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T.Hoffman
05-25-2012, 06:24 PM
We have a 20 watt, 36v DC solenoid valve we use for a particular application that includes timing cycles.

We have some prototyping to do, and would like to try a new low-power version of this valve that is only available in 24v and uses 5watts.
We would still need the same timing cycles as the previous valve, so would like to run it off the 36v timed source.

What's the best way to wire up this 24v valve to run off that 36v output?
Voltage divider circuit?

armedandsafe
05-25-2012, 07:32 PM
I would first try a series resistor. Measure the current drawn by the new relay, calculate the resistance for 12 VDC across your dropping resistor and get that size and wattage.

R = E / I (Resistance = Voltage divided by Current)
R = 12 / I (Make sure you pay attention to the decimal point when you record the current.)

Power (wattage) = E*I (Voltage times Current)

You only have to be close on the resistance. +- 10% is not going to make that much difference.

I had a similar problem once, and had to get a pump running NOW!. I had 24 VDC available to the burned out relay and 40 miles to town on a Sunday. I grabbed a couple of 12 VDC automotive relays from the mechanics and wired them in series. Then I paralled the contacts to insure they would not burn too badly before I could acquire the proper relay tomorrow.

Pops

darryl
05-25-2012, 08:04 PM
That would take a 56 ohm resistor, power rating 5 watts would be good. There's about 2.5 watts of heat to dissipate. For a cooler running resistor use a 10 watt.

armedandsafe
05-25-2012, 11:36 PM
That would take a 56 ohm resistor, power rating 5 watts would be good. There's about 2.5 watts of heat to dissipate. For a cooler running resistor use a 10 watt.

You are assuming the new relay consumes 5 watts. Which is what he said ("Uses 5 watts.") I was assuming, (I know!) that it was rated at 5 watts, which would mean that it actually draws less than 208mV. That was why I wanted him to get the actual specs or measure the draw.

I suspect the difference is going to be pretty slight. ;)

Pops

tdmidget
05-25-2012, 11:48 PM
Since 110V AC controls are pretty much standard in the US and 24vDc in the rest of the world How did you wind up with 36 vDC and why would you want to stay with it?

T.Hoffman
05-26-2012, 12:09 AM
Thanks for the info gentlemen.

Since 110V AC controls are pretty much standard in the US and 24vDc in the rest of the world How did you wind up with 36 vDC and why would you want to stay with it?
I have to deal with the cards I'm dealt. It's going into an application that has been in production for many years, using a number of 36v solenoids.

They're looking for the cheapest solution working with the current arrangement. If a simple valve type change can make a big improvement vs minimal costs, I'm sure there is no motivation to be changing anything else.

armedandsafe
05-26-2012, 12:17 AM
Thanks for the info gentlemen.

I have to deal with the cards I'm dealt. It's going into an application that has been in production for many years, using a number of 36v solenoids.

They're looking for the cheapest solution working with the current arrangement. If a simple valve type change can make a big improvement vs minimal costs, I'm sure there is no motivation to be changing anything else.

That being the case, I'm sure the cost of a 10 watt resistor is not going to be a breaker. ;) Check sizes and be sure to include mounting for the resistor, rather than just hanging off the end of the connecting wire.

Pops

T.Hoffman
05-26-2012, 12:21 AM
That being the case, I'm sure the cost of a 10 watt resistor is not going to be a breaker. ;) Check sizes and be sure to include mounting for the resistor, rather than just hanging off the end of the connecting wire.
This is just for prototype experimentation.
If it turns out that this is a decent idea and worth pursuing, the valve company says with the amounts we need, they would make a 36v version...

armedandsafe
05-26-2012, 12:50 AM
For prototype operation you can get by with the resistor inline with the connecting wire. I've discovered over the years of designing machine controls that it is not good in the field when the left-footed maintenance people get to messing around with your pretty machine. :D

One trick I did, which won't be applicable in your case, was build the resister into the relay mount. That one didn't work, because the techs insisted on using Tandy relay mounts as replacements because they were cheaper/easier to find. :(

Pops

T.Hoffman
05-29-2012, 03:25 PM
I would first try a series resistor. Measure the current drawn by the new relay, calculate the resistance for 12 VDC across your dropping resistor and get that size and wattage.

