Ohms test on solenoid mystifying


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Old 08-20-08, 06:13 PM
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Ohms test on solenoid mystifying

I was testing to diagnose why a power vent gas water heater was not working. I tested everything - including a couple of gas valve solenoids, that mount right on the longer style gas valve. This system is employed also on gas dryers.

Well, at first I thought I had a solenoid issue (common as cause on dryers when they do not fire up) because I got a 'no-reading' through the thread like spool of wire (solenoid). You can remove these solenoid spools right off the rod they sit on that protrudes from the gas valve. To see these spools - they look only like the wire enters the spool into a spade, coils around and exits at another spade next to the incoming wire.

When I put my ohms meter on the bottom solenoid, I got a reading of 180 ohms. But on the top identical-looking solenoid I got no reading at all. Dead. So I thought. This led me to all sorts of unnecessary phone calls and trips.

Lo and behold, back at the address, the lady is home next door in the duplex rental. I ask her if I could test her identical water heater. I test HER solenoids. I got a reading. Then I tested the bottom. I got a reading. Both 180 ohms. Then I retested her top one again and this time I got no reading. I exclaimed under my breath, "What?!", thinking the gods were toying with me. I kept touching my probes together to make sure I wasn't having meter problems. Nope.

Then I made a discovery. On the bottom solenoid, test, if you touch red probe to + terminal and black probe to - solenoid spade terminal, you get 180 ohms. But on the top solenoid, you get no reading if you do it that way. But you WILL get the 180 ohms reading by reversing the probes on each spade.

??????????

Never heard of having to reverse probes on an ohms test. Only for DC volts.

??????????

Have at it.

................................

Came back to say these solenoids are like spools of thread, even in size and size of wire, and are rated 24 VAC and 12 VDC .3 and .4 amps each. -if this helps you.
 
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Old 08-20-08, 06:23 PM
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It is possible to see what you are experiencing if there is a diode or rectifier in the solenoid.
 
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Old 08-20-08, 06:41 PM
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Came back to say these solenoids are like spools of thread, even in size and size of wire, and are rated 24 VAC and 12 VDC .3 and .4 amps each. -if this helps you.
HHmmm. If I install a single diode in a circuit, 24 volts AC will become around 12 volts DC and since a diode will only allow current in 1 direction, that would explain the polarity dependant reading.

180 ohms seems like a lot of resistance though but if you have gotten similar readings on multiple solenoids, I would suspect it is proper.

So, at the problem house, did you recheck the solenoids? Did they both read 180 when using correct polarity?

Did you check for voltage when they should be energized?
 
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Old 08-20-08, 07:41 PM
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If this gas valve is used with a "flame rectification" flame proving system then having a polarized solenoid on the main gas valve makes perfect sense. The 24 volts AC is rectified by passing through the flame rod to the main valve coil. This is a safety feature in that applying straight AC (via the flame rod shorting to ground) will NOT open the main valve.
 
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Old 08-21-08, 06:20 AM
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Originally Posted by nap View Post
HHmmm. If I install a single diode in a circuit, 24 volts AC will become around 12 volts DC and since a diode will only allow current in 1 direction, that would explain the polarity dependant reading.

180 ohms seems like a lot of resistance though but if you have gotten similar readings on multiple solenoids, I would suspect it is proper.

So, at the problem house, did you recheck the solenoids? Did they both read 180 when using correct polarity?

Did you check for voltage when they should be energized?
Both water heaters now work. Both have 180 ohms (18 x 10 on my x10 range scale) on analog meter.

Reason for water heater outage was bad control module even though I thought it was testing good. I tested for that from the get go, and never thought that was it, based on my results of the energizing I was showing. Each one, (the one at each duplex, before I changed out the bad module) allowed HSI to come on and gas valve to click and hold 24 volts at + at solenoid for 2 seconds and then HSI turned off (slowly dies down in brightness). Both water heaters did this and timed out the same - at least I thought they did, while I was counting outloud.

But my mistake may have been in using a digital meter during this. Maybe the numbers do not change as fast as they would have if I had used an analog. (My analog's volt function burned out and I need a new analog meter - maybe just for this very reason!)

I got 73 ohms on my digital meter off the brand new IDENTICAL HSI, and 33 on the other duplex's HSI, so I swapped out burners with HSI's attached, thinking I'd now get fire = no.

The only thing I can figure with being fooled by all my tests with all the inline 24 volt safeties (that all showed good) and the control module itself, was that there was say a 1/10th of a second differential in time when the transition occurs between gas valve staying on during HSI glow and when the HSI goes out.

The only reason I ended up swapping out the control module, and not hastily ordering the new gas valve from out of state, direct (I had already talked to the company LD, and they said they'd ship to me direct - not all companies will) was based on my gut. I knew that gas valve failure (now that I knew the solenoids were acting like the other duplex's water heater that worked = good) is like 20 times more rare than having the control module fail.

