hybridfree/eevblog-posts huggingface dataset:
Current handling spec for rotary switch - Page 1
Forum section: electronics / beginners

author: JuiceKing
author_rank: Regular Contributor
timestamp: August 02, 2012, 03:27:48 pm
content: Hi,I'm looking at 2P4T switches. The maximum load it needs to switch is 25V at 0.25A VRMS.Holy smokes, these switches are expensive. And, they don't even meet this spec. For example:SWITCH, ROTARY, DP4T, 200mA, 220VNo. of Poles: 2No. of Switch Positions: 4Angle of Throw: 45°Contact Current AC Max: 150mAContact Current DC Max: 200mAContact Voltage AC Max: 220VContact Voltage DC Max: 28VCircuitry: DP4TThis one is rated at 150mA, but I don't need 220V. Can I expect it to handle 250mA at 25V without a problem?- Ken

author: mariush
author_rank: Super Contributor
timestamp: August 02, 2012, 03:49:17 pm
content: Is that DC or AC? There's 2 different limits for them...Contact Voltage DC Max: 28VContact Current DC Max: 200mA250mA is a bit higher than 200mA, the max current listed. If in average you're going to have less than 200mA on it, I would say maybe they'll work, I'm not sure.

author: JuiceKing
author_rank: Regular Contributor
timestamp: August 02, 2012, 03:54:58 pm
content: Quote from: mariush on August 02, 2012, 03:49:17 pmAC. My current requirements are 250mA and the switch is rated at 150mA. However, I will only put 25VAC across it, and the switch is rated much higher. So, if power is the real concern, it's fine. But if current rating must be respected even at a lower voltage, I need a higher spec part--more money and none in stock (at least none at Newark).- Ken

author: SeanB
author_rank: Super Contributor
timestamp: August 02, 2012, 04:17:44 pm
content: No, it can handle 150Ma AC only, and any voltage from zero to 220VAC. So 25 VAC is in the range for voltage, but the current is rather over the limit. That being said, if you are going to have few operations of the switch, or are prepared to have a life considerably less than rated, you can use it at this overload. If the current is only to be carried but the switched current is low then it will work.

author: T4P
author_rank: Super Contributor
timestamp: August 02, 2012, 04:27:09 pm
content: I think you'll need to settle to use a lower rated switch to switch on a transistor ...I THINK of course.

author: HardBoot
author_rank: Regular Contributor
timestamp: August 02, 2012, 05:54:44 pm
content: I wouldn't worry about it's spec too much, just gotta check it's contact resistance when running your worst-case current through it.

author: JuiceKing
author_rank: Regular Contributor
timestamp: August 03, 2012, 03:31:55 pm
content: Many thanks for the help here.I looked around some more and found to my surprise that I could save money by getting a switch with exactly the capacity I need at lower cost by filtering for one with MORE poles than I need. I guess I happened to filter by a odd-ball configuration that they don't stock or don't buy in quantity.It is harder than it should be to shop for parts on line.
Current in a charging lipo cell - Page 1
Forum section: electronics / beginners

author: ppTRN
author_rank: Regular Contributor
timestamp: February 05, 2025, 11:13:31 am
content: Hi everyone,I designed and built a simple lipo charger (TP4056 based) and I am currently testing it.- Without battery it outputs 4.27V and STBY led is on. It outputs 70mA since it is supply the circuitry connected tho the battery.- If I short the output to ground it shuts the voltage off and does not draw any current (integrated short circuit protection I guess)- If I put a Lipo charged at 3.97V, the STBY led is still on but the CHRG led does not light up. This also happended when the battery was more discharged. The current drown is about 500mA, as per the programming resistor that I choose (2 4.7k in parallel).The thing that worries me tho is the fact that even at this state of charge the circuit still drows 500mA.When is it supposed to go into into constant voltage mode and stop feeding current to the lipo?I really do not want to find myself with an overloaded battery.The only suspect that I may have is about a polyfuse in series with the lipo, that cause a 100mV voltage drop across it, and a shunt resistor used to evaluate current consumption, that only drops a few mVIn the schematic attached you can find 2 lipos, but actually only one is connected.I am still leaving it connected since I guess the lipo should reach the 4.2V programmed, just wanted to check with you guys.

author: kjr18
author_rank: Regular Contributor
timestamp: February 05, 2025, 01:52:25 pm
content: Standby LED usually green or blue) should be lit when no battery is connected, and charge LED should be flashing briefly. When charging, only charge LED should be lit. And when charge is finished, only standby LED should be lit.As for CC/CV curve, probably around 4,12V should start lowering current, and keep constant voltage at 4,2V until current drops below 0,1 C, 50mA in your case.With fuse it would simply start dropping earlier, because TP4056 monitors it's own pin, not battery terminal. In most cases, there is no noticeable difference in voltage, but in your case it will be some (because of mentioned fuse.

author: ppTRN
author_rank: Regular Contributor
timestamp: February 05, 2025, 02:14:19 pm
content: Thank you.It actually started falling after a while.Now I feel more confident about the correctness of the circuit.
µCurrent instead of current range on multimeter - Page 1
Forum section: electronics / beginners

author: jts
author_rank: Newbie
timestamp: December 26, 2013, 07:48:12 pm
content: I'm looking to buy my first real multimeter and I'm leaning towards the Agilent U1232A, but after watching the µCurrent video on kickstarter I'm thinking that it would be smarter to get the U1231A, a cheaper model without the current range, and just use the µCurrent for current measurements since multimeters are horrible at it anyway. What would you recommend?

