Measuring your multimeter while attempting to measure your circuit
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By the way, did you check the battery voltage when it was actually connected to the resistor?
I measure approx 2.2 volts across the 10 Ohm resistor in my circuit. I'm assuming that would mean something like:
2.2 / 10 Ohm = 220mA
However, I know when I measure that I only see 157mA of current. So...
220mA - 157mA = 63mA (which are missing somewhere).
I plug in the missing 63 like so:
.8V (voltage drop) / .063 = 12.69 Ohms
So I'm still seeing a extra ~12 Ohms that I can't account for. That is obviously higher than my 9-10 Ohm extra I was calculating but could be due to rounding. Does that seem at least somewhat correct? I'm just noodling my way through this to learn more. Thanks,
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That's a great point and helps me to learn. However I am using 2 AA in series to obtain 3v. I still thought 300mA might be a bit too much to pull but in the spirit of experimentation I rolled on. :)
Carry on! :-D Electronics remains my first love, despite nearly 40 years since school! It's fascinating, always...
Will Rogers never met me.
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I measure approx 2.2 volts across the 10 Ohm resistor in my circuit. I'm assuming that would mean something like:
2.2 / 10 Ohm = 220mA
However, I know when I measure that I only see 157mA of current. So...
220mA - 157mA = 63mA (which are missing somewhere).
I plug in the missing 63 like so:
.8V (voltage drop) / .063 = 12.69 Ohms
So I'm still seeing a extra ~12 Ohms that I can't account for. That is obviously higher than my 9-10 Ohm extra I was calculating but could be due to rounding. Does that seem at least somewhat correct? I'm just noodling my way through this to learn more. Thanks,
Combining a cheap multimeter and an overloaded battery may lead to all kinds of silly conclusions. If you want to comment on a multimeter, make sure you use a lab-grade power supply and resistor both capable to act nominally in your circuit; and likewise when checking a battery, make sure to use a lab-grade multimeter and resistor. And make sure none of the components involved heats up (which indicates operation outside the nominal operating range), as currently your resistor probably and your battery most likely does. :)
Luc Pattyn [My Articles] Nil Volentibus Arduum
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I'm working on Chapter 4 of Practical Electronics For Makers and I'm working the reader through some very simple circuits so we can see how resistors affect (lower) current. So I set up the simplest circuit that is simply a 3V battery supply and one 10 Ohm resistor. Fist I calculate the expected current value I will see using Ohm's law (E = IR).
3V / 10 Ohm = 0.3A (300mA)
I hook up the circuit to run through my multimeter so I can measure (and confirm) the value. My meter displays:
157.7 mA
What? Google... Apparently meters have their own internal resistance. I decide to turn Ohm's law around and calculate internal resistance.
3V / 0.157 = 19.10 (Ohms)
So it looks like my meter has around 9-10 Ohms of its own resistance. Hmmm... Measuring the measurements of a measuring device.:~ Of, course if I had higher resistance in my circuit I'd probably not notice this, because it is only 10 Ohms. I posted a question about this on electronics stack exchange where you can see pictures and the circuit. Multimeter, measuring current calculates less. Can I calculate meter's internal resistance this way? - Electrical Engineering Stack Exchange[^]
In addition to the other sources of error (I'd wager battery internal resistance is the biggest), you've run into ammeter burden[^]. In most modern digital multimeters, the maximum burden is approximately the full scale on your lowest voltage range, probably something like 200mV. (It measures current by the voltage drop across a shunt.) Welcome to real-world electronic engineering! Cheers, Peter
Software rusts. Simon Stephenson, ca 1994. So does this signature. me, 2012
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I watched a YouTube that explains it very well #194: What is ammeter burden voltage, and why you should care. - YouTube[^] And someone else pointed out a special port on my meter (marked 20A) which has a smaller burden and I was was able to read 270mA on the circuit. Interesting that it was the meter and not the battery.
Next time you'll measure the voltage over the resistor. :)
Wrong is evil and must be defeated. - Jeff Ello
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I'm working on Chapter 4 of Practical Electronics For Makers and I'm working the reader through some very simple circuits so we can see how resistors affect (lower) current. So I set up the simplest circuit that is simply a 3V battery supply and one 10 Ohm resistor. Fist I calculate the expected current value I will see using Ohm's law (E = IR).
