hybridfree/eevblog-posts huggingface dataset:
Current Probe deskew evaluation - Page 1
Forum section: electronics / beginners

author: abhinav_937
author_rank: Newbie
timestamp: June 21, 2023, 11:15:31 am
content: Performed a Double Pulse Test for a GaN-based Half Bridge.I noticed some delay in the current waveform from the TRCP0600 Rogowski probe.There are no deskew values on the datasheet or on their website.How to evaluate the deskew value?Scope used: MDO3054

author: David Hess
author_rank: Super Contributor
timestamp: June 21, 2023, 12:52:25 pm
content: Built a test circuit with a low value non-inductive resistor and measure the delay between the voltage across the resistor with the current through the resistor.

author: JohnG
author_rank: Frequent Contributor
timestamp: June 21, 2023, 01:33:26 pm
content: I would strongly consider the possibility that your current probe does not have enough bandwidth to properly estimate the switching loss of a GaN half-bridge. With a 30 MHz BW, to first order the probe transition time will be on the order of 11-12 ns. The error will become significant once the measured transition time is below 30-40 ns, depending on your definition of significant. This is probably not enough to accurately capture the current waveform for switching loss estimation for GaN devices, unless they are very large or you have slowed them down a lot.John
Current requirement for a microwave switch - Page 1
Forum section: electronics / beginners

author: medical-nerd
author_rank: Regular Contributor
timestamp: February 08, 2017, 11:12:37 pm
content: Hiya allI have a number of presumably pin diode SPDT microwave switches originally from a military receiver.MVS P/N 2S3020-93 SC00131-01These require +/-5VI have a low noise supply that provides +/- 6V that I want to use.I have done the usual searches over the past few months and have found nothing regarding these.Other microwave switches have a supply voltage tolerance of 5% so I was going to use a 5.1V zenner as a shunt regulator to drop down the voltage, presuming that the current requirement would be relatively low.However I can't find information on current requirements - can anyone give a ballpark figure??I have considered using the voltage drop from a silicon diode but since I don't know what this beasties tolerances are do not want to risk breaking it since they are obviously irreplaceable. (and I don't want to go down the LDO regulator route, I'm already having difficulties since the beastie I'm building has voltage requirements of +/-5V, +6.5V, +8.5V +/-12V, +15V, +/-24V, +30V).CheersSteve

author: TimFox
author_rank: Super Contributor
timestamp: February 08, 2017, 11:27:25 pm
content: A reverse-biased PIN diode (switch OFF) needs only the low leakage current. The reverse voltage needs to be larger than the peak AC voltage.When forward biased (switch ON), the device requires sufficient DC current to pass the AC current at the operating frequency.The stored charge is the product of the DC bias current and the storage time (diode parameter), and the stored charge must be larger than the integral of the AC current over a half cycle at the lowest operating frequency.

author: medical-nerd
author_rank: Regular Contributor
timestamp: February 08, 2017, 11:31:59 pm
content: I have other pin diode rf switches that I know have integrated electronics for control of switching - thus my noobie question.cheersSteve

author: medical-nerd
author_rank: Regular Contributor
timestamp: February 08, 2017, 11:52:22 pm
content: Application notes such as MACOM AN3021 PIN Diodes for Microwave Switch Designs give an impression of a shunt bias of 10-20mA per diode plus current for the driving electronics for the device- or am I misinterpreting thischeersSteve
current sense led circuit - Page 1
Forum section: electronics / beginners

author: Flump
author_rank: Frequent Contributor
timestamp: June 01, 2015, 11:34:43 pm
content: can anyone suggest a way of making a led light up when current drawn by a device drops below 300mA ?it is for a ocxo, when the heaters on it draws 400mA but when its up to temp it settles to about 250mAi would like visual confirmation of that.an example circuit to look at would be great if anyone has something like it.

author: V8Snail
author_rank: Newbie
timestamp: June 01, 2015, 11:58:25 pm
content: Do you have a link to the model of OCXO you are using?

author: Flump
author_rank: Frequent Contributor
timestamp: June 02, 2015, 12:41:24 am
content: yep its one of thesehttp://www.sungwhatech.com/product/pdf/aging/8663.pdf

author: TerminalJack505
author_rank: Super Contributor
timestamp: June 02, 2015, 02:03:02 am
content: Here's one way to do it. Assuming that you can put a resistor between the heater and ground, that is.

author: V8Snail
author_rank: Newbie
timestamp: June 02, 2015, 04:40:03 am
content: Quote from: TerminalJack505 on June 02, 2015, 02:03:02 amThis is along the lines I was thinking, except instead having just the heater monitored, which looks to be impractical with this package, connect pin 4 and GND of the 8663 package between VCC and R1 of this circuit.Bear in mind, the unit he referred to is 12V not 5V, but this is a good way to ensure the voltage drop across the current sense resistor stays well within the required 5% of 12V VCC.

author: TerminalJack505
author_rank: Super Contributor
timestamp: June 02, 2015, 12:56:39 pm
content: Quote from: V8Snail on June 02, 2015, 04:40:03 amWhoops. I didn't look at the OCX's datasheet before drawing the schematic.It might be better to monitor the current from the high-side. This way the sense resistor doesn't affect Vref. That would be a little trickier.

author: TerminalJack505
author_rank: Super Contributor
timestamp: June 03, 2015, 01:41:58 am
content: Here are a couple of circuits that should work for sensing the current from the high-side.One uses an INA138 high-side current monitor IC.The other uses more jellybean parts. The P-channel MOSFET shown in the circuit can be pretty much any part.

