EEVblog Electronics Community Forum
Electronics => Beginners => Topic started by: raspberrypi on November 16, 2016, 01:01:30 am
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I have an led hooked up to a power supply that I built. The sole purpose of this project is to learn E=IR and mess with different resistors etc etc. Super simple: power supply is set to 5 volts goes into 330 Ohm resistor then led (the kind you can't just plug into to 3 volts without it burning up, do these have a name?). So I put the meter on amps, plug the leads into the amp and common, hook up in series, and nothing! LED doesn't light and no reading on the screen. What is happening? Does the meter have too much resistance to let the current flow? It's not like I'm trying to measure micrro amps this circuit would use milliamps. Its a basic Fluke multimeter with seven functions.
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You may have a blown fuse in your meter.
Does the LED light without the meter in the circuit?
Does the meter display a current without the LED in the circuit?
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Just to be clear: You are connecting everything in series, right?
Positive terminal of Power Supply > Positive Input of DMM
Negative Input of DMM > One end of 330R resistor
Other end of 330R resistor > Anode of LED
Cathode of LED > Negative terminal of Power Supply
If you have everything connected like this and you still aren't seeing the LED light up or any reading on the meter, it is probable that your meter's "10A" fuse is blown. It is unlikely that the resistance of the meter itself is high enough to cause your problem.
What happens when you use the "mA" mode of the meter instead of the "A" mode?
What model Fluke are you using?
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The fuse read 6.5MOhms. The last guy that owned it was a complete idiot, so dumb I'm surprised he even tried to use it. I bet he plugged in the amps setting and tried to measure a car battery and blew it up, them blamed the meter for his mistake. If you're careful, any harm in just shorting out the fuse? I can't see myself just ordering one fuse, there are no electronic part stores around here. Its the good old fluke 73. This thing looked brand new, I guess not.
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The meter will certainly work just fine if you jumper the fuse holder.
BUT.... the fuse is there to protect you, and the meter, from unintentional misuse. Sure as shootin, if you don't replace the fuse with a proper one, somewhere down the line, maybe years from now when you've forgotten all about it, someone will do something that _should have_ blown the fuse, and will blow up the meter instead, and maybe the user too.
After all... we know the fuse was needed for protection at least once.
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I have blown the fuse in my meter because I was trying to compare it's current readings with my own current readings by measuring the voltage drop across a resistor. When going back and forth between the two measurements, it's really easy to forget to change the probe from the current input to voltage input and vice/versa. Best you keep a fuse in there.
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BTW, in this situation it is much easier to just measure the voltage drop across the 330R resistor and use ohms law. No need to keep switching probe sockets and you don't need to disconnect anything to put the meter in series.
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BTW, in this situation it is much easier to just measure the voltage drop across the 330R resistor and use ohms law. No need to keep switching probe sockets and you don't need to disconnect anything to put the meter in series.
how about this http://www.rapidtables.com/calc/electric/power-calculator.htm (http://www.rapidtables.com/calc/electric/power-calculator.htm)
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Multimeters also have a burden voltage, caused by the current shunts inside the multimeter, used to measure the current.
If you configure a fixed voltage like 3v and put the multimeter in series with the led, the led will actually see a voltage slightly lower than 3v... the actual value will vary depending on the current amount and the multimeter setting you use (burden voltage will be smaller on the Amps range as the current shunt has lower resistance, and higher on the mA range, but you get higher precision on that range)
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BTW, in this situation it is much easier to just measure the voltage drop across the 330R resistor and use ohms law. No need to keep switching probe sockets and you don't need to disconnect anything to put the meter in series.
how about this http://www.rapidtables.com/calc/electric/power-calculator.htm (http://www.rapidtables.com/calc/electric/power-calculator.htm)
Well if you really need to, :-\ but I=V/R is very simple... as is V=IR and R=V/I. It's really best to get used to get into the habit of just doing the sum rather than using a website as a 'crutch'. It is something you're going to be doing a lot. .
Edit: There are actually very few occasions that you actually need to use a DMM in current mode, there's usually a resistor somewhere suitable that you can measure the drop across. As mariush points out, having to take into account shunt+fuse+lead resistance is a pain (as is accidentally blowing the meter fuse).
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:scared: This makes no sense why this doesn't work here are pics of the worlds 2nd simplest circuit other then a dead short: :scared: :scared: :scared:
Do you think this set up would make for a really clean 5 volts if you put a scope where the LEDS are? From the wall the it goes into that power conditioner which has a gigantic toroid coil in it, then a linear power supply then that LED driver with that huge smoothing cap... Waaaaaaaay too much circuitry to light an LED.
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Measure the actual resistance of the 330ohm resistor.
