adding a dummy load to a circuit OR seperate dummy loads - Page 1
Forum section: electronics / beginners

author: lots o tots
author_rank: Contributor
timestamp: March 24, 2018, 04:07:05 pm
content: I have two (2) separate temperature control relayshttps://www.creatroninc.com/product/12v-10a-temperature-control-relay-55-to-110c/. the relay can control AC or DC power, i have been using the AC method using a ac/dc power adapter so far but can change to DCi am going for similicity and cost here if possible.I want to add a load to both circuits, but would like to try to match up the total power consumption of each of these loads as close as possible. the actual power consumption in Watts does not matter as long as it is the same. i thought of a few different options, but i realize there may be some better options. The relay switches turn on and off every one minute or so.which options do you think would provide the best results?1. using a 5V to a heating padhttps://www.creatroninc.com/product/5v-heating-pad-5x10cm/?search_query=heating+pad&results=2and a 5 volt ac/dc power supply. I realize that the power draw of 2 different circuits will be similar, but not the same. I was wondering if it was possible to add a lower power resistor to one of the circuits to increase the power consumption. I see many resistors available that are 1/8 or 1/4 watt. Is it possible to under power a resistor that it will consume less?2. same method as number one, but using two (2) separate, new and identical low power 12v fans 80mm or 50mm fans.3. dummy load (could this just be a simple resistor with a high accuracy)?4: other option?accuracy is highly desired. i would prefer to have at least 0.1%, but i don't know if that is a pipe dream.

author: ebastler
author_rank: Super Contributor
timestamp: March 24, 2018, 04:36:15 pm
content: Why do you feel you need to match the load? (At all, or even to 0.1%??)Maybe you have a good reason, but I would like to understand your use case first, before we all start to give advice on more or less costly and complex solutions.

author: lots o tots
author_rank: Contributor
timestamp: March 24, 2018, 05:44:06 pm
content: QuoteBefore the explanation, if you think there is a much better way of doing this, i am all ears as you have stated.This is not the most efficient design due to me being a beginner, but i am going for simplicity too and doing my best. This is for a home project, so accuracy or efficiency and reliability of design does not have to be perfect. If this was for my workplace, i would do this in a much more efficient manner such as using a proper PID controller among many other things, but want to do the best i can within reason of time and cost.i am going to be putting an ac load inside a 2 separate enclosed boxes and want to measure the power consumption of both using two different loggers. through some quick calculations, i figure i need about 12 watts continuous (25 watts minimum with relay) or so for each box and will use a light bulb for this (i figured the light energy will be absorbed by the opaque surfaces, since most of the consumption will be heat and using AC due to high wattage. i will measure the exact electrical consumption of the bulb to calculated heat loss. i would like to use a heat emitted of some sort ceramic heater instead of a standard incandescent bulb, but need something inexpensive). This may be a weakness in my logic, but i cant think of something better.The goal is to measure the power consumption using an arduino and a current sensor. The circuit is for the arduino is from an online tutorial, so that part is covered.the relay i mentioned will control the bulb. but on the same circuit, i plan on adding an AC/DC adapter (it will have inefficiencies in unit and power factor issues) to add a much smaller DC load. It is the DC load that i will measure and apply a (calculated multiplication factor in my logger) to the DC load value to calculate the A/C power consumption.some of my electronic gear is not high end, so i would like to rely on good component tolerances where possible. That is why the relative value is important. If both components read 0.XXXX W of power consumption that is fine, i will just change my (calculated multiplication factor).even though there are inefficiencies in the system, i think being able to control an accurate load will help.Please ask me any questions for clarification if needed

