Question regarding Ohm's law and a Digital Multimeter.

[Spork Schivago]

Hello,

I have a few questions. I have a simple circuit I made using a breadboard. It's 4 AA batteries hooked up to a potentiometer and an LED. The potentiometer is a 2K one. My DMM shows the voltage from the batteries as 6.46VDC. The resistance on the potentiometer is currently set for 986 Ohm (according to my DMM). I try predicting what the current should be for the circuit using Ohm's law. I have I = ΔV/R. I = 6.46 / 986, I = 6.551724 mA. When I use my multimeter to test the current though from the positive terminal that goes to the LED, I get 4.31mA.

The way I test it is I place the multimeter probe between the wire going to the red post (the positive terminal) and I place the other probe on the positive side of the LED (the anode side I believe). Why does the multimeter show the current to be 4.31 mA but Ohm's law says it should be roughly 6.55 mA? I've replaced the battery in my multimeter. Does it mean it's failing? Thanks.

[mariushm]

Re: Question regarding Ohm's law and a Digital Multimeter.

There's a thing called burden voltage.

Basically, the multimeter has a very small resistor inside and measures the voltage drop on that resistor to tell you how much current flows through the circuit.
The resistor value can be one for a particular scale (0-200mA for example) and another one for the 1A or 10A scale, so the burden voltage will change.

So when you put the multimeter in the circuit, you no longer have 986 ohms, you have 986 ohms plus the internal resistance of the meter (shunt resistor to measure current) PLUS the resistance of your leads (put your meter in resistance measurement and short the leads, you may be surprised to measure 0.2-0.5 ohms for the leads alone).

In your particular case, the measurement may be more accurate if you force your multimeter on the smallest range possible.

See these for better/more detailed explanations:

#194: What is ammeter burden voltage, and why you should care.
https://www.youtube.com/watch?v=BH4RpvXAcRs

Burden Voltage, How a Multimeter Takes Measurements and the uCurrent
https://www.youtube.com/watch?v=fRP98k3Rh1E

http://forum.allaboutcircuits.com/th...rection.51060/


There are specialized tools to measure very low currents, like uCurrent (mentioned in the video above as well) . Its documentation explains burden voltage in detail :

https://cdn.badcaps-static.com/pdfs/...dfce09d2ef.pdf

The above tool isn't very cheap because it uses expensive (precise, low noise) opamps to amplify the very small voltage drops to more meaningful values, easy to understand.

[Spork Schivago]

Re: Question regarding Ohm's law and a Digital Multimeter.

Thank you very much for the articles and videos. I will watch them shortly. So this is normal then. I was also curious as to why I couldn't check the resistance of the potentiometer while the circuit was live. I think I figured that one out though. I believe it's because the multimeter sends a small amount of voltage and current through the resistor to calculate the resistance. With my multimeter, it has no way to know what the current and voltage is that's going through the circuit. Am I on the right track here? Thanks again for the help.

[kaboom]

Re: Question regarding Ohm's law and a Digital Multimeter.

Hello,

I have a few questions. I have a simple circuit I made using a breadboard. It's 4 AA batteries hooked up to a potentiometer and an LED. The potentiometer is a 2K one. My DMM shows the voltage from the batteries as 6.46VDC. The resistance on the potentiometer is currently set for 986 Ohm (according to my DMM). I try predicting what the current should be for the circuit using Ohm's law. I have I = ΔV/R. I = 6.46 / 986, I = 6.551724 mA. When I use my multimeter to test the current though from the positive terminal that goes to the LED, I get 4.31mA.

The vf of the LED subtracts from your supply voltage.

That 986 ohm resistance is dropping 4.25V from the ~6.5V supply (maybe less under load), leaving the 2.25V LED forward voltage.

Measure the voltage across the LED, with another meter, if you're curious.

There's a thing called burden voltage.

Burden voltage at 4.3 mA?? I don't think so...

I'm surprised you're making the OP "chase ghosts" with those links, without even considering the well-known effect of junction drop. It's misleading, to say the least.

For your erroneous claim of "burden voltage" to be correct, that meter would need an impedance of ~520 ohms.

[retiredcaps]

Re: Question regarding Ohm's law and a Digital Multimeter.