Then measure the voltage across it while the LED is lit.
The current through the resistor is the same as through the diode.
I=U/R
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I suggest you get two of these and keep one on current and the other one for voltages. Helps avoiding blowing fuses. Downside is they use 9v batteries, but the fuses are common and cheap.
http://www.harborfreight.com/7-function-digital-multimeter-90899.html (http://www.harborfreight.com/7-function-digital-multimeter-90899.html)
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:scared: This makes no sense why this doesn't work here are pics of the worlds 2nd simplest circuit other then a dead short: :scared: :scared: :scared:
Do you think this set up would make for a really clean 5 volts if you put a scope where the LEDS are? From the wall the it goes into that power conditioner which has a gigantic toroid coil in it, then a linear power supply then that LED driver with that huge smoothing cap... Waaaaaaaay too much circuitry to light an LED.
I've been staring at your photos for ten minutes trying to figure out what you mean. What is "not working"? Are the LEDs not lighting up?
Why doesn't the DMM show a reading? Well, it's turned off, and I only see one of its probe leads connected to the circuit.
Why aren't the LEDs lighting up? I can't tell from the photo. Perhaps you have their polarities the wrong way around. Perhaps your LED driver has something to do with it. What is the part number/specifications of that module? Is it supposed to be a constant current driver?
The way to trouble shoot such a circuit starts with the voltages. Is your wall-wart providing the correct input voltage for the LED driver module? Test this using the voltmeter function of your DMM. Are the LEDs good and have you identified which pin is anode and which is cathode? Use the Diode check function of the DMM to test the LEDs. Is the LED driver doing what it is supposed to do? Check its output voltage.
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I suggest you get two of these and keep one on current and the other one for voltages. Helps avoiding blowing fuses. Downside is they use 9v batteries, but the fuses are common and cheap.
http://www.harborfreight.com/7-function-digital-multimeter-90899.html (http://www.harborfreight.com/7-function-digital-multimeter-90899.html)
Works for me. Battery included, too, iirc. And they can be had for free sometimes, during HF sales events.
I velcroed a pair of them to a little backboard and made up some appropriate dedicated jumpers to make the hookup easy-peasy.
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:scared: This makes no sense why this doesn't work here are pics of the worlds 2nd simplest circuit other then a dead short: :scared: :scared: :scared:
It is impossible to tell what you are doing based on a photo. Draw a diagram of exactly how you have things hooked up.
I wrote this beginners primer to include with the multimeters (ADM02) that I sell here in NZ, maybe it will help you...
http://sparks.gogo.co.nz/assets/_site_/downloads/multimeter-notes.pdf (http://sparks.gogo.co.nz/assets/_site_/downloads/multimeter-notes.pdf)
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:scared: This makes no sense why this doesn't work here are pics of the worlds 2nd simplest circuit other then a dead short: :scared: :scared: :scared:
Do you think this set up would make for a really clean 5 volts if you put a scope where the LEDS are? From the wall the it goes into that power conditioner which has a gigantic toroid coil in it, then a linear power supply then that LED driver with that huge smoothing cap... Waaaaaaaay too much circuitry to light an LED.
Please try to calm down! You're not making sense.
Your original problem of trying to measure the LED current was explained by the blown meter fuse. Several replies have explained how to measure the current by looking at the voltage drop across the 330R resistor. Now you have posted a photo with LEDs apparently lit. What's the problem? Please try to explain clearly.
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So the photo is to show that every thing works. The problem is when I put the meter in series with the circuit. The red probe is to show where the meter was inserted in series although I tried many points.
Yes:
the meter was on
the fuse problem was solved
the leads were plugged into the current setting
the leds and circuit light up, but not when the meter is in series anywhere in this or any other circuit.
voltages were measured at every point and all = my math less tolerance/loss (the point of not just taking math short cuts, this is for TEACHING real world vs pure math).
point of this is to teach people how e=ir and p=vA
No:
I'm not retarded although this circuit would suggest other wise
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the fuse problem was solved
How? You were reluctant to buy a new fuse, did you jumper the fuse holder?
If so, have you tested the current measurement and verified that it works, before trying your LED circuit again?
As an example of how to test it, remove everything except the 5V power supply and a 330R resistor from your circuit. Connecting your meter in series should give a reading of around 0.015A. (5/330, as you are obviously aware).
If it doesn't, then you haven't solved the fuse problem. Maybe the "complete idiot" who blew the fuse did so with enough force to damage the meter in some way?
If it does, then there is another problem with your circuit.
Whatever the issue, if things appear confusing it is often a good idea to keep simplifying what you are doing until you are sure you understand it, and then build back up again, step by step.
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You clearly need a cheaper meter.
