Current discrepancy

[ABCD]

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?

[baljemmett]

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?

Not 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

[ABCD]

Thanks, that explains it excellently.

[Zero999]

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.

[vk6zgo]

Actually,baljemmet answered the question,the forward voltage drop of the LED has to be subtracted
from the 9 volts supply,so:-
 
                  I is not=  9/680,

                  I = (9-LED volt drop)/680,-----------hence the reduced current.

All the other comments are red herrings.

VK6ZGO

[Zero999]

All the other comments are red herrings.

No, 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.

[Zero999]

Personally I think given the circuit; a 5% resistor shunt is more then adequate.

I 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.

[ABCD]

In the future just place your current liminting resistor at LED cathode and the other leg of the resistor at ground. You can then use the resistor to sample the circuit with negligible impact on the circuit. The voltage across the resistor is proportional to current E = IR ---> I=E/R.

Do you mean like this: 

x--LED--Meter--Resistor-x

Where x's are the + and - power strips on the breadboard.

Or

x--LED--Resistor--Meter--x

[vk6zgo]

All the other comments are red herrings.

No, 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.

No,it was an injection of commonsense into a discussion which had introduced a lot of extraneous matter which
was 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 the
value 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 great
accuracy 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 general
technical 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.

For most practical 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

[vk6zgo]

All the other comments are red herrings.

No, 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.

No,it was an injection of commonsense into a discussion which had introduced a lot of extraneous matter which
was 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 the
value 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 great
accuracy 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 general
technical 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.

For most practical 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

While I agree that sometimes to much information can be confusing, the OP mentioned burden voltage so a cursory explanation is warranted. While for the circuit in question the current reading should provide more then enough accuracy. It never hurts to explain typically one takes voltage measurments for determining currents both for practical reasons and its less invasive.

True,but it might have been a good idea to let the OP know that this would be very unlikely to be the cause of the
disparity 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 much
general electronics,either the supply voltage is sufficiently high,or the current drawn sufficiently low,that the effects of
burden 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

[Zero999]

But--- This is an LED operating from a 9 volt supply!

No, 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.

[Neilm]

You cant read SMD resistor codes or TH color codes and you've been doing this for how long?

Considering 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

[ABCD]

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.

[vk6zgo]

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

You cant read SMD resistor codes or TH color codes and you've been doing this for how long?

For a guy who whines about to much information and nitpicking you sure dont have difficulty in doing it yourself.
Using a meter or current probe is a heluva lot more likely under most circumstances to lead to errouneous measurments then Reading the value of a resistor and calculating current that way.

How do you propose to take a current measurment DC in circuit without breaking the circuit to insert your meter?

Of course burden voltage is affecting only low voltage and high current measurments they use a shunt "R" to take the measurment I=V/R and there is also the fuse to consider this will all reduce overhead.

Ouch!!!

The point I was making was that your concern about burden voltage in series current measurement seemed to indicate that you were
really fussy about accuracy,& that the possibility of error in resistor values was likely to lead to greater error in
the 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 the approximate current,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

[vk6zgo]

"With such a simple circuit yea its easy to take a current measurment and the meter and its leads will have little impact on the reading."

This is exactly what I have been trying to convey!

VK6ZGO