@Spork Schivago, what is the make and model of your multimeter?

[eccerr0r]

Re: Question regarding Ohm's law and a Digital Multimeter.

You can't measure resistances of a resistor in circuit because there are things "parallel" to the resistor, which will give an false reading... Left as an exercise to the reader.
You'll have to calculate the value of the resistor by the voltage across it and the current through it.

[mariushm]

Re: Question regarding Ohm's law and a Digital Multimeter.

@kaboom :

My Uni-T UT61E (22000 count multimeter, see here: https://www.youtube.com/watch?v=Bz6CYDRl1M8 ) uses about 1kohm shunt resistor for uA (I think 0-2.2mA), a 10 ohm resistor when set in 0-220mA range and a much lower resistor for 0-10A range.

So you do the math, see how 10 ohm in series in the circuit would affect the measurement with a multimeter like mine.

If I need to measure low currents, I often find it easier to just put a 0.1 ohm resistor and measure the voltage drop across it with the multimeter set on 0..220mV range.
The voltage drop across a 0.1 ohm resistor is very small in this case.

Back to questions .. yes spork, you can't measure resistors while circuit has electricity flowing through it, because that electricity interferes with the multimeter's own test signal. You need to at least pull one lead of the resistor off the prototyping board, or desolder it, or disconnect the power source.

(note: I'm saying the numbers for shunt resistor values from a review page for the meter and i think they're correct from memory, but i didn't check the values again right now, they could be slightly incorrect)

[budm]

Re: Question regarding Ohm's law and a Digital Multimeter.

+1 with KABOOM,
"I have I = ΔV/R. I = 6.46 / 986, I = 6.551724 mA. When I use my multimeter to test the current though from the positive terminal that goes to the LED, I get 4.31mA."
if you just have the resistor and 6.46V Voltage source only, but you introduce another element is series with the resistor so you will no longer have 6.46V drops on the resistor any more, if you measure the V drop on the resistor then you can calculate the current flow in the circuit.
Just like this guy that I try to explain how the series circuit function and he still does not understand, at least you do:
https://www.badcaps.net/forum/showth...911#post553911

[kaboom]

Re: Question regarding Ohm's law and a Digital Multimeter.

@kaboom :

My Uni-T UT61E (22000 count multimeter, see here: https://www.youtube.com/watch?v=Bz6CYDRl1M8 ) uses about 1kohm shunt resistor for uA (I think 0-2.2mA), a 10 ohm resistor when set in 0-220mA range and a much lower resistor for 0-10A range.

Don't give me a weak-willed youtube link. Prove to me, and everyone else reading this thread, that you can follow along. This is basic stuff, so much so, it was once taught in every Jr. High science class.

You're trying to rebut my post with a comment about a microamps range, with it's associated and necessary high impedance?! Nice try- this is the bullshit that pushed Kiriakos too far. Someone gets all "maker" like, and will not think outside the project box, at least the one kipkay (the reckless laser user) gives you to "work with." And alluding to "uCurrent," as in that maker-whore sellout kit, which claims to be the only way, since the discovery of electricity, to measure low currents??

You allege that the 10A range uses a "much lower value resistor;" care to tell us, approximately, what such a shunt typically measures? Or do you have to defer, again, to copy&paste bullshit, because you've got no idea how these things work, and have never bothered to figure it out? Anyone can pull stuff from yootoob, but it doesn't mean they've ever troubleshot a DMM divider chain, or calibrated one.

Don't pee on my leg, then tell me it's raining; and don't cry to me about comparing calibration to the "simple task" of measuring that stupid 4 mA, or how the two have no bearing on each other.

Just try me, sucker- you're just an amateur.

So you do the math, see how 10 ohm in series in the circuit would affect the measurement with a multimeter like mine.

No. You calculate it, right here in your reply; show us all, you supertech.

I'll try to make it easy. You obviously haven't paid attention, as evidenced by your missing the numbers right in front of you, for the calculation you think you'll "show me up" with.

Here we are: The OP's load current is 4.31 mA, and the value you suggest for a burden is 10 ohms. Hint: the result will be far short of the 2.25V that "went missing," according to you, due to "burden voltage."

Now then, what is this alleged burden voltage, with 4.31 mA across a 10 ohm shunt?