author: idpromnut
author_rank: Unknown
timestamp: December 26, 2013, 08:21:20 pm
content: Since I have the U1232A, I figure I could answer this!This was actually my first "high quality" meter that I bought as well, and well worth it I think. I don't have a uCurrent (yet, but it will be in the mail by this spring it looks likebut I can't say that I have run into many issues when measuring A, mA or uA for my purposes. In truth, most of my work is not in the low voltage ranges, so I don't tend to notice the burden voltage of my meter that much. Some quick 4-wire measurements of the resistance of the A and uA ranges for the Agilent (using my HP 3478A):Amps Range: 0.0191 Ohm (6A range)uAmps Range: 3.0183 KOhm(note that the Agilent was measuring ~ 100uA, so this was the 600 uA range)So the uA range has a fairly hefty resistance (and I guess that would translate into a fairly high burden voltage). Picking through the manual for the 1232A, we can see on page 107 Agilent lists the burden voltage for the 600 uA and 60 uA ranges as up to 2.5V.All that said and done, I come back to my original observation that for my typical uses, it's not a big deal for me.

author: Unknown
author_rank: Unknown
timestamp: December 26, 2013, 10:06:40 pm
content: The µCurrent lacks any fuses or other forms of input protection and is obviously limited to currents under about 1 A. Imagine a review of a multimeter without any input protection? I would only use the µCurrent in known safe circuits powered from a current-limited power supply, like a bench supply or small battery. My DMM has beefy HRC fuses that can safely interrupt high-energy circuits, so I'd much rather use my DMM near high-energy circuits or in unknown circuits.

author: idpromnut
author_rank: Unknown
timestamp: December 26, 2013, 10:14:16 pm
content: And actually, since I'm a beginner as well, is it correct to measure to the resistance of a multimeter's current range(s) to gauge how much impact it will have on a circuit under test as I did above? Of course The Fine Manual had some numbers for the burden voltages on the various ranges so that would probably be the best basis for impact analysis of a multimeter.

author: EEVblog
author_rank: Unknown
timestamp: December 26, 2013, 11:12:28 pm
content: No, get a mutlimeter with a proper current range. Don't think that the uCurrent is substitute for that, it's not.

author: jts
author_rank: Newbie
timestamp: December 27, 2013, 04:33:06 pm
content: Thanks for all the replies, I'll be going with the U1232A or the U1233A!
Current latching button? - Page 1
Forum section: electronics / beginners

author: Muffins
author_rank: Regular Contributor
timestamp: July 19, 2022, 09:20:23 am
content: Hi guys.I need some help with some terminology here. I have a machine where there is a push button that when pressed remains on while there is current through it, another button in series cuts the circuit and resets the state of the first button.The circuit is like the attached image.I've searched around and am not getting what I want. I need a 22mm diameter industrial type button, like available on RS and all that. I don't need a mechanical latch, it needs to be like a contactor and latch with current. What are these things called?

author: BeBuLamar
author_rank: Super Contributor
timestamp: July 19, 2022, 09:45:41 am
content: If you talk about the entire circuit then it's generall called "Start/Stop latching relay circuit" or something like that.The pushbuttons are just normally closed for the stop button and normally open for the start button.

author: vu2nan
author_rank: Contributor
timestamp: July 19, 2022, 12:33:16 pm
content: Quote from: Muffins on July 19, 2022, 09:20:23 amItisan electromagnetic contactor or relay.The following schematic is for a contactor / relay with a DC coil. It would be the same for one with an AC coil, in which case the freewheeling diode would be replaced by an RC snubber.Industry standard 22.5 mm diameter push button switches with 1NO + 1NC contacts could be used.When the 'On' / 'Start' button is actuated, the coil 'K1' would get energised and the contact 'K1' would close, effectively bypassing the 'On' / 'Start' button. Hence the coil would remain energised even after the button is released.When the 'Off / 'Stop' button is actuated, the coil 'K1' would get de-energised and the contact 'K1' would open and remain so even after the button is released.This circuit is known as a 'contactor / relay - based, non-remnant latching circuit' for 'start / stop' functions.
Current limit by LM358 in buck-boost circuit - Page 1
Forum section: electronics / beginners

author: Vindhyachal.takniki
author_rank: Frequent Contributor
timestamp: June 15, 2020, 07:15:49 am
content: 1. I have a buck-boost IC, where input is 12V or higher and output I need is 12.6V to charge a lithium -ion battery 3 cells series2. Need to limit charge current of battery to 0.650A, so according to datasheet I have selected 0.150ohm resistor.3. If I dont connect D4, and connect battery to charge, it start taking 4Amps of current. But when I connect D4, current starts falling but it falls at very slow rate. Takes rough around 10 min to fall, meantime IC gets super hot and sometimes shut down itself.What I am doing wrong here?4. Also IC temperature goes to 98C sometimes, I placed a sensor on top of IC, Tj max allowed is 125C as per datasheet so very close operating I am.
Current-limit circuits for 78XX voltage regulators - Page 1
Forum section: electronics / beginners

author: khatus
author_rank: Regular Contributor
timestamp: March 15, 2019, 04:54:04 pm
content: Hello guys i Want to find the value of RSCand R1in the following circuitIn this circuits, Q1is the high-current pass transistor, and its emitter-to-base resistor has been chosen to turn it on at a given load current. In the circuit, Q2senses the load current via the drop across the resistor RSCcutting off drive when the drop exceeds a diode drop(i.e 0.7V). Now i want to calculate the value of RSCand R1But How can i solve this problem???For linear voltage regulator