author: rx8pilot
author_rank: Super Contributor
timestamp: June 03, 2015, 02:21:57 am
content: I think there may be some of the INA current sense amplifiers with fault outputs. Could have a lower part count. Set it to reverse indicate anything over 300ma or whatever threshold you need.

author: Flump
author_rank: Frequent Contributor
timestamp: June 05, 2015, 07:36:09 am
content: thank you all for the advise
Current sense resistor and dmm reading  - Page 1
Forum section: electronics / beginners

author: Stuartambient
author_rank: Regular Contributor
timestamp: August 31, 2016, 02:19:32 pm
content: I've done a few tests with these resistors and they were all spot on, but the latest is driving me a bit battery.Here's my problem. I have the voltage range on the dmm set to 200mV. The reading I get off this simple circuit is .2 which I'm interpreting as 2 micro Volts.With the circuit I have 4 resistors (including the shunt) in serial with a battery that has 6.11V on load.R1- 984 OhmR2- 2.160kR3- 21,500kCSR - 1.018 OhmAside from the CSR, the voltage drops added up correctly and using resistance I calculated to see about 247mA in current.I'm confused because if I use 2 micro Volts and divide by the resistance (1.018) I don't seem to get the right answer. I do understand that if I am in fact reading micro volts, accuracy beyond even one number is not going to happen on my DMM.

author: Stuartambient
author_rank: Regular Contributor
timestamp: August 31, 2016, 02:31:41 pm
content: Wait, I take it back. Using the 197mA's i get with E=IR, 197ma*1.018 = .2So it's correct outside of lack of precision, which I'm looking into. Maybe not another DMM but something I can build.

author: Dave
author_rank: Super Contributor
timestamp: August 31, 2016, 02:59:34 pm
content: You're getting 247uA, as in microamperes, not milliamperes. Multiply that by 1R and you get 247uV, which your meter will correctly read as 0.2mV.

author: Stuartambient
author_rank: Regular Contributor
timestamp: August 31, 2016, 03:31:55 pm
content: Ok, I guess my math is still a bit fuzzy.So how do calculations get performed, meaning, the voltage drop on R1 (984 ohms) was 242mV. I was dividing 242/984. Which is why I was thinking mA. If I do a .242/984 then it looks like it is micros, except I need to be better at scientific notation. Looks like this number would be the micros, 2.459349593495934959349593495935e-4

author: Dave
author_rank: Super Contributor
timestamp: August 31, 2016, 03:37:18 pm
content: 2.459E-4 = 245.9E-6 = 245.9 microI'd suggest you practice until it becomes second nature.

author: Stuartambient
author_rank: Regular Contributor
timestamp: August 31, 2016, 03:39:08 pm
content: Quote from: Dave on August 31, 2016, 03:37:18 pmYep, that is what I was thinking. I flew through that chapter a bit to fast.

author: Stuartambient
author_rank: Regular Contributor
timestamp: August 31, 2016, 09:17:22 pm
content: Sorry, I had confused exp button for exponent button.
current sense resistor location - Page 1
Forum section: electronics / beginners

author: gogoman
author_rank: Regular Contributor
timestamp: November 19, 2017, 07:59:33 pm
content: Hi the circuit uses an H bridge to drive a 15v DC motor. The current sense resistors can be placed in series with the motor or between the bridge and ground of each legWhat are the advantage/disadvantage of of one location over the other.Thanks

author: rstofer
author_rank: Super Contributor
timestamp: November 20, 2017, 01:02:48 am
content: If you put the resistor in the ground side of the H-bridge, you shift slightly the voltage level required to turn the MOSFETs on.The best place to put the resistor is in the + supply lead to the H-bridge and then use a High Side Current Sense Amplifier.https://www.maximintegrated.com/en/app-notes/index.mvp/id/746

author: T3sl4co1l
author_rank: Super Contributor
timestamp: November 20, 2017, 01:27:00 pm
content: Quote from: rstofer on November 20, 2017, 01:02:48 amThe difference, in the first case, is minuscule (and, indeed, may have the benefit of providing short-circuit current limiting -- because it is a source degeneration resistor as well), while saving the cost of a current sense amplifier.Some controllers provide the current sense amp function internally, allowing the resistor to be placed anywhere. Or it's internal (integrated switch controller). This is probably more common among SMPS controllers than motor controllers, not sure.Tim

author: rstofer
author_rank: Super Contributor
timestamp: November 20, 2017, 04:18:42 pm
content: If you were to run the current sense voltage into a 5V ADC, wouldn't you need an amplifier either way? The voltage drop across the resistor probably shouldn't be anywhere near 1V and that just doesn't fill up much of the ADC input range.ADCs vary, of course, but still, I would think some kind of amplifier is required. The advantage of the high side amplifiers I have used is that they also include some selectable internal gain. Then again, so do the low side amplifiers...When I built a boat with a 24VDC motor, I used a 50 mV shunt in the high side. It's been a long time but I think it was rated for 50A full scale. By itself, the 50 mV wouldn't scale well with the ADC input. 50 mV shunts are available in a wide range of full scale ratings. These are typically used in industrial applications.Here is a paper that discusses MANY approaches to current monitoring:http://cds.linear.com/docs/en/application-note/an105fa.pdf
Current Sense Transformer guidance - Page 1
Forum section: electronics / beginners

author: danners430
author_rank: Regular Contributor
timestamp: March 29, 2018, 03:51:56 pm
content: Hey folks,I was wondering if anyone knew where to find some guidance regarding current sense transformers?I am considering them as one method of detecting model trains in a block section, since they'd be using DCC (AC square wave). My only issue is that in a short circuit condition, there is a potential for a short pulse of 5A or more, where a normal current might be in the order of mA magnitude...Cheers guysSent from my ONEPLUS A3003 using Tapatalk

author: Zero999
author_rank: Super Contributor
timestamp: March 29, 2018, 04:10:09 pm
content: Presumably the normal current is hundreds of mA? That's smaller than most current transformers are designed for. You'll need to wrap several turns through the middle of a higher current rated unit.The surge shouldn't be a problem, so long as it isn't prolonged.How about a Hall effect current sensor? Again the nominal current is on the low side, but they're not damaged by overloads.

