Главная > Analog circuits, In English > Dummy load, now truly resistive

Dummy load, now truly resistive

This article was inspired by this Hack a Day post, which is about building a dummy load using a FET and op. amp. No doubt, this build is good, but according to the schematic it turns out that it is a constant current load. In some cases it can be useful, but one can also be in need of a constant resistance load. So in this article I’ll show such design.

Let’s take a look at the main part of a Re:load schematic:


Here the voltage from shunt resistor is compared against some constant value, which we get by dividing constant power rail voltage, produced by three-terminal regulator which can be seen on full schematic. Theory says that when operational amplifier is used with a feedback, it drives the output in a way that is aimed to equalize the voltage applied to its inputs. So MOSFET’s gate is driven in a way that results in a constant voltage on a shunt resistor, which in turn gives us a constant current through it and consequently through all our load. As this reference level is independent from the input voltage, resulting load current is also independent from it and this means that our load violates Ohm’s law, i.e., it is not resistive load.

To make our load behave as a resistor and obey Ohm’s law we can throw away power regulator and connect divider directly to an input:


LM358 can handle power supply voltage up to 32 volts, so zener diode is needed mostly to avoid IRL540‘s gate breakdown which can possibly be caused by an output of an op. amp at supply voltages above 10V, which is maximum for this FET’s gate.

The resistive appearance of this schematic can be shown by calculating it’s equivalent resistance (the first equation represents a fundamental op. amp’s property mentioned above):



Sorry I have no time to test this circuit in real life, but here are the results of simulation in Proteus:






It can be clearly seen that the circuit behaves like a resistor with equivalent resistance of 10 Ohms. This perfectly matches the value we can get by substituting appropriate resistor values into given formula.

Рубрики:Analog circuits, In English
  1. Michael
    08/02/2013 в 15:31

    (for instance, a low pass RC network with a time constant of a second)

  2. Michael
    08/02/2013 в 15:30

    I’ve built a circuit like this an in reality you need an RC network between the output of the opamp and the gate of the MOSFET. Otherwise you will still get the correct average current through the load resistor, but the current will be very very noisy.

    • YS
      09/02/2013 в 01:09

      Thanks for the remark. I definitely missed this in a post (partly because what I discuss is more an example than real circuit).

  3. petro-ew
    07/02/2013 в 13:34

    yes )

  4. 07/02/2013 в 10:25

    In English I can read it in Hack A Day / original post. Where is Russian lost?

    • YS
      07/02/2013 в 12:23

      On HaD you can read about constant current load, while what disussed here is a constant-resistance load.

      I’ve just checked HaD main page — they didn’t post a link to my article yet. 🙂 So you can’t read it on Hack a Day. 🙂

      There’s no Russian version yet. As this was created in a response to an english post, I originally wrote it in English.

  5. Vga
    07/02/2013 в 10:08

    Quite interesting. But… Why so English?)

    • YS
      07/02/2013 в 12:25

      That’s ’cause it was created as a response to an article which is written in English and targets English-speaking readers of HaD. 🙂

  1. 08/02/2013 в 15:47
  2. 08/02/2013 в 15:21
  3. 08/02/2013 в 15:20
  4. 08/02/2013 в 15:08
  5. 08/02/2013 в 15:02

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