author: RES
author_rank: Regular Contributor
timestamp: March 15, 2019, 05:44:04 pm
content: check this application note:http://www.ti.com/product/LM340/technicaldocuments(direct link)http://www.ti.com/lit/gpn/lm340on page 16 you see your example.(figure 25, top right) two formulas:RSC = 0.8 / ISCR1 = βVBE(Q1) / IREG Max (β +1) – IO Max

author: iMo
author_rank: Super Contributor
timestamp: March 15, 2019, 06:27:07 pm
content: It will not work as a current limit.When the Rsc creates 0.7V the Q2 starts to conduct and the excessive current will flow through 7805.You have to put a resistor R2 into collector of the Q2.Set R1 such you get enough current for 7805 (10mA).For example 50ohm.The R2 for example 50-100ohm.Edit: I would put R3=100ohm into the Q2 base.With Rsc 1ohm it will limit to 1A, with Rsc=4ohm to 250mA, and so on..
Current limit potentiometer problems - Page 1
Forum section: electronics / beginners

author: niflheimer
author_rank: Contributor
timestamp: June 16, 2013, 01:24:08 am
content: Hello. I've built a simple PSU based on the two regulator design - with a L78S12 in current limit configuration and a low dropout , 3 amp CS5203 for voltage regulation . It works quite well , but the trimpot I'm using for current limiting heats up a lot.Is there any way to reduce the current through it ? At anything over 150mA it heats up a lot (as it should actually )I don't have any simple opamps at hand - the only ones I have are some dual ones which need a two rail supply. And I'm out of quality wirewound pots as well . I do have a lot of triacs , mosfets and transistorsThank you!Edit - forgot to add the schematic

author: Skimask
author_rank: Super Contributor
timestamp: June 16, 2013, 01:52:28 am
content: http://www.eevblog.com/2010/08/01/eevblog-102-diy-constant-current-dummy-load-for-power-supply-and-battery-testing/Replace to pot with this circuit maybe?

author: don.r
author_rank: Frequent Contributor
timestamp: June 16, 2013, 02:13:54 am
content: Have the wiper of the pot drive the base of a NPN power transistor with the collector connected to the regulator's output and the emitter to the far side of the pot. Add a further resistor between the output and the pot to lower the current flowing through it.

author: niflheimer
author_rank: Contributor
timestamp: June 16, 2013, 12:02:23 pm
content: Ty don.r , will try it tonight - I wasn't sure if it would work when I thought about doing it this way .@skimask - I plan on doing a CC load after Dave's design , but at the moment I don't have the parts at hand - and I'm too lazy to drive 10 km to the nearest part storeI'm using this supply to test a box of assorted laser diodes I got for free - and some of the DPSS ones pull up to an amp
Current limited AC bench supply - Page 1
Forum section: electronics / beginners

author: KJP
author_rank: Newbie
timestamp: April 25, 2021, 05:27:05 pm
content: I have two variacs for ac things connected to isolation transformer, the small variac has a little 3 amp circut breaker on it. I am wondering what is a good way to keep my accidents less explosive than 3 amps. Light bulbs seem to be the answer from google, but I wonder if there is a better way? My dc supply has the constant current dial, that would be my goal, to dial in a limit to how bad things go boom.

author: Brumby
author_rank: Unknown
timestamp: April 26, 2021, 03:20:44 am
content: DC is easy. AC becomes a bit more complex.First you have to define how you are wanting the current limiting to operate - on peak, average or RMS values. Then you have to define the response and recovery times as instantaneous limiting will simply clip the waveform ... unless that is what you want.You will need to spell out how you want the supply to respond to various overload situations and then you can move forward.

author: bdunham7
author_rank: Super Contributor
timestamp: April 26, 2021, 03:45:28 am
content: Designing an constant current supply that appears to work for AC the way it does for DC would be hard and probably unnecessary. Lightbulbs are good, but they have foldback characteristics and have to be just right for the load. A better option, IMO, would be an adjustable electronic circuit breaker that would sense the current and then just turn off a relay if that limit were exceeded. You'd leave the circuit breaker in the circuit just in case.A TMCS1101 current sensor, a comparator, a latch circuit and a relay would be all you need.

author: Gyro
author_rank: Super Contributor
timestamp: April 26, 2021, 08:44:28 am
content: Whatever else you do, make sure your have the breaker protection on theOutputside of the variac (ideally with additional fuse on the input side).The most common failure of Variacs is trying to output excessive current at low output voltage settings. Under these conditions, a breaker on the input will not trip because there is no overall excessive input current, but the small section of winding in use and the wiper brush will quickly burn out.

author: KJP
author_rank: Newbie
timestamp: April 26, 2021, 02:46:43 pm
content: Thanks for all the sugestions, food for thought. I will do the breaker on output, and try the ajustable breaker idea

author: rstofer
author_rank: Super Contributor
timestamp: April 26, 2021, 03:36:36 pm
content: Maybe run the variac output through a separate box with a fuse holder. You can then use a number of fuse types: Fast blow, standard, slow blow. You can stock a variety of amperages in each type. Here's a kit of standard fuses:https://www.amazon.com/HUAREW-Values-Glass-Fuses-Assortment/dp/B085YCS3Q5Not universally true but circuit breakers are slow. They generally have an instantaneous trip feature but it is usually set for 10 times full load.
Current limiting a buffer IC's outputs - Page 1
Forum section: electronics / beginners

author: ultrageek.lloyd
author_rank: Contributor
timestamp: August 21, 2012, 08:53:56 pm
content: I'm currently interfacing a microcontroller to a router via EJTAG, and per the EJTAG standard, the current draw on the EJTAG port is not to exceed 8mA per pin. I've placed an sn74hc244 buffer IC between the microcontroller and the router to reduce loading on the microcontroller pins, but I'm wondering if I need to do any current limiting on the buffer output pins, and if so, what's a good way to do so. Any help would be appreciated.Thanks,ultrageek

author: Stephen Hill
author_rank: Regular Contributor
timestamp: August 21, 2012, 10:19:14 pm
content: A simple resistor in series with the opamp output would do the trick. Assuming 5V, that would be 625ohm so a standard 680ohm resistor would do.