author: David Hess
author_rank: Super Contributor
timestamp: March 30, 2018, 05:42:25 pm
content: High current which saturates the current transformer is not a problem especially if the output is clamped like with diodes unless you manage to drive so much power into the secondary resistance that it overheats. Usually saturation is avoided to preserve linear measurement but that does not matter in your application.Current transformers are sometimes used to drive a bridge rectifier producing a DC output for a sensed AC input. The forward voltage drop of the diodes increases the required volt-second product for linear operation but could make your detection circuit easier. Do *not* half wave rectify the output of a current transformer; it requires a low output impedance under all conditions. If the secondary is ever open, then the current transformer becomes a voltage step up transformer and may produce damaging voltages across its secondary.
Current sensing and overcurrent protection circuitry with LT1999 - Page 1
Forum section: electronics / beginners

author: Kryptychon
author_rank: Contributor
timestamp: May 22, 2022, 03:05:32 pm
content: Hi,I'm trying to design a current sensing and overcurrent protection circuit based in the LT1999-20 (20V/V gain)The requirements are:1. Must be able to sense current in both directions2. Overcurrent shutdown threshold must be adjustable and it must pull the EN1/EN2 lines to GND as fast as possible (should protect MOSFETS on a h-bridge by disabling the driver IC)3. Provide a scalable output signal for further processing with an external MCU's ADC to realize current depending acceleration limiting etc. Scaling the output seems desirable so the supply voltage of the LT1999 can be idependent from the external ADCs ref voltage)The circuit I got so far measures two voltages from two shunt resistors with two LT1999.The following circuit should amplify / attenuate the LT1999's output signal for further processing and it should pull the EN1/EN2 lines to GND if the overcurrent threshold is reached.I've a couple of questions:1. Is my approach to realize an adjustable gain via an OpAmp okay?2. The LM358 seems unsuitable to me because of it's slow slew rate and the low GBW (compared to the LT1999). Am I right?3. What's you opinion regarding the "external processing"? Should I integrate the ADC directly into the current sensing circuit and use e.g. I²C for communication with the MCU? I imagine transmitting digital signals is less error prone than doing so with analog signals.Besides my concrete questions, I really appreciate any feedback from you.Thanks!Kryp

author: Terry Bites
author_rank: Super Contributor
timestamp: May 24, 2022, 02:14:18 pm
content: INA300?see page 24. You can set the trip point with a resistor or with a control voltage on the LIMIT pin.

author: Kryptychon
author_rank: Contributor
timestamp: May 28, 2022, 01:11:00 pm
content: Hi,thanks for your reply.The INA300 looks quite nice as over current protection device. First I thought it cannot be used in a bi-directional current flow setup, but reading the data sheet shows this application is possible with two oINA300 which is nice.To clarify my requirements: I want this current sensing and over current protection board to protect a double h-bridge (See my other post here:https://www.eevblog.com/forum/projects/double-brushed-motor-controller-feedback-appreciated/).So, with the INA300, the over current protection part is much easier to implement than on my original design. Thanks for that!Regarding the LT1999 output signal scaling, I just realized, that digitizing it directly on this board might be the better approach. As the LT1999 and the ADC share the same Vcc/Vref, the output of the ADC is always ratiometric. Ensuring that with a "distributed" setup might be a little harder.To sum it up:- I'll ditch the LM339 in favour of the INA300 for the protection part.- I'll use a ADC to digitize the LT1999's outputs and do the communication between ADC and MCU via a busWhat would be a appropriate ADC for this purpose?Thanks in advance!RegardsKryp
Current Sensing circuit - Page 1
Forum section: electronics / beginners

author: brian_mk
author_rank: Regular Contributor
timestamp: July 03, 2023, 08:27:13 am
content: I have been looking at an OCXO 'oven ready' current sensing circuit. See attached schematic.The basic idea is that when the OCXO reaches temp, the supply current drops.The circuit senses the drop in current and turns on a LED. Simple.I want to do something similar.However, I'm struggling to understand what limits the current through Q5 when Q6 turns on and the LED switches off.It looks as if when Q5 turns on, it virtually shorts the power rail through the Q6 base-emitter junction.I expected a current limiting resistor beween Q5 collector and Q6 base.I tried creating an LTSpice model to investigate. It confirms there is no large current through Q5 when it turns on but I don't understand why.What am I missing?