author: ebastler
author_rank: Super Contributor
timestamp: March 24, 2018, 07:38:47 pm
content: Thanks for the explanation. But I think I am still missing something; I am not sure I understand what the ultimte goal is and how you want to approach it. Three questions in particular:(a) Do I understand correctly that you want to ultimately build a two-channel power measurement device for AC loads, where the two light bulbs just serve as "test" or "demo" objects? (I.e. later you want to connect unknown loads?)(b) I understand that the "DC loads" are meant as reference loads, which remain permanently attached to you measurement device, and you want to use those to calibrate your measurements of the unknown "AC loads" -- right?(c) I struggle with your use of "AC load" and "DC load", and with understanding the purpose of using both in your system. A lamp, as well as a resistor, can be connected to either AC or DC, so the load per se is not an AC load or DC load. You seem to say that you want to connect AC voltage to your "real" test objects, and DC voltage to your "reference" objects. If I got that right: Why use two different voltage sources? Why don't you just switch between either your unknown load (light bulb), or a reference load (e.g. a higher value precision resistor, e.g. 1 kOhm), connecting either of them tothe same AC voltage?Precision voltage and current measurements are an art in its own right, and you can develop that to high degrees of perfection. The "Metrology" secton of this forum is the place to go when you want to get serious about this.But I think there may be some fundamental misunderstanding (just as likely on my side as on yours) which we should sort out first.

author: lots o tots
author_rank: Contributor
timestamp: March 24, 2018, 08:37:40 pm
content: Quotenot exactly. there will be two (2) separate chambers with two (2) separate relays, (2)two seperate arduinos.... so it is not two channel because everything is separated. the parts are so inexpensive that i thought it would be easier to keep everything seperate and easier for a novice. both loads will be connected at the same time, i just want to measure the one with low power (DC) as the current measuring device is DC.QuoteyesQuoteI dont know how to connect a high power item to DC, i figured it was easier just to control/measure an AC load (using the AC and DC control device in my original link). i also didnt know how to measure that high power device. both loads would go in the champber.QuoteI didn't realize i could put that resistor in A/C at all or and measure the refernece load. didn't know i could out a light bulb in DC. I am just so used to DC. the current sensor i have is a DC sensorhttps://www.adafruit.com/product/904.also how would i be able to sense something using an arduino if it was in AC?

author: ebastler
author_rank: Super Contributor
timestamp: March 24, 2018, 09:13:23 pm
content: Quote from: lots o tots on March 24, 2018, 08:37:40 pmAh, then I had misunderstood your earlier post. I thought you already had an AC sensor circuit from a tutorial, and wanted to add a separate reference for higher accuracy.If you have a DC sensor, the easiest solution would be to base your whole sensing on DC voltage and current. Have a relay to switch between the unknown load and a reference load, as you had planned (I believe), and use DC for both.(This assumes that all loads you want to measure can be used with DC. Light bulbs certainly can; some motors e.g. are AC only. I am still not quite sure what the ultimate goal of your circuit is. Do you want to measure other objects besides the lamps? Or do you want to measure the lamps under different conditions, e.g. different operating temperatures?)

author: lots o tots
author_rank: Contributor
timestamp: March 24, 2018, 09:41:34 pm
content: Quotethe purpose of the relay is to turn off the circuit once it reaches a certain temprature and turn it back on if it gets too cold. both loads will be on at the same timethe ultimate goal is to change the insulation of the housing box. the interior conditions will stay the same. when having reduced insulation, then the power ddraw increases. measure an accurate DC power draw and log it. I only planned on measuring the "refrence load".30 Watts seems like alot in DC, meaning higher cost power supplies, but i will see what i have lying around in my collection.I also want to know if can i increase the power draw of one circuit just by adding a low power resistor. if a resistor is rated at 1/4 watt, is that the maximum draw or always at 1/4 watt? It would be awsome if i can make fine adjustments just by adding a resistor far less than 1/4 wattin additoin, what are your thoughts about my original 3 questions?