Don't run off now to your maker friends; why not look in your "Engineers' Mini Notebook?" Remember those?

If I need to measure low currents, I often find it easier to just put a 0.1 ohm resistor and measure the voltage drop across it with the multimeter set on 0..220mV range.
The voltage drop across a 0.1 ohm resistor is very small in this case.

Damn. Right.

And it's even smaller with a paltry 4.31 mA flowing, quite possibly below the resolution of a common 2000 count DMM.

Back to questions .. yes spork, you can't measure resistors while circuit has electricity flowing through it, because that electricity interferes with the multimeter's own test signal.

More like the meter cannot distinguish from V*R during a resistance check, or I*R when measuring "ohms" across resistor(s) in a live ckt.

(note: I'm saying the numbers for shunt resistor values from a review page for the meter and i think they're correct from memory, but i didn't check the values again right now, they could be slightly incorrect)

So the OP is using that exact meter? You must be there with him, and tore his meter apart before he built the circuit! We're talking about his meter.

FWIW, my harbor freight 830 clone's shunt values are 3.3 ohms on the 200mA scale, and 12.2 ohms on the 20mA scale; the highest value of which is a mere 1.2 percent of the OP's dropping resistor value.

This is not about me; you gave false information to the OP, which he gratuitously thanked you for, even though it was incorrect. That's worse than youtoob comments; where someone asks a question about an unfamiliar subject, and are given "an answer," which, more often than not, is wrong, but still thank the answer-giver, as if they were an all-knowing entity. This is bad- we end up with two people, now, who are in the dark.

I have a simple question for you, super tech. If I simply copy&pasted exactly what I had in my reply to the OP from a supposed "authority" such as wikipedia, or a "maker" site, you'd believe it w/o question. Am I correct? Because all we're supposed to do is robotically, or perhaps Google-otically, mindlessly copy, paste, and repeat, without ever thinking or learning, right?

If you can't handle or understand such simple electronic principles, you've got no business misleading others with your so-called answers.

[Spork Schivago]

Re: Question regarding Ohm's law and a Digital Multimeter.

@Spork Schivago, what is the make and model of your multimeter?

It's a Craftsman 82357 DMM. It's an older model but has served me well over the years. I cannot find the manual to it, neither the physical copy or a digital copy. There used to be one on Sears website but the link is no longer valid.

[kaboom]

Re: Question regarding Ohm's law and a Digital Multimeter.

It's a Craftsman 82357 DMM.

I doesn't matter.

Don't overthink this, it's a simple series ckt.


Take the 6.5V supply, subtract 2.25V from that (LED junction drop- all diodes/PN junctions have this), then divide into your 986 ohm pot.

(6.5-2.25)/986= 4.31 mA

To double check, remove pot from ckt and set/confirm 986 ohms. Reconnect ckt, then measure voltage across 986 ohm pot. Divide this voltage into 986- should be very close to 4.25V.

And, again, 4.25/986= 4.31 mA.

Hope that helps, OP.

[budm]

Re: Question regarding Ohm's law and a Digital Multimeter.

Or just measure the V drop on the resistor to calculate the current flow as I suggested.

[Spork Schivago]

Re: Question regarding Ohm's law and a Digital Multimeter.

+1 with KABOOM,
"I have I = ΔV/R. I = 6.46 / 986, I = 6.551724 mA. When I use my multimeter to test the current though from the positive terminal that goes to the LED, I get 4.31mA."
if you just have the resistor and 6.46V Voltage source only, but you introduce another element is series with the resistor so you will no longer have 6.46V drops on the resistor any more, if you measure the V drop on the resistor then you can calculate the current flow in the circuit.
Just like this guy that I try to explain how the series circuit function and he still does not understand, at least you do:
https://www.badcaps.net/forum/showth...911#post553911

Thank you Budm! It was a flaw in my understanding of how it all worked. Once I measured the voltage drop across the resistor, I got a much, much closer value for current compared to what the DMM had. My math showed a voltage drop across the potentiometer of 4.31VDC. I had the potentiometer set to 988 Ohm. That gives me a current of roughly 4.36 mA. The DMM gives me 4.31 mA. I'm assuming the .05 mA is probably from the burden voltage or my DMM not being perfectly accurate.