author: fourtytwo42
author_rank: Super Contributor
timestamp: July 03, 2023, 08:47:15 am
content: What you are missing is the part played by R21/R22 these introduce a schmitt action (hysteresis) but also control the collector current of Q5 by starving Q5 of base current once Q6 starts to turn on. It is somewhat dependent on the current gain of the transistors used and indeed you could add a small resistor to Q5's collector to provide a certain limit (say 100R).Oops my mistake

author: brian_mk
author_rank: Regular Contributor
timestamp: July 03, 2023, 09:25:51 am
content: I was aware that R22 provides hysteresis.However, when Q6 turns on, R22 pulls the voltage on Q5 base lower so Q5 base emitter voltage and Q5 base current increases.That will increase Q5 collector current rather than decrease it as you suggest.I am starting to think that Q5 collector current must be limited by R21.I agree that will depend on the gain of Q5.It means that Q5 never turns fully on and does not go into saturation.If I remove R22 from my LTSpice simulation, it removes the hysteresis (as you would expect) but has little effect on the current through Q5 when Q6 turns on.

author: magic
author_rank: Super Contributor
timestamp: July 03, 2023, 09:45:22 am
content: That's correct, R22 resistance is much too high to have much effect.Q5 collector current is simply β times R21 current, and the latter depends on R18 voltage hopefully not being too large.

author: ledtester
author_rank: Super Contributor
timestamp: July 04, 2023, 01:35:20 pm
content: Quote from: brian_mk on July 03, 2023, 09:25:51 amCan you share your .asc file?(Unfortunately the forum doesn't allow .asc attachments so just change the suffix to .txt or put it in a zip file and attach the zip.)
220V AC Current Sensing With Shunt Resistor - Page 1
Forum section: electronics / beginners

author: Bskitter
author_rank: Regular Contributor
timestamp: June 06, 2017, 07:50:43 pm
content: Hello Guys,I am doing a project with A WiFi controlled Triac circuit for dimming some light bulbs over a web server.I want to implement current sensing on the output between the triac and the load.I have used ASC712 sensors in the past but if there are magnetic relays near the sensor it will effect the reading. (Hall Effect)So I would like to make a circuit with a Shunt Resistor in series with the load and then read the voltage of the Shunt Res with some op amp or integrate package from TI or some other company.Could anyway provide me link to circuit diagrams or some papers on these types of circuits. (Or theory about current sensing in multi meters),Byron

author: Bskitter
author_rank: Regular Contributor
timestamp: June 07, 2017, 04:20:29 pm
content: Anybody ?

author: Vtile
author_rank: Super Contributor
timestamp: June 07, 2017, 04:59:52 pm
content: Kelvin resistance measurement and current transformers.. I don't know if it is lost in translation or do you really have competence dealing with remote mains custom power electronics.Don't burn your house or customers house.
Current sensing with ULN2003A - Page 1
Forum section: electronics / beginners

author: ZeroResistance
author_rank: Frequent Contributor
timestamp: November 21, 2016, 05:23:00 pm
content: I have to drive a bunch of relays with a ULN2003A at any time only 1 of the 7 outputs would be on but I also need to measure the current of the relay coil which would be around 100mA but my intention is not to measure steady state current I want to measure the current rise and fall in the relay coil.For this I am thinking to connect a 1E resistor to the ULN2003A emitter pin. and then sense the voltage across the pin by feeding it to an adc input of a microcontroller.Does this setup sound OK? I mean would that resistor connected on ULN2003A emitter pin cause any issues with switching of the ULN?Can I increase the value of the resistor to 2E, would a op-amp after this be a better idea or just the resistor is OK?The other issue is I'm going to drive the ULN2003A with 3.3V output from the microcontroller this seems to be ok as per the datasheet which says the Vin(on) is around 2.4V for Ic of 200mA but I just want to confirm this with the experts on this forum.Thanks

author: Zero999
author_rank: Super Contributor
timestamp: November 21, 2016, 07:32:56 pm
content: Yes, it should be fine.The only disadvantage of not using an amplifier for the current sense resistor is that, unless you have a high voltage drop across the sense resistor, you wouldn't be using very many counts on the ADC, which would reduce the resolution.

author: HwAoRrDk
author_rank: Super Contributor
timestamp: November 22, 2016, 02:15:25 am
content: If your microcontroller has the option to use a 1.1V internal ADC reference (e.g. like Atmega328), that will provide better resolution. Depending on your expected peak current (i.e. not >500mA), a 2 Ohm resistor would probably be ideal then. It'd need to be at least a 1/2W one, though.
Current sensor output changes with temperature? - Page 1
Forum section: electronics / beginners

author: Jay112
author_rank: Regular Contributor
timestamp: November 23, 2015, 03:30:14 am
content: I'm using an ACS712 to tell me when a 250 watt infrared heat lamp isn't running when it should be. Technically everything is working fine, and it does allow me to accurately detect when the bulb burned out or isn't running.But I'm curious about one anomaly I've seen:When the ambient temperature drops by about 10 degrees F (from 50 to 40), the current reading drops by about 25% of its former output.I've read that the ACS712 is very temperature stable, and shows the same readings from -40C to 150C (like on this page here:http://embedded-lab.com/blog/a-brief-overview-of-allegro-acs712-current-sensor-part-1/). So I'd be surprised if it was the current sensor that was causing the fluctuation.Is it possible that the IR heat lamp itself is using less current as the temperature drops?Or is it possible that the 10' wires (26 gauge solid copper) connecting the sensor to my microcontroller are affecting the values as they get colder? The MCU itself is kept at controlled room temperature (75 F).Do you have any ideas why this would be happening?

