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Confusing Volt measurements (newbie) Utility Plugins
Old 26th December 2010
  #1
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datune's Avatar
Confusing Volt measurements (newbie)

I just started learning electronics, and while doing a little experiment, I stumbled upon something which confuses me.

The potentiometer has 2.5k resistance, the LED has listed (not that I understand what those values mean, but I am guessing it is supposed to be operated at maximum those values?):

lv:typ.3000mcd (how bright it shines?)
I(f): 20mA (what does the f stand for anyway?)
U(f): 3,2V

If the potentiometer is turned fully counter clockwise, and I measure directly at the battery pack, I get a value of 4,9Volt (batteries have been in use for a while). If I than measure at the two points displayed in the picture below, I can measure that the LED takes 2,69Volt, and the potentiometer takes 2,23Volt. So far so good.

Here is what confuses me:

If i now turn the potentiometer fully clockwise, and I measure directly at the battery pack, it displays only 3,59Volt (huh?????). Again measuring at the two points displayed in the picture below, I can measure that the LED takes 3,56Volt, whereas the potentiometer takes 0,02Volt.

What happened to the rest of the Voltage available from the battery? There's roughly 1,3Volt missing, who or what is using those?



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Old 26th December 2010
  #2
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Quote:
Originally Posted by datune View Post
lv:typ.3000mcd (how bright it shines?)
I(f): 20mA (what does the f stand for anyway?)
U(f): 3,2V
Yup, lv is light intensity. The f stands for "forward" i.e. current flowing from anode to cathode.

Quote:
Originally Posted by datune View Post
What happened to the rest of the Voltage available from the battery? There's roughly 1,3Volt missing, who or what is using those?
My best guess?

Batteries have output impedance, which rises the flatter they get. Since you are drawing more current with the pot turned down, more voltage will be dropped across the battery's resistance as per Ohm's Law V=I*R.
Old 26th December 2010
  #3
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Hi
OUCH!!
You must ALWAYS have a current limiting resistor for an LED. Turning the pot to minimum will be (almost) zero ohms and the battery will try to destroy the LED.
I (f) the f