You are probably spot on about your polarity and flame rectification theory, as each solenoid's spades are marked + and -. But I still cannot visually SEE this, and have to have it explained to me how resistance can be had one direction (when I had one wire detached from spade, and unit turned off), and yet not the other direction. ??? I cant't envision a spool of wire thread that has it one way but not another.

Does a law of magnetism come into play, regarding this?

......................................

furd,

You are probably right. But I still need more of an explanation to understand how wires can seemingly seem 'disconnected' one way, but not the other way. ?????
 
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Old 08-21-08, 06:27 AM
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Originally Posted by GregH View Post
It is possible to see what you are experiencing if there is a diode or rectifier in the solenoid.
I welcome any DETAILED explanation about this, as I cannot visualize. Visualization is perhaps my biggest ally in my ability to fully understand and repair all the stuff I have worked on through the years. I can "see" things. Like x-ray vision, in a way. THIS I cannot "see".
 
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Old 08-21-08, 08:58 AM
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I do not know for a fact, never worked with the equipment, but from what was posted here this is how I visualize it. At the beginning of the coil on the solenoid there is a diode (rectifier). It is hidden by the winding but connects one end of the coil to the spade lug. This means current can travel only one way. Your average multimeter uses a battery to check for resistance. Since it is DC your measuring with and DC only flows one way through a diode it won't work if the leads are reversed.
 

Last edited by ray2047; 08-21-08 at 11:51 AM.
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Old 08-21-08, 03:25 PM
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Okay. Thanks. I can follow that. But now explain how current can get through one direction in a diode and not the other way. Or is there some site I can go to that shows an exploded view of some diode where I can better visualize this? Maybe this will be my first baby steps towards becoming an electronics expert.
 
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Old 08-21-08, 04:07 PM
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You have a good example of how a diode works in your automotive battery charger.

A transformer reduces 120 AC volt line voltage to 12 volts AC.
The diode in the charger allows the current to pass only in one direction creating modified direct current power.
If you were to reverse the connections on the diode no current will flow.
 
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Old 08-21-08, 05:23 PM
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Well. I am at that point, stuck in my thinking, that we now know that it can only go one way through a diode. But my problem is I guess I do not know the insides of diodes to understand how that can be. So, since electricity usually flows like water, the best way I can come to understand this, is in thinking with one polarity, the water can flow downhill. But reverse the polarity and it can't flow uphill.
 
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Old 08-21-08, 05:54 PM
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Just do a google on "semiconductor". The diode is THE basic semiconductor and many other devices derive from its characteristics.
 
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Old 08-21-08, 06:35 PM
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here is a site that explains things a bit but very simply put, a diode is a one way check "valve" for current.

http://www.allaboutcircuits.com/vol_3/chpt_3/1.html

the single diode simply stops 1/2 of the current (1/2 of the sine wave) so that effectively cuts the voltage in 1/2 so using 24 volts AC and a diode ends up passing 12 volts DC. (not an exact resulting voltage but that becomes another lesson).
 
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Old 08-22-08, 04:44 AM
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You may want to check that, Nap. A diode isn't a voltage divider. A single diode on a 24 volt AC transformer will pass 24 volts through the diode (less the .7 or .3 volts used by the diode), but current will flow only in one direction.

http://en.wikipedia.org/wiki/Rectifier
 
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Old 08-22-08, 06:23 AM
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This weekend when I have a bit more time, I will look this up. I am no electronics type man. I find it fascinating how current can flow one way but not another. So I have learned something quite revealing here.

I wonder what person was able to figure out something like that could even be done? We live in an amazing world.
 
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Old 08-22-08, 03:40 PM
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Originally Posted by Rick Johnston View Post
You may want to check that, Nap. A diode isn't a voltage divider. A single diode on a 24 volt AC transformer will pass 24 volts through the diode (less the .7 or .3 volts used by the diode), but current will flow only in one direction.

http://en.wikipedia.org/wiki/Rectifier
so, lets see; to calculate the voltage of an ac waveform, you utilize the peak to peak voltage, correct?

So, if I remove 1/2 you are saying you will still have the full 24 volts?


Oh, ok, but just for fun, visit the following websites.

http://www.wisc-online.com/objects/i...?objID=SSE4203



http://www.wisc-online.com/objects/i...p?objID=SSE402

this is the other lesson I referred to previously.
 
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Old 08-22-08, 07:17 PM
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Originally Posted by nap View Post
so, lets see; to calculate the voltage of an ac waveform, you utilize the peak to peak voltage, correct?
No, you use zero-to-peak (half the peak-to-peak) voltage divided by the square root of two (approx. 1.414). The peak voltage of a "24-Volt" AC circuit reaches almost 34 Volts briefly each cycle. A half-cycle later it is nearly -34 V.
Originally Posted by nap View Post
So, if I remove 1/2 you are saying you will still have the full 24 volts?
Yes. The peak is still 34 Volts (less the .7V diode junction potential) above ground (or below if you reverse the diode). Crazy isn't it? It gets even better when you start trying to do something useful with the rectified AC.

The resistor in the "lab" - your first link - dissipates the same amount of power (that is, it heats up the same amount) as it would if it were across a DC voltage equal to the peak AC voltage (minus that .7 V JP) divided by Pi, or about 10.6 VDC for your 24 VAC example.