author: Vgkid
author_rank: Super Contributor
timestamp: November 23, 2015, 04:42:50 am
content: I wonder if the AC is playing trouble with the sensor.Found this, might help.https://forum.arduino.cc/index.php?topic=179541.0
Current shunt for Home Made DC Bench power supply - Page 1
Forum section: electronics / beginners

author: dave_birdi
author_rank: Contributor
timestamp: April 02, 2014, 10:45:20 pm
content: Hi all,I am currently in the process on designing a bench power supply (which I intend on doing a proper write-up of at some point in the future).The initial design does not have scope for digital control as of yet (control is derived from a 10-turn pot feeding in to an LM338)However for my initial specification the design does need to be able to sense current and output voltage. Output voltage can be sensed by a microcontroller easily enough with a potential divider (and can be fairly accurate as I intend on powering the microcontroller from a precision 5V source)I have read up on the theory surrounding the use of a shunt resistor and an op-amp in an inverting amplifier configuration that supplies an output voltage from the op-amp which can be fed in to an ADC on microcontroller (an Atmel in this).The output voltage from the op-amp can be used to infer the voltage drop against the shunt resistor and hence the current flowing through the circuit.My design as it stands involves the rectification from a mains transformer (output 20 V RMS) to produce a D.C. output.My question is:How important is it that I have split supply when using an op-amp for the above purpose? I have seen some op-amps that are designed to be used in a single-supply basis but these seem comparatively expensive.I have found this link to be quite useful:http://www.eng.yale.edu/ee-labs/morse/compo/sloa058.pdf

author: Fank1
author_rank: Regular Contributor
timestamp: April 03, 2014, 11:42:38 pm
content: Just get a single supply op-amp.
Current shunt location question - Page 1
Forum section: electronics / beginners

author: Chris Wilson
author_rank: Super Contributor
timestamp: August 02, 2015, 11:51:31 pm
content: I think I may be doing something daft!I have Chinese made 0 to 300V / 0 to 50 Amp dual digital meter.The DC volts reads fine, red wire to + 100V DC at the bridge rectifier / electrolytic caps , but no current is displayed. If I disconnect the shunt and feed the yellow and black wires with a few mV the current section of the display starts working. I attach the schematic of the circuit I am using it with, and I have the shunt in a break in the 100V DC ground wire. In other words the bridge rectifier, smoothing caps negative wire is not taken directly to chassis ground, it now goes to chassis ground through the larger of the shunt terminals. I know the device is drawing a good few Amps when the current meter section remains at zero. Thanks for any help.

author: tautech
author_rank: Super Contributor
timestamp: August 03, 2015, 12:25:24 am
content: Chris, ensure the Com and meter supply negative are not commoned.IME with similar meters, they require their supply is isolated from the rails to be measured.

author: Chris Wilson
author_rank: Super Contributor
timestamp: August 05, 2015, 10:22:41 am
content: Hi again, and thanks for the reply. I am not sure I can actually use this dual volts / amps digi meter in this circuit. I managed to blow one up tryingI think the issue is having the 100V DC supply negative tied to ground *AND* the 12V DC regulated supply negative also tied to ground. I can measure amps by putting the shunt in the 100V DC positive supply wire, and running the meter off a battery, but it appears even with the black negative power supply wire to the meter floating trying to *ALSO* read the 100V DC + wire won't work.For whatever reason the shunt in the negative 100V DC wire, before it is grounded, didn't seem to give an amperage reading.I have a couple of amp only digi meters that read 0 to 50 amps, using the same 50 amps / 75mV shunt, but they need a 5V DC supply. I am now wondering how to test one to see if the 5V supply negative needs isolating from ground. I can use a 5V regulator on the 12V supply in the amp, but do I need to float it somehow above ground?These things are neat, cheap as chips, but have complexities I hadn't considered over and above good old analogue meters! And hole sawing a round hole is a lot quicker than setting up to mill a rectangular hole with tight corner radii

author: PSR B1257
author_rank: Frequent Contributor
timestamp: August 05, 2015, 04:14:50 pm
content: QuoteWhy???As your schematic shows, there is no connection between the auxiliary supply and the supply to be measured
Current shunt Question - Page 1
Forum section: electronics / beginners

author: dekra54
author_rank: Unknown
timestamp: September 13, 2016, 05:14:26 pm
content: Hello forumI'm currently trying to fix an old Daub audio amplifier with more ore less success.This amplifier has 2 Channels and dedicated to each is an inrush/overcurrent protection Board and one is totally broken.I'm now thinking of running booth transformers of one of this boards since i have relays that fit in the footprint and can handle about 2.5 the current but obviously the current sensing has to be upgraded too.If i'm right i only have to solder the shunt resitor from the faulty board in parallel so it will trip at double the current it would normally do.Attached a picture of the schematic and the shunts in question in the blue circle R36 R37Greetings DennisEdit: The traces on the PCB are also thick enough to handle the current (about 10-25mm in the relevant area)
Current Source Design Issue - Page 1
Forum section: electronics / beginners

author: IuriC
author_rank: Contributor
timestamp: May 05, 2014, 03:41:55 pm
content: I'm trying to design a simple current source to bias a PTC sensor. This is a learning exercise using just parts I got in my bench.I almost have no experience with OpAmps, so please don't laugh on me.What I have to do to compensate the poor output swing of this amplifier in single-supply? And why this circuit is not working?Thank you.

author: tszaboo
author_rank: Super Contributor
timestamp: May 05, 2014, 03:49:57 pm
content: Inverting and not inverting input is reversed (+ and -). Don't worry, this happens.MOD: And the TL072 needs at least 7V power supply. Go for the TL972 for similar thing.

