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Advice to speed up the decay time of an electromagnet when it is turned off please.

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Triumph Herald

New Member
Hello All,

I have an electromagnet made from a Microwave oven transformer.
I am only using the coil with the thicker wire.
I've separated the transformer core into the "E" shaped bit and the "keeper" shaped bit and removed the other coil.
When I turn the current off, I would like the attractive force between the two halves of the core to decay more quickly than it does at present.

The magnet is powered by a 6V lead acid battery.
I turn the magnet off using a Crydom DC60S3 SSR controlled by the DTR line of the serial port on a laptop. I've written a simple Perl programme to turn it off.

Following advice I've found online and in the local electronics shop I currently have a diode across the coil to allow the back emf to do its thing and not damage the Crydom. The diode says: GW506 IN5404 on it.

I have a 7 ohm resistor in the circuit to reduce the strength of the magnet to what I want.
My multimeter measures the resistance of the coil as zero, so I assume that it is less than 1 ohm (or less than 0.5 depending on how it does its rounding).
The Crydom data sheets says:
Maximum On-State Voltage Drop @ Rated Current=1V (rated current is 3A).
Maximum Turn-Off Time [msec] =0.3 (but that is under the heading "Input Specs")
I can't find information about how long the Crydom takes to go from "conducting" to "not conducting" on the output side.
My meter is telling me that the coil current is 0.7 A.


During use, the coil is energised for approximately 45 seconds, then turned off.
It isn't re-energised for about 30 seconds after that.

My priorities are:
- the attractive force between the two parts of the core to fall away as quickly as possible.
- not damaging the Crydom

I don't care about:
- energy efficiency
- time for the magnet to gain its attractive force (if less than 5 seconds)

I measure the decay of attractive force of the magnet by hanging it vertically with a known weight on it (15kg) and measuring the time taken for the two parts of the transformer core (the "E" shaped bit and the "keeper) to separate after the PC turns the Crydom off.
I determine "separation" using a second circuit on the same PC to measure when a current can no longer be passed through the 2 parts of the transformer core with a crude attempt to debounce in software.

At the moment separation times are of the order of 0.25 to 0.35 seconds.
I would like to get it down to less than 0.05 seconds if possible!

I'm looking for advice on circuits and component values for a snubber circuit which might help me get there (or any other means!)
I did "A" level physics 35 years ago so the theory comes back to me when I read stuff on the internet, but I'm just too rusty to sensibly calculate values.

Thanks in advance for any help you can give me.

Crydom DC60S3
www.crydom.com/en/products/catalog/d_c60.pdf

If you get this far and are curious as to why I'm doing all this, this is the reason:
I am building a Track Cycling start gate for my son.
The "E" shaped part of the transformer core is (indirectly) attached to a fixed post and the "keeper" bit is (indirectly) attached to the bike.
The Laptop generates the countdown "beeps" and triggers the Crydom at time=T0, then records the separation time with the other circuit.
In the future I want to measure the forward displacement of his hips during the "start" phase (pre time=T0) and of the bike (post time=T0) so that we can plot velocity, acceleration jerk and momentum and compare it to frame by frame video from the side and the "separation" time above.
Should be fun.

Cheers
Geoff.


 

Les Jones

Well-Known Member
Most Helpful Member
Hi Geoff,
I think it is unlikely that the turn off time of the SSR is the cause of your problem. I think the problem is that the diode across the coil to protect the SSR from back emf from the coil works by maintaining the current in the coil until the stored energy in the inductance of the coil is lost. (Via heat in the coil resistance and the forward volts drop of the diode.) Adding a resistance in series with the diode should help but the resistor value must not be so high that the voltage developed across is would be enough to exceed the voltage rating of the SSR. The voltage rating of the SSR is 60 volts so I would limit the voltage to about 45 volts. As you say that the current is 0.7 amps then the maximum safe value of resistance would be 45/0.7 = 64 ohms. Although the average power disipated in the resistor will be very low the peak power will be about 31 watts. I would choose at least a 5 watt resistor. (probably wire wound.).

