Measure ac amps and resistance with LT1078 opamp Precision rectifier. - Page 1
Forum section: electronics / projects

author: anishkgt
author_rank: Frequent Contributor
timestamp: September 11, 2017, 04:11:08 pm
content: Hi all,Is it possible or maybe determine the amps put into nickel strip when spot welding with a microwave oven transformer (MOT) controlled with an Arduino Uno? The timing is short for about 50ms and the open voltage is about 2vac and amps is about 1000 - 1500A. I know that possible with the ohm's law but not sure where to start from.V = 2vI = 1000 to 1500A - that is at dead short. Question is for a time period how much amps is put into it ?

author: Kleinstein
author_rank: Super Contributor
timestamp: September 11, 2017, 07:01:50 pm
content: It is possible to measure. This could be as little as tapping of the wire at two points and go to a scope. With the copper wire of known cross section one can calculate the approximate resistance.

author: anishkgt
author_rank: Frequent Contributor
timestamp: September 12, 2017, 12:54:16 am
content: Thanks.I've tapped at the electrodes and connect them to a osc. pic attached, the diameter of the wire is 9.5mm. The pulse is 20ms

author: Tomorokoshi
author_rank: Super Contributor
timestamp: September 12, 2017, 01:48:28 am
content: You could also wrap a few turns of wire around the high-current cable and measure the induced voltage. Calibrate on a load of known current.

author: ejeffrey
author_rank: Super Contributor
timestamp: September 12, 2017, 02:42:58 am
content: Oscilloscopes are not floating, so without a differential amplifier you are just measuring the voltage at the high side of the shunt with an unknown total resistance to ground. The high dI/dt will also generate large magnetic fields which you can pick up with other parts of your probe. This will make it hard to calibrate your amplitude, since the signals will sum. Your bandwidth is not that great, so this shouldn't be a hard application for a simple differential amplifier, which can also help keep your loop area small.This is a classic example of an application for a current transformer. Tektronix makes current probes that do exactly what you want, but they cost several hundred dollars. If you don't mind winding your own you should be able to do that much cheaper, with calibration left as an exercise to the reader. It won't be DC coupled if you care about that.

author: NiHaoMike
author_rank: Super Contributor
timestamp: September 12, 2017, 04:09:39 am
content: Measure the turns ratio, then measure the primary current.

author: Zero999
author_rank: Super Contributor
timestamp: September 12, 2017, 08:25:44 am
content: Quote from: NiHaoMike on September 12, 2017, 04:09:39 amI agree. If isolation from the mains is required, then use a Hall effect sensor or current transformer.

author: anishkgt
author_rank: Frequent Contributor
timestamp: September 12, 2017, 10:58:51 am
content: i appreciate everyone's reply but i am trying to keeps the BOM low hence the simplest form to measure it. Searching for the same idea i came accross this formulaWeld current measuringYou can determine the weld current by measuring the voltage across a certain distance of the weld cable.Calculate the weld current as follow:I = U * diameter  / (0.0175 * length )For the measurement of the weld current, two wires are attached to a weld cable at a distance of 44.5cm. The voltage at short circuit is 0.34V; so the maximum weld current = 0.34V * 25mm2 / (0.0175 * 0.445m) = 1100A.posted here.http://www.avdweb.nl/tech-tips/spot-welder.htmlWould the current formula be true here ?

author: Zero999
author_rank: Super Contributor
timestamp: September 12, 2017, 06:13:05 pm
content: Quote from: anishkgt on September 12, 2017, 10:58:51 amIt seems reasonable. I take it that the formula assumes the cable is made of copper, which has a positive temperature coefficient of resistance.The formula can be verified, using the figures given for copper, in the table linked below:https://en.wikipedia.org/wiki/Electrical_resistivity_and_conductivity#Resistivity_and_conductivity_of_various_materials

author: anishkgt
author_rank: Frequent Contributor
timestamp: September 12, 2017, 07:40:41 pm
content: Thanks hero999.Yes the cable is a flexible copper cable of 9mm diameter. From the table I can see that there several properties of it. But what does the 0.0175 mean in the formula ?Www.Georgehobby.wordpress.comEquipments: DSO104z, Hakko FX888D

author: Zero999
author_rank: Super Contributor
timestamp: September 12, 2017, 10:36:59 pm
content: Quote from: anishkgt on September 12, 2017, 07:40:41 pmIt looks similar to the figure for given for annealed copper, with the decimal point in a different place.The formula is confusing. It says cable diameter, given in mm2, which makes no sense. Diameter is measured in mm. Cross-sectional area is measured in mm2.I'd use the figure from the table 1.72×10-8Ohm×m2, which gives the resistance of a cube of copper 1x1x1m. 1mm2= 1×106m2, so the resistance of a 1m length of 1mm2copper is 1.72×10-2Ohms.Now if you know the cross-sectional area of the cable, which can be calculated from the diameter, area = pi(1/2d)2, you can work out the resistance, then the current given by a certain voltage drop, using Ohm's law.

