Current discrepancy - Page 1
Forum section: electronics / beginners

author: ABCD
author_rank: Contributor
timestamp: May 21, 2011, 10:12:00 pm
content: I was attempting to measure current on a 9VDC circuit powering an LED (in series with a 680-ohm resistor). I understand that the meter is placed in series with the circuit under test. By Ohm's law, the current flowing out of the resistor should be 13.23 mA (9/680), right? When measuring with two different meters, I get 10.61 and 10.81 mA. So at least the meters are in agreement. I have crocodile clip patch cable attached to the resistor, to the + probe, and another one connected to the - probe and the positive terminal of the LED. I'm not sure if that's a factor (the cable), and I don't think burden voltage is either at 9V.I measured the battery and resistor values with each meter and using those numbers with Ohm's law I get 13.72 and 13.77 mA. The meters in question are the EX330 and AM220 from Dave's reviews. I got burden voltage data from the PDF comparison sheet.Since both meters are in close agreement, I don't think that they're damaged, so I'm thinking there's some theory I haven't grasped yet and/or technique.Am I missing something here?

author: baljemmett
author_rank: Unknown
timestamp: May 21, 2011, 10:35:04 pm
content: Quote from: ABCD on May 21, 2011, 10:12:00 pmNot quite -- the voltage across the resistor will be 9 volts minus the forward voltage drop of the LED. In this case, your LED appears to be dropping about 1.7 volts (which is a pretty typical sort of figure), which would give a current of (9-1.7)/680 = 10.7mA -- corresponding with an average of your readings. If you measure the voltage across the resistor you should see that it's around 9-1.7 = 7.3 volts, and Ohm is once again satisfied

author: ABCD
author_rank: Contributor
timestamp: May 22, 2011, 12:50:38 am
content: Thanks, that explains it excellently.

author: Zero999
author_rank: Super Contributor
timestamp: May 22, 2011, 09:21:00 am
content: What's the battery voltage with the LED connected?The battery voltage will drop below 9V as it discharges and under load so the actual battery voltage when the load is connected needs to be used in the calculation.

author: vk6zgo
author_rank: Super Contributor
timestamp: May 23, 2011, 03:04:44 pm
content: Actually,baljemmet answered the question,the forward voltage drop of the LED has to be subtractedfrom the 9 volts supply,so:-I isnot= 9/680,I = (9-LED volt drop)/680,-----------hence the reduced current.All the other comments are red herrings.VK6ZGO

author: Zero999
author_rank: Super Contributor
timestamp: May 23, 2011, 05:02:09 pm
content: Quote from: vk6zgo on May 23, 2011, 03:04:44 pmNo, all the other comments are perfectly valid and have contributed more to the discussion than yours which was just a put everyone else's down and repeated the first post.The DVMs burden resistance can affect readings, especially at lower current settings and as I said a 9V battery won't necessarily be be 9V, after its discharged a bit it'll be a little less and when it's brand new it'll be slightly more.It's true that the LED voltage drop should be taken into account, no one said it shouldn't but they're other things to be taken into account. I think a drop of 1.7V is a little on the low side for a red LED, it's probably more like 2V and the remaining 0.3V difference is due to the battery voltage being a bit higher than 9V and the burden resistance.

author: Zero999
author_rank: Super Contributor
timestamp: May 23, 2011, 06:03:29 pm
content: Quote from: AcHmed99 on May 23, 2011, 05:26:58 pmI agree, a precision resistor would be of no benefit.If I wanted constant brightness as the battery discharges, I'd go for a constant current source but would keep it simple such as a transistor with a couple of diodes or a JFET with the source connected to the drain, not very accurate but good enough given the application.

author: ABCD
author_rank: Contributor
timestamp: May 24, 2011, 12:29:00 am
content: Quote from: AcHmed99 on May 23, 2011, 05:26:58 pmDo you mean like this:x--LED--Meter--Resistor-xWhere x's are the + and - power strips on the breadboard.Orx--LED--Resistor--Meter--x