author: ultrageek.lloyd
author_rank: Contributor
timestamp: August 21, 2012, 11:48:12 pm
content: Quote from: Stephen Hill on August 21, 2012, 10:19:14 pmThanks for the reply. I thought of that, but I was afraid that the voltage drop across the resistor would cause the router to see less than Vcc, and hence never see a logic high. Now that I think about it though, for inputs, the router is the current source, meaning that for a logic high current shouldn't be flowing, and for a logic low, current will flow from router to buffer, meaning the voltage drop doesn't matter. I guess it's just a case of me not understanding how digital signals work =]. Thanks for your input, I think that solution should work.ultrageek

author: poptones
author_rank: Frequent Contributor
timestamp: August 22, 2012, 01:49:26 am
content: You put the output resistor inside the feedback loop. Voila! It compensates for the voltage drop and you still get protection.
Current limiting adjustable - Page 1
Forum section: electronics / beginners

author: danman1453
author_rank: Newbie
timestamp: September 14, 2014, 03:42:25 pm
content: I am trying to develop a current limiting circuit to be powered by a li-ion battery. I want to limit the current available to a resistive wire. Approx 1.8 ohms. Max current of about 4 amps.I have found a circuit to limit the current, but I want to make it adjustable.The upper circuit is the one I found. The lower is my try at making it adjustable. I feel I missed something, and I am looking for input/suggestions/corrections.Thanks.
Current limiting battery charger supply - Page 1
Forum section: electronics / beginners

author: davegravy
author_rank: Regular Contributor
timestamp: March 16, 2024, 11:49:58 pm
content: I have an MPPT solar charge controller which has a fixed input current limit of 5A. Unsurprisingly it's fed by solar panels.I have an AC-DC converter that produces 15VDC and just under 5A and has overcurrent protection (hiccup mode shutdown). I need to charge the battery from this supply when it's available (and disable solar charging). I think I can manage power path with an LTC4412, the bigger question is whether I should:1) feed the solar charger with the AC-DC converter using some intermediary current limiting circuitry so its protection doesn't trip, or2) add a second parallel charger (without MPPT) designed with a lower input current limit, dedicated to the AC charging path.It's not an option to change the AC-DC converter or solar charger.It seems redundant to have two chargers butcomments like thismake me think I'm barking up the wrong tree by adding current limiting. If current limiting's a viable option, are there any simple circuit options?
Current limiting for COB LED - Page 1
Forum section: electronics / beginners

author: 4kruby
author_rank: Contributor
timestamp: August 19, 2021, 12:50:17 pm
content: Hello,I have a COB LED strip - precisely this one -https://makerbazar.in/products/11-1v-12v-strip-7-inch-cob-led-light-color-cool-white?utm_source=globoRelatedProductsin the website, there is no information about the power rating of the LED. Just the voltage is given.I want to run these LEDs from a Lithium cell(3.7v). So I use a boost converter to step up to 11.5 volts. I tried upto 11.7 volts and the strip was drawing 100mA of current.I didnot go upto 12 volts because, in the website where i bought this, they mentioned that above 12v, the LEDs will be damaged. But I guess that won't happen if we limit the current draw of the diodes.So, my question is, how to find out the maximum brightness I can get from these LEDs (maximum current they can consume) without ruining them or attaching a extra heatsink?Thanks,
Current limiting from my UPS battery - Page 1
Forum section: electronics / beginners

author: pranavAppu
author_rank: Newbie
timestamp: March 12, 2017, 05:51:45 pm
content: I have made a single transistor flyback driver.. and looking for the best power source for hot arcs.I've found a battery from my old UPS(about:600VA) but a direct connection would kill my transistor.Any ideas to limit the current to the highest safe zone..?(I WANT HOT ARCS)Forgot to mention; it's a 2N3055 transistor.Sent from my Lenovo S850 using Tapatalk
Current Limiting Resisters - Page 1
Forum section: electronics / beginners

author: TheHippySeal
author_rank: Contributor
timestamp: March 16, 2012, 11:23:55 pm
content: Hi guys. How do I figure out if a device needs a current limiting resister on VCC or GND? Atmel micros don't seem need one on either, only the inputs or outputs. Is this the same for all ICs? The device I need to know at the moment is this Schmitt trigger.http://www.ti.com/lit/ds/symlink/sn74ls14.pdfIn my circuit the trigger input comes from an external source, then the trigger output goes to the micro input. Should I put the resister before or after the Schmitt trigger?Cheers.

author: Psi
author_rank: Super Contributor
timestamp: March 16, 2012, 11:39:02 pm
content: I dont quite understand why you'd want a current limit on vccThe micro will only draw the current it needs under normal conditions. So you don't need a limiting resistor in series with vcc.If you're worried about overloading the micro controller outputs then you can current limit individual pins with a resistor so the short circuit current cannot get above the maximum.

author: TheHippySeal
author_rank: Contributor
timestamp: March 16, 2012, 11:48:30 pm
content: Thanks for the reply. So micros are like that, but what about other ICs? Particularly that Schmitt trigger.

author: Psi
author_rank: Super Contributor
timestamp: March 17, 2012, 12:10:26 am
content: Quote from: TheHippySeal on March 16, 2012, 11:48:30 pmI cant say i've ever seen a IC that requires a resistor in series with VCC.You just supply the IC with a voltage that is within its valid supply range given in the datasheet.

