ergodic Posted March 20, 2010 Posted March 20, 2010 My usage into what SCE called "Tier 5" marginal pricing: about 34 cents per KWh. It's a motivator to try and trim power consumption. So, for two weeks I've been trying to figure out a branch circuit I have that's drawing about .8A when nothing on it is on - at least as far as I and the electrician who wired it can find. The circuit is kitchen/dining overhead and garage area lighting. There are about 12 Insteon controllers on the circuit, a mix of KPLs, dimmers and relays. Totally frustrated at not being able to find where the power is going, I put an old KPL on the bench. I connected power and neutral and clamped the current at 2 - 3ma - about what I'd expect. So then I pulled the cover off a garage box with two live controllers in it. When I disconnected the two switchlincs (which were off), current on the branch dropped from 800ma to 680ma. One device in this box is a SL relay that is just used as a 2-way and the load isn't even connected. The other is an outside door light dimmer. Both were in the off state. Each of the two switchlincs appears to be dead-drawing about 60ma (about 7W!). That seems awfully high, but multiplied across all the controllers on that circuit it would explain the off-state loading. So has anyone ever measured the overhead loading for any of the Insteon controllers? The SH data sheets I have say nothing specific about it. 60ma at 120V is about 7W which seems way too much. Is it possible that some way a miswiring or defective controller somewhere in the branch could cause this to happen?
Illusion Posted March 20, 2010 Posted March 20, 2010 In my studies, the off state draw of most the devices I tested were in the area of 7W. I have found a few that are far less than that, but 7W is not unusual in my experience. I do wish they were less.
IndyMike Posted March 20, 2010 Posted March 20, 2010 Hello Ergodic and Illusion, The current measurements that you've observed appear in line with what I've seen in the past. The good news is, you're not being charged for this. This is because the devices are drawing power "out of phase" - it's referred to as apparent power. Your utility company only charges you for the "real" component (in phase portion) of this power consumption. In order to perform this measurement you need a device with power-factor correction. It measures both the current and voltage (or current and phase) to compute the real component of the power. Wiki: Power Factor As an example, I have a lamplinc plugged into my UPM energy meter. Current consumption is 80 ma. Corrected power consumption is 1 W. This device is drawing current nearly 90 degrees out of phase from the voltage waveform. Here's a list of devices that Dave Houston baselined some years ago. Dave improved the accuracy of his meter through long term tests: http://davehouston.net/x10-power.htm
Illusion Posted March 20, 2010 Posted March 20, 2010 Wow! Hi IM, I had not even considered Power Factor. These things are crazy. I just tested 6 modules and am seeing a PF of .2 on all the standard appliance lincs and lamp lincs. I think that is the lowest PF I have ever seen on a consumer product. 6VA 1W PF .18 Interestingly, the new dual band lamp lincs are very different: 1VA 0W PF .8 So I will correct my earlier statement. I do see about 7VA not Watts. The power company does indeed charge residential customers for watts, not Volt-Amps. If you were in a commercial or industrial building that may be different though. IM is absolutely correct and I retract my statement about wishing they drew less power.
Brian H Posted March 20, 2010 Posted March 20, 2010 Yes the power line derived power supplies use a capacitor for the voltage drop and mess up many meter readings because of the phase shift. I also use a Kill A Watt when I do tests.
IndyMike Posted March 20, 2010 Posted March 20, 2010 Hi Illusion, You're absolutely correct about commercial applications. Power companies penalize "big users" with poor power factors (causes problems for their distribution systems). It's interesting that the dual band LL's are showing better power factors. There may be "incentives" for oem's to improve power factors on small devices as well. Most older CFL's exhibited horrible power factors. I was surprised that this was allowed. Obviously a "green" trade that was made at the time. My recent dimmable CFL's are showing PF's of ~ .9 so things are improving. Interestingly, the new dual band lamp lincs are very different: 1VA 0W PF .8 So I will correct my earlier statement. I do see about 7VA not Watts. The power company does indeed charge residential customers for watts, not Volt-Amps. If you were in a commercial or industrial building that may be different though.
ergodic Posted March 20, 2010 Author Posted March 20, 2010 I've found in the past that the Kill-A-Watt (at least the basic one I have) not too useful. While it seems to handle the power angle compensation OK (at least it says it does), it's totally confused by switching dimmers. Mine basically shuts down or reads 0 as soon as the dimming level goes below about 95%. Does the UPM meter handle this better? Is there anything I could easily rig to clamp-and-tap in the panel without actually pulling wires out of breakers that would give a true power reading?
