Electronic Load diy - questions about current peaks in the DUT - Page 1
Forum section: electronics / projects

author: Avelino Sampaio
author_rank: Frequent Contributor
timestamp: January 22, 2024, 07:06:16 pm
content: To avoid a high initial current peak, extracted from the DUT, I initially keep the Mosfet in cut-off. This is well resolved. The issue is that, when activating the relay, a large peak current (16A) passes through the Mosfet, even though it is in cut-off. However, it is short-lived as you can see in image 1. I believe that this current is due to the high capacitance of the IRFP460. The Cds capacitance is around 520pf. The test was carried out with my PSU at 10v. The shunt resistor has a value of 0.125RI ask for help with my following questions:1) Do you consider this peak current harmful for any specific type of DUT?2) Could this level of current cause problems for the relay contacts?3) Any suggestions to minimize this peak current, caused by the Mosfet capacitance?4) I'm very curious about how this is handled by a professional electronic load project, since several MOSFETs are used. If you have one of these beauties, could you show an image capture from the oscilloscope, from this initial moment?

author: Kleinstein
author_rank: Super Contributor
timestamp: January 22, 2024, 07:44:52 pm
content: The pulse is so short (< 1 µs), that it would not hurt the relay. It should also not hurt a power supply - it's current from an output capacitor. With that speed it is also the question how much of the signal is from shunt inductance and not just the actual resistance. Some parasitic inductance could also explain the ringing.I don't think the other electronic loads care about that time scale. They usually not even care about turning the FET off, if the voltage is too low and this way avoid the much larger peak from starting with the fet turned on all the way.

author: Avelino Sampaio
author_rank: Frequent Contributor
timestamp: January 23, 2024, 09:50:20 am
content: Quote from: Kleinstein on January 22, 2024, 07:44:52 pmThank you very much for clearing my doubts. It's good to know that the project is behaving as expected. But my curiosity about how a professional electronic load behaves remains.

author: Avelino Sampaio
author_rank: Frequent Contributor
timestamp: January 24, 2024, 12:46:38 pm
content: I did an experiment using a 3k9 resistor (R36), with the aim of carrying out a controlled charge of the C11 capacitor and the Mosfet capacitance. I changed the PIC programming to delay the relay activation by a few milliseconds, to allow the capacitors to charge. The result was very satisfactory, achieving a reduction in peak current by 10x. See in image 1 that I lowered the scale from 1v to 100mV.

author: Avelino Sampaio
author_rank: Frequent Contributor
timestamp: January 28, 2024, 01:51:15 pm
content: I did a test with an LED, putting it in series with the DUT, to analyze whether it would resist activating the relay. It gave a small blink, more due to capacitor C11. I decided to remove the 3k9 resistor and accept this short pulse that passes through the mosfet, as there doesn't seem to be enough energy to damage anything. Completing my project and very satisfied with the learning.

author: Martin72
author_rank: Super Contributor
timestamp: January 28, 2024, 10:35:10 pm
content: Hi,QuoteJust to make sure that I have understood it correctly:I take a supply and connect it to my load.The load is set to 0A, so it doesn't have to supply anything.Then I press the load on/off button and use a current clamp to see if a corresponding "inrush current" appears, right?I can do that, but I don't think you can see anything.

author: Avelino Sampaio
author_rank: Frequent Contributor
timestamp: January 29, 2024, 07:44:40 am
content: Quote from: Martin72 on January 28, 2024, 10:35:10 pmMy curiosity was about the inrush power generated when connecting the PSU, given that there must be ten to twelve mosfets in parallel, in addition to the RC shock absorber. An output capacitor from a psu seems to be able to deliver this short duration power without any problems.I have a question related to the RC shock absorber. Note the image of an HP electronic load, where the 3R9 resistors are 5w. Why did the designer use 5w resistors? I'm using a 1/4w resistor in my project and now I have this question as to whether it is undersized.

author: jan28
author_rank: Contributor
timestamp: January 29, 2024, 04:36:00 pm
content: Quote from: Avelino Sampaio on January 29, 2024, 07:44:40 amYou don't need a big power resistor for a single spike like turn-on or hot plug. The power of the spike might be high, but the time duration is very short. This results in a small amount of total energy that only needs to be dissipated occasionally.You might need it if you connect a very noisy DUT (e.g. bad switching power supply). The RC filter will filter smaller individual spikes, but these spikes are repeating constantly so the resistor will dissipate more energy. I assume that is the reason for the HP electronic load to have 5W power resistors.

author: Avelino Sampaio
author_rank: Frequent Contributor
timestamp: January 30, 2024, 07:19:03 am
content: Quote from: jan28 on January 29, 2024, 04:36:00 pmUse will be more focused on batteries and linear sources. Eventually I will be able to use it with low voltage switched sources. I decided to change R14 (2R2), which was 1/4w, to 1w.