Les.
 

crutschow

Well-Known Member
Most Helpful Member
For fast turnoff of a coil you need to rapidly dissipate the inductive coil energy.
The fastest way to do this is to add a zener in reverse series with the 1N5404 diode (anode to anode).
The zener generates a higher reverse voltage across the coil when the switch opens, to more rapidly dissipate the inductive energy in the coil and drop to zero coil current.
With just the 1N5404 diode, the energy is dissipated only in the diode forward drop and the coil resistance.

The zeners value should be such as to not exceed the relay's 60V rating.
With a 6V battery, this means the zener voltage should be no more than about 50V or so.
Here's a 51V, 5W zener that should handle the 0.7A current pulse when the switch opens.
 

Triumph Herald

New Member
Thank you both very much. I think the penny has just dropped!

Would I be correct in thinking that both the Zener and the resistor are both doing the same thing (providing a resistance), but while the resister behaves according to Ohm's law the Zener has a high effective resistance until we get to 51V but then the voltage doesn't need to increase much as the current increases considerably.
In either case we don't care about current flow in the other direction, because the standard diode prevents it.

I'll try to get these components and try them out and report back the separation time at 15kg.
I understand that these 2 solutions work separately and I won't try them together!

Thanks again for your help.
Cheers
Geoff.
 

Les Jones

Well-Known Member
Most Helpful Member
Hi Geoff,
The zener will be slightly better than the resistor. If you wanted to make it even faster you would need a switching device (SSR mosfet or transistor.) with a higher voltage rating which would allow you to use a higher value resistor or higher voltage zener. You are correct that you should use either of these solutions but not both.

Les.
 

spec

Well-Known Member
Most Helpful Member
Hi TH,

I have traveled in many Triumph Heralds and Vitesses.

Here are some observations about your system- do not take any of what I say personally- I only mean to help.

Using an RS232 serial interface DTR line for accurate timing can have problems- RS232 is not a real time interface and is specifically designed to have slow edges.

The other problem with RS232 lines is that they are only required to provide 5V/3K= 1.66mA and the Croydon solid state relay (SSR) type DC6053 requires a minimum turn on current of 2.2mA. But, a typical PC (personal computer), including laptop, RS232 interface line would probably be capable of supplying more current than the RS232 specification minimum limit.

So that sets the scene: you say that the solenoid (microwave oven transformer [MOT]) is taking in the order of 0.3 Secs to de-energize. The problem is that your method of measuring the de-energize delay may not be accurate due to the non real-time nature of the laptop PEARL/RS232 route.

My advice, to assist a resolution to your problem, is to measure the signal across the solenoid and compare that with the time that the solenoid bar actually separates from the rest of the transformer core. An oscilloscope would be ideal to display the signal across the solenoid but, failing that, a moving coil multimeter (set to the volts range), or a high efficiency LED and suitable series resistor may suffice.

Incidentally, along the lines that Les described in post #2, you could try this: instead of connecting the stubber diode directly across the solenoid, you could try connecting the snubber diode across the solenoid and the seven-Ohm current limiting resistor (anode connected to the SSR terminal and cathode connected to the 6V battery positive terminal).

But, Les provides a more effective solution.

spec
 
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MikeMl

Well-Known Member
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Telling simulation:

28.png

A 5H inductor with an initial current of 5A. The diode represents the typical diode snubber. Vz represents adding a Zener of various voltages. The first case, (Green traces), is when Vz=0, or no Zener at all; just the snubber diode. Note that with these values, it takes almost 30s for the current I(L1) to decay. It is hard to see on trace, but the voltage V(c) is clamped at ~1V (forward drop of a Si diode).