IndyMike Posted March 20, 2010 Posted March 20, 2010 Hello ergodic, I'm a bit surprised that your Kill-a-watt gets confused with dimmers. I had understood it to be equivalent to, or better than, my UPM EM100. Are you measuring incandescent or CFL loads? If CFL, the Kill-a-watt may have some problems with the poor power factor on some CFL's. I can say that my UPM appears to measure both incandescent and dimmable CFL loads under full-on and dimmed conditions. My dimmable CFL is a Sylvania 15W. It measures 19,52 VA and 15 W (PF ~. with the UPM. Hopefully someone with a Kill-A-Watt can comment on their use with dimmers. IM
markens Posted March 20, 2010 Posted March 20, 2010 This is fascinating. In retrospect, it does make sense that these devices have measurable current draw with such low PFs when off given the capacitance coupling. Hopefully someone with a Kill-A-Watt can comment on their use with dimmers. I dug out my Kill-a-Watt and tested a bunch of relatively recent devices (both dimmer and relay), using a 60W incandescent light bulb as the load. I saw stable readings for all dim values I tested (0/22/44/50/65/87/100%), so the Kill-a-Watt did not seem confused by dimmers with an incandescent load. I didn't try a CFL load. My results mirror what others have posted WRT OFF values. --Mark
IndyMike Posted March 20, 2010 Posted March 20, 2010 Mark, Thank you for the confirmation. My understanding was that the Kill-o-watt was pretty good in determining device consumption and PF. At one time I actually considering buying one as a backup. IM
ergodic Posted March 21, 2010 Author Posted March 21, 2010 Well, I just tested mine again. The old and crappy (=$20) Kill-a-watt shuts off its display at somewhere just about 90% dim level when connected through to my highly-sensitive, tightly-engineered, resistive test load: an old kennel pad heater from a dog of long ago. When I looked, the UPM 130 meter spec indicates it measures power factor. Is it able to display it, or does it just incorporate it in the true power computation?
IndyMike Posted March 21, 2010 Posted March 21, 2010 ergoic, Sorry to hear about the Kill-a-watt. It should handle a resistive load easily. Don't mean to be insulting but - you aren't connecting the Kill-a-watt to to output of the dimmer are you? The UPM EM130 does appear to display power factor (My older EM100 does not): EM130 Manual
ergodic Posted March 22, 2010 Author Posted March 22, 2010 Not insulting - it's exactly the sort of stupid thing I seem to do all the time. But, no. Still it seemed an interesting question. So I tried it on the load side, after the lamplinc. Connected there it shuts off almost instantly - as soon as the level goes below 100%. So I toddled over to my local Home Depot this afternoon and picked up a new 4460. And it works perfectly. Go figure. For years I've just assumed it was the nature of the design.
apostolakisl Posted March 22, 2010 Posted March 22, 2010 Hello Ergodic and Illusion, The current measurements that you've observed appear in line with what I've seen in the past. The good news is, you're not being charged for this. This is because the devices are drawing power "out of phase" - it's referred to as apparent power. Your utility company only charges you for the "real" component (in phase portion) of this power consumption. [/url] Let me see if I understand what you are saying here. By drawing power out of phase with the 120V ac the switch is effectively drawing less voltage. Since power is volts time amps, with lower volts, there is less wattage despite the high current load. Yes? If so, how does the insteon draw out of phase? Seems like everyone else out there uses a transformer to drop the voltage.
markens Posted March 22, 2010 Posted March 22, 2010 Let me see if I understand what you are saying here. By drawing power out of phase with the 120V ac the switch is effectively drawing less voltage. Since power is volts time amps, with lower volts, there is less wattage despite the high current load. Yes? If so, how does the insteon draw out of phase? Seems like everyone else out there uses a transformer to drop the voltage. The issue is AC power theory vs DC power theory. For DC, your understanding is correct that P=VI. AC power calculation is somewhat more complicated because of its frequency component. A power factor of 1.0 means the load is purely resistive (incandescent light bulb, heater element, etc), and the current flowing is exactly in phase with the voltage frequency. Power factor goes towards 0.0 as the load's reactive elements increase (motor winding, capacitor coupling) and the current flowing shifts away in phase from the voltage. Watts is the measure of real power (in phase with the voltage), and is what residential electric meters measure. The VA (volt-amp) is a measure of apparent power, which is what the wires must actually carry. Generally you'll see appliances such as heaters (PF=1) rated in Watts and those with motors (PF<1) rated in VA. Wikipedia has a good article with more info about power factor. --Mark
oberkc Posted March 22, 2010 Posted March 22, 2010 Perhaps another way of looking at this is to recognize that household AC power varies in voltage from +110 to -110. It is not a constant 110 volts. At any given point in time the voltage could be anywhere in between. Current, is also alternating between extremes. The power factor recognizes that peak current (amps) may, or may not, occur at peak voltage. If peak current occurs when voltage is only 60, then power would be amps x 60V. The power factor is a way to define how far peak current shifts away from peak voltage. So, even with AC, power = volts times amps. It is just that you cannot assume that volts and amps are simultaneously at their peaks.
apostolakisl Posted March 22, 2010 Posted March 22, 2010 I was just reading that wikipedia page again and am sort of getting this. It looks like you need a lot calculus to figure the area under the curve. It would appear that there is an ebb and flow of power as capacitors are charged from the utility and then discharged back to the utility as the utility voltage drops and crosses 0. So I don't know how a meter measures current and volts in a situation like this. Seems like you probably need a fancy one to correctly measure all of this stuff. I can see why the electic company wouldn't like this as it would mess with the wave form and voltages if done on a large scale. So does anyone then actually know how many watts the insteon switches actually consume and therefore what role it would have on our electic bill? Incidently, typical 110 (or 120) is the average voltage. The sine waves peak much higher. The equation is 110 (or 120) times square root of 2. For 120 household current the peaks are 170 volts.
oberkc Posted March 22, 2010 Posted March 22, 2010 So does anyone then actually know how many watts the insteon switches actually consume and therefore what role it would have on our electic bill? You may want to go re-read some of the early posts. It appears that there has been some attempt to answer this question, before we all hijacked the discussion.
