Constant current load Help needed again - Page 1
Forum section: electronics / beginners

author: LukeB
author_rank: Contributor
timestamp: September 28, 2018, 10:12:34 am
content: Hello again,I have another really newbie question, sorry, I have this constant current load that I am building as a learning experience and as a tool. I think most of the circuit is all good but I want to know what you think about C2. Is that the best way to smooth out the load?? As you can tell I am still learning but I would like some criticism on my project here so that I can learn. Hopefully this doesn't waste anyone's time but I am here to learn.Thanks for the help in advance.BTW, VCC is probably going to be 12v or 9v and the load will probably be for a lead acid car battery. The max current really only needs to be 1A but up to 4 would be nice. Probably depends on mosfet temp.

author: Prithul0218
author_rank: Contributor
timestamp: September 28, 2018, 02:58:00 pm
content: I am new here. I am not an expert, just a young hobbyist. So judge my words before applying in your project.It seems to me like what you're doing is smoothing the feedback signal. But that's not what you want to do. You want a fast changing feedback but your capacitor is actually slowing it down. Maybe a tiny capacitor would be ok, like, a few pFs but not necessary.However, in order to make the load current smooth, what you should do is make the output of the op amp stable. Check the attached photo. It is from the part 1 of Scullcom Hobby Electronics DC load project which I highly recommend you take a look at, here's the full YouTube playlist with 10 videos:https://www.youtube.com/playlist?list=PLUMG8JNssPPzbr4LydbTcBrhoPlemu5DtYou would probably want to use that type of RC filter configuration.Correct my mistakes and help me grow.Thanks.

author: Dave
author_rank: Super Contributor
timestamp: September 28, 2018, 05:03:29 pm
content: Quote from: Prithul0218 on September 28, 2018, 02:58:00 pmNo mistakes, your advice is solid.

author: Prithul0218
author_rank: Contributor
timestamp: September 28, 2018, 06:00:30 pm
content: Thank you Dave.

author: LukeB
author_rank: Contributor
timestamp: September 29, 2018, 12:40:32 am
content: So why do you need to tie it back to the inverting input?With my method wouldn't smoothing the feedback out a bit cause the output to be smooth anyway?

author: Prithul0218
author_rank: Contributor
timestamp: September 29, 2018, 03:05:45 am
content: Quote from: LukeB on September 29, 2018, 12:40:32 amThe op amp configuration with resistor and capacitor you see in the picture is probably called "Inverting Integrator". I suggest you to look it up in order to understand why the resistor is connected to the Inverting input.Making the feedback smooth, the op amp will not be able to detect the noises. It will "fool" the op amp into thinking that it's already noiseless. But your op amp needs to detect the noise and control the output accordingly in order to get rid of the noises. That's why making the feedback smooth will not work.Somebody correct me if I'm wrong.Thank you.

author: Kleinstein
author_rank: Super Contributor
timestamp: September 30, 2018, 10:34:01 am
content: The capacitors in the feedback loop are not primarily about noise, but to avoid oscillations. The capacitor around the OP provides local feedback that is not effected by the possibly slow MOSFET. The actual feedback from the shunt is only for the lower frequencies, were the power part is well behaved.The capacitor in parallel to the shunt does the wrong thing, it slows down the feedback signal and this way causes a phase shift that promotes oscillation.One can also see the OP with local FB via the RC as a kind of PI regulator. The capacitor sets the I term while the series R (not always found in current sinks) sets the P term. So the capacitor is there to slow down the regulator enough that it does not oscillates. If chosen right the addition P term can help with stability and the response to a changing set current.

author: Prithul0218
author_rank: Contributor
timestamp: September 30, 2018, 10:51:44 am
content: Quote from: Kleinstein on September 30, 2018, 10:34:01 amI have used this method before but never thought of it that way. It really is like a PI controller.I have a side question. How does one select the right values for the capacitor and the resistor? I just used the same values as the schematic.