author: ebastler
author_rank: Super Contributor
timestamp: March 24, 2018, 10:21:37 pm
content: Aaahh, now I get what you want to achieve (or I think I do). Let me recapitulate -- please confirm whether I got it right:So, the purpose of this device will be to evaluate the effectiveness of various thermal insulation options. You want to build two identical copies of the same circuit, and of the same heater + insulation box, for comparative studies. As part of making them "identical", you want to ensure that the heaters consume the same electrical power (and convert it into heat). Right?If that is indeed the goal, I don't see why your Arduino device needs to measure the electrical power consumed by the heaters (lamps) at all. Why don't you just measure thetimefor which the heaters are turned on or off? If the insulation is weaker, the duty cycle (on time vs. off time) will become higher, and vice versa. Or maybe that is what you intended to do anyway?If you are worried that your two heaters (lamps) do not have identical electrical properties, and hence electrical power, you can calibrate them once, using a multimeter, before you install them in the circuit. I don't think you need to check this continuously via the Arduino.I assume that you want to run both heaters from the same supply voltage. To make their powers match, you can add a small resistor in series with the lamp which has the lower resistance, to bring it to the same total resistance as the other lamp. Mount this resistor inside your insulated box together with the lamp. This may be what you are describing as option (1) in your original post?The power rating of a resistor specifies themaximumpower it can tolerate before it overheats. You can "under-power" a resistor by driving it with a voltage which results in less electrical power being converted to heat in the resistor. (And youshouldunder-power it by at least a factor of 2 in this application, better a factor of 5, since it can't easily get rid of the heat.)I am not sure what the fans in your original option (2) are supposed to do, or where you would position the dummy load (your option 3). But the series resistor option should be fine. It works with wither AC or DC (across the lamps and resistor), by the way; use whatever voltage source is available.

author: lots o tots
author_rank: Contributor
timestamp: March 24, 2018, 11:36:32 pm
content: Quotei honestly couldn't have said it better myself. yes.Quotei actually thought of that. the issue is i dont have the skills to (log) time in the arduino, at least the tutorial didnt cover that. i could use an electro mechanical minute counter, but they are not easy to find. I do have two identical "kill a watt" logging devices which would be pefect, but i want something that i can modify better.Quotei am only using the arduino because the temperature draw will change will time as the system reaches steady state. my calculations incidcate is it may take hours. i did plan on calibrating them, but how would i change the properties? Buck or boost coverter?Quoteassumption is correct. I know this can be done for DC, but if i choose AC (undecided) load, can a resistor be added to AC too? i thought resistors handled only minimal current.Quoteperfect! I just need to calculate the resistor value. i will probably run at 12V or 5V depending on my available power supplies. V=IR or P=V2/R? i have never sized a resistor before. i suppose this will need to be done later after i find out the difference.Quotethe fans are a loads themselves. I wondered if brand new 12V computer fans with the same spec would draw almost identical power.It would seem that a precision resistor may still be the best choice.

author: ebastler
author_rank: Super Contributor
timestamp: March 25, 2018, 08:02:01 am
content: Quote from: lots o tots on March 24, 2018, 11:36:32 pmIt will work in the same way for AC. As long as you work only with resistive elements (heaters, lamps, resistors), not capacitors or inductive coils, any DC calculations can directly be transferred to AC, using the "effective" (root mean square) values for AC voltage and current.AC is not automatically higher current than DC. But yes, if you want to power a heater and run it on a low voltage, you will probably deal with currents of several Ampere, and need to size your resistors appropriately (see below). The standard resistors used in small-signal circuits will indeed be specified for somewhere between 1/8W and 0.6W only, but you easily can get resistors specified at 10W (and much more, but eventually they start to get expensive). See e.g.https://www.reichelt.de/11-Watt-axial/2/index.html?ACTION=2&GROUPID=3120https://www.reichelt.de/from-50-Watt-axial/2/index.html?ACTION=2&LA=2&GROUPID=5274(Switch to English in the upper left, if the site doesn'a already come up that way.)By the way, if you already have the current sensor modules for DC, I would still suggest powering the heaters by DC for simplicity. If you need 2*25W of heating power and use 12V heaters, that's less than 5A total. You can get 12V 5A power supplies cheaply, for use with LED strips or halogen spots. Higher powers, e.g. 12V 10A, are also available. The Adafruit current sensor will limit your power, due to its 3.2A limit.QuoteYes, that's how it works.You mentioned earlier that you want at least 25W heating power, and your Adafruit current sensor can only handle 3.2A. So you will need to run at 12V, since 5V * 3.2A will not give you enough power.Assuming a heater module specified at 12V and 30W, that would run at 2.5A, and would hence have a resistance of R = U/I = 4.8 Ohm. Assuming that the variation between units is no more than 20%, your matching resistor will be anywhere between 0 Ohm and 1 Ohm. For the power rating, the highest resistance is the worst case. P = I² R = (2.5A)² * 1 Ohm = 6.25W.I would use a 15W to 25W resistor in this case, because it will run hot inside its insulated box. If the resistance you need to match your two heaters/lamps is smaller, the required power rating will also be lower.