R = E / I (Resistance = Voltage divided by Current)
R = 12 / I (Make sure you pay attention to the decimal point when you record the current.)

Power (wattage) = E*I (Voltage times Current)

You only have to be close on the resistance. +- 10% is not going to make that much difference.

Turns out the new solenoid only uses about 2.4watts, measured at ~100ma with 24v DC supply.

So my resistor should be around a 120ohm 5watt (or 10watt) resistor inline to my new solenoid with the 36v supply? sound right?

05-29-2012, 03:36 PM
120ohm is correct but you should only need a 2w resistor at that current?
Max.

armedandsafe
05-29-2012, 03:38 PM
A 12VDC drop at 0.100 amps would be 12/0.1, thus 120 Ohms. Yup, you got it. ;)

The 5 Watt will be sufficient, but check the heat given off by that component in operation, with concern to other items near/touching it. The 10 watt ones might be needed for that reason, while not needed for operational reliability.

This sounds like a good solution. Keep us posted, eh?

Pops

darryl
05-29-2012, 03:43 PM
The resistor is only going to be dissipating 1.2 watts, but go with a 5 watt to keep the temperature rise down. No need to start melting insulation, etc.

armedandsafe
05-29-2012, 03:45 PM
Max, 2 watt resistor will be fine, for the steady state math, but you have to take into consideration the flash heating at pull-in. You are driving an inductive device and the initial in rush through that resistor is going to be pretty hefty. I ran into that when designing trigger circuits for cannons. In that case, the relay was tripping and releasing many times per second, so the demand was much higher that I think we will see on this application. However, I would still start with the 5 watter. (Yes, I do tend to get anal, at times. :D )

Pops

05-29-2012, 04:04 PM
Max, 2 watt resistor will be fine, for the steady state math, but you have to take into consideration the flash heating at pull-in. You are driving an inductive device and the initial in rush through that resistor is going to be pretty hefty.
Pops

But that would only apply to AC solenoids, with DC the inrush is not seen, only the coil resistance.
With AC it is not only the inrush which would be less than one cycle duration there would also be high current until the armature shifted over.
AC solenoids have a very low resistance and the current is not limited until the inductive effect takes place, (after 1 cycle and pull in occurs).
This is why DC solenoids and relays are superior, if the armature/spool on an AC solenoid does not shift, or someone pushes it over manually, a burn out results.
Max.

T.Hoffman
05-29-2012, 04:05 PM
This will be buried in some heat shrink, and around some other cabling bundles so I'd rather be on the safe side.

Plus I don't have any 120 ohm 5 watters here at the moment.
But I DO have several 250 ohm 5 watts here.

How 'bout a couple of them in parallel, wouldn't that be equivalent to 125 ohms at 10watts?

darryl
05-29-2012, 04:12 PM
That would be perfect.

T.Hoffman
05-29-2012, 04:18 PM
cool- thanks guys for the help!
I'll be trying this out tomorrow.

vincemulhollon
05-30-2012, 10:32 AM
So... did it work?

Wanna try something fun? Assuming you have a low current 36 volt lamp laying around (key is low current, like a small fraction the current the solenoid draws...) put that lamp across the resistor. Or three 12 volt lamps in series.

Now you have a test indicator as seen here:

1) lamps are dead = you got no power
2) lamps are incredibly dull red glow barely visible because you're running 36 volts of lamp off 12 volts = all is well
3) lamps are bright full brightness = your solenoid is shorted out and you're running 36 volts of lamp off a 36 volt drop. This is also why guys are telling you to dramatically over-spec the dropping resistor... if the coil shorts that "12 volt drop" instantly changes to "36 volts" and thats gonna smoke a low rated resistor for sure maybe start a fire or something. I think the other guys know that but I didn't see them mention the "WHY" reason.

You could stick an old fashioned analog voltmeter with color coded red and green ranges there instead of a lamp. Been there done that. Of course an old fashioned analog voltmeter movement is like \$30 now.