author: TheHippySeal
author_rank: Contributor
timestamp: March 17, 2012, 03:14:17 am
content: Ok. Thanks for the info.

author: andyg
author_rank: Regular Contributor
timestamp: March 17, 2012, 03:34:32 am
content: If you really want to limit the current draw on VCC due to say a fault, then you can use a PTC Resettable Fuse at VCC's source.

author: wkb
author_rank: Super Contributor
timestamp: March 17, 2012, 07:37:12 am
content: Quote from: andyg on March 17, 2012, 03:34:32 amIf you need fuses in the supply line of a single digital IC or MCU it is dead already anywaySeries resistors in supply lines of digital ICs I have never seen. You would not like the end result anyway, as it limitsthe speed of the IC and it could also make the logic levels go haywire.
Current limiting resistor for buck converter? - Page 1
Forum section: electronics / beginners

author: Pariston
author_rank: Contributor
timestamp: November 25, 2020, 04:03:40 pm
content: Hi,I have a quick question. Should I use a current limiting resistor for 102-VXO7803-500-M-CT-ND (https://www.digikey.ca/en/products/detail/cui-inc/VXO7803-500-M-TR/11569239). Input voltage 24V.

author: Manul
author_rank: Super Contributor
timestamp: November 25, 2020, 07:33:54 pm
content: On the input? No, you should not use any resistors. You may use a fuse if you want. You may also use noise filters, but you probably do not need them.

author: Zero999
author_rank: Super Contributor
timestamp: November 25, 2020, 07:41:08 pm
content: Don't use a series resistor. It will reduce the efficiency and could cause instability. The output is short circuti protected, but it's a good idea to add a fuse to the input, in case a fault develops inside the switching regulator.

author: Pariston
author_rank: Contributor
timestamp: November 27, 2020, 12:16:23 am
content: Thanks for replying!I have a question on the current rating for this fuse. Circuit is powered from the wall through SDI36-48-U-P5 (48V DC) power supply which has Barrel Plug on the output side. Is there any danger to the circuit, the moment plug is connected? I am afraid of initial spark to cause damage in some way.Additional information:48 V power supply is required for my high power led and can not be avoidedBuck converter operating voltage is 5V to 36 V, so, I decided to power this buck converter with 24V zener diode on the line, PDZVTFTR24BCT-ND‎ has maximum zener current of 40 mA. Should be enough for this tiny buck converter.

author: Manul
author_rank: Super Contributor
timestamp: November 27, 2020, 02:31:40 pm
content: Fuse for that buck converter? Roughly, if it is 3.3V 500mA max output, and the input voltage is 24V, max input current should be around 500mA / (24V / 3.3V) * 1.2 = 82.5mA. I choose 1.2 multiplier, because likely efficiency could be around 80-95%.About 48V and zener. In my opinion it is bad idea. You should use a converter.

author: Zero999
author_rank: Super Contributor
timestamp: November 28, 2020, 04:52:11 pm
content: Quote from: Pariston on November 27, 2020, 12:16:23 amThe zener diode would have to do in series and its current rating would depend on the load connected to the buck converter, but it's not the right way to do this, which is to use a DC:DC converter rated to 48V in.https://www.digikey.ca/en/products/detail/mornsun-america-llc/URB4803MT-3WR3/13168103
Current limiting resistors between logic chips or micro - Page 1
Forum section: electronics / beginners

author: Moriambar
author_rank: Unknown
timestamp: April 29, 2021, 11:24:33 am
content: Hello.I am currently building (for fun) some project with counters ('161 or '163 (datasheet)) and a micro (currently using an arduino uno, but an attiny/atmega will be the final choice).In order to interface the arduino with the counters I'll probably use a shift register ('164).The question is: between logic chips or logic and micro do I need current limiting resistors for changing signals or is there some kind of "built-in" current limiting logic?I hope I made myself clear.Cheers

author: Kleinstein
author_rank: Super Contributor
timestamp: April 29, 2021, 11:38:48 am
content: The CMOS logic chips have a kind of current limit. Depending on the series and supply voltage the chips may survive a short.Somthing like 74HC at a low voltage like 2.5 or 3 V may be OK. The old CD4xxx (= HEF4xxx or MC14xxx) series is in most cases quite limited in current at 5 V, but it is slow and slightly different chips.Depending on the supply and functions one normally gets away without current limiting resistors. Just make sure not to program thise µC pins as output that see an logic ouput.With logic inputs there is no need for current limiting resistors.

author: Benta
author_rank: Super Contributor
timestamp: April 29, 2021, 05:54:44 pm
content: If MCU and logic run on the same supply voltage, there's no reason for further precautions. The issues may come when you go "off board".

author: Moriambar
author_rank: Unknown
timestamp: April 29, 2021, 05:59:57 pm
content: Quote from: Benta on April 29, 2021, 05:54:44 pmsame supply.Thanks

author: David Hess
author_rank: Super Contributor
timestamp: April 29, 2021, 11:10:15 pm
content: The output resistance of the CMOS drivers will protect them against shunt capacitance so no external series resistance is required when diving transmission lines and other gates.
Current limiting to an inductive load - do I include its resistance in calcs? - Page 1
Forum section: electronics / beginners

author: dsegel
author_rank: Contributor
timestamp: March 31, 2019, 08:15:44 pm
content: I'm using a transistor to drive a coil with 18V DC. The coil has a resistance if 45 ohms. I want to limit current to about 350ma. Do I need to add a 50ohm resistor in series, or does the coil resistance already limit the current?