Zick Posted March 22, 2010 Posted March 22, 2010 So does anyone then actually know how many watts the insteon switches actually consume and therefore what role it would have on our electic bill? You may want to go re-read some of the early posts. It appears that there has been some attempt to answer this question, before we all hijacked the discussion. I reread the whole thread and still somewhat confused too. I was always under the impression that I was getting charged for all my insteon devices (which I have a lot of!) but does it matter how many of these are in a house? Sure 1 or 10 of these things isn't much to the power company but what if you have say 100 of these devices? That's a possible 700 (watts or VA (still little confused about which one it is)). Would that many make any difference? Sorry electricity is not my strong suit.
apostolakisl Posted March 22, 2010 Posted March 22, 2010 I saw some stuff from indymike where he listed lamplincs at 1 watt and another at 0 watt. I think something is up with that 0 watt, maybe a rounding error. Anyway, I only have a couple of lamplincs and appliance links but have tons of switchlincs. I didn't see anything about switchlincs. 1 watt seems reasonable which would mean that my entire insteon setup would still be under 100 watts. Also, what about the ISY? It gets power from the PLM. Does the PLM draw different when it has the ISY connected?
markens Posted March 22, 2010 Posted March 22, 2010 I reread the whole thread and still somewhat confused too.I was always under the impression that I was getting charged for all my insteon devices (which I have a lot of!) but does it matter how many of these are in a house? Sure 1 or 10 of these things isn't much to the power company but what if you have say 100 of these devices? That's a possible 700 (watts or VA (still little confused about which one it is)). Would that many make any difference? Sorry electricity is not my strong suit. Bottom line is that the typical residential electric meter measures (and you are charged for) Watts, not VA. Measurements of insteon devices, including switchlincs, show they consume no more 1W each, which is what you are charged for. So 100 devices would total less than 100W. Don't know about usage of PLM/ISY. --Mark
Illusion Posted March 22, 2010 Posted March 22, 2010 My ISY draws about 1.5W. I measured this when I was building my UPS power supply for it. http://forum.universal-devices.com/view ... 4309#24309 Not sure how much the PLM pulls.
IndyMike Posted March 22, 2010 Posted March 22, 2010 I was just reading that wikipedia page again and am sort of getting this. It looks like you need a lot calculus to figure the area under the curve. It would appear that there is an ebb and flow of power as capacitors are charged from the utility and then discharged back to the utility as the utility voltage drops and crosses 0. So I don't know how a meter measures current and volts in a situation like this. Seems like you probably need a fancy one to correctly measure all of this stuff. The old electro-mechanical meters were a work of art (spinning dial meter). Fortunately, with modern microprocessors, we don't need a strong physics/e-mag background to do this anymore. Most of the current devices simply sample the Voltage and Current waveforms at a high rate. Average power is then a simple multiplication of the two over a time period (T). Note: I have seen my UPM get "confused" in the presence of voltage spikes (CFL's, etc). This normally occurs at low power consumption/near 90 degree phase shift ( PF near 0). It will sometimes alternate between 0 and 8W on my LL (0w being the correct reading). Adding a filter to the input of the UPM typically corrects this fluctuation. So does anyone then actually know how many watts the insteon switches actually consume and therefore what role it would have on our electic bill? The Dave Houston link that I provided earlier lists measurements for a number of devices. I trust both Dave's methods and the numbers he provided. Since there wasn't much data on current Insteon devices, I've just kicked off a long term (kWh) test on a spare LampLinc I have. I'll report back in about a week. IM
ergodic Posted March 22, 2010 Author Posted March 22, 2010 I actually found a clamp power meter I can justify: $149! Displays power factors down to 30% - or at least so it claims. God knows if it's any good at that price but I'll post back when I've got it in hand and tried it.
IndyMike Posted March 22, 2010 Posted March 22, 2010 I actually found a clamp power meter I can justify: $149! Displays power factors down to 30% - or at least so it claims. God knows if it's any good at that price but I'll post back when I've got it in hand and tried it. Hi Ergodic, An inductive power meter (clamp style) may have problems measuring the low current draw of your Inseon devices. Also the 30% power factor number - is that a typo? Most HA devices consume power (off power) with power factor well below that (15 - 20%). If your intent is to measure circuit currents to major devices and lighting loads, these items may not be a concern.
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