Supply = 5VDC
LED fwd voltage = ~1.7 V
Resistor = 330 \$\Omega\$
Math:
ILED = (Vsupply - Vfwd)/R
ILED = (5 - 1.7)/330 = 0.010 A or 10 mA.
Iresistor = Vdrop/R
Iresistor = 3.44/330 = 0.0104 A or 10.4 mA
10 mA = 10.4 mA to within experimental error.
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You clearly need a cheaper meter.
Supply = 5VDC
LED fwd voltage = ~1.7 V
Resistor = 330 \$\Omega\$
Math:
ILED = (Vsupply - Vfwd)/R
ILED = (5 - 1.7)/330 = 0.010 A or 10 mA.
Iresistor = Vdrop/R
Iresistor = 3.44/330 = 0.0104 A or 10.4 mA
10 mA = 10.4 mA to within experimental error.
Horrible Freight? Whats wrong with buying another fluke? Thanks for the math but I threw in a curve ball: 3 100 ohm chinese resistors that all have real values in the top 98 ohms range. Must be the meter.
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Horrible Freight? Whats wrong with buying another fluke?
I'd be tempted to get your head around the basics first before paying out on more expensive gear. With regard to the resistors, they have tolerances, 98 ohms is only 2% off, well within spec for a 5% resistor. Nothing to indicate that the meter resistance range is wrong.
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1) Did you measure the actual voltage drop of the LED or just guess 1.7V? Some of my red LEDs are 1.5V and others are over 1.9V.
2) Did you measure the actual resistance of the resistor or just guess than the stripes are orange and not red or brown? I have never bought Chinese resistors (that have the wrong color stripes hand-painted on).
3) Did you measure the 5V or just guess it is accurate?
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The accuracy of a multimeter when measuring resistance can be something like 0.5-1% +/- 1-2 digits. For example, the Fluke 83v is +/- 0.4% + 1 digit on the 6000 ohm range, you can see in the specs here: http://support.fluke.com/find-sales/Download/Asset/2161164_6116_ENG_B_W.PDF (http://support.fluke.com/find-sales/Download/Asset/2161164_6116_ENG_B_W.PDF)
So if you measure a resistance that's exactly 1000 ohm, a multimeter like the Fluke 83v may read it as low as 0.4% less + 1 digit, or 996+1= 997 ohms, or may read it as up to 1004+1 = 1005 ohm. Anything between 997..1005 would be considered valid, and the actual value shown would also vary with the ambient temperature and humidity in which the multimeter is used. In general, most multimeters would show values much closer to the reality, not such extremes.
After taking this into account, keep in mind resistors have a tolerance value.. they may be 5% or 1% .. resistors are measured at the factory and binned into +/- 5% or they maybe be +/-1% - so with our 1000 ohm resistor, depending on what resistor you use, you may have a resistor with a value that's between 950 ohm and 1050 ohm for the 5%, or a resistor with 990 .. 1010 ohm value, for the 1%.
For cheap carbon film resistors, the actual resistor value may also drift a bit as the temperature of the resistor changes... so your 1000 ohm resistor may be 995 ohm when you start the circuit and may be 996 ohm after 5 minutes of electricity flowing through the circuit and lightning the led.
And I've said it a few posts above .. burden voltage.
If you put the multimeter on the current range and you put the multimeter in series with your resistor and led, the multimeter looks like a resistor to the circuit, and the resistor value that appears in the circuit may change depending on what range of current the multimeter may stay on.
So for example if your multimeter is a 2000 count multimeter, it may have a range of 0.. 19.99 mA where the multimeter appears to the circuit as one resistor value (let's say 100 ohm) , and it may have a current range of 20mA .. 199.9 mA where it shows up in the circuit as a 10 ohm resistor
So you suddenly no longer have just 330 ohm or whatever, you may have 430 ohms , or you may have 340 ohms, which could screw up your math.
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The HF meter + clipleads I used in my demonstration above has a burden resistance of about 3 ohms in the 2000 mA range ( as measured with a Fluke 87-III ).
The LED's forward voltage, again measured with the Fluke 87-III, is 1.743 V. The 330R resistor measures 328 ohms on the Fluke.
So go ahead and apply whatever corrections you like to my numbers above; the facts remain that my demonstration works and shows the validity of Ohm's Law. And also demonstrates that there is nothing wrong with using the HF "el cheepo" meter for this kind of work.
Feel free to provide your own demonstrations of this simple problem.
Subbing 3 x 100R resistors for the 330R isn't going to make the demonstration fail, nor will it make the circuit not function when the meter is connected in series. It will however allow slightly more current through the LED with a 5 volt DC supply, IF the meter is in good order and all connections are properly connected.