me.

 

[mxfrank]

I don't know where you got the above quote that supports your argument but to me a diode that is forwarded biased by the counter EMF of a collapsing magnetic field is a dead short to voltage and current when connected across the coil.

Actually, a better bet would be a combination of a resistor and a diode in series. The diode would block forward flow, but allow the reverse kick to dissipate. The resistor would dissipate energy faster.

Describing this as happening at the speed of light is misleading, it's a relatively slow process. Energy is gradually dissipated as heat, due to the internal resistance of the coil. For example, I've used a diode alone to suppress sparking in a fuel pump, and the slowing of the pump is acutally audible. In a clock mechanism, delayed release could have implications to timekeeping.

You need to remember that ohm's law doesn't explain everything going on here. The reason you get sparking is that flyback voltage=L dI/dT, where L is the inductance of the coil and dI/dT is the rate of current change. dI/dT is determined by the time constant, T=L/R, where R is the serial resistance of the circuit. Increasing R decreases discharge time. When the switch opens what is the resistance of the circuit? It's not quite infinite, but it's a very large resistance which grows as the air gap opens. If R grows, then the time constant is reduced, and dI/dT becomes larger. Voltage rises towards infinity, creating a spark and limited only by the electrical resistance of the air gap. It stops when the energy stored in the coil is dissipated. as heat, which is what causes pitting.

The same is true for your flyback circuit. The diode alone provides a short circuit for discharge current. But since the only resistance is the serial resistance of the coil itself and the diode, the time constant is very large. By the simple addition of a resistor, the time constant can be reduced. This will result in faster discharge, but voltage will rise. The trick is finding a resistor that limits voltage rise to something low enough to prevent spark erosion., say 36v. There are calculators for this, if you hunt around.

 

[Toughtool]

I was being cute about the speed of light thing.

Similar to the time constant of a capacitor-resistor circuit, t=R*C, the inductance (the coil) has a time Constant. t=L/R, where t is time in seconds, L is the inductance in Henry, R is resistance in Ohms. We don't know what the inductance or time constant is in this direct current circuit. We do know the approximate resistance of the coils. Based on the voltage of 24 volts DC and the typical ITR/IBM time control system, the maximum of 2 amps per circuit is assumed. This includes a master movement wind coil of 43 Ohms (.132 amps), a program magnet coil of 22 Ohms (.545 amps), and sixty-six (66) model type 561-2 secondary coils of 240 Ohms each (* 0.024 amps ea). The lengthy process is to calculate the total resistance of 68 resistances in parallel using the formula of 1/Rtotal = 1/R1+1/R2+1/R3+1/R4+...1/R68. Or we can take a short cut and use Ohms law (R=E/I) that tells us that the total load resistance in this master clock system is the Voltage (24V) divided by the total current, in amps (I=2.0 amps) and is equal 12 Ohms.

Type Description ___________ 12 Volt 24 volt 12V to 24 V Conversion Resistor
25 Master's Winding Magnet .273 amps .132 amps 43 ohm 5 watt (actual 3.36 watts)
25 Impulse Accumulator .280 amps .140 amps 43 ohm 5 watt (actual 3.36 watts)
805-2 Program Magnet .545 amps .279 amps 22 ohm 10 watt (actual 6.54 watts)
561 Indicating Clock Movement (20V) *.028 amps. Use R1, 150 ohm 1 watt resistor to
(*20 volt coil measured with 20V applied: I=28 mA) convert coil from 20 to 24 volts.
561-2 Indicating Clock Movement .050 amps .024 amps 240 ohm 1 watt (actual .6 watts)
563-2 Indicating Clock Movement .333 amps .171 amps 36 ohm 5 watt (actual 3.9 watts)
565-2 Indicating Clock Movement .170 amps .085 amps 68 ohm 2 watt (actual 2.1 watts)
569-2 Tower Clock Movement .050 amps .024 amps 240 ohm 1 watt (actual .6 watts)

When the master relay is opened, the magnetic fields of all the coils suddenly collapses, at the speed of light minus the time constance, in seconds, of L/R. (t=L/C) It has been said the counter electromotive force (CEMF) voltage rise is 10 times the applied voltage (so >240 volts?). But I think this CEFM voltage could be higher because of the fast rise time due to the contacts opening. The diode starts conducting when it's junction reaches 0.7 volts and holds the voltage at 0.7 volts, not >240 volts. This is now a two voltage crop series circuit, a voltage drop of 0.7 volts and a voltage drop of >240 volt drop across the 12 Ohm total load resistance. This looks to me to be better way of protecting the contacts than the suggested threshold voltage of 30 volts from the scheme that mxfrank mentioned. So how is the resistor scheme wired? If the resistor is 10 times the load resistor, it would have to be in series to make a difference. A series resistor would certainly slow the decaying CEMF induced, and probably upset the current to all the coils. If the resistor is in parallel to all the coils, the total resistance would be closer to 12 Ohms. Rt-R1*R2/R1+R2 (or 1/Rt=1/R1+1/R2)

The point is; the contacts are subjected to a lower voltage after the the diode switches on, thous reducing the damage from arcing. The slow decay of the CEMF energy does not matter to the contacts because this happens after the contacts are opened. As long as the time constant is shorter than one second. Why? The two second contact, starting at the 59' 10” pulse from the master clock's movement happens every other second so all this starts over and over for about 21 pulses.

 

[Toughtool]

Actually, a better bet would be a combination of a resistor and a diode in series. The diode would block forward flow, but allow the reverse kick to dissipate. The resistor would dissipate energy faster.

I agree about the reverse kick dissipating with a forward biased diode. Inductor time constant formula is t=L/R.

See Inductor lesson: Magnetic Fields and Inductance | Inductors | Electronics Textbook

Algebraically, if L remains the same, then "t" goes down if R goes up because they are inversely proportional. The text below seems to support mxfrank's remarks that, "the resistor would dissipate the energy faster."

"Hypothetically, an inductor left short-circuited will maintain a constant rate of current through it with no external assistance:" and ...

"Practically speaking, however, the ability for an inductor to self-sustain current is realized only with superconductive wire, as the wire resistance in any normal inductor is enough to cause current to decay very quickly with no external source of power."

"When the current through an inductor is increased, it drops a voltage opposing the direction of current flow, acting as a power load. In this condition, the inductor is said to be charging, because there is an increasing amount of energy being stored in its magnetic field. Note the polarity of the voltage with regard to the direction of current:"

"Conversely, when the current through the inductor is decreased, it drops a voltage aiding the direction of current flow, acting as a power source. In this condition, the inductor is said to be discharging, because its store of energy is decreasing as it releases energy from its magnetic field to the rest of the circuit. Note the polarity of the voltage with regard to the direction of current."

As you can see, inductors are very complex electronic components. I do know that the open circuit ac voltage of my "Buzz box" Lincoln 250 amp arc welder is only about 25 volts. So it doesn't take much voltage to maintain an arc capable of melting a lot of steel together.