Load at the output of a power supply board - Page 1
Forum section: electronics / beginners

author: Yac
author_rank: Newbie
timestamp: September 03, 2019, 06:21:19 am
content: Hello everybody,I have to set up a test bench to test two power supply boards, the first one provides a voltage of 24V (50W) and 48V (200W) and the second one provides a 5V (65W), +12V(7W) and -12V (7W).I have to measure these voltages with a voltmeter at nominal load.So, what load can I put on each output ?Should I use a thermal radiator ? If so, how to choose the type of radiator.Thank you in advance.

author: ledtester
author_rank: Super Contributor
timestamp: September 03, 2019, 03:58:53 pm
content: Quote from: Yac on September 03, 2019, 06:21:19 amUsually DC power supplies are tested with an electronic load.Otherwise any resistive load will work, such as power resistors and incandescent lamps.Given the voltage and nominal wattage of a power supply, do you know how to compute the resistance required to draw the nominal power?

author: Yac
author_rank: Newbie
timestamp: September 03, 2019, 07:16:45 pm
content: Thank you for your help.An electric load is a little expensive for what I want to do.To calculate the resistance value I think it is necessary to use the ohm law.But I don't know if the resistance will heat up a lot and if I have to use a heat sink to dissipate the heat.For example, for the 48V output, I have a current of 4.2A, if I use a power resistor of 12 Ohms, 200W, Do I need a heat sink ?

author: ledtester
author_rank: Super Contributor
timestamp: September 03, 2019, 07:37:21 pm
content: Quote from: Yac on September 03, 2019, 07:16:45 pmYou can get inexpensive electronic loads circuits on ebay/aliexpress.This is a good one that works up to 30V and meets its advertized specs:https://www.aliexpress.com/item/32776310672.htmlThis is another popular one which works at higher voltages, but you'll need to get two to test your 42V, 200W supply:https://www.aliexpress.com/item/32916595565.htmlQuoteA power resistor is a resistor mounted on a heat sink. If the resistor is rated for 200W then there shouldn't be any problem.How long do you have to test the supplies for? The power rating on a resistor is for continuous use. If you are only connecting the power supply for a short time you can get by with a lower rated resistor.

author: Yac
author_rank: Newbie
timestamp: September 03, 2019, 08:04:30 pm
content: Thank you for these ideasThe tests must be done automatically, I think these loads are not appropriated because I have to control the measuring devices via ethernet.In fact, I don't know how long the tests will last, so I can't take the risk of choosing a lower rated resistor.

author: james_s
author_rank: Super Contributor
timestamp: September 03, 2019, 09:21:24 pm
content: I typically use incandescent lamps. Automotive bulbs are inexpensive and widely available in sizes ranging from less than a Watt to 100W or more. 12V bulbs of the same type can be wired 2 in series for 24V or 4 in series for 48V.

author: ledtester
author_rank: Super Contributor
timestamp: September 03, 2019, 10:25:42 pm
content: Quote from: Yac on September 03, 2019, 08:04:30 pmHow are you going to switch in different loads via ethernet? Seems like you would need some ethernet controlled relays.The second electronic load circuit (the 150W one) above has its current set by a voltage at a terminal which can also be set by relays.DC electronic loads are in principle very simple circuits, e.g.:https://www.codrey.com/electronic-circuits/simple-electronic-dc-load/or Dave's own explanation:

author: Yac
author_rank: Newbie
timestamp: September 04, 2019, 07:02:41 am
content: Hi,Quote from: james_s on September 03, 2019, 09:21:24 pmIf I use an incandescent lamp how do I measure the voltage and current in this lamp?Quote from: ledtester on September 03, 2019, 10:25:42 pmThe bench is equipped with a Keysight 34972A Data Acquisition, which will allow me to measure the voltage of each resistor using the 34901A multiplexer.With the 34972A Keysight, I can control the multiplexer relays via Ethernet.

author: Jwillis
author_rank: Super Contributor
timestamp: September 04, 2019, 07:18:36 am
content: Use an ammeter in series with the incandescent lamps . For testing the 5V 65W you'll need at least a 20amp meter.65W/5V=13A

author: Yac
author_rank: Newbie
timestamp: September 04, 2019, 07:33:48 am
content: Quote from: Jwillis on September 04, 2019, 07:18:36 amIt is required to measure the voltage of a resistor. I don't think that an incandescent lamp will be accepted.So, I have another question.If I use a programmable electronic charge for the 48V output only, how will the other outputs be charged?

author: ledtester
author_rank: Super Contributor
timestamp: September 04, 2019, 08:35:47 am
content: Quote from: Yac on September 04, 2019, 07:33:48 amI'm not sure I understand the question.How many outputs does each power supply board have? Does each board have 5 outputs (a 24, 48, 5, 12 and -12)? And if so, do you need to test the supply while each output is delivering its nominal power? Maybe you want to have an load attached to each output. Or if you only need to test one output at a time you can use a relay to select which output gets tested in which case you can get by with just one load (either electronic or resistive).

author: Yac
author_rank: Newbie
timestamp: September 04, 2019, 10:30:47 am
content: Quote from: ledtester on September 04, 2019, 08:35:47 amExactly, one board has 2 outputs and the other has 7.The idea of using a load with a relay to select the output to be tested is interesting.But the other outputs will not be loaded, is that a problem?

author: ledtester
author_rank: Super Contributor
timestamp: September 04, 2019, 10:51:15 am
content: Quote from: Yac on September 04, 2019, 10:30:47 amIt all depends on the power supply and what the objective of the test is.Some multi-output power supplies require a minimum load on one or more of the outputs in order to meet its specifications. You're just going to have to ask someone who knows the answer to that question.If an output needs a minimum load, just connect a fixed resistor to that output which will draw that load. Then use a relay to switch in an additional load as required by the test.

author: Yac
author_rank: Newbie
timestamp: September 04, 2019, 12:55:02 pm
content: Thank you very much for the time you took to answer me,thank you all for your helpI'm going to explore the different possibilities.