author: lots o tots
author_rank: Contributor
timestamp: March 25, 2018, 04:34:46 pm
content: I think i am going to stick will A/C because the extra cost of the resistor with high tolerance needs and i dont have two good power supplies as needed.that means i will need two resistors. one for the AC load to balance it, and one precision DC load to measure with low toleranceI think i need to figure out the AC resistor later.Regarding the DC resistor which i am considering mounting outside the box,I was on the newark element 14 site and navigated to the web page below. i really want to get a 0.01% tolerance resistor (maybe 0.1%?) between 0.12W to 0.6W as your recommended. please correct me if i am wrong. if i run it at 5V and aim for 0.25W, then the current would be 0.05A which requires a resistor of 100 Ohms.if i use the following resistor which has a 0.6W rating. you said it was a good idea to run a resistor below its rating to ensure it doesn't break which makes sense.http://canada.newark.com/vishay-foil-resistors/s102c-100r00-0-01/resistor-metal-foil-100-ohm-600mw/dp/35H5066how to ensure it actually runs at 0.05A as i am aiming for?the following url is just a list of resistors available at newarkhttp://canada.newark.com/w/c/passive-components/resistors-fixed-value/through-hole-resistors/prl/results/2?resistance-tolerance=posneg-0.005pc|posneg-0.01pc|posneg-0.02pc|posneg-0.025pc|posneg-0.05pc|posneg-0.1pc&range=not-exc-nic&sort=P_ATT_BASE_VALUE_1000001_EN_US

author: ebastler
author_rank: Super Contributor
timestamp: March 25, 2018, 04:55:43 pm
content: Sorry; I must still be missing something. You stated earlier that you don't want to measure just the "on" time of the heater, because you don't have a reliable way to measure time. Instead, you want to measure the actual heater power (current and voltage, I assume). If you drive your lamp or heater with AC, and only have a DC current sensor, how do you plan to do that?I would recommend that you- either measure time only (it's easy with an Arduino, really),- or use a DC supply for the heaters,- or find a ready-made AC sensor, or instructions how to build one.It is not that difficult to measure AC voltage or current (you need to rectify it and apply a correction factor), but you don't seem comfortable enough with this stuff to design your own AC sensing solution.EDIT: I re-read your earlier posts and found this passage, which I had not remembered in full detail before:QuoteSo you will be measuring DC voltage and current (with a calibaration factor which will need to be determined separately). But how are you going to get the duty cycle information?! The Adafruit current sensor you linked to does not integrate the current over time, but provide the instantaneous current only, right?So you are back to the same problem of having to determine the duration of the ON and OFF periods. Do you plan to do that via time measurements in the Arduino? I still think you would simplify your circuit very much if you just measure the time during which the thermostat switches the heater on an off, respectively.

author: lots o tots
author_rank: Contributor
timestamp: March 25, 2018, 11:43:10 pm
content: Quoteusing the method from Great Scott. it measures, logs power consumption (Wh), current and voltage and uses a small display:https://youtu.be/lrugreN2K4w.also i am thinking about measuring time from your recommendation, but it wont be as full featured as shown in the video. i would have to add a time function to Great Scott's code.Quoteyes. i do rely quite a bit on online tutorials and tutorial videos (especially when coding is involved) to help me through the learning process, and am generally not comfortable unless i know and understand what i am doing. I have less knowledge about AC circuits than DCQuotethe original plan was the duty cycle information will be logged (except for time) using GreatScott's method in the above listed video. His videos are excellenti would like to modify the code in the video to add a time variable, then i think i will be set and dont need a precision resistor for DC (just any simple lower power load). this is at a stage before buildng a circuit so i don't know how easy it will be.i would only need a resistor later on when i find out the difference between the two A/C load values.do you by chance the answer to the question about resistor sizing?