author: magic
author_rank: Super Contributor
timestamp: March 31, 2019, 08:24:47 pm
content: Of course it does. 18V/45Ω=400mA so you only need to add a little more resistance in series.

author: MrAl
author_rank: Super Contributor
timestamp: March 31, 2019, 08:56:23 pm
content: Quote from: dsegel on March 31, 2019, 08:15:44 pmHi,Be prepared to catch the back EMF from the coil when you turn it off. The voltage can shoot up very very high and cause a bad shock.Typically a diode is used or zener or something else to absorb the energy.

author: dsegel
author_rank: Contributor
timestamp: April 01, 2019, 04:02:08 am
content: I was planning on using something like a TVS diode because the coil is driven in both directions (i.e. the current direction could be either way). How do I choose a voltage rating? Is 30V breakdown enough for an 18V coil? I measured the voltage spike with my scope to be around +/- 150-200V.And I can just put the diode across the coil terminals, right?

author: MrAl
author_rank: Super Contributor
timestamp: April 01, 2019, 06:53:37 am
content: Quote from: dsegel on April 01, 2019, 04:02:08 amHi,Well it partly depends on how often you have to turn on the coil and how long it stays off for, and the total energy that has to be dumped. The energy will partly dictate the diode power rating because it has to dissipate energy.Do you know the inductance of the coil?And how often do you turn it on and turn it off, like the frequency?

author: Rerouter
author_rank: Super Contributor
timestamp: April 01, 2019, 07:09:03 am
content: The diode voltage is based upon what will damage the transistor, equally the higher the voltage of the TVS diode, the faster the energy in the spike is dumped, With the trade off of the diode being heated much more,The max current will always be the same as the max current through the inductor, (ignoring shoot through) its just how quickly you want to dissipate that spike.

author: dsegel
author_rank: Contributor
timestamp: April 01, 2019, 09:01:52 pm
content: The coil will only be going on for 200ms and off for at least 900ms each time.Don't know the inductance; I'll see if I can find that or figure it out.
Current-limiting transformer using capacitor - Page 1
Forum section: electronics / beginners

author: FHR
author_rank: Contributor
timestamp: May 04, 2015, 07:27:04 pm
content: Hi!I have a microwave oven transformer connected to the mains (240V 50Hz) with 4uF capacitor in series with phase/live wire.The thing I don't understand is, why primary current with not-connected secondary wiring is 0.40A and when I short-circuit the secondary it actually lowers to something like 0.25A.Does anyone have any clue why this happens?

author: T3sl4co1l
author_rank: Super Contributor
timestamp: May 04, 2015, 08:37:20 pm
content: If you measure a few voltages and currents in the circuit, you should be able to come up with a reasonable value for the primary inductance. If you draw up the phasor diagram of circuit impedances (complex values), you should see just what you're looking at.Probably it's above resonance, so you'd get considerably more power with a larger capacitor. Eventually, you'll apply so much voltage to the primary that it starts to saturate, pushing it back above resonance. This is essentially how a ferroresonant transformer regulates voltage (or more precisely, flux == V/freq), although the circuit isn't quite the same.A more linear example of resonance would start with a low voltage, maybe 6.3 or 12.6V, and running the primary in series with the capacitor. Then you'll see a good bit of voltage and current multiplication.As far as I know, the general operating curve is elliptical -- whereas a pure series resistance or inductance has a linear relationship (between O/C voltage and S/C amperage).Tim

author: FHR
author_rank: Contributor
timestamp: May 05, 2015, 02:46:35 pm
content: Didn't know this ferroresonant transformer existed. It kinda makes sense now. Thanks
Current Limiting/Control  - Page 1
Forum section: electronics / beginners

author: SkrillBill
author_rank: Regular Contributor
timestamp: November 10, 2017, 08:28:36 pm
content: Hello,i'm in the process of designing and building a power supply for my workbench. For details on that project, referthis thread.In that build im using a few LM-series regulators to control voltage. Maximum output will is planned to be 1A, but realistically 500mA is a closer goal. At this time im not sure about variable current, but i do want to limit the current to keep the magic smoke inside the components.How would i go about limiting current with a fixed voltage from the regulator without inducing more strain than necessary? Would i limit it before or after the regulators? And, should i worry about having a constant current of some kind (via a choke i think?)?This is all new to me, and i'm not 100% sure how current control or limiting works or what i need to make it work. The more in depth the explanation the better, i want to understand the concept not be told "do this with this and it will work".Many thanks!

author: danadak
author_rank: Super Contributor
timestamp: November 10, 2017, 11:27:51 pm
content: Google "current limit circuits power supply", quite a few hits, but many circuitsfairly simple.Normally limit is placed after voltage source. Its done by sensing load currentand using that in a feedback loop, sometimes as simple as a transistor, a coupleof diodes, and a rsense. Other better aopproaches for greater range and accuracyusing an OpAmp.There are some circuits that foldback the current.Regards, Dana.
Current measurement and on/off control using Arduino - Page 1
Forum section: electronics / beginners

author: rthorntn
author_rank: Frequent Contributor
timestamp: April 04, 2013, 06:20:36 am
content: I am interested in measuring the current and being able to switch on/off equipment remotely.Ideally I would like an enclosure with an IEC male connector and an AU power socket on the panel, so mains to IEC and then the device plugged in, say for example it's a 240v 1hp pool pump.I would like to have a circuit to monitor the current and a relay to switch it on/off.Power for the circuit (ACS758?), Arduino and relay should be stolen from the mains.Has anyone seen anything like this to buy, obviously it's dealing with mains so a DIY job for a beginner may be a bad idea?Thanks.Richard