author: LvW
author_rank: Frequent Contributor
timestamp: May 05, 2014, 04:31:10 pm
content: Quote from: NANDBlog on May 05, 2014, 03:49:57 pmI don´t think so. We have one single sign inversion within the feedback path (common emitter configuration).Hence, we havenegativefeedback.Another issue is the question of single supply restrictons.

author: IuriC
author_rank: Contributor
timestamp: May 05, 2014, 04:50:20 pm
content: Quote from: NANDBlog on May 05, 2014, 03:49:57 pmI need a "positive" feedback because there is already a sign inversion like LvW said.Quote from: NANDBlog on May 05, 2014, 03:49:57 pmDidn't notice that in datasheetQuote from: AcHmed99 on May 05, 2014, 03:53:50 pmThank you man!Quote from: LvW on May 05, 2014, 04:31:10 pmHow can I calculate the boundaries of my load? I have to maintain the linear operation in the transistor and CM range/output swing of the OpAmp?Thank you all for the replies!

author: LvW
author_rank: Frequent Contributor
timestamp: May 05, 2014, 05:52:16 pm
content: Quote from: IuriC on May 05, 2014, 04:50:20 pmJust for clarification: When there is a sign inversion within the feedback loop you always have NEGATIVE feedback. - even in case the feedback path is connected to the non-inv. opamp input.

author: tszaboo
author_rank: Super Contributor
timestamp: May 05, 2014, 07:26:09 pm
content: Quote from: LvW on May 05, 2014, 04:31:10 pmNow, that you mentioned it, that is a PNP transistor. Silly me.On the other hand, is it neccessary to have the transistor at all? Cannot you put the PTC in the feedback, or drive it directly by the opamp?

author: IuriC
author_rank: Contributor
timestamp: May 05, 2014, 07:43:06 pm
content: Quote from: NANDBlog on May 05, 2014, 07:26:09 pmI Can't find in the datasheet if it can source at least 1mA in its output.

author: tszaboo
author_rank: Super Contributor
timestamp: May 05, 2014, 08:16:45 pm
content: Quote from: IuriC on May 05, 2014, 07:43:06 pmhttp://www.ti.com/lit/ds/symlink/tl071a.pdfFigure 9 tells you the output load. You might need to use Figure 10 to modify it for your power supply, If you use this. But if I were you I would just chose a peak-to-peak 5V amplifier.
Current Source for DIY Multimeter - Page 1
Forum section: electronics / beginners

author: akowalczyk
author_rank: Contributor
timestamp: May 15, 2019, 07:02:45 am
content: Hello!I'm working on a 4.5 digit multimeter design and am not sure what the best current source would be for the ohms range.I have +/- 5V, but that can change if a better solution is out there if I need to bump the voltage a bit (it's battery powered but I'd be OK boosting to get higher voltages)Right now I am looking at using this reference design with a REF200 current source providing a regulated 100uA from this app note:http://www.ti.com/lit/an/sboa046/sboa046.pdfAlthough the app note mentions being able to program the output current, I'm running into issues with trying to source anything other than 100uA, which only really works up to about 40k, after that I need a 1uA and 10uA current source to test larger resistor values with 5VAny recommendations are helpful!

author: MrAl
author_rank: Super Contributor
timestamp: May 15, 2019, 01:24:25 pm
content: Hi,Do a search for "precision current source".
Current source? - Page 1
Forum section: electronics / beginners

author: rexxar
author_rank: Frequent Contributor
timestamp: August 15, 2013, 05:58:51 am
content: I bought a multimeter kit, thinking it'd be fun to build andanythingwould be better than what I already had. It didn't occur to me that I'd have to calibrate the thing, and now I'm stuck.I figured I'd buy a DMM Check to get the resistance and volts ranges "close enough". While they have versions with current sources, they're more expensive, and are only in the milliamps range. I need to calibrate the 20A range on this meter, and I don't have anything like a precision current source.I was thinking that once I got resistance and volts in the general area of calibrated, I could get one of those huge power resistors, stick it across my power supply, and apply Ohm's law to figure out how much current should be passing through.Is that sensible, or are there ICs or something that will do a better job at low cost?

author: Simon
author_rank: Unknown
timestamp: August 15, 2013, 06:34:56 am
content: No, you can't do that. Your meter will have it's own resistance causing burden voltage and so many things to account for. Not exactly precision. Have a look at Daves videos on his "dummy load" you need to make something that works on the same principle with precision resistors.

author: ve7xen
author_rank: Super Contributor
timestamp: August 15, 2013, 07:20:24 am
content: You can probably get in the ballpark if you measure the current shunt, measure your power resistor, add the two and then use Ohm's law. Power resistors don't tend to have very good temperature coefficients though, so do the measurement quickly before it has a chance to heat up.If you want more precision you need a reference and a feedback loop of some sort.

author: sub
author_rank: Regular Contributor
timestamp: August 17, 2013, 10:40:28 am
content: If you have a working ohms range, perhaps you could derive the high-current calibration from the resistance of the shunt+connectors.

author: Zero999
author_rank: Super Contributor
timestamp: August 17, 2013, 01:12:21 pm
content: You could use another volte meter to measure the voltage across the resistor and calculate the current.

