I not equal E/R; what's happening - Page 1
Forum section: electronics / general

author: gbynum
author_rank: Newbie
timestamp: October 21, 2020, 01:52:20 am
content: I've a friend who sells machines with heaters in them. He tries to match the heating elements (cal-rod style) closely for some unknown (to me) reason.To measure the (nominal 0.43-0.47 ohm) elements, he uses 2 procedures, a 4 wire resistance meter, VD480C+, which he checks with a "precision" 0.500 ohm resistor. He gets good data.As a check, he runs a moderate current from a Lincoln AC-225-S welder (the tombstone) through an element. He reads voltage across the element and current in the welding cable.Voltage has been read with 3 voltmeters, a Fluke 87V, a Fluke 179, and a Chinese import. They agree within 0.2%. All are advertised as "true RMS". Indication is 30.35 volts.Current has been read with an Amprobe clamp meter, and a Fluke clamp meter. Visual interpolation of the analog suggests 59.94 amps. The Fluke indicates 60.0 amps.If I accept the resistance, 0.45 ohm, and voltage, 30.35 volts, current is calculated at 67.4 amps ... but the meters indicate ~60 amps.It is my GUESS that the waveform from the welder is sufficiently nasty to confuse the current instrumentation; he has a Fluke Scopemeter with RMS calibration, and a 5-100 kHz current probe (80i-1000S) which he hasn't tried ...I'm of the opinion that the voltage and resistive element give the most accurate current of the 3, but am frankly confused at the magnitude of the difference. I tried using Google to search for a transformer waveform ... no luck (or inadequate search skills). Could the welder waveform be such that the true RMS instrumentation gives bad data?

author: WattsThat
author_rank: Frequent Contributor
timestamp: October 21, 2020, 02:33:41 am
content: An arc welder is not a pure dc supply. A big transformer feeding some big diodes in series with a big inductor. Lots of ripple there so are you measuring the welder output voltage as DC or AC?Then you’ve got the negative temperature coefficient of resistance of the heaters. Probably not high enough to account for 10% but that needs to be taken into account as well. Because of the tempco, I would expect voltage and current to be more accurate than measuring resistance which is guaranteed to change with the temperature rise.BTW, I agree, matching heaters makes no sense if there is any thermal mass. But hey, what do I know, I’m just an unknown clown on the Internet.

author: CatalinaWOW
author_rank: Super Contributor
timestamp: October 21, 2020, 02:47:44 am
content: The only reason I can see for matching heaters is to try to match element life. I don't know how much the elements vary but the low ones will run a little hotter and presumably fail somewhat sooner than their higher resistance mates. If this follows an Arrhenius relationship like many things do the life difference will be exponential with temperature difference, could be quite noticeable.While matching element life is arguably a good thing, it isn't obvious to me that it is good enough to be worth the extra effort.

author: Tom45
author_rank: Frequent Contributor
timestamp: October 21, 2020, 03:00:40 am
content: It's a heater. So it is meant to get warm. When it heats up, the resistance will rise.Is the measured 0.43 to 0.47 resistance measured at ambient temperature?Running 30.35 volts and 60 amps means the heater is dissipating 1800 watts. So it has to be getting warm and so its resistance is going to rise from that at ambient temperature. Actually I would have expected the hot resistance to rise more than seems to be the case from your numbers.

author: bob91343
author_rank: Super Contributor
timestamp: October 21, 2020, 03:46:42 am
content: It depends on the material of which the heater is made. Some alloys have very low temperature coefficient of resistance and others have higher values.Ohms law works perfectly when you don't try to compare apples with oranges. The voltage and current relationships in the vicinity of 1 kW will be different from those at low stress due to temperature rise. Even the best alloys change significantly with temperature. And it's not even a linear relationship. The shape of the wave, within reason, is immaterial as long as you are measuring rms.Even an apparently small detail such as the torque used on the fasteners to mount the elements will have a major effect on thermal transfer and, thus, temperature.

author: gbynum
author_rank: Newbie
timestamp: October 21, 2020, 03:25:58 pm
content: To WattsThat, it is an AC welder, no diodes.To the rest of you, the consensus seems to be that the resistance is changing with temperature. He buys these SUPPOSEDLY to not change R much with temperature ... and he looks at them dynamically from startup. We don't THINK it is a thermal effect.They are LARGE heaters ... the 30V on them for his receiving inspection just barely gets them warm. They will have 277 across them in the real application which involves a 2400F oven.We're trying to understand if, and how, the very inductive welder transformer supply, with perhaps E and I out of phase (?) would give these results. He's reading E across the heater terminals with a sense connection ... if the heater is a resistive load, won't E and I be in phase?Puzzled ...