author: Rerouter
author_rank: Super Contributor
timestamp: April 04, 2013, 06:29:29 am
content: use relays for switching, and current transformers for the measurement, being i don't know your background, by keeping everything as isolated as possible it should work well,

author: rthorntn
author_rank: Frequent Contributor
timestamp: April 17, 2013, 03:37:37 am
content: Thanks Rerouter!
Current measurement - Page 1
Forum section: electronics / beginners

author: zitman
author_rank: Contributor
timestamp: August 14, 2013, 08:41:56 am
content: Hi all,First post and probably a dumb question but one I am struggling to get my head around nonetheless.I have just bought a new multimeter (Extech EX330) as my Fluke 114 did not have current measurement. I was checking out the current measurement ranges and have connected a 68 ohm resistor across a 5v power supply and measured the current through the resistor.I connected the Fluke across the resistor and get a voltage of 4.55V without the Extech in series. When I put the Extech in series on the mA range I get a voltage of 3.9V and a current of 57.4mA which makes sense (via Ohms law).What I don't understand is why connecting the Extech in series on the mA range drops the voltage across the resistor by 0.65V.My assumption is that the Extech is adding some more resistance into the circuit. Is this correct or if not can someone please explain this in reasonably simple terms please.TaZ

author: nukie
author_rank: Frequent Contributor
timestamp: August 14, 2013, 08:55:22 am
content: It's called Burden Voltage. Check out Dave's uCurrent while your at it.

author: ve7xen
author_rank: Super Contributor
timestamp: August 14, 2013, 09:08:51 am
content: Quote from: zitman on August 14, 2013, 08:41:56 amExactly. What the meter is actually doing is putting a known resistance into the circuit and measuring the voltage drop across it. Because of Kirchoff's current law and Ohm's law, if you know V and R you can derive I.Likewise, there will always be a voltage drop across the shunt, which is usually related to the current through it (ie. it is purely resistive), referred to as burden voltage.

author: zitman
author_rank: Contributor
timestamp: August 14, 2013, 09:16:34 am
content: Thanks both,I have measured the resistance of the Extech on the mA range using the fluke and got 11.1 ohms which when I plug the numbers into Ohms law I get the result I expected. I will check out Daves uCurrent video and Kirchoff's current law too.TaZ
Current measuring in DC-Link of Inverter - current signal riddle - Page 1
Forum section: electronics / beginners

author: HGruen
author_rank: Newbie
timestamp: December 19, 2019, 10:00:36 am
content: Hello,for a permanent magnet synchronous motor I designed an inverter (based on the TI designhttp://www.ti.com/tool/TIDA-00774?jktype=design), though simplified to a lower power level. It is based upon three halfbridges CSD88584Q5DC (23nC Qdiodereverse) and a smart driver DRV8323 with integrated differential shunt amplifiers.The idea is, replacing two current sensor by one shunt in the dc link as shown in the attachment. Depending on the switches' state, the measured voltage drop across the shunt is - in theory at least - equal to one of the absolute values of the phase currents. The phase currents change slowly in comparison to the shunt signal, therefore they can be seen as constant during the sampling time.In my setup though, I have an effect that I cannot explain. For low currents (~1A) my reconstruction method works fine. As the currents are increasing, the reconstructed waveforms become strongly distorted. I guess it's due to the exponential falling / rising of the shunts voltage right after the switching instants. The moments for sampling the shunts voltage is situated almost at the 'end' of each possible time'slot'. If the signal's not yet settled, it results in distortions in my reconstructed phase currents.I have no idea how to explain that exponential rising/falling of the signal. I cannot reproduce it using LTSpice by adding various parasitic elements to the circuit. Guessing the charge of the exponential area results in several hundred nC - the reverse recovery charge of the halfbridge diodes is 'only' about 20nC.The recorded waveforms where recorded using DC-coupling at the osci. I get a similiar signal if I use the integrated differential amplifier of the driver (due to the limited slew rate it gets smoothed but the effect is still there).I'd be glad about hints for solving this puzzling issue. Is it a measuring error?(similiar version for german readers can be found herehttps://www.mikrocontroller.net/topic/486380- i updated the sketch)
Current measuring pref a currentclamp - Page 1
Forum section: electronics / beginners

author: flaxo
author_rank: Newbie
timestamp: March 08, 2021, 03:26:41 pm
content: Hi!I need some help choosing the right sensor to measure current.The idea is to be able to measure a single phase from a 3-phase motor with rated current of 75A ( bonusstar if it can handle up to 445A) and the signal from it should be received by a DAQ.The input card can take care of 20mA (-20 to +20mA) or 10V (-10V to +10V) and has 50Ω impedance on the current mode and 107kΩ on the voltage mode.The crux of the matter is that the installation should preferably not be invasive as the idea is that you can easily move around the measuring points. I started by looking around a little on current tongs but was limited by the low maximum current on the card. Also looked at solutions where the signal was a voltage such as fluke i200S but where the requirement for 1MΩ as input impedance put a spanner in the works.Will there be major errors if I use the i200S?Or is the best alternative a split core current transducer and run a current divider, worst case.Excuse my bad English..

author: Terry Bites
author_rank: Super Contributor
timestamp: March 08, 2021, 03:53:39 pm
content: http://www.pemuk.com/products/rct-industrial-sensor/rctrms-3-phase.aspx?You need to be careful if you have a VFD most current transformers are rated for 50 or 60Hz. Rogowski based sensors have wide bandwidth.