As Vz is stepped to 12V (red traces), the current I(L1) decay time is cut dramatically to ~2s, while the voltage V(c) is clamped at 12V+one diode drop.

As Vz is stepped to progressively higher voltages (25, 50, 100; blue, violet, yellow traces), the decay time is cut even more, but at the expense of V(c) going higher and higher. The voltage rating of whatever is switching the current in the inductor must accommodate the peak V(c).
 

ericgibbs

Well-Known Member
Most Helpful Member
hi TH,
This clip from a datasheet covers SSR's with DTR drive.
E
 

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MikeMl

Well-Known Member
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Before there is too much jubilation, I reran the sim for 2s, and plotted the power in the simulated Zener V1 (top plot pane, same color scheme), and in the diode D1 (bottom plot pane). The energy stored in the inductor has to go some where! Where are you gonna find a 25W to 100W Zener?

28p.png

Note that the area under all of the Zener dissipation curves is constant...
 
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MikeMl

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Here is a way of shifting the power dissipation into a power transistor. I am showing the 12V Zener case. The top pane is the power dissipation in the Zener itself, and the bottom pane shows the power that the transistor has to handle.

29.png

The examples I used are a bit extreme compared to Geoff's actual case, but just scale the currents appropriately...

Oh, and I am sure that you all can see that simulation is totally worthless...:D
 
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crutschow

Well-Known Member
Most Helpful Member
To go along with Mike's simulations, here's one that shows the difference in decay time for a zener and a resistor that gives the same peak voltage.

upload_2017-2-26_10-17-46.png
 

Les Jones

Well-Known Member
Most Helpful Member
Hi crutschow,
You simulation show that your zener solution is a lot better than my resistor solution.

Les.
 

crutschow

Well-Known Member
Most Helpful Member
...................
You simulation show that your zener solution is a lot better than my resistor solution.
It somewhat depends upon the dropout current of the solenoid.
 

Triumph Herald

New Member
Thank you all.

I'll start with a 56 ohm 5W resisistor (because that's what I can get hold of).

MikeMI
I think that I understand your simulations.

Your initial simulation is for 5H and 5A.
My current is less than 1A so the 5W resistor might survive.
Ah, but I have no idea what my coil is. Was 5H an educated guess, or just a value to start with?

If a Zener is damaged by the power level, does it fail to high resistance or low resistance?
I assume that if it fails to (very) high resistance, then next time I will have no diode protection at all?
And if it fails to low resistance then I'll be back to a slow release with just the original diode.
Is that correct?


Can I put 5 off 10V Zeners in series to get an effective 50V Zener?
They would each see 1/5th of the power dissipation then wouldn't they? (manufacturing tolerances aside)


MaxheadRoom.
I assume that you mean deliberately reversing the polarity of the supply.
I wondered if that was possible.
Is there a single component that I could fit to achieve this?


spec.
Thanks for the appraisal of my system.
It was one of those things that just grew and grew.
I kept going down this route, because a previous version with a much smaller magnet recorded a disconnect time in my drop tests of a a few milli seconds, and this was good enough. But that magnet wasn't strong enough, so I got the bigger magnet and all this started !


Thanks again.

Cheers
Geoff.
 

crutschow

Well-Known Member
Most Helpful Member
If a Zener is damaged by the power level, does it fail to high resistance or low resistance?
I think they usually fail shorted.
The high power damages the junction and it ceases to act as a diode, but just becomes a conductor.

Given the time to release you observed of 0.25 to 0.35 seconds with just a diode, I would expect your inductance is less than a Henry.
It that case a 5W, 45V zener should be able to dissipate the energy of 0.7A through that inductance without damage..
 

MaxHeadRoom78

Well-Known Member
MaxheadRoom.
I assume that you mean deliberately reversing the polarity of the supply.
I wondered if that was possible.
Is there a single component that I could fit to achieve this?
It is normally either done electronically, H bridge etc, or mechanically with either a switch or a relay.
Max.
 
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