Power calculation is off - Page 1
Forum section: electronics / beginners

author: ghamille
author_rank: Newbie
timestamp: April 01, 2024, 12:35:00 am
content: Hi all, very beginner question hereI have a DC power supply which I set to 9V. I build a straightforward circuit to see if I understand P, so I don’t blow up my resistors in the future. The circuit is as follow:Power Supply + (set to 9V) —> 330ohm (measured at 324.9ohm) —> Power Supply -Now I was under the impression that this would draw 92/324.9 ~= 0.25W. The problem is, my power supplies shows 0.27W. In fact, it shows 0.25W when supplying 8.5V.So the question is, what variable am I missing here? I just want to make sure I understand how to calculate the limits properly, so I can build reliable circuits!

author: xrunner
author_rank: Super Contributor
timestamp: April 01, 2024, 12:42:10 am
content: Well from what it's indicating9v * .03 = .27WSo it's calculating it properly (from the numbers it's using).Now, whether those numbers areexactlycorrect, well, that's a whole other story.

author: ghamille
author_rank: Newbie
timestamp: April 01, 2024, 12:49:20 am
content: That’s trueSo it would just be a limitation of the instrument/display?And there’s no phenomenon where it would use 0.27W at 9V, with a 324.9ohm resistor?

author: xrunner
author_rank: Super Contributor
timestamp: April 01, 2024, 12:56:01 am
content: Quote from: ghamille on April 01, 2024, 12:49:20 amThe accuracy of the PS measurement system / display yes.QuoteNo. Did you verify with a known accurate voltmeter that the PS is actually holding at 9V when you attached the resistor?I bet it isn't exactly 9.000 V.

author: ghamille
author_rank: Newbie
timestamp: April 01, 2024, 01:01:54 am
content: Perfect! I did and you’re right, it’s not exactly 9V, it’s 8.97V.I also tried measuring the current but I think it’s so low, that the resistance from my multimeter is messing up the values. The power decreases to 0.18W when I try to measure the current.

author: Faranight
author_rank: Unknown
timestamp: April 01, 2024, 05:52:09 am
content: It's within the margin of error. Some instruments have a burden voltage, which is basically a voltage drop across the instrument due to its internal resistance. When you're adding a current meter in series it will affect the measurement to a degree. Also, how accurate is your multimeter when you measured the resistance of the 330 ohm resistor? Was the multimeter calibrated? My suggestion would be to get some more precise element i.e. a 0.1% resistor and use *that* in your circuits. Then see how different instrument arrangements affect your measurements.

author: ejeffrey
author_rank: Super Contributor
timestamp: April 01, 2024, 06:02:37 am
content: Most power supplies have rather poor accuracy on the built in current meter and power is just calculated by multiplying voltage * current. From the display it has only 10 milliamp resolution. That's an easy 20% error -- 25 milliamp rounding up to 30 and the spec is probably worse

author: ghamille
author_rank: Newbie
timestamp: April 01, 2024, 12:49:58 pm
content: QuoteGood question! It’s a relatively good one (Fluke 87) and I assume it was calibrated at the factory, but I’ve had it for more than 5 years and did a lot of work with it on my RV/solar panel installation. Is calibration something that has to be done regularly?QuoteThat’s a really good idea, I will try that.

author: golden_labels
author_rank: Super Contributor
timestamp: April 02, 2024, 03:47:47 am
content: Don’t think of power supplies as precision instruments. That’s not their job. See the voltage and current settings as rough limits. If you need a precise voltage or current measurement, attach multimeters.On theoretical layer, accuracy of the power indicator is even worse. It’s not measured directly, but calculated from voltage and current. And if you multiply two values, their error margins add up. So if your PSU offers 1% accuracy, the power measurement may be worse than 2% off.

author: Zero999
author_rank: Super Contributor
timestamp: April 02, 2024, 08:14:37 am
content: P = V2/R, so a % change in the voltage results in roughly double the % change in power dissipation in a resistor.

author: Rick Law
author_rank: Super Contributor
timestamp: April 03, 2024, 01:08:57 am
content: Quote from: ghamille on April 01, 2024, 12:49:20 amIt is not just the limitation of the instrument or the display. It is the limitation of reality.Your 330ohm resister is NOT and never will be exactly 330ohm except for the brief moments as it transits between just below to/from just above 330. In reality, changes in temperature changes the resister. You, as a human being, you would be a rather good space heater. Your entering the room affects airflow and the room temperature and thus changes the resister's resistance.Generally,day to day electronic stuff, you probably will see variations around 1% to 5%. Of course, very high precision stuff do exist and a corresponding higher price. The more precision you want, the more exacting the environment and condition must be. You will need a well equipped laboratory (and careful work) to measure down to 1/1000000 (1 part per million).So, inaccuracy is something we must live with. You choose the accuracy (and cost) in accordance with your need. If you feel that's a bit of cheat. Nature cheats. Below certain resolution, physics itself breaks down and measuring it is not possible as dictated by our present understand of physics. No one can tell you the length of a rod in meters down to 35 decimal places. At that resolution, our ability to understand what "length" is breaks down. So nature itself "rounds off". Living with limited accuracy is just living within reality.Edit: corrected a line ("...You choose the accuracy (and cost) in ..."). Originally with text that represented an escape sequence causing the text to display incorrectly.