A pity incandescent lamps have such a ridiculous temperature coefficient that you can't just stick a 2 watt 12 volt lamp in series with the 2 watt 24 volt solenoid and skip the resistor.

This ends my episode of "dumb tricks with old fashioned light bulbs"

T.Hoffman
05-30-2012, 10:49 AM
So... did it work?

Well... I dunno yet.

I designed and fabricated a fluidics mounting block for the new solenoid in clear acrylic with input/output ports, turned out great.

Made up a new cable for powering the new solenoid with appropriate connector housings and pins, and found my 5w resistors to put in there as well.

Then, just as a precaution I figured I'd better check the actual power in the instrument that we are connecting this to.
The one that is currently powering a 36v DC solenoid, at least that's what the label says.

I measure the output across the pins with nothing connected and get around 17 volts DC......
I sent some emails out to people more familiar with this instrument than I am, wondering what is going on here.

05-30-2012, 11:38 AM
What type of equipment are you connecting to? PLC etc?
If you are testing at the output and it happens to be solid state device that is operating the output, you could be getting strange reading because of this.
A better explanation of what you are connecting to and the nature of the output would be better.
Max.

T.Hoffman
05-30-2012, 11:58 AM
What type of equipment are you connecting to? PLC etc?
If you are testing at the output and it happens to be solid state device that is operating the output, you could be getting strange reading because of this.
A better explanation of what you are connecting to and the nature of the output would be better.
Max.

Due to proprietary issues with this prototype instrumentation, I'd better not get into that too much.
Better safe than sorry for my job.....

But I checked the output of other 36v sol valves on that instrument, and they actually ARE outputting 36v DC . So I know it's not my meter or how I'm measuring them....
I can measure the output with the valve connected or with no load and both ways get 36v DC.

The 17.5 output for that particular valve is not an isolated case. I checked several other instruments, and all measured the same.

I connected the new valve without the dropping resistors in-line to see how it would work at the 17.5v DC.
Worked fine, so I'll tinker with that until I get an answer from others here at work about that 17.5v output.

05-30-2012, 12:25 PM
If this is a custom piece of equipment then I would think whoever designed the output functions should be involved in order to qualify what exactly is happening here?
Max.

T.Hoffman
05-30-2012, 12:37 PM
If this is a custom piece of equipment then I would think whoever designed the output functions should be involved in order to qualify what exactly is happening here?
Max.
Precisely the emails sent yesterday.
But knowing how slow the wheels turn at times, it might be a while before I get a knowledgeable response.

vincemulhollon
06-04-2012, 08:53 AM
But I checked the output of other 36v sol valves on that instrument, and they actually ARE outputting 36v DC . So I know it's not my meter or how I'm measuring them....
I can measure the output with the valve connected or with no load and both ways get 36v DC.

The 17.5 output for that particular valve is not an isolated case. I checked several other instruments, and all measured the same.

I connected the new valve without the dropping resistors in-line to see how it would work at the 17.5v DC.
Worked fine, so I'll tinker with that until I get an answer from others here at work about that 17.5v output.

Pulse width modulated output? Put a 'scope on the other outputs see whats there. 18 times 2 is suspiciously close to 36 volts...

T.Hoffman
06-04-2012, 08:57 AM
Nope, nothing fancy about the output for this solenoid vavle.

It operates a drain opening....

Still waiting for a response to my inquiring emails.

Paul Alciatore
06-04-2012, 01:14 PM
Do keep in mind that the driving circuit is presently providing 36 Volts at a certain current. When you substitute a 24 Volt coil with a series resistor, it may require more current. Generally speaking, a particular design of a relay or solenoid requires a certain strength of magnetic force from the coil. When various coil Voltages are provided, the lower Voltage coils do need higher current (current X number of turns needs to be constant).

Perhaps this is what you are talking about when you talk about the power, I don't know. But if the transistor or other device that is providing that current to the coil, then it will need to be able to provide the current needed by your substitute device with a 24 Volt coil which is likely higher.