Load for big power supply - Page 1
Forum section: electronics / projects

author: stoica adrian
author_rank: Contributor
timestamp: November 30, 2014, 02:15:45 pm
content: Hi, i want to make a load (30A.5500W) for a big power supply (55V 100A). I want to make this load using resistor. Please a idea for me how too make this load. Best Regards

author: KM4FER
author_rank: Regular Contributor
timestamp: November 30, 2014, 03:46:52 pm
content: Is this just for testing the power supply, or a more permanent device?I once tested a repair job on a 12V DC power supply by using a coil of iron wire used for tying steel reinforcing rod together. I measured the resistance of the wire and then cut to length so that at 12V I would get 20A of current. Worked a treat. The wire got very warm and started smoking but the power supply held up just fine.

author: krivx
author_rank: Frequent Contributor
timestamp: November 30, 2014, 04:58:09 pm
content: Go to the dump and start pulling heating elements out of old ovens.

author: timb
author_rank: Super Contributor
timestamp: November 30, 2014, 05:04:19 pm
content: Quote from: krivx on November 30, 2014, 04:58:09 pmSent from my Tablet

author: stoica adrian
author_rank: Contributor
timestamp: November 30, 2014, 05:07:35 pm
content: I need for testing only, but i don't know how to calculate that load because it will warm up, and i need to be sure that at the end it will have 30A and sustain 5500W

author: stoica adrian
author_rank: Contributor
timestamp: November 30, 2014, 05:12:53 pm
content: I will make 3 of these to get out from power supply 90A

author: stoica adrian
author_rank: Contributor
timestamp: November 30, 2014, 05:18:37 pm
content: Heating elements ..... i like this

author: Seekonk
author_rank: Super Contributor
timestamp: November 30, 2014, 05:35:30 pm
content: Buy a roll of strapping for HVAC ducts, the steel stuff with all the holes in it. And a fan. What you want is a lot of surface area to get the heating density down.

author: max_torque
author_rank: Super Contributor
timestamp: November 30, 2014, 11:44:58 pm
content: Buy a cheap roll of 1/4 inch (6mm) copper brake pipe tubing. Unwind the roll and spiral wrap it around something non metallic, like a length of wood, or plastic drain pipe. Use Metal P - clips as movable clamps to attach wires to the copper pipe, by moving these clamps up / down the length of the pipe to set the resistance and hence the current / load. Now use some flexible plastic pipe to connect the pipe to your household cold water tap as cooling.You can absorb 10s of kW this way very cheaply indeed, akthough obviously the system is NOT isolated, so make sure you check your PSU output, bot h in terms of isolation and max voltage etc!

author: stoica adrian
author_rank: Contributor
timestamp: December 01, 2014, 08:43:04 am
content: how do you calculate the resistance of the load, using ohm law? Or you using a shunt resistor for measure live the amp, and you go from a large resistance to a smaller one, and you stop when you have your amps at the shunt?

author: m100
author_rank: Regular Contributor
timestamp: December 01, 2014, 04:01:29 pm
content: In years gone by I used home made load banks made with lots and lots of aluminium clad wirewound resistors like this bolted to 4mm aluminium and providedlotsof airflow.Doing the same today I'd probably choose thick film on steel Welwyn WDBR's bolted to a big RSJ and just stand it outside.http://uk.farnell.com/webapp/wcs/stores/servlet/Search?catalogId=&langId=44&storeId=10151&gs=true&st=welwyn%20WDBRhttp://www.farnell.com/datasheets/1851672.pdf

author: stoica adrian
author_rank: Contributor
timestamp: December 01, 2014, 04:56:44 pm
content: I'm glad for all this type of solution, i will think witch tip is suited for me. Thanks all

author: Seekonk
author_rank: Super Contributor
timestamp: December 01, 2014, 05:10:51 pm
content: Going back to your first question..... If you want a 30A load on a 55V power supply that would be 1.83 ohms at 55V for a total of 1650W. You stated 30A.5500W. I think you will have trouble doing this test if you don't know ohms law because many resistors will have to be in parallel to obtain this wattage. Just a 5500W oven will not draw enough current if the element is rated at 240V. Building your own resistor is the only cost effective way.

author: stoica adrian
author_rank: Contributor
timestamp: December 01, 2014, 07:24:38 pm
content: Sorry, my mistake, i calculate the power out of the source 55x100, not the power at the 30A load.

author: tautech
author_rank: Super Contributor
timestamp: December 01, 2014, 07:52:49 pm
content: 2 big industrial diodes in series for ~ 1.2 V conduction.The sort they use in welders.Maybe stud mount on a water jacket for cooling. Or as it is only 55 V.... submerse them in water.If they are on the same water jacket one will need to be anode stud the other cathode stud.Thats if your PSU can handle such a load.Before you all raise your hands in horror, we have used this principle on 30 V 5 A PSU's to test fan cycling, overload protection, foldback, and as a general "hard" load.I also know of welders that have been tested using coils of fencing wire submersed in a drum of water.55 V @ 100 A is only equivalant to a small welder and should be no big deal to load test.

author: jlmoon
author_rank: Unknown
timestamp: December 01, 2014, 08:22:20 pm
content: Quote from: krivx on November 30, 2014, 04:58:09 pmOr you can go buy Ni-chrome heater element replacement kits at your local appliance repair depot. Saw one the other day 5500W .. amost bought it.. $5.95 with ceramic insulators and all. They work great for static loads and such, just be sure to use proper techniques in mounting and installation in fire proof enclosures.

author: SeanB
author_rank: Super Contributor
timestamp: December 02, 2014, 06:52:32 pm
content: The generator test bed was literally a bed, with a cut in the short side of the frame and a big block of SRBP as a join. The steel slats were the resistor to test the unit could deliver a minimum of 2kA of current at 28V.The 3 phase alternator was tested using 3 steel rods inserted into a 55 gallon drum filled most of the way with very salty water, and kept on a concrete block for insulation.Test till the bed frame glows visibly red in daylight, and the drum boils vigorously.

author: Seekonk
author_rank: Super Contributor
timestamp: December 02, 2014, 11:19:49 pm
content: A 5500W 240V electric hot water heater element is 10.5 ohms and cost about $9. Buy five of them and put them in parallel. That will give about 2 ohms. Close enough or sub in a couple 3500W elements. Then stick them in a tub of water. 2000W 120V water heater elements are also common. Four of these in parallel would give you 1.8 ohms, exactly what you want.Never done the calculation, but at my camp I heat water with PV. With a 10 gallon tank 1 KWH of energy gives me a really hot shower.