3V / 10 Ohm = 0.3A (300mA)
I hook up the circuit to run through my multimeter so I can measure (and confirm) the value. My meter displays:
157.7 mA
What? Google... Apparently meters have their own internal resistance. I decide to turn Ohm's law around and calculate internal resistance.
3V / 0.157 = 19.10 (Ohms)
So it looks like my meter has around 9-10 Ohms of its own resistance. Hmmm... Measuring the measurements of a measuring device.:~ Of, course if I had higher resistance in my circuit I'd probably not notice this, because it is only 10 Ohms. I posted a question about this on electronics stack exchange where you can see pictures and the circuit. Multimeter, measuring current calculates less. Can I calculate meter's internal resistance this way? - Electrical Engineering Stack Exchange[^]
Years ago, I had a weird problem with a prototype board and slapped a Tektronix scope on to start looking for spurious signals and the like. And the problem went away. Finally worked out that the scope probe was smoothing out a high frequency nasty so I contacted Tektronix to find out the probe capacitance - and their suggestion was "ship one of our scopes with each board" ... funny guys, very funny. (They did tell us, and a capacitor added to the PCB cured the problem).
Bad command or file name. Bad, bad command! Sit! Stay! Staaaay... AntiTwitter: @DalekDave is now a follower!
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I measure approx 2.2 volts across the 10 Ohm resistor in my circuit. I'm assuming that would mean something like:
2.2 / 10 Ohm = 220mA
However, I know when I measure that I only see 157mA of current. So...
220mA - 157mA = 63mA (which are missing somewhere).
I plug in the missing 63 like so:
.8V (voltage drop) / .063 = 12.69 Ohms
So I'm still seeing a extra ~12 Ohms that I can't account for. That is obviously higher than my 9-10 Ohm extra I was calculating but could be due to rounding. Does that seem at least somewhat correct? I'm just noodling my way through this to learn more. Thanks,
That's strange. I would espect about
3.0Vin a circuit with only the batteries and the resistor (the multimeter providing negligible effect with its high parallel resistance). [update] According to the answer to your post on electronics stack exchange it looks you are asking too much to your batteries (I'm not an expert). So I would suggest you to make a test within limits of both resistor and the batteries (e.g. use a100 Ohms resistor). [/update] What's the power rating of your resistor? With2.2Vyou are still well out-of-range on a1/4Wresistor. -
I'm working on Chapter 4 of Practical Electronics For Makers and I'm working the reader through some very simple circuits so we can see how resistors affect (lower) current. So I set up the simplest circuit that is simply a 3V battery supply and one 10 Ohm resistor. Fist I calculate the expected current value I will see using Ohm's law (E = IR).
3V / 10 Ohm = 0.3A (300mA)
I hook up the circuit to run through my multimeter so I can measure (and confirm) the value. My meter displays:
157.7 mA
What? Google... Apparently meters have their own internal resistance. I decide to turn Ohm's law around and calculate internal resistance.
3V / 0.157 = 19.10 (Ohms)
So it looks like my meter has around 9-10 Ohms of its own resistance. Hmmm... Measuring the measurements of a measuring device.:~ Of, course if I had higher resistance in my circuit I'd probably not notice this, because it is only 10 Ohms. I posted a question about this on electronics stack exchange where you can see pictures and the circuit. Multimeter, measuring current calculates less. Can I calculate meter's internal resistance this way? - Electrical Engineering Stack Exchange[^]
As an electronic engineer I welcome you to this world! There is a proverb in German about measurements "Wer misst misst Mist". My freehand translation: If you measure you measure bullshit. There are so many factors affecting your result and you already realized: A breadboard circuit doesn't represent the ideal world of Ohms Law! Resistors have tolerances (do you know the exact value?), your meter has tolerances, your probe has resistance, your wires etc. And you are only in the low DC voltage area! The real fun starts with high frequency, where every cable, every socket etc is a combination of capacitor, induction and resistor. I suggest you measure the voltage drop over the resistor and then calculate your current. Also try to use a regulated power supply or similar that gives you a constant current. With a 3V battery, your won't have a lot of fun for long and your measurements will be different every time. There are also online circuit simulators (https://www.circuitlab.com/) and SPICE ([Online Circuit Simulator with SPICE](https://www.partsim.com/simulator)). SPICE is around for a long time and open source.
