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I was inspired by Dave Jones blog about making a simple dummy load for testing stuff (power supplies, batteries, etc.) ...



Not having a pre-printed circuit board around I assembled one on a small piece of protoboard:



Schematic:



I didn't bother with a load read-out like Dave did. All you need to do is clip a multimeter across the big current sense resistor, and as it is 1 Ohm, then 100 mV on the meter will represent 100 mA current.

It seems to work pretty well. I could only get the lower limit down to around 20 mA but the op-amp wasn't a rail-to-rail one. The pot is a 10-turn wirewound, which gave reasonably sensitive control.
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Looks good...I'm getting tired of slapping together resistors to simulate random loads...
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Interesting circuit. I will simulate this circuit and see what happen. I was thinking of building a variable dummy load just my ex-employer ( when I was working as a technician fixing / testing Switching Power Supply ), they where usign a bigger variable load system.

Thank Nick.
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All credit to Dave Jones for describing it. Although I put the thing on paper, with his blogs you tend to have to copy stuff down from the screen. But he does explain what everything is doing which is great.

You can buy these of course, but there is a certain satisfaction in doing it yourself.

Out of interest I measured the current consumed by the 9V battery powering the circuit. It was about 1.5 mA. So, small consumption makes it nice and portable.
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It's hard to resist over-egging a design like this. smiley

I added an indicator LED so I would realize I had my battery plugged in. For an extra milliamp drain a high-power LED is very bright.

Also I marked two test points on the circuit:

  • TP:A - a voltmeter between here and Gnd shows the input voltage to the second op-amp. Effectively this becomes the "set current" (whether or not this is achieved). So for example, 500 mV would be a set current of 500 mA. So you could attach a meter, and dial-up the current you want, before powering up the (main) circuit.
  • TP:B - a voltmeter between here and Gnd shows the actual voltage dropped by the 1 ohm current-limiting resistor, and thus the current actually being drawn by the circuit under test.
« Last Edit: February 05, 2012, 12:06:21 am by Nick Gammon » Logged


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If you wanted to have a computer-controlled dummy load you could simplify the circuit like this (untested):



Now you just use analogWrite to set up the required voltage (and thus, load current). Since analogWrite (255) outputs 5V roughly then you would need to scale it. For example, to have a 2.5 A load you would analogWrite (127).
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If you wanted to have a computer-controlled dummy load you could simplify the circuit like this (untested):



Now you just use analogWrite to set up the required voltage (and thus, load current). Since analogWrite (255) outputs 5V roughly then you would need to scale it. For example, to have a 2.5 A load you would analogWrite (127).

Not going to work with that wire going from pin 1 to pin 2 of that op-amp. Pin 2 should wire to top of the 1R resistor. I think the main disadvantage of this circuit is the inefficiency due to the 1 ohm feedback resistor which leads to lots of wasted power. I would redesign it using a .1 ohm resistor and scale the feedback op-amp to compensate.

Lefty
« Last Edit: February 05, 2012, 04:32:50 am by retrolefty » Logged

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Not going to work with that wire going from pin 1 to pin 2 of that op-amp. Pin 2 should wire to top of the 1R resistor.

Oops. Major blunder there with reworking the circuit. Thanks! Fixed schematic now.
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I think the main disadvantage of this circuit is the inefficiency due to the 1 ohm feedback resistor which leads to lots of wasted power. I would redesign it using a .1 ohm resistor and scale the feedback op-amp to compensate.

Thanks for the feedback. You are probably right in terms of efficiency, weighed against the difficulty of finding a 0.1 ohm resistor with high accuracy. I think Dave did his 1 ohm resistor by putting 10 x 10 ohm in parallel, which tends to indicate that he found the likelihood of finding a 0.1 ohm resistor rather low.

I've got one here, I wonder if the heat would make 0.1 ohm drift faster than 1 ohm? I suppose there are always trade-offs in this sort of thing. For situations where you might just want to test a "wall-wart" to see if it delivers the promised voltage at the promised rating, a few seconds' test would probably suffice, in which case you don't really care if you waste power or not.
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Certainly if the circuit's function is for just using as a programmable 'dummy load' then worrying about efficiency is silly, it's all going to be converted to heat anyway. However what you have drawn is really a programmable constant current source useful for continuous operation for say driving high power LEDs or other constant current applications where increased efficiency is always welcomed. I guess these days one could look at high current hall effect current shunt sensors, their costs are higher but they would result in the most efficient operation.