I also did a test just to see if I was understood where I went wrong. I figured out the resistance that LED was using by using Ohm's law. I new because it's a series circuit, current should be the same everywhere. I checked the voltage drop on the LED, it was 2.023VDC. I divided that by the current, 4.36 mA, and got a resistance of around 464 ohm. I added that to the resistance of the potentiometer (988 ohm) and 1452 ohm. Using Ohm's law again, 6.46VDC (from the battery) divided by the total resistance of the LED + potentiometer gives me roughly 4.45 mA.

I figured the reason I have a bigger number is because of the resistance and everything from the breadboard itself and the wires. When resistance goes up, I was under the impression current goes down.

Thanks!

[budm]

Re: Question regarding Ohm's law and a Digital Multimeter.

"That gives me a current of roughly 4.36 mA. The DMM gives me 4.31 mA. I'm assuming the .05 mA is probably from the burden voltage or my DMM not being perfectly accurate."
What is your meter spec for the current measurement?

[Spork Schivago]

Re: Question regarding Ohm's law and a Digital Multimeter.

I doesn't matter.

Don't overthink this, it's a simple series ckt.

Take the 6.5V supply, subtract 2.25V from that (LED junction drop- all diodes/PN junctions have this), then divide into your 986 ohm pot.

(6.5-2.25)/986= 4.31 mA

To double check, remove pot from ckt and set/confirm 986 ohms. Reconnect ckt, then measure voltage across 986 ohm pot. Divide this voltage into 986- should be very close to 4.25V.

And, again, 4.25/986= 4.31 mA.

Hope that helps, OP.

I don't understand what an LED junction drop is or what a PN junction is yet. I believe I know what a forward voltage drop is though. It's just the voltage drop across the anode and cathode of a diode, right? I'm still really new to this. I've finished reading all the tutorials at the Physics Classroom website. Now I'm working through the experiments in this digital book and kit I have. Make: Electronics. Thanks.

[mariushm]

Re: Question regarding Ohm's law and a Digital Multimeter.

Dude, you seriously need to chill and stop being so aggressive.

I was simply trying to educate the guy because he was going all theory, without realizing some of the pitfalls.

It's 3 AM here now, it was about 1-2 AM when I wrote the previous message and it took some time to find the videos in those authors' channels, so I listed the shunt values from memory and listed all three to be as complete as possible, not to confuse anyone (for example by listing the uA range).
I didn't double check the actual values not because I was lazy but because it is late and not in the mood to go to my workbench and spend 10-20 minutes setting everything up.

Again, my point was simply to educate him about the fact real world ... test leads have resistance, current shunt has resistance, potentiometer has resistance.
Yes, I missed the point about the LED's forward voltage but again, excuse the late hours.

For extra education, I could have continued by mentioning that the forward voltage of the led will vary with the heat. At 4mA, it won't happen but with higher current leds (like 10-20mA for example) after some time as the led heats up its forward voltage will change and so will the current going through the circuit, naturally.

Also note that potentiometers usually suck when it comes with power dissipation. The resistance of the potentiometer will also change with its temperature, usually potentiometers are much more sensitive compared to regular resistors.
The power dissipated can be measured with formula P = IxIxR ... for 5mA and 986 ohm the power would be P = 0.005 x 0.005 x 986 = 0.02465 watts if you use or 24 mW

A small trimmer potentiometer like this one for example has a power rating of 100mW so 24mW would not cause problems, but things would be different with higher currents.

[Spork Schivago]

Re: Question regarding Ohm's law and a Digital Multimeter.

"That gives me a current of roughly 4.36 mA. The DMM gives me 4.31 mA. I'm assuming the .05 mA is probably from the burden voltage or my DMM not being perfectly accurate."
What is your meter spec for the current measurement?

I don't have a manual or anything. But I have it set to mA when I measure it. The meter is set to auto range. If I set it manually, it only moves the decimal over one place. It's either 00.00 or 000.0 and it doesn't change the reading at all for current either way. I still get 4.31 mA. Where the probes plug in, for the mA and micro amp settings, it says 400 mA MAX FUSED. How do I tell what the meter spec for the current measurement is?

[kaboom]

Re: Question regarding Ohm's law and a Digital Multimeter.

