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I am beginning an electronics design project to build a constant current load to drive testings for batteries, bench power supplies etc. I have been hard trying to locate some black-on-white information over the Net to get the basics: what is the main operation principles of PWM controlled constant current dummy loads and, what do the particular parts like operational amplifiers of the circuit do. I think, I have been trying to find these basics with wrong search string but can't help myself anymore to invent some other words than "dummy load basics" or so. I hope you guys are able to help me in finding the answers.

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When they're sold commercially, it looks like they're called "Electronic loads". I doubt if the term "PWM" will be relevant to your search. The principle of the electronic load won't depend on how the D/A conversion is done to generate the control signal. – The Photon Dec 19 '11 at 16:51
Welcome to Stackexchange! Just as a note for future questions, there is no need to add "Hi" and "-S. J. L." to your questions. Your post is "signed" in the bottom right corner of your post and people can easily click to your profile if they want to learn more about you. – Kellenjb Dec 19 '11 at 20:56


I got a great source for you, from Dave again! Here is the EEVblog #102 - DIY Constant Current Dummy Load for Power Supply and Battery Testing. And here is some info from that:

The circuit uses two OP-AMPs to set the constant current.

The first OP-AMP (IC1A), is configured as a voltage follower. What it does is; it takes the voltage at its non-inverting pin (pin 3) and shows it in its output (pin 1). This configuration is used because of its very high input impedance and very low output impedance.

Second OP-AMP (IC1B), is configured as a voltage follower too, however in the negative feedback there is Q1. So in this case (as its general behavior), OP-AMP will try to equalize the voltages in its non-inverting and inverting inputs (pin 5 and 6). In order to do this, OP-AMP will do whatever it can do with its output. So, in the end, voltage at pin 5 and pin 6 are going to be equal.

Saying that they are equal, let's continue. Upper pin of R3 is connected to the inverting pin of the second OP-AMP (IC1B:6). Lower pin of R3 is connected to the ground. So, voltage across R3 is equal to the voltage at pin 6, which is equal to the voltage at pin 5.

Let's say we have 3V at the output of IC1A. We will have a voltage of 1.5V at the non-inverting terminal of IC1B (pin 5), because of the voltage divider resistors R1 and R2. The voltage at IC1B:5 is going to be equal to IC1B:6 and that will be equal to R3.

With the ohm's law:

1.5 = I * 1
I = 1.5

So, as you see, voltage across R3 is equal to the current going through it which means that the current you are sinking from PS-1 pin which is connected to the battery or power-supply you want to test.

Let's say you want to have 1.2 A from your battery, then you will set your pin 5 to 1.2 V and that means pin 3 to 2.4 V.

To conclude, we can say that the voltage you set with your potentiometer (R4), is going to be the constant current source value.

Schematic This is my first time drawing a schematic with Eagle, if I've made a mistake, I'm sorry.

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lol ;) the community is quite tiny – qdot Dec 19 '11 at 12:04
Can you write an answer that actually includes content in the answer itself? It is fine to link to other sources to supplement your answer, but should not be the entire answer. – Kellenjb Dec 19 '11 at 20:57
@Kellenjb, Added the content. – abdullah kahraman Dec 20 '11 at 8:44

David L. Jones's video blog covered that particular subject really well, with a sample design, some explanation, etc.


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Same time, same thing! lol – abdullah kahraman Dec 19 '11 at 12:03
Can you write an answer that actually includes content in the answer itself? It is fine to link to other sources to supplement your answer, but should not be the entire answer. – Kellenjb Dec 19 '11 at 20:57

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