 I recent completed building a 100+ watt oil cooled 50 ohm dummy load used in ham radio transmitter testing. I used 20 1,000 1% 2 watt resistors ordered on E-bay for just a couple of bucks. I wired them in a circular parallel manner and soldered them to a BNC connector mounted on the lid of a 1 quart paint can, filled the can with mineral oil and have been able to pump 100+ watts of RF into it for many mins at a time. After about tens mins the can is warm to the touch but handles the load nicely.

http://k4eaa.com/dummy.html

Lefty
« Last Edit: February 05, 2012, 04:50:57 pm by retrolefty » Logged

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@retrolefty

Nice site. Good work and nice idea using mineral oil for cooling and a recycle paint can. Very "McGyver".  smiley-lol

@Nick Gammon

I did the simulation using Circuit Wizard. http://www.new-wave-concepts.com/

During the simulation, the pot was set around 65 %, the current going through the MOSFET and the resistor was 2.8 A. A saturated value. I increace the R value, the current max was reduced, but the pot at low setting ( under 10 % ) the current at R was under 100 mA. Decreasing the R, will increase the I across the R. The low setting of the pot will be increase ( start around 250 mA ). Also, it depend the MOSFET type ( resistance across the MOSFET ) , the voltage at the gate ( to optain saturation value )  and the power resistor ( controlling the max current ).

I will figure out a better way to design and make the variable dummy load. It a matter of tweeking the circuit so I can get a low current through the R ( at low pot setting ) and set a max value ( at max pot setting ), so I can test my PSU I built / test.

 I may use the Arduino to monitor the current going through the R. Right now, a current meter wil do the job.
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I used 20 1,000 1% resistors ordered on E-bay for just a couple of bucks.

I initially read that "I used 20,000 resistors". Wow, I thought, you have a lot more patience than me!

Interesting article, thanks.

BTW, when measuring the AC component of the dummy load it seems quite high (a few hundred millivolts) when the MOSFET kicks in, at least at the "edge" point. Would you recommend a capacitor somewhere to smooth that out? If so, what one and where?
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I used 20 1,000 1% resistors ordered on E-bay for just a couple of bucks.

I initially read that "I used 20,000 resistors". Wow, I thought, you have a lot more patience than me!

Interesting article, thanks.

BTW, when measuring the AC component of the dummy load it seems quite high (a few hundred millivolts) when the MOSFET kicks in, at least at the "edge" point. Would you recommend a capacitor somewhere to smooth that out? If so, what one and where?

Hum, well the mosfet rather high gate capacitance coupled with the limited output impedance (current drive) of the op-amp could very well make that circuit somewhat unstable during initial transition. Not sure if adding more capacitance would help or make it worst.

Lefty
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@Nick Gammon

I just finish a simulation using Circuit Wizard. I re-do the circuit of your "Dummy Load" circuit, And the circuit can be control be Arduino using a D/A for the 5 V section ( Vin ) and the current monitor, the voltage at the Power resistor can be scale down ( to 5 V ) and going into an analog pin.

The analog circuit follow this formua.

Vout = 9/5 * Vin + 3

First Op-Amp section :  Vout-a = - ( Vin * 1.8 )

Second Op-Amp section :  Vout = -( Vout-a + -3 V )

Therefore :  Vout = - ( - ( Vin * 1.8 ) + - 3 )

Here the schematic.


* DummyLoad.jpg (38.59 KB, 1024x501 - viewed 278 times.)
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I've been messing around with lead acid battery chargers and working on a load to test capacity.  I have a heck of a lot of trouble with batteries here in the desert and I want to get it under control.  So, I stumbled across this thread in passing. 

Nick, that is a cool idea.  I've had several instances where a wall wart didn't come close to supplying it's rated power and it really annoyed me.  In one instance it was a USB cable that had enough resistance to limit a 5V 2A wall wart to around 200ma; that was a pain to find.  This little device would solve that troubleshooting problem for me.  It could also help isolate the bad NiCd batteries that turn up in solar accent lights bordering my walkway.  They seem to die too often; this could help me figure that one out too.

Lefty, I have a couple of questions on your RF load.  You have folks adjust the voltage up by .4 to allow for the diode, doesn't it have a .7V forward drop?  Also, you show two diodes in the picture and only talk about one in the description.  I realize that .3V or even 1V with two diodes isn't statistically significant in the scheme of things, but it might confuse someone (besides me).  I also wonder about the voltage reading since you only use a half wave to charge the capacitor and RMS calculations usually use the entire wave form, but then again this would only be a small error at HF and could probably be ignored.  The other question is the value of the capacitor.  At 3MHz it's impedance should be (if my math isn't too rusty) less than an ohm rising to around 5ohms at 50Mhz.  What am I missing?
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