Limit current on capacitor - Page 1
Forum section: electronics / beginners

author: joeyjoejoe
author_rank: Frequent Contributor
timestamp: October 01, 2023, 10:54:08 pm
content: I'm trying to design a device that will work off parasitic power from 1-wire system.I'm putting a capacitor in front of my device, and the voltage of the system is 12V.I need to keep the current draw under 45mA (absolute max), or else the source power (fairly high impedance) will cutout.My first thought was to put a 680ohm resistor in front of this. But now I'm thinking, doesn't this just mean the resistor will be turning that into heat and not keeping my power draw down?(I am assuming an ideal eSR of 0 on the cap for simplicity)

author: ozkarah
author_rank: Regular Contributor
timestamp: October 01, 2023, 11:32:33 pm
content: With that 680 Ohm resistor, the current will be limited to 17mA. If you reduce the resistor value to 220 Ohm then that will let you draw 54 mA most.However, you need at least a 1W resistor.For a simple solution, you might wanna look for a current-limiting NTC resistor whose resistance will drop when higher currents are needed.

author: golden_labels
author_rank: Super Contributor
timestamp: October 02, 2023, 04:40:04 am
content: Do you meanthe1-Wire® (it’s a word owned by Maxim referring to a particular solution) or just any attempt to communicate and send power over the same wire?If the former, the values are already given by specs: ≤ 800 pF capacitor, 1.5 kΩ – 2.2 kΩ pull-up resistor and supply voltage of a bit over 5 V. And the resistor is not present on the downstream side. Power comes through the pull-up resistor on the upstream side and that already limits current to about 3.3 mA. And that is way more than the device can actually consume: it must exhibit resistance like ten times that of the pull-up resistor, so currents are below hundreds of µA.(1)If the latter, than — if I get your idea right and there is no pull-up resistor on the upstream end — you will not be able to send data. If you wish to do that, you will need the resistor to be at the upstream end and with much higher value.(2)With all the limitations that entails.If you are fine with one-way communication only (upstream to your device), tell us something more about its electric requirements. Voltages required, margins, current consumed on average and peak etc. A few things that are expected in general are:The capacitor is likely to be much smaller, unless there are particular requirements requiring a larger one. The role of this capacitor is to provide power during short periods of communication. 1-Wire itself uses sub-nanofahrad capacitors.Required power rating of the resistor is unlikely to be 1 W. The only high current it sees is the inrush current, which is a short event. Resistors can safely withstand such infrequent overloads.If the capacitor happens to be of higher value, your device must be able to deal with slowly rising power supply voltage.A side note: you may add attachments to your posts, below the post form. This way the entire post is held on the forum.(1)To see why, notice that the pull-up resistor and the load (device) form a voltage divider. That divider determines (and limits) voltage the device can get even without any communication.(2)Higher value is needed, because both ends use open-drain outputs for sending data. They are shorting the pull-up resistor to ground.

author: joeyjoejoe
author_rank: Frequent Contributor
timestamp: October 02, 2023, 12:37:15 pm
content: Now we're talkingSo, I don't think it's THE 1-Wire as you say. That said, I am reverse engineering a garage door opener wired-remote. So I know very little of the other end, but based on an existing project and on running my logic analyzer, it seems to burst at 9600bps and does about 12 bytes.I can pull about 45mA - anything more, and the voltage drops to 0. (There is quite a bit of voltage drop over the very long remote wire which is thin - maybe 22 or 24ga)My hope is to run a custom device based on an ESP32-MINI-C3 to signal on this wire. I've plugged in a previous design using the same buck converterhttps://www.rohm.com/products/power-management/switching-regulators/buck-step-down/integrated-fet-nonsynchronous/bd9g101g-product, and it works. The next challenge now is sustaining power while getting it to pull down the wire.I'll plug in my device and measure the load a bit better off 12VDC to see the numbers. My hope is that at 12V, while I will be close to 45mA, I will get away with it.My first attempt I'll put in an easily replaceable resistor in order to tune things. It will be a simple THT 1/4W, but it sounds like based on your comments moving to a 0805 resistor eventually would be fine from a power point of view?

author: Terry Bites
author_rank: Super Contributor
timestamp: October 02, 2023, 08:15:42 pm
content: Use the reverse leakage from a nasty Schottky as the limitter?

author: Terry Bites
author_rank: Super Contributor
timestamp: October 02, 2023, 08:23:47 pm
content: How about putting a big Schottky in reverse. Choose for leakage current ?

author: golden_labels
author_rank: Super Contributor
timestamp: October 02, 2023, 08:25:01 pm
content: Once again: please do not spread post content across multiple services. If it’s your own work, the forum provides an option to attach images (below the message body editor). In the case of your last post you even used a wrong URL, so most people aren’t even aware there is any image.I am still missing a complete picture of the situation, so I will be making some guesses. Corrent me, if I am wrong. You have a remote for a garage door opener. Both the door opener and the remote are acting as black boxes. You want to replace the remote. The opener provides power and receives (but doesn’t send) data. You still didn’t say, what are requirements of your device, which is what I asked about. So, once again: what does your device require to run? ESP32-C3-MINI requires 500 mA minimum (IVDD, tab. 4.2, p. 14), based on the datasheet. That’s over ten times more than values mentioned so far and also over maximum current from BD9G101G (IOUT, p. 5). Do you have any information, if there are pull-up resistors present in the original devices (and which one)? How much current is the original pilot drawing?As for the rest: 12 V / 45 mA ≈ 267 Ω. The wire can’t be causing the drop, unless you run it across a town.(1)Your description resembles more source’s internal resistance (if the voltage falls to 0 smoothly) or an overcurrent or thermal protection kicking in (if it falls abruptly). In the former case it may not be possible to draw current that high continuously.As for a 1/4 W resistor being sufficient: without knowing the details it’s hard to tell with any certainity. It may and,with the original assumptionsit sounds plausible. The point was that a 1 W rated resistor is not needed, because it’s just a short, rarely occuring burst of high current. As for using 0805: I have no opinion or practical experience with how these handle exceeding limits.(1)See tab. 10 (p. 23) inCopper Wire Tables (1966) from National Bureau of Standards(archive.org), orp. 27 in the 1914 release.