author: ebastler
author_rank: Super Contributor
timestamp: March 26, 2018, 04:02:46 am
content: Quote from: lots o tots on March 25, 2018, 11:43:10 pmThere already is a timing function in Scott's code. It measures the voltage and current at regular intervals (100 ms), and adds up (integrates over time) the measured instantaneous power. You can use the same approach to measure the duty cycle: Every 100 ms, add "1" to a counter if the heater is on, add "0" if it is off. Count the total number of samples (= time intervals) in a second counter. Divide the two counter values to obtain the duty cycle.That video is indeed nicely done, btw!QuoteThe remaining open question was about sizing the reference resistor, assuming you want to use a 0.6W resistor, right? You have already done the math to calculate the maximum operating current: Imax = P/U, so for P = 0.6W and U = 12V, Imax = 50mA. Add a safety margin, let's say a factor of 2 as this resistor will sit out in the open, so I = 25mA is the target.Now simply calculate R = U/I = 12V/0.025A = 480 Ohm. That exact value is not in the E12 series of commonly available resistor values, but 470 Ohm is (and will only give you a 2% higher current, which is not a problem).As a shortcut, you could also directly calculate the value from P = U²/R, so R = U²/P. With U = 12V and a target of P = 0.3W, you get R = 480 Ohm again.

author: lots o tots
author_rank: Contributor
timestamp: March 27, 2018, 12:05:23 am
content: Quoteone thing that is not clear to me me, is if i were to aim for 0.025 amps, how do i ensure that the resistor draws that exact power. the voltage i can control with the power supply, but how to i control the current to ensure the safety factor is maintained?

author: ebastler
author_rank: Super Contributor
timestamp: March 27, 2018, 03:00:03 am
content: Quote from: lots o tots on March 27, 2018, 12:05:23 amOhm's law will take care of that for you. If you set the voltage U on your power supply, and connect a resistor with resistance R, the current I = U/R will flow through the resistor. So, you select a suitable value R to get just the current I which you want.Think of it like a section of tubing for water which has a small diameter to restrict the flow. If you connect that to a pump or an elevated reservoir which provides a certain water pressure (voltage), and the tubing provides a certain obstruction (resistance) to the flow of water, those two values will define how much water flows through the tubing (current). The larger the pressure, the more flow. The larger the resistance (restriction), the less flow.

author: lots o tots
author_rank: Contributor
timestamp: March 28, 2018, 12:34:28 am
content: Quotereally what i am getting at is if the resistor dictates the current, how do i build in a safety factor. wouldn't that be a safety factor of 1. am i missing something?

author: ebastler
author_rank: Super Contributor
timestamp: March 28, 2018, 04:34:08 am
content: Quote from: lots o tots on March 28, 2018, 12:34:28 amYou mean a safety factor with regard to the resistor's power rating, to ensure it does not run hot? We had figured that in when determining the desired target current for the resistor.In the example above, the 0.025A current was chosen such that theactualpower which is converted into heat in the resistor will be 0.3W. If you then buy a 0.6W resistor with the appropriate 470 Ohm resistance, you will have a safety margin of 2.EDIT: Trying to clarify further:The resistor'sresistancevalue (R) determines how much current (I) actually flows through it, when a given voltage (U) is applied across the resistor. I = U/R.This, in turn, determines what amount of power (P) isactuallyconverted into heat in the resistor. P = U*I.The resistor'spower ratingis independent from the above. It doesnotdetermine how much current actually flows or how much power is converted into heat; it only states what the resistor cantoleratebefore burning up. Choose a resistor with a sufficient power rating so it will tolerate the actual operating conditions, which are determined by U and R.