author: IanB
author_rank: Super Contributor
timestamp: October 21, 2020, 03:34:48 pm
content: If you want a good measure of resistance, it's best to use pure DC for the measurement. Removes a whole lot of uncertainty.One possibility would be to use a stable DC supply like a laboratory power supply, or even a car battery to pass current through the element. While doing this, measure voltage drop across the element and current through the element. Whatever measurements you get from doing this will give the actual resistance under these conditions.If this resistance is different from what you get with a 4-wire resistance meter when measured between the same two terminal points, then the difference is due to temperature change. There is no other reasonable possibility.

author: KaneTW
author_rank: Frequent Contributor
timestamp: October 21, 2020, 03:38:48 pm
content: Current and voltage will be in phase for a resistive load, regardless of source.Where are you measuring the voltage? Is there possibly voltage drop in the measurement setup? 50 milliohm is very easily contact resistance somewhere.

author: Tom45
author_rank: Frequent Contributor
timestamp: October 21, 2020, 03:58:58 pm
content: gbynum, thanks for the further details.With the modest warming when driven by 30 volts, the 10% change in resistance (0.45 to 0.50 ohm) isn't that surprising. What is the resistance material for the heater elements?What is the expected current when an element is driven with the full 277 volts? And what would the calculated resistance be at that full voltage?

author: Conrad Hoffman
author_rank: Super Contributor
timestamp: October 21, 2020, 04:49:21 pm
content: Temperature is the likely culprit. If the welder waveform is really bad, you might be hitting the crest factor limit of the meter(s) or might need to handle AC and DC separately as in root sum of the squares. DC makes life a lot easier.

author: bdunham7
author_rank: Super Contributor
timestamp: October 21, 2020, 05:31:05 pm
content: Your RMS voltage measurements are likely accurate because you have multiple instruments and we know 2 of them to be models that are reliable and actual TRMS.I don't know what your resistance meter is, didn't find it on Google. Without knowing the specs and the standard and without a second instrument, you just don't know. A single calibration test point is inadequate at those levels and being 50mohms off isn't hard to imagine. I would measure the resistance with a lab power supply at a known current--say 1 amp--and then measure the voltage with your 179 and 87V. Or, measure the current with one and the voltage with the other and do the math.As for current, you didn't mention the model numbers of the current clamps. The welder transformer may well be somewhat saturated at that low of a current and may have distinctly non-sinusuoid output. If your clamp meters are not TRMS, then that alone could account for the difference.If you want to figure out the thermal effect on the element resistance, you can simply repeat your experiment at lower and higher currents, depending on what you have available. Plot out the results on a linear graph and if you have a straight line, the resistance isn't changing. If it curves, it is changing. Just saying they are not supposed to change is rather weak tea--a real spec would be more to the point. How hot are they getting when you test them? A low-TCR element might change 0.05%/C, which means 1% for every 20C. You are putting 1800 watts into them so they have to be warming up at least a little bit.EDIT: I meant to include: Perhaps they have some inductance as well? 120uH would be ~10% of the DC impedance.

author: babysitter
author_rank: Frequent Contributor
timestamp: October 21, 2020, 07:25:34 pm
content: Go for a wheatstone bridge - e.g. one arm (a voltage divider) of fixed 1:1 well-matched resistors, the other arm the series of two heaters to be compared, voltmeter between the arms centerpoints.The heater combination with the least voltage will be the best-matched, at any current.

author: gbynum
author_rank: Newbie
timestamp: October 22, 2020, 01:45:58 am
content: OP here. We'll close it out. A guy on the NEC forum said what we expected; the series inductance of the welding transformer and resistance of the load easily explain what we have.his numbers, For AC linear circuits, Z = E/I, and Z = sqrt(R2 + X2). So if your resistance meters are truly measuring R, what you've shown (for the simplest model) is that the welder - heater circuit has a reactance X = sqrt(0.505 - 0.450) = 0.23 ohms.It's the L as we suspected, but couldn't explain.Thanks!George

author: bdunham7
author_rank: Super Contributor
timestamp: October 22, 2020, 01:59:55 am
content: Quote from: gbynum on October 22, 2020, 01:45:58 amYou're going to have to review some basic laws. The reactance of the source can't make the current flow out of phase with the voltage in a purely resistive load. It could be 'L', as I mentioned above, but it would have to be the inductance of the heating element, not the source.

author: KaneTW
author_rank: Frequent Contributor
timestamp: October 22, 2020, 02:17:37 am
content: The inductance of the welding transformer doesn't matter. The series inductance of the load does.If you drive the heating elements with AC power, you will have the same issue. Due to series inductance your impedance will increase, so they will dissipiate less power.

author: gbynum
author_rank: Newbie
timestamp: October 23, 2020, 04:06:35 pm
content: OP here; thanks all. With your collective input, I understand that we don't understand.He's decided to do nothing, just take the readings he has on 2, one in 2018 and this one (which were very similar) and use that as his baseline in the future.I'd like to understand better, but he's ok as it is.THANKS!