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I'm working on Chapter 4 of Practical Electronics For Makers and I'm working the reader through some very simple circuits so we can see how resistors affect (lower) current. So I set up the simplest circuit that is simply a 3V battery supply and one 10 Ohm resistor. Fist I calculate the expected current value I will see using Ohm's law (E = IR).
3V / 10 Ohm = 0.3A (300mA)
I hook up the circuit to run through my multimeter so I can measure (and confirm) the value. My meter displays:
157.7 mA
What? Google... Apparently meters have their own internal resistance. I decide to turn Ohm's law around and calculate internal resistance.
3V / 0.157 = 19.10 (Ohms)
So it looks like my meter has around 9-10 Ohms of its own resistance. Hmmm... Measuring the measurements of a measuring device.:~ Of, course if I had higher resistance in my circuit I'd probably not notice this, because it is only 10 Ohms. I posted a question about this on electronics stack exchange where you can see pictures and the circuit. Multimeter, measuring current calculates less. Can I calculate meter's internal resistance this way? - Electrical Engineering Stack Exchange[^]
Hi, The picture on your electronics stack exchange post shows that you are using a solderless breadboard. This is very likely contributing significantly to your "phantom" resistance. I count at least 6 breadboard connections in the picture any one of which could contribute several ohms if you are unlucky! The contacts in these things are only spring loaded so any oxidisation on the contacts or the wires will contribute additional series resistance. Make sure that all the wire ends, including the resistor are freshly cleaned and I would also try to remove some of the oxidation from the breadboard contacts by inserting and removing each wire several times. Better still, take the breadboard out completely and solder the battery and one end of the resistor directly together. Even then you will still have spring loaded connections from the crocodile clips on the ends of the meter leads. I understand that you are trying to produce something that is reproducible by people without an electronics workshop, but for you tests I would also try to eliminate the battery by using a bench power supply so you can eliminate voltage droop.
Graham
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Hi, The picture on your electronics stack exchange post shows that you are using a solderless breadboard. This is very likely contributing significantly to your "phantom" resistance. I count at least 6 breadboard connections in the picture any one of which could contribute several ohms if you are unlucky! The contacts in these things are only spring loaded so any oxidisation on the contacts or the wires will contribute additional series resistance. Make sure that all the wire ends, including the resistor are freshly cleaned and I would also try to remove some of the oxidation from the breadboard contacts by inserting and removing each wire several times. Better still, take the breadboard out completely and solder the battery and one end of the resistor directly together. Even then you will still have spring loaded connections from the crocodile clips on the ends of the meter leads. I understand that you are trying to produce something that is reproducible by people without an electronics workshop, but for you tests I would also try to eliminate the battery by using a bench power supply so you can eliminate voltage droop.
Graham
Excellent advice! I hate breadboard - but nowhere near as much as I hate wirewrap ... :laugh:
Bad command or file name. Bad, bad command! Sit! Stay! Staaaay... AntiTwitter: @DalekDave is now a follower!
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Excellent advice! I hate breadboard - but nowhere near as much as I hate wirewrap ... :laugh:
Bad command or file name. Bad, bad command! Sit! Stay! Staaaay... AntiTwitter: @DalekDave is now a follower!
OriginalGriff wrote:
but nowhere near as much as I hate wirewrap
Just when I hoped you would wrap up my new boards for me...
I have lived with several Zen masters - all of them were cats. His last invention was an evil Lasagna. It didn't kill anyone, and it actually tasted pretty good.
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OriginalGriff wrote:
but nowhere near as much as I hate wirewrap
Just when I hoped you would wrap up my new boards for me...
I have lived with several Zen masters - all of them were cats. His last invention was an evil Lasagna. It didn't kill anyone, and it actually tasted pretty good.
Oh I do it from time to time - but I don't like it. The amount of time you spend chasing bad contacts because you wiggled it slightly... :mad:
Bad command or file name. Bad, bad command! Sit! Stay! Staaaay... AntiTwitter: @DalekDave is now a follower!
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In addition to the other sources of error (I'd wager battery internal resistance is the biggest), you've run into ammeter burden[^]. In most modern digital multimeters, the maximum burden is approximately the full scale on your lowest voltage range, probably something like 200mV. (It measures current by the voltage drop across a shunt.) Welcome to real-world electronic engineering! Cheers, Peter
Software rusts. Simon Stephenson, ca 1994. So does this signature. me, 2012
-
Years ago, I had a weird problem with a prototype board and slapped a Tektronix scope on to start looking for spurious signals and the like. And the problem went away. Finally worked out that the scope probe was smoothing out a high frequency nasty so I contacted Tektronix to find out the probe capacitance - and their suggestion was "ship one of our scopes with each board" ... funny guys, very funny. (They did tell us, and a capacitor added to the PCB cured the problem).
