Dummy load

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.

Looks good...I'm getting tired of slapping together resistors to simulate random loads...

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.

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.

It’s hard to resist over-egging a design like this. :slight_smile:

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.

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).

[quote author=Nick Gammon link=topic=90343.msg678752#msg678752 date=1328417809] 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). [/quote]

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

retrolefty: 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.

retrolefty: 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.

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

@retrolefty

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

@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.

retrolefty: 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?

[quote author=Nick Gammon link=topic=90343.msg679499#msg679499 date=1328470539]

retrolefty: 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? [/quote]

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

@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.

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?

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).

The author of the article told that he used two series diode because that is what he had on hand and needed the PIV value that two series diodes provides. I assume if you were to use the one in the drawing it has a high enough PIV rating. And yes the diode drop will limit both the total accuracy possible and the minimum amount of RF power that can be measured. But for the purpose of measuring the nominal RF output power of a typical 100 watt ham transmitter, it is simple, cost effective, and that method has been used for many many decades.

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?

Well to review, a half wave filted rectifier with charge up to the peak value (not peak to peak) of the RF signal (ignoring the diode voltage drop), the fact that it’s half wave Vs full wave does not effect the measurement value as the meter used to read the value is required to have high (10megohms+) input impedance, so as to not ‘load down’ the measured value. The value of the cap has no bearing on accuracy as it will charge to the peak value of the RF voltage. Power can be calculated with this by multiplying the peak value by .707 to give the RMS voltage value. Power is then calculated by: (Vrms X Vrms) / 50 ohms. This assumes that the RF being measured is a sine-wave. Keep in mind this kind of measurement circuit is mostly used to just allow a visual reference to be watched as one makes tuning adjustments on a transmitter looking to adjust things for a maximum value. The fact that one can calculate the approximate power is usually just a secondary purpose or goal. Many ham operators certainly use other more accurate watt meters, but it’s hard to beat the bang for the buck that this simple diode method allows. I personally didn’t build that measurement circuit into my version of the dummy load. I rather just hook my scope to the power input and ground via a BNC T adapter and measure the RF signal directly on my analog oscilloscope as a AC signal.
**Lefty **

But, the capacitor is directly across the terminals from center through the diodes to ground, so it's in parallel with the resistors. Measuring across it would see the combination of the impedance of the capacitor (around 1 to 5 ohms depending on frequency) and the reisistance of the load resistors. Something a lot lower than 50 ohms.

I now realize that you didn't write the article, so you can't speak for the author, and I understand that you don't use his measurement techniques; just the load itself. I started wondering when he missed on the diode forward bias drop value and cited how accurate it was. Which it could be at higher wattage levels where a lower value won't matter, but not for a 1W handheld. Now, I'm trying to understand how it can work. Every calculation I do show essentially a dead short through the .01 microfarad capacitor. Even when I consider that the frequency is cut in half by the diode in series.

I bet he meant to type .01pF or something similar, which would make more sense. Or maybe I'm missing something.

You're right about how simple the load portion of this project is though.

draythomp: But, the capacitor is directly across the terminals from center through the diodes to ground, so it's in parallel with the resistors.

Measuring across it would see the combination of the impedance of the capacitor (around 1 to 5 ohms depending on frequency) and the reisistance of the load resistors. Something a lot lower than 50 ohms.

No, the resistors (the 50 ohm load) wire to the BNC connector's center pin and ground. The cap is wired to the output side of the diode to ground. The cap just filters the halfwave rectified signal so as to hold a charge equal to the peak of the RF signal voltage. The diode effectivly isolates the cap from the load resistor(s).

I now realize that you didn't write the article, so you can't speak for the author, and I understand that you don't use his measurement techniques; just the load itself. I started wondering when he missed on the diode forward bias drop value and cited how accurate it was. Which it could be at higher wattage levels where a lower value won't matter, but not for a 1W handheld. Now, I'm trying to understand how it can work. Every calculation I do show essentially a dead short through the .01 microfarad capacitor. Even when I consider that the frequency is cut in half by the diode in series.

I bet he meant to type .01pF or something similar, which would make more sense. Or maybe I'm missing something.

Why don't you check out this link about building and using simple low power measurement circuits using diodes. It's better written then that first link I used. The circuit there called 'peak voltage detector' is the exact equivalent of the first one I linked. http://epic.mcmaster.ca/~elmer101/rfpower/rfpower.html and a construction page at http://epic.mcmaster.ca/~elmer101/rfpower/projvi.html Now if your only interested in measuring low power RF signals then you might look into the chip I built a small RF milliwatt meter with. It has a frequency range of DC to 500Mhz and a power measurement range from -75dbm to +15dbm and is very accurate, which is total of 90db of range, a feat just short of amazing. It's based on the Analog Devices AD1307 logrithmic amp in a 8 pin DIP package and is powerd with just +5vdc. I paid about $15 maybe 10 years ago for the chip, but in my opinion worth every cent as accurate RF power levels to 500Mhz is not a trivial thing. It's output analog DC voltage would be perfect for wiring to a arduino analog input pin for display or logging functions. http://www.analog.com/static/imported-files/data_sheets/AD8307.pdf

Lefty You're right about how simple the load portion of this project is though.

Wait, I finally figured out where I was having problems. I was thinking impedance which is a result of an AC current flowing through the capacitor. Not really flowing through since the alternate voltage will suck the power off the cap as the wave form varies. The diode doesn’t really isolate the cap, it changes the RF into pulsed DC and there is no bottom side of the wave form to affect the cap. This means that the cap will charge up to peak voltage in a few cycles and essentially stay there.

None of the sites clarifies this simple little item. I even looked at a bunch of power supply calculation examples trying to understand why the impedance of the capacitor wasn’t a factor other than filtering the pulse by discharging along the current path. My kind of thinking (related to AC impedance) comes from calculating the proper capacitor value to control a ceiling fan since a resistor would have to be too large and a triac introduces way to much noise into the motor assembly (causing a buzz).

Sigh

I was right about the diode forward voltage drop though :slight_smile:

I was right about the diode forward voltage drop though

There is a pretty easy method to 'compensate' for the diode voltage drop using a simple op-amp circuit that uses a second 'matched' diode in it's feedback circuit to factor out the voltage drop and allow small millivolt measurements. One of those last links shows and tells about it.

Lefty