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Anyone have a schematic for a Constant Current DC load tester?

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Hi everyone,
I've been surfing the web trying to find a schematic for a constant current DC load tester for use with a battery bank (48VDC) Ideally something that can handle 50amps or more would be fantastic! So far I've only found either low current or useless schematics online with nothing really suitable.

TIA for any help :)
 
Welcome to ETO!
Ouch! 50A or more @ 48V is gonna be expensive. A 1Ω load resistor, for example, is ~£75. What is your budget?
How accurate does the 'constant' current need to be?
Do you want a fixed 50A (or whatever) or must it be adjustable?
If adjustable, over what range?
I'm not surprised you haven't found a suitable circuit, but I'm sure the guys here could design something for you.
 
Fortunately I'm not working on a budget, whatever it costs will be my budget.

Adjustable would be a nice feature to have, as for accuracy I could tolerate a + or - of 2amps. And the range hmm well I'm not too fussed but maybe adjust between 10~50amps.
 
Hi everyone,
I've been surfing the web trying to find a schematic for a constant current DC load tester for use with a battery bank (48VDC) Ideally something that can handle 50amps or more would be fantastic! So far I've only found either low current or useless schematics online with nothing really suitable.

TIA for any help :)

Welcome to ETO!
Ouch! 50A or more @ 48V is gonna be expensive. A 1Ω load resistor, for example, is ~£75. What is your budget?
How accurate does the 'constant' current need to be?
Do you want a fixed 50A (or whatever) or must it be adjustable?
If adjustable, over what range?
I'm not surprised you haven't found a suitable circuit, but I'm sure the guys here could design something for you.

Hi EF,

Nice that you have joined ETO. I see you are from New Zealand; care to put it next to 'Location' on your user page so that it shows in the window next to your posts.
This helps us to know your component access and your mains voltage. It is also just plain interesting to know where people are.:)

I have done a few battery discharge circuits along the way, but not on such a grand scale; as Alec says, 'ouch'.:D

Basically you have these practical options, as I see it:

(1) Fixed Load
(1.1) 1 Ohm 2.5KW resistor: https://www.mouser.co.uk/ProductDetail/TE-Connectivity/TE2500B1R0J/?qs=iJVBkmKXgrlgth4ME9/cMw==
(1.2) 25 off 100W resistors in parallel (100W seems to be the maximum power before resistor prices rise disproportionately).
(1.3) Connect an inverter to the battery and connect the inverter mains output to a 2.5Kw (approx) load: domestic heater, kettle element, oven element, oven cooking ring. This approach could be converted into variable constant current (if that makes sense) by adjusting the inverter output voltage.

(2) Variable load switched steps
(2.1) Say, 25 off 100W resistors that could be switched in or out to change the battery load.

(3) Continuously variable
(2.1) Variable resistor. This is the simplest variable approach but would be costly
(2.2) 48V input 2.5KW switch mode power supply, with output voltage adjustable from 0V upwards, connected to a 2.5KW load. This could be modified to generate a variable constant current load.
(2.3) Custom variable electronic. This could be designed to draw an adjustable constant current. The circuit would use basic electronic techniques and would be simple to design. But it would involve a lot of circuitry, mainly repetition, and large heat sinks. I hope you choose this approach because, not only will it give the best performance, but it will be the most fun.:D

One thing to bear in mind, whichever approach you choose, is that the load would be dissipating 2,500 W (powerful domestic heater) and that heat would need to be conducted away from the electronic components and also out of the room that the equipment is located in. As a result, the load is liable to be quite physically large. Even with the simple single fixed-load resistor you would be wise to provide fan cooling to get the heat away from the resistor.

I do hope that you get back and tell us which approach you favor.:)

spec
 
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What is the purpose of the load tester? I ask because, if it's to assess the battery discharge profile, then a fixed load resistor and a Coulomb counter might be a cheaper alternative to using a constant current load. On the other hand, if it's for assessing something like battery peak current capability then ripple on the 'constant' current may be an issue.
 
You also could make a continuously variable by setting the coarse current by switching in and out the 25x 100W resistors and have a 150W linear current sink to get the fine adjustment.
 
You also could make a continuously variable by setting the coarse current by switching in and out the 25x 100W resistors and have a 150W linear current sink to get the fine adjustment.
Good idea.:cool:

spec
 
A large number of smaller resistors probably will be less costly and definitely more easy to cool. How do you plan to remove 2500 W?

ak
 
This is the easiest and cheapest solution for an adjustable constant current load that I could initially come up with:

2016_10_08_Iss1_ETO_CONSTANT_CURRENT_LOAD_VER1.png
NOTES
(1) The switch mode power supply units (SMPSUs) (£8.33UK from ebay) can generate a stabilized output voltage of between 12V and 80V from an input voltage of 48V. Each PSU can output a maximum of 10A
(2) The output voltage is adjustable by a potentiometer on the PSU printed circuit board (PCB). (3) Thus the output power of each power supply will be adjustable from 12V * 12V/8R Amps = 18W, and 80V *80V/8R Amps = 800W. **broken link removed**
(3) Thus, with three PSUs, the current range from the 48V battery would be adjustable from, (3* 18W)/48V = 1.125A and, (3 * 800W)/48V = 50A.
(4) The 8R resistor dissipate 800W and can be made up of eight 100W resistors (around £24UK total for a set of eight)
 
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Since that provides a max load current of 18A, did you envisage adding further PSU's to the chain, Spec?
 