Bad command or file name. Bad, bad command! Sit! Stay! Staaaay... AntiTwitter: @DalekDave is now a follower!
-
That's strange. I would espect about
3.0Vin a circuit with only the batteries and the resistor (the multimeter providing negligible effect with its high parallel resistance). [update] According to the answer to your post on electronics stack exchange it looks you are asking too much to your batteries (I'm not an expert). So I would suggest you to make a test within limits of both resistor and the batteries (e.g. use a100 Ohms resistor). [/update] What's the power rating of your resistor? With2.2Vyou are still well out-of-range on a1/4Wresistor. -
As an electronic engineer I welcome you to this world! There is a proverb in German about measurements "Wer misst misst Mist". My freehand translation: If you measure you measure bullshit. There are so many factors affecting your result and you already realized: A breadboard circuit doesn't represent the ideal world of Ohms Law! Resistors have tolerances (do you know the exact value?), your meter has tolerances, your probe has resistance, your wires etc. And you are only in the low DC voltage area! The real fun starts with high frequency, where every cable, every socket etc is a combination of capacitor, induction and resistor. I suggest you measure the voltage drop over the resistor and then calculate your current. Also try to use a regulated power supply or similar that gives you a constant current. With a 3V battery, your won't have a lot of fun for long and your measurements will be different every time. There are also online circuit simulators (https://www.circuitlab.com/) and SPICE ([Online Circuit Simulator with SPICE](https://www.partsim.com/simulator)). SPICE is around for a long time and open source.
-
Hi, The picture on your electronics stack exchange post shows that you are using a solderless breadboard. This is very likely contributing significantly to your "phantom" resistance. I count at least 6 breadboard connections in the picture any one of which could contribute several ohms if you are unlucky! The contacts in these things are only spring loaded so any oxidisation on the contacts or the wires will contribute additional series resistance. Make sure that all the wire ends, including the resistor are freshly cleaned and I would also try to remove some of the oxidation from the breadboard contacts by inserting and removing each wire several times. Better still, take the breadboard out completely and solder the battery and one end of the resistor directly together. Even then you will still have spring loaded connections from the crocodile clips on the ends of the meter leads. I understand that you are trying to produce something that is reproducible by people without an electronics workshop, but for you tests I would also try to eliminate the battery by using a bench power supply so you can eliminate voltage droop.
Graham
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I think that is a great idea and will try that and let you know. I also learned a bit more about this see : This thread the Lounge[^]
Thank you for your Feedback. Thinking again (and again and again ...) about this: In case you have a second multimeter available, I think the most easy Thing is to measure the voltage over the ammeter with the second meter. Having the current displayed by the ammeter and the voltage over the ammeter makes it easy too judge wheter the ammeter's internal shunt is the reason ;) I like your experiments and the articles. Best regards Bruno
It does not solve my Problem, but it answers my question
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I'm working on Chapter 4 of Practical Electronics For Makers and I'm working the reader through some very simple circuits so we can see how resistors affect (lower) current. So I set up the simplest circuit that is simply a 3V battery supply and one 10 Ohm resistor. Fist I calculate the expected current value I will see using Ohm's law (E = IR).
3V / 10 Ohm = 0.3A (300mA)
I hook up the circuit to run through my multimeter so I can measure (and confirm) the value. My meter displays:
157.7 mA
What? Google... Apparently meters have their own internal resistance. I decide to turn Ohm's law around and calculate internal resistance.
3V / 0.157 = 19.10 (Ohms)
So it looks like my meter has around 9-10 Ohms of its own resistance. Hmmm... Measuring the measurements of a measuring device.:~ Of, course if I had higher resistance in my circuit I'd probably not notice this, because it is only 10 Ohms. I posted a question about this on electronics stack exchange where you can see pictures and the circuit. Multimeter, measuring current calculates less. Can I calculate meter's internal resistance this way? - Electrical Engineering Stack Exchange[^]
You can't monitor a system without affecting its behavior. :) I'm glad I don't do electronics (my father did).
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You can't monitor a system without affecting its behavior. :) I'm glad I don't do electronics (my father did).