If you replace the resistor in the "lab" circuit with a capacitor, the cap will charge up to the peak voltage (about 34V in your example - actually almost 33.3V due to that pesky .7V drop in the diode) and stay there. If you place a resistor (or resistive load of some kind) in parallel with the diode, the capacitor's charge will bleed off through the load and the voltage will drop slowly or quickly depending on the values of R and C, until it charges up again the next cycle. If C is big enough and your 24-V transformer and wiring have no losses (fat chance of that in the real world, but in an arm-waving theoretical example, why not?), the resistor will dissipate nearly the same amount as if it was across a 33.3 V DC source.

Welcome to AC circuit analysis.

Originally Posted by nap View Post
Oh, ok, but just for fun, visit the following websites.

http://www.wisc-online.com/objects/i...?objID=SSE4203

http://www.wisc-online.com/objects/i...p?objID=SSE402

this is the other lesson I referred to previously.
Good stuff (except they misspelled 'sine' as 'sign' early in the second. Oops...)!
 

Last edited by ByteWrangler; 08-22-08 at 07:35 PM.
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Old 08-22-08, 07:42 PM
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They also misspelled "occilliscope" (oscilloscope).

Remember, nap, that 120VAC is the RMS voltage, not peak-to-peak. (Household peak-to-peak voltage is actually RMS divided by .707, or 169.7 volts.)

If a 24-volt AC transformer is used as the source, it too is rated at its RMS value.

AC RMS voltage of a pure sine wave into a resistive load is the same as the circuit's DC voltage into that same load. This does not apply to other waveforms (sawtooth, square, etc.).

One more tidbit: Use two diodes in the circuit in opposite orientation and the output voltage will double. Fascinating.
 

Last edited by Rick Johnston; 08-22-08 at 08:00 PM.
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Old 08-22-08, 08:11 PM
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ok. Here is a real simple formula for half wave voltages:

Half rectified wave Vrms= Vpk/2

so, if we are dealing with 24 volts ac

33.94/2=16.97 volts. or more correctly 33.94-.7/2=16.62 (rms) or if you want average 7.632

Still not the 24 you want to claim.

That is why you can run a 12 volt DC solenoid on 24 volt 1/2 wave rectified voltage.

bytewrangler; you are getting into totally different areas not germane to the situation.
 
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Old 08-22-08, 08:36 PM
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Originally Posted by nap View Post
ok. Here is a real simple formula for half wave voltages:

Half rectified wave Vrms= Vpk/2

so, if we are dealing with 24 volts ac

33.94/2=16.97 volts. or more correctly 33.94-.7/2=16.62 (rms) or if you want average 7.632

Still not the 24 you want to claim.

That is why you can run a 12 volt DC solenoid on 24 volt 1/2 wave rectified voltage.

bytewrangler; you are getting into totally different areas not germane to the situation.
Actually, for pure sinusoidal voltages (and only for pure sinusoids, as Rick notes),
Vrms = Vpk / sqrt(2) = Vpk / 1.414 (approx).

33.94 V / 1.414 = 24.00 V.

Rick - Missed the misspelled oscilloscope. I guess I'm better with small words.

Isn't the peak-to-peak of "household 120 V" +169.7 to -169.7, or 339.4 Volts? 169.7 is the maximum difference from ground, or zero to peak.

Lest everyone be totally confused, note that multiplying by 1.414 (the square root of two) is the same as dividing by .707 (half the square root of two). Similarly, dividing by the square root of two is the same as multiplying by half the square root of two. It works only in this case.
 
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Old 08-22-08, 09:13 PM
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from your own post bytewrangler:

The resistor in the "lab" - your first link - dissipates the same amount of power (that is, it heats up the same amount) as it would if it were across a DC voltage equal to the peak AC voltage (minus that .7 V JP) divided by Pi, or about 10.6 VDC for your 24 VAC example.
without getting too specific (which I was trying to stay away from because it makes for these uneeded discussions given the original post), you , in that statement above, supported my original claim as to the voltages available to the solenoid based upon a dc current and a half wave rectified current of the originally indicated levels. (12 VDC and 24 VAC)



but since you want to get picky.

Isn't the peak-to-peak of "household 120 V" +169.7 to -169.7, or 339.4 Volts? 169.7 is the maximum difference from ground, or zero to peak.
there is no true relation between ground and the line voltage unless there has been such a relation due to a connection to ground made. Also, 120 volts is merely the nominal voltage. Your actual voltage may vary and generally does so if you guys want to be picky, how about calling it nominal grounded household voltage? Anything else would be presuming facts not stated.

so, even after all this, the voltage available to the solenoid from a 24 volt ac half wave rectified supply is still not 24 volts as Rick stated.

Rick Johnston: You may want to check that, Nap. A diode isn't a voltage divider. A single diode on a 24 volt AC transformer will pass 24 volts through the diode (less the .7 or .3 volts used by the diode), but current will flow only in one direction.
enough?
 
 

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