author: flaxo
author_rank: Newbie
timestamp: March 09, 2021, 06:44:43 am
content: I missed to mention that they are all direkt drives (for now), so no VFD. I got my hands on a i400S is there any idea to order parts to be able to go from BNC to screwplint or will the impedance missmatch screw things over for me and in what way?

author: Terry Bites
author_rank: Super Contributor
timestamp: March 10, 2021, 09:30:25 pm
content: If you make up your input impedance to 1M by adding a series resistor you can set your calibration error to a known figure. Of course the reading will be 5% ish of its true value. (10700000/(1000000+1070000)). But 50% ish on the 10mv/A setting. Which you may be able to compensate for with a simple gain factor in your DAQ.You can bypass the big resistor with a small cap to reduce noise. Again your DAQ probably has filters you can set. Keep the bandwidth as low as possible.Some kind of calibration will be needed. This will be the hard part I expect.

author: The Electrician
author_rank: Frequent Contributor
timestamp: March 11, 2021, 07:51:42 am
content: Even though the spec for the i200s:https://www.fluke-direct.com/pdfs/cache/www.fluke-direct.com/i200s/datasheet/i200s-datasheet.pdfmentions a load impedance of >1 MΩ // 100 pF, It says the output impedance for the 20A range is ≤ 20 Ω @ 1 kHz, and for the 200A range it's ≤ 10 Ω @ 1 kHz. I would think your card input impedance of 107k should work just fine.
Current Mirror on the wall ... - Page 1
Forum section: electronics / beginners

author: PerranOak
author_rank: Frequent Contributor
timestamp: February 08, 2016, 10:37:57 am
content: I had just about got to grips with the simple current mirror.Then I saw an example with one input but multiple outputs - excellent!However, this suddenly included another BJT (Q1 on the attached) in the input side WITH NO EXPLANATION AT ALL!Why do they do this just as I am getting it!

author: EPTech
author_rank: Regular Contributor
timestamp: February 08, 2016, 11:10:16 am
content: Hi There,This looks more like an equivalent circuit or a circuit for simulation to me. I suggest you find a better circuit.http://sub.allaboutcircuits.com/images/03129.png

author: Andy Watson
author_rank: Super Contributor
timestamp: February 08, 2016, 12:07:40 pm
content: One input transistor, three output transistors, the output current will be three times the input current. 1:3Two input transistors, three output, the current will be mirrored 2:3Although, in practice, variation in Vbe/temperature will probably lead to very disappointing results - unless you can get transistors on the same die.

author: PerranOak
author_rank: Frequent Contributor
timestamp: February 08, 2016, 01:07:13 pm
content: Thanks both.It was really just to understand the concept.So the "extra" transistor is just to alter the ratio?

author: Ian.M
author_rank: Super Contributor
timestamp: February 08, 2016, 01:37:20 pm
content: In real life, as the transistors need to be on the same die to match their Vbe, (both process variations and temperature), the current ratio is controlled by the ratio of their emitter junction areas.

author: w2aew
author_rank: Super Contributor
timestamp: February 09, 2016, 02:45:24 am
content: Quote from: PerranOak on February 08, 2016, 01:07:13 pmYes. This is often done on an IC because it's generally easy to match and ratio transistor sizes/areas etc. one of the 'arts' in IC design is to recognize the component tolerance are pretty poor, but devices match and ration very well - so you design circuits to take advantage of ratios and scaling rather that absolute values.
Current mirror with  stabilizing emitter resistors - Page 1
Forum section: electronics / beginners

author: selotelefon
author_rank: Newbie
timestamp: December 13, 2023, 11:20:55 am
content: When i used 3.3k resistor for Rl i observed major change in the current. What accounts for this change?

author: magic
author_rank: Super Contributor
timestamp: December 13, 2023, 11:23:48 am
content: Saturation. Q4 has too low collector-emitter voltage to work normally.

author: mawyatt
author_rank: Super Contributor
timestamp: December 14, 2023, 05:15:19 pm
content: Since the current mirror is derived from the same voltages with same emitter resistors and same device multiplicity, then RL can't be any larger than Rref, otherwise Q4 will saturate. With RL equal to Rref, then Q4 has a similar Base to Collector voltage of ~0 volts as forced by Q3's Base to Collector connection. As RL increases beyond Rref then Q4 enters Quasi-Saturation, then full Saturation.If one requires RL to be higher in value (with lower current), then changing the ratio of RE1 and RE2 allows scaling, ultimately becoming a Widlar Current Source as RE1 approaches 0 ohms, or changing the active device ratios allows the output current range to be higher or lower than the current supplied by Rref.Many options with resistor and/or device ratios.Best,

author: ArdWar
author_rank: Frequent Contributor
timestamp: December 14, 2023, 05:15:59 pm
content: Quote from: MichaelB on December 14, 2023, 04:53:44 pmI guess you mean lowering Vee to -15V? Not really since you'll also change the ref current that way. And the load side transistor will stay saturated. As a rule of thumb assuming transistor and supplies are identical you can't put larger load resistor than the set resistor (minus some base current so actually slightly larger).However you may supply the load with different voltage. (i.e. "lifting" the RLground)
Current output signal from DAC to differential voltage - Page 1
Forum section: electronics / beginners

author: DanioIO
author_rank: Contributor
timestamp: December 23, 2014, 09:15:23 am
content: Hi,I have a DAC which outputs it current signal and i want to convert it to differential voltage signal so i can process it later. I would like to use differential amplifier and i figure out to circuits which one is the best for my application so should i use something else?

author: bktemp
author_rank: Super Contributor
timestamp: December 23, 2014, 09:29:06 am
content: The second circuit should work.This application note can help:www.ti.com/lit/pdf/sbaa150