Since that provides a max load current of 18A, did you envisage adding further PSU's to the chain, Spec?

Whoops: have I got it wrong? That's what happens when your missus has 'Songs of Praise' blasting on the TV while you are doing schematics.:p
Here is my reasoning:
(1) The maximum output power from each PSU is, (80V * 80V)/8R = 800W
(2) The required total power is, 48V * 50A = 2,400W. Thus 2,400W/800W = 3 power supplies needed.

The PSUs are assumed to be 100% efficient, but are more likely to be 70% to 85% in practice, depending on input voltage, output voltage, and output power.

spec

PS: you could probably find a single PSU to do the job, but a single PSU would probably cost more than three smaller PSUs, which are only around £8.33UK each.
 
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An interesting technique there spec.
It may be wise to be cautious of the maximum current specifications of those power supply modules, but the idea is easily scaleable for more or less load current and avoids the necessity of controlling low resistance values when trying to use just one element for the load.

JimB
 
An interesting technique there spec.
It may be wise to be cautious of the maximum current specifications of those power supply modules, but the idea is easily scaleable for more or less load current and avoids the necessity of controlling low resistance values when trying to use just one element for the load.

Yes, you are right Jim. Four PSUs would probably be needed in practice. The schematic is only meant to be an outline to show the principle though.

The other approach I had a look at was to use pulse width modulation (PWM). You could probably do the whole thing with four man-sized NMOSFETs, four power resistors, and a 555 or microcontroller.:)

spec
 
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(1) The maximum output power from each PSU is, (80V * 80V)/8R = 800W
Where does the 80V figure come from? Battery is 48V.
 
I have placed power resistors in a tub of water. As long as you replenish the water before it boils away, a nominal 100W wire-wound ceramic power resistor will dissipate ~1000W while boiling water around it.

I also found some diesel-electric train braking resistors that rated at 0.15Ω @ 1200W ea. that I used to make a battery tester.
 
Out of curiosity: such a test should last for how long?
 
Where does the 80V figure come from? Battery is 48V.
Hi Alec,

The power supplies can output 12V to 80V. Remember that power in equals power out in any system.

(By the way I haven't checked that these particular power supplies will actually output any voltage lower than 48V when they have a 48V input, but as this is an outline circuit that is not important; it is the principle that counts. In practice a buck converter would probably be better.)

spec
 
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Below is an outline schematic of the constant current load approach mentioned in post#13
See post #24 for detailed schematic
2016_10_08_Iss1_ETO_CONSTANT_CURRENT_LOAD_VER2.png

NOTES
(1) The microcontroller adjusts the pulse width modulation to maintain a constant voltage drop across R29. This in turn defines the average total current flowing through the four NMOSFETs.
(2) The battery will smooth out the current pulses.
(3) An Arduino (UNO) single board computer would be suitable for the microcontroller function.
(4) The load current is set by RV1. The microcontroller reads the voltage on the wiper of RV1 to determin what current load is set.
(5) The DRIVER is a single chip NMOSFET gate driver to provide enough current to quickly charge and discharge the NMOSFET's high effective gate capacitance.
(6) The Voltage regulator is just a high voltage three terminal voltage regulator (TL783).
(7) A simple sketch (program), loaded into the microcontroller, would implement this function.
(8) R41, R42, R43, and R44 are 600W types.
 
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We know that Inrush current limiters have a later PTC to tend to maintain a constant current, but not suitable for this app.

Tungsten lamps also have a similar PTC of > 8:1 from room to 3200'K and make excellent constant current loads at half power.

Since you require about a 2000 W load I would suggest get enough 150W halogen or 300W rated at 120V and run them at 48V. This can be calibrated with a 100mV shunt and DMM.

OTherwise if you need precision, use a current shunt with a MOSFET to regulate the voltage difference between 49 and 51 mV across a 50 mV shunt or some magnet wire that is 1 milliohm , twisted in half so that it is non-inductive.

The MOSFET needs to be low RdsOn in the low mOhm range and can be driven by a high speed comparator with a fast pull-up R using Nch on the low side to switch all your tungsten lamps. Cost $100
 
Wow that's an impressive response! Went to bed with 3 posts, wake up and we're up to 19! Thank you to all who are interested in helping me :)

I appreciate the efforts to keep costs to a minimum, however the very reason I'd like to build something myself is not to save money (well maybe a little) But more to the point it's about building something rock-solid reliable. And also I enjoy the learning process along the way rather than buying manufactured end products. So don't worry so much on the most cost effective solution, rather I'd like the best solution to build a constant current adjustable DC load tester.

For clarity I'll quote the post # first, then followed by a Q or A

#4, Method (3) Continuously variable is probably the best solution? This will be used for testing a bank of deep cycle gel batteries. It's very important to build a load tester that can maintain a fixed current draw until the batteries are drained.

#5, I guess I already answered this but it's purpose is to drain a bank of deep cycle gel batteries

#8, The only practical heat management for my application will be the use of very large heatsinks and fans

#16, That does depend on the size of the bank I'll be testing, It should be designed in such away that it can handle hours of abuse.

#18, That looks like a real nice idea you have there! While I'm not knowledgeable enough to create a program for an adruino for such an application, I certainly have experience editing and know all the basics when it comes to working with arduino's.

Spec, I could be wrong but I think you're liking this man-sized project aren't you? :p After reading my latest post I imagine you have a fair idea of what I'm after design wise. Do you think the idea you posted in #18 is the best solution? It sure looks great to me!
 
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