Once I measured the voltage drop across the resistor, I got a much, much closer value for current compared to what the DMM had. My math showed a voltage drop across the potentiometer of 4.31VDC. I had the potentiometer set to 988 Ohm. That gives me a current of roughly 4.36 mA. The DMM gives me 4.31 mA.

Too little to matter.

I'm assuming the .05 mA is probably from the burden voltage or my DMM not being perfectly accurate.

NO! It's a series ckt. The same current flows thru all components, including your meter.

Stop thinking about this "mysterious" "burden voltage," which was misrepresented by mariushm. There was no reason for him to mention that.

I also did a test just to see if I was understood where I went wrong. I figured out the resistance that LED was using by using Ohm's law. I new because it's a series circuit, current should be the same everywhere. I checked the voltage drop on the LED, it was 2.023VDC. I divided that by the current, 4.36 mA, and got a resistance of around 464 ohm. I added that to the resistance of the potentiometer (988 ohm) and 1452 ohm. Using Ohm's law again, 6.46VDC (from the battery) divided by the total resistance of the LED + potentiometer gives me roughly 4.45 mA.

Alkaline cells, right? They've got a high-ish source impedance, meaning the supply voltage drops once you load them; actual voltage under load results in less current than no-load calculations suggest.

That's another thing: people taking their supply voltages for granted- it's more real and observable than the effects of burden voltage a four-point-anything milliamperes.

I figured the reason I have a bigger number is because of the resistance and everything from the breadboard itself and the wires. When resistance goes up, I was under the impression current goes down. Thanks!

That's a stupid assumption, unless your breadboard was recovered from a flood. The current that LED draws is to that breadboard what a transistor radio is to a central aircon (35mA vs 25A).




Not coming back, mariushm?

Is it because you've got no argument, or are you googling the answer to what I asked you? How about apologizing to the OP for misleading him, and making him look for "ghosts in the machine"?

[Spork Schivago]

Re: Question regarding Ohm's law and a Digital Multimeter.

...Also note that potentiometers usually suck when it comes with power dissipation. The resistance of the potentiometer will also change with its temperature, usually potentiometers are much more sensitive compared to regular resistors.
The power dissipated can be measured with formula P = IxIxR ... for 5mA and 986 ohm the power would be P = 0.005 x 0.005 x 986 = 0.02465 watts if you use or 24 mW

A small trimmer potentiometer like this one for example has a power rating of 100mW so 24mW would not cause problems, but things would be different with higher currents.

Ahh! So that probably explains why the resistance of the potentiometer is changing from day to day. The temperature is fluctuating a good bit in this room. In the morning, nice and cool. At night, sweltering hot! It doesn't change much, but enough to make me question some things.

[Spork Schivago]

Re: Question regarding Ohm's law and a Digital Multimeter.

Too little to matter.



NO! It's a series ckt. The same current flows thru all components, including your meter.

Stop thinking about this "mysterious" "burden voltage," which was misrepresented by mariushm. There was no reason for him to mention that.

Ok...

Alkaline cells, right? They've got a high-ish source impedance, meaning the supply voltage drops once you load them; actual voltage under load results in less current than no-load calculations suggest.

That's another thing: people taking their supply voltages for granted- it's more real and observable than the effects of burden voltage a four-point-anything milliamperes.

Yes, they're Alkaline cells. Just so I understand this, you're saying when I put a load on them (for example, use them to provide energy to a circuit with a resistor), the voltage that they're providing will actually be less than if I measured the voltage from the cells out of circuit?

That's a stupid assumption, unless your breadboard was recovered from a flood. The current that LED draws is to that breadboard what a transistor radio is to a central aircon (35mA vs 25A).

Keep in mind, I'm new to this. I have no teacher or anything. Everything I've learned I've learned by reading and trying some experiments. I don't know hardly anything yet. That's why I'm asking the questions. I'm okay with being wrong. I just like to know when I'm wrong so I can correct my thinking. I don't understand the analogy you provided about the transistor radio and a central aircon. I just started learning this stuff maybe a week ago. I've spent probably a good 3 days on it though. Had some other stuff to do around the house. I am going to make mistakes. And I'm sure there will be more "stupid assumptions".