author: Marco
author_rank: Super Contributor
timestamp: August 17, 2013, 05:25:13 pm
content: Quote from: rexxar on August 15, 2013, 05:58:51 amA high current version of Dave's constant current load is relatively cheap component wise. A 20 Ampere shunt is a couple of bucks, a low offset voltage opamp is a couple of bucks (Dave's LM324 is a bit dodgy, maybe an AD8638 or something) a 100 Watt MOSFET is a couple of bucks, a MOSFET driver less than a buck (zetec totem pole) ... so less than 10 bucks of components.
Current sources in small signal model analysis - Page 1
Forum section: electronics / beginners

author: Silenus
author_rank: Contributor
timestamp: May 30, 2015, 04:46:52 am
content: I've been working on some electronics questions regarding BPJ transistors and circuit gains and cant seem to find an answer to a question that affects my analysis. When you have an indpendentcurrent source that is connected between a transistor and voltage rail in a given circuit topology, what do you do in terms of a small signal model? I understand these currents are there to bias the circuit to a particular operating point. Would that mean they are off (open circuit) in small signal model since we are dealing with small signal (not DC conditions)?

author: mathias
author_rank: Regular Contributor
timestamp: May 30, 2015, 07:59:03 am
content: Exactly!You might take into account their impedance, for example when you use a BJT current source to bias a differential pair. In that case, the finite impedance will cause common mode gain.

author: LvW
author_rank: Frequent Contributor
timestamp: May 31, 2015, 11:02:42 am
content: Quote from: Silenus on May 30, 2015, 04:46:52 amYes - for a small-signal analysis based on a corresponding (small-signal) model all dc current sources are considered as an open circuit and all dc voltage sources are replaced by a short to ground. However, dont forget that the VALUES of these sources play an important role for calculating the small-signal parameters used within the model.

author: jpb
author_rank: Super Contributor
timestamp: May 31, 2015, 12:19:16 pm
content: Just to add to what every one else has already said, when looking at the small signal impedance of sources you just need to think about what changes (nothing) when you apply small changes of voltage or current.A fixed current will not pass any more (or less) current what ever small signal change in voltage is applied so it looks like an open circuit. A fixed voltage doesn't change in voltage what ever change of current passes through it so it looks like a short circuit.Most small signal current sources in transistor models are modeled as voltage or current controlled sources, e.g. the transconductances gm in a FET model where a voltage change in one place (on the gate) leads to a current change in a different place (the drain current).
Current through ground plane - Page 1
Forum section: electronics / beginners

author: Jr.Maxwell
author_rank: Contributor
timestamp: February 11, 2020, 10:27:49 am
content: Hello amazing peopleI've searched the interwebs and this amazing forum but found a lot of confusing answers.My question is:"Does current flow through the ground plane?" I would assume so for completing a closed circuit current needs to flow, but does that mean that if I have a ground plane, through it current flows?Thank you for your time and help

author: EEEnthusiast
author_rank: Frequent Contributor
timestamp: February 11, 2020, 10:31:49 am
content: Just think of the ground plane as a flat current conductor. Then your question is answered..Yes it does carry current.

author: Jr.Maxwell
author_rank: Contributor
timestamp: February 11, 2020, 10:41:42 am
content: So hypothetically, if I were to touch the ground plane and the floor I/m standing on had a different voltage referenced to ground, I would be 'schocked'?

author: EEEnthusiast
author_rank: Frequent Contributor
timestamp: February 11, 2020, 10:52:28 am
content: Possibly.. unless the ground plane is earthed (connected to the mains ground/earth terminal).However note that carrying current does not mean that it is at a higher potential w.r.t. floor or earth.But for safety reasons, any ground plane inside a high voltage circuit should be treated as hot.

author: SL4P
author_rank: Super Contributor
timestamp: February 11, 2020, 11:07:52 am
content: The ground plane is often/usually tied back to 0V or ground - also as a ‘shield’ for the adjacent circuitry.

author: Jr.Maxwell
author_rank: Contributor
timestamp: February 11, 2020, 11:55:28 am
content: Thanks guys!this helped a lot.
Current to Flux  to current - Page 1
Forum section: electronics / beginners

author: J4e8a16n
author_rank: Regular Contributor
timestamp: June 30, 2015, 11:25:37 am
content: Hi,About the toroid as a coil.In a Joule Thief, when the current comes in from 0V, the coil résist the current by storing it in the flux (and magnetic field). When the voltage induced is equal to the Vcc, the flux is full and let the current pass like a wire. The flux is still there.When the incoming source current stop or slow down for any reason, the flux will be transform again in a current flowing trying to maintain the flow of a current.

author: DmitryL
author_rank: Regular Contributor
timestamp: June 30, 2015, 02:06:08 pm
content: I think that Maxwell, Heaviside and co. explained it slightly better.https://en.wikipedia.org/wiki/Maxwell%27s_equations

author: J4e8a16n
author_rank: Regular Contributor
timestamp: June 30, 2015, 02:50:10 pm
content: Quote from: J4e8a16n on June 30, 2015, 11:25:37 amIt is a simple question. The flux is still there? It wait around the coil?JP
Current-to-voltager amplifier (TIA): dual vs. single supply OpAmp - Page 1
Forum section: electronics / beginners

author: Bigman
author_rank: Regular Contributor
timestamp: April 02, 2015, 04:02:22 pm
content: Hi,in the attachment "TIA_dial_supply" you see my first circuit for a transimpedance amplifier. In this circuit the reference voltage at the non-inverting input is just in between the negativ and postive supply voltage of the OpAmp. To get this reference voltage, the circuit compromises a rail-splitter.Then, I heard about Input Common-Mode Voltage range, and I was wondering: if an OpAmp even accepts voltages below his negative supply voltage, I may not need the rail-splitter anymore.So, I came up with the circuit "TIA_single_supply". The difference to the 1st circuit: the non-inverting input and the negative supply voltage are now connected with GND.But according to the simulations, this is not working. Interestingly, the Output-voltage is even lower than the input voltage in case of the single supply.(The choosen OpAmp has an Input Common-Mode Voltage range of V- - 0.2V, and V+ + 0.2V.)Perhaps somebody can give me a hint, where my thinking is wrong.Thanks!

