Constant Current Dummy Load


Stephen Coates' Website - stevecoates.net

Created: 17th November 2012
Updated: 15th December 2013


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Martin Lorton's Forum Topic

Dave Jones' Video

Introduction

Following some inspiration from both Dave Jones' and Martin Lorton's sites, I decided I will build a constant current dummy load. This is a nice simple circuit, which is easy and cheap to build, and very useful, especially if you want to design power supply circuits. Basically, you dial in a current level, and it will draw that amount of current, regardless of the voltage. Much better than just sticking some power resistors in as your dummy load!

How It Works

Basically, a voltage is set by a multiturn potentiometer (a single turn pot will work if you prefer). The op-amp buffers this and sets a voltage on the gate of the MOSFET. This causes the MOSFET to allow some current through the drain to the source. The 1 Ohm power resistor helps share the power with the MOSFET and assists in providing feedback to the op-amp. This feedback enables the current to stay constant if the voltage on the 'Load Input' changes. 

Update (December 2013)

I have revised the circuit to use a logic level MOSFET. This means you can use a power supply much smaller than the 24V shown below in the circuit diagram. Any logic level MOSFET with an appropriate voltage/current rating should do, but I used an F12N10L, which is available from Farnell. The power supply is simply a 9V PP3 battery. Simply swap out the IRF540 for the F12N10L, and swap the 24V supply for a 9V supply, and it should work fine.

The Circuit

Circuit diagram of the dummy load

An LM324 op-amp was used here. An LM358 will also work. Any other bog standard op-amp will also probably work provided you get the power supplies correct.

The voltage input in this circuit goes up to 24V. This is due to the use of an IRF540 MOSFET. Ideally, a logic level MOSFET should be used instead, as this will allow a much smaller voltage to be used (around 5V).

The 1R resistor must be able to dissipate quite a bit of power. In my brief testing, I found that it could dissipate around 10W, but even the 50W resistor got hot, so you may want to attach it to a heatsink, especially if you use a smaller one.

The IRF540 MOSFET was also attached to a heatsink. In my brief testing, this can get very hot, and I did in fact kill one transistor when using a heatsink which was too small, so I swapped it for a bigger one. I used a 2.7°C/Watt heatsink which is available from Farnell as product code 171-0614 for about £4.

Breadboard Build

I built it on a breadboard. I had to attach the MOSFET to some wires as there is no room on the breadboard for the heatsink.

Breadboard Photograph

Working Photo

In this photo, the multimeter shows the dummy load drawing 1.02A from a 5V power supply. The oscilloscope shows that there are no oscillations, hence the circuit is stable. I have used it up to about 3.4A.

Photo of the dummy load working.

Oscillations

It is quite easy to get this circuit to oscillate. This is unwanted and can make it difficult to set the current accurately. If the capacitor in the above circuit diagram is removed, the circuit will oscillate. I got some fun looking waveforms on my oscilloscope, which vary as the current is adjusted :).

I tried the capacitor in other locations and still got oscillations, so settled on having it in its current location.

Click here to see the oscillation problem.

A slightly more permanent build

I built the circuit up on some strip board and added some banana connections. This was only intended to be temporary till I make a PCB, but it actually works OK.

A photograph showing the dummy load circuit built on stripboard