author: anishkgt
author_rank: Frequent Contributor
timestamp: September 13, 2017, 01:11:00 am
content: So i did some basic calculation the know valuesI = 0.29 * 25 / (0.0175 * 0.13)= 3,186A for 20ms??0.29is the voltage at two points 13cm apart3AWG wire is25mm2as per the table herehttp://www.rapidtables.com/calc/wire/awg-to-mm.htmand with ohms law i get0.09milliOhmsis that really that much amps that is being put into for 20ms ?does the milliohms mean the resistance between the points that are 13cm apart ?

author: Zero999
author_rank: Super Contributor
timestamp: September 13, 2017, 03:17:43 pm
content: Quote from: anishkgt on September 13, 2017, 01:11:00 amI get 3242A, using the figures, in the table I linked to.R = (0.13×1.72×10-2)/25 = 89.44×10-6I = 0.29/89.44×10-6= 3242A.Our figures are close enough.Have you considered the inductance? That might also matter, at such a high current, but 20ms may still be long enough, for that not to dominate.

author: anishkgt
author_rank: Frequent Contributor
timestamp: September 13, 2017, 03:21:49 pm
content: inductance as in, in the wire itself or inside the transformer core ? I've no clue how it would have an effect here nor know how to measure it.Note:I am just trying get the formula correct to have it added in the code to display the resistance and current after a spot weld.

author: IanB
author_rank: Super Contributor
timestamp: September 13, 2017, 04:19:35 pm
content: I think (guess?) that the inductance of a short length of straight wire is negligible, so the differential voltage measured across two nearby points on the wire will mainly reflect the resistive voltage drop?

author: Kleinstein
author_rank: Super Contributor
timestamp: September 13, 2017, 04:32:29 pm
content: In principle there is a little contribution from the straight wire, but not much. One can compensate for much of it by running the taping wire close to the main wire until the two sensing wires meat. I don't think it will make a big difference for low frequency (like 60 Hz).

author: anishkgt
author_rank: Frequent Contributor
timestamp: September 13, 2017, 04:40:56 pm
content: So the straighter the cable more accurate would be the reading, would that be correct. Is there a minimum distance that the taps should be at ?Www.Georgehobby.wordpress.comEquipments: DSO104z, Hakko FX888D

author: Zero999
author_rank: Super Contributor
timestamp: September 13, 2017, 05:56:38 pm
content: Yes, the inductance should have a negligible effect.According to the calculators, linked below, the inductance of a 13cm piece of 5.64mm diameter (25mm2) wire is just 98nH, so assuming a constant voltage, the time it will take for the current to reach 3242A is just 1.1ms. di/dt= V/L = 0.29/98×10-9= 2.959184×106A/s. Time to reach 3242A: t = I/di/dt= 3242/2.959184×106= 1.096×10-3s.https://www.eeweb.com/toolbox/wire-inductancehttp://chemandy.com/calculators/round-wire-inductance-calculator.htmEDIT:Quote from: anishkgt on September 12, 2017, 07:40:41 pmQuote from: anishkgt on September 13, 2017, 01:11:00 amThat doesn't add up, unless you're taking about the thickness of the insulation. 25mm2cable will have a diameter of over 5.64mm, but not 9mm!

author: anishkgt
author_rank: Frequent Contributor
timestamp: September 13, 2017, 11:26:47 pm
content: So would that be true ? and going forward I would need to rectify the output if need to feed it to an arduino correct ? The max time that would be used is 1000ms. So to get a bridge of a rating would 1kA that would cost more. Or is there a better way to use feed the voltage to an arduino ?Www.Georgehobby.wordpress.comEquipments: DSO104z, Hakko FX888D

author: anishkgt
author_rank: Frequent Contributor
timestamp: September 16, 2017, 03:32:56 pm
content: Quote from: Hero999 on September 13, 2017, 05:56:38 pmWell the thickness was with the insulation which is 9.5mmWww.Georgehobby.wordpress.comEquipments: DSO104z, Hakko FX888D

author: Zero999
author_rank: Super Contributor
timestamp: September 16, 2017, 10:49:52 pm
content: Quote from: anishkgt on September 13, 2017, 11:26:47 pmI wouldn't consider connecting the Arduino directly to such a high powered circuit. Rectifying the high current, just so the voltage can be measured by an ADC is crazy.Ideally a Hall effect or current transformer should be used. Failing that, use a differential amplifier, with the reference, at half the ADC's full range voltage, to convert it to a suitable voltage.The gain is simply R2/R1.http://www.eetimes.com/document.asp?doc_id=1272328If REF is simply half the supply voltage, then it can be arranged as below.For example, if the ADC's range is 5V, you could select a gain of 4.R1 = 3kR2 = 12k2×R2 = 24kWith an input voltage of 0.29VRMS, the peak to peak voltage would be 0.29×root(2) = 0.54V. A gain of 4 would make the peak voltage 0.54×4 = 2.16V and biased at 2.5V, the ADC's voltage would range from 2.5 - 2.16 = 0.34V to 2.5 + 2.16 = 4.66V. The op-amp needs to be a rail-to-rail type, such as the MCP602.

author: anishkgt
author_rank: Frequent Contributor
timestamp: September 24, 2017, 09:31:06 pm
content: I am sorry for not replaying in a while. I was actually trying to understand the opamp thing and allaboutcircuits.com had suggested to have a look at LT1078 precision rectifier. I can understand the technical side of how its done but not sure about the components to use like the resistors and if anything more would be needed before i order them.http://www.linear.com/solutions/1608

author: anishkgt
author_rank: Frequent Contributor
timestamp: September 25, 2017, 07:20:06 pm
content: after a bit of understanding how to work with LTSpice and experimenting with resistor values i've got precise rectificationThat diode seems to be outdated as per digikey what would be a replacement to it ?

author: Kleinstein
author_rank: Super Contributor
timestamp: September 25, 2017, 07:48:46 pm
content: The 1N4148 is about the most common small diode - no need to replace it. Maybe choose an SMT variant.Anyway, if you go with the signal to an µC, I would not rectify the signal, but directly sample the AC signal and do RMS calculation in software. The rectifier will add offsets and will not work very well with small signals.For measuring such a high current an indirect sensor, like hall effect might be better, as it give protection to the µC. With high current sparks there can easily be high induction voltages. So any circuit (even the hall effect sensor variant) will need protection for the amplifier / ADC.

author: anishkgt
author_rank: Frequent Contributor
timestamp: September 25, 2017, 11:56:59 pm
content: QuoteThank you for the suggestion, appreciate that.Will add a RMS calculation in the software. I've checked on all the halla effect sensors but all are bulky and cost more.QuoteWhat would be the ideal way to avoid those induction voltage at high current/voltages ? a tvs diode ? or an mov ? would it be possible to simulate these spike on LTSpice ?

author: Kleinstein
author_rank: Super Contributor
timestamp: September 27, 2017, 03:49:25 pm
content: To avoid induction voltage spikes, the first thing is to avoid large area circuit loops. So keep the wires and return wire close together (maybe twisted pain were useful).The other point is more or less normal over-voltage protection, like a series resistor and diodes towards the supply and maybe a TVS diode. A MOV will not help very much they work at 100's of Volts and thus not sufficient to protect a µC's input. It sometimes also help to have an amplifier between the signal source and the ADC. For the rather low shunt voltage, just back to back diodes should be OK too.

author: anishkgt
author_rank: Frequent Contributor
timestamp: September 27, 2017, 03:54:48 pm
content: Thanks that was very useful. Appreciate it.Www.Georgehobby.wordpress.comEquipments: DSO104z, Hakko FX888D

author: anishkgt
author_rank: Frequent Contributor
timestamp: September 28, 2017, 08:06:26 am
content: Quote from: Kleinstein on September 27, 2017, 03:49:25 pmA little clarification, since the voltage at the pins 1 and 8 are in milli volts, how would a resistor help as it just reduces the current to the opA or should it be kept at minimum and not very high to alter it, like 800or 1K to start with ? and a TVS such as thishttps://www.digikey.com/product-detail/en/wurth-electronics-inc/824501600/732-9940-1-ND/5871094i was looking at the 'Peak Pulse (10/1000µs)', would that be correct ?

author: Zero999
author_rank: Super Contributor
timestamp: September 29, 2017, 12:43:09 pm
content: There isn't need for any rectification. Bias the signal, so it centres around half the full scale reading of the ADC, then any rectification becomes a programming exercise.