author: TimFox
author_rank: Super Contributor
timestamp: April 02, 2015, 04:17:04 pm
content: In your second circuit as drawn, with the input current (with zero DC component) to the inverting input and a feedback resistor from output to inverting input, an ideal op-amp (ignoring power supply issues) would have an output that swings positive and negative from ground, to balance the input current through the feedback resistor. Even though the input common-mode range goes below the negative power supply pin (which you have grounded), the output cannot swing negative with respect to the negative power supply = ground.If you want to see the variation on the input, you need to connect the non-inverting input to an intermediate positive voltage, or add negative current (resistor from a supply negative to ground) to bias the output positive (between the two supply pins).
Current Transformer design - Page 1
Forum section: electronics / beginners

author: Helix70
author_rank: Unknown
timestamp: February 19, 2024, 05:16:44 am
content: Hi Guys,I am looking for a book, website or other help with the design of a current transformer.I have a single primary turn and size of the conductor, frequency (50Hz), a minimum core inside diameter requirement, and a turns ratio, and want to specify a core material (probably GOSS, but not sure), a cross sectional area of the core, and secondary wire diameter to achieve 100mA secondary at 10A primary, ramping up to 1A secondary at 100A primary before saturating.Any tips on where to look?

author: Ugur
author_rank: Regular Contributor
timestamp: February 19, 2024, 06:42:33 am
content: Take a look here, maybe it will give you an idea.https://www.electronics-tutorials.ws/transformer/current-transformer.html

author: Helix70
author_rank: Unknown
timestamp: February 19, 2024, 09:48:22 pm
content: Thanks, I did find that, but I am looking for more detail about core size and material selection.
Current transformer on DC input (100A) - Page 1
Forum section: electronics / beginners

author: theoldwizard1
author_rank: Regular Contributor
timestamp: December 24, 2014, 01:31:06 am
content: I am researching building a power inverter. I think the answer is "No", but can a current transformer work on a DC (battery) cable ?The obvious second choice is a shunt.

author: macboy
author_rank: Super Contributor
timestamp: December 24, 2014, 01:39:03 am
content: You are right, no current transformer for DC.Shunt or Hall effect.

author: Andy Watson
author_rank: Super Contributor
timestamp: December 24, 2014, 01:39:54 am
content: No. A standard current transformer won't work with DC, but a Hall-effect sensor will. There are ready-made sensors available for current sensing - I can't recall the details at present, but searching for Allegro Microsystems should return some interesting results.

author: theoldwizard1
author_rank: Regular Contributor
timestamp: December 24, 2014, 03:39:53 am
content: Okay, someone is going to have to explain how a "hall-effect sensor" measures current !The one I am familiar with are for measuring rotational speed by turning a transistor on pr off in the presence/absence of a metal shudder.

author: AG6QR
author_rank: Frequent Contributor
timestamp: December 24, 2014, 04:55:29 am
content: Hall effect sensors sense magnetic fields. A current (DC or AC) produces a magnetic field. The biggest complication is how to make the sensors reject stray external magnetic fields while being sensitive to the induced fields from the current.Wikipedia does a better job of explaining it than I could in a single post.http://en.wikipedia.org/wiki/Hall_effect
current & voltage requirements for pcbs - Page 1
Forum section: electronics / beginners

author: Adhith
author_rank: Frequent Contributor
timestamp: October 02, 2016, 06:12:59 am
content: hello everyone..most of the time when i make circuit boards i notice some over heating issues related to ics or with regulator ics or in some cases, all the leds in the circuit wont work properly but it may work fine with lower voltages than specified in the circuit diagram. Recently when i plugged a usb cable from my pc to the board (for 5v supply) the pc was automatically shutdown. i still can't figure out about the basic current & voltage requirement or its effects on my circuits. so, can anyone help me or direct me to solve my problem.

author: danadak
author_rank: Super Contributor
timestamp: October 02, 2016, 12:50:19 pm
content: If you are using USB from LapTop to power external boards.In the USB 1.0 and 2.0 specs, a standard downstream port iscapable of delivering up to 500mA (0.5A); with USB 3.0, it movesup to 900mA (0.9A).Sounds like you are overloading USB, although most USB is overloadprotected, so not sure why your laptop is sutting down. Unless youare using its brick side to power LEDs and the brick is shutting downdo to overload.Regards, Dana.
Current, Voltage, resistance, and Ohm's law (video) - Page 1
Forum section: electronics / beginners

author: Aaron D. Parks
author_rank: Contributor
timestamp: August 14, 2018, 03:37:21 pm
content: Hey, EEVblog peeps! Happy Tuesday. It's been a while since I posted here, but I just posted a video that I thought might be valuable to some of the folks who come by the Newby forum here.Over the past months my wife and I put together an online course to study for the technician class ham radio test here in the US and in that process I think a lot of good ideas for videos came up, so I hope to be making more in the near future.As always, I'd be delighted to hear your feedback on the video. I've been working on it, but I'm by no means an expert at making these yet