author: vk6zgo
author_rank: Super Contributor
timestamp: May 24, 2011, 04:28:49 am
content: Quote from: Hero999 on May 23, 2011, 05:02:09 pmNo,it was an injection of commonsense into a discussion which had introduced a lot of extraneous matter whichwas not necessary to explain the original problem.Current through the resistor can be determined by I=V/R where V is the voltage across the resistor & R is thevalue of the resistance,without the LED or the battery being part of the calculation.This can then be compared with the measured result,& that calculated by the other method.Obviously,in this case,the accuracy of the volts,& ohms functions of the DMM,could be questioned,if greataccuracy is needed.But--- This is an LED operating from a 9 volt supply!If burden resistance was of concern in everyday circuits like this,it would be widely discussed in the generaltechnical community,&I would have come across it as a problem during my over 40 years working in Electronics.It is plainly obvious to just about anyone who has worked in the real world of Electronics,that a current meter in series with a circuit will affect the current in that circuit.Formostpractical purposes,however, the Amps range on a DMM can be regarded as an ideal Ammeter.Although it is laudable to bring the various effects to the notice of beginners,it can send them "chasing their tails",after things that have only a very peripheral bearing on the problem.VK6ZGO

author: vk6zgo
author_rank: Super Contributor
timestamp: May 24, 2011, 12:39:23 pm
content: Quote from: AcHmed99 on May 24, 2011, 11:09:19 amTrue,but it might have been a good idea to let the OP know that this would be very unlikely to be the cause of thedisparity of readings in this case.Since I wrote this,I had a look at Dave's Blog on burden voltage.It seems to be of more importance for high current, low voltage supplies where the volt drop across the meter becomes an appreciable part of the original supply voltage,affecting the regulation & the actual voltage of the supply as fed to the active devices.This of course will produce current readings which are not as expected.This is an important point IF you are working with low voltage supplies where substantial current is drawn,but in muchgeneral electronics,either the supply voltage is sufficiently high,or the current drawn sufficiently low,that the effects ofburden voltage may be ignored.The practice of determining current from I+=V/R ,requires an accurate knowledge of the value of R,which normally requires disconnecting one leg of the resistor.You can measure it in circuit,or take the markings as correct,in each case,there are possible errors in your calculations.VK6ZGO

author: Zero999
author_rank: Super Contributor
timestamp: May 24, 2011, 05:37:52 pm
content: Quote from: vk6zgo on May 24, 2011, 04:28:49 amNo, it's running from a 9V battery which could be as low as 6V if it's nearly dead or >10V if it's a NiMH battery hot off the charger, so just blindly using 9V in the calculation is wrong.

author: Neilm
author_rank: Super Contributor
timestamp: May 24, 2011, 06:39:44 pm
content: Quote from: AcHmed99 on May 24, 2011, 03:25:11 pmConsidering all the different marking patterns SMD resistors have (3 digit plain, 4 digit plain, EIA-96 1%, 3% and 5%) as well as the different styles through holes have (4 band with and without tolerance, 5 band + special markings) I would still use a multi-meter to check.Neil

author: ABCD
author_rank: Contributor
timestamp: May 24, 2011, 11:00:19 pm
content: The battery was a Duracell 9V alkaline. I measure it to be 9.25V and 9.23V with two different meters (EX330, AM220, respectively). The current readings I got didn't account for the voltage drop across the LED (which according to a guy on RadioShack.com, where I got the LED, measures to be a little over 2V) when I made my calculation. I think that was the main issue.

author: vk6zgo
author_rank: Super Contributor
timestamp: May 25, 2011, 10:36:10 am
content: Quote from: AcHmed99 on May 24, 2011, 03:25:11 pmOuch!!!The point I was making was that your concern about burden voltage in series current measurement seemed to indicate that you werereallyfussy about accuracy,& that the possibility of error in resistor values was likely to lead to greater error inthe resultant calculated figure in your preferred method than in the series case.Of course I can read resistor codes,but the value is not always as marked.For instance,5% tolerance,value change with age,etc.I also read voltages across resistors & calculate theapproximatecurrent,as it is very much more convenient,& certainly the method of choice for measuring inside circuits.If you are measuring the current of a complete unit, you have no resistor to hang across,& you have to use an ammeter in series,or a clamp meter.Interestingly,while we have been chasing our respective "hobby-horses" ,the OP,ABCD.has come to an understanding of the original error,&even accomodated Hero999's concern about varying battery voltage,by measuring & recording the actual voltage.VK6ZGO

author: vk6zgo
author_rank: Super Contributor
timestamp: May 25, 2011, 12:24:48 pm
content: Quote from: AcHmed99 on May 25, 2011, 11:33:44 amThis is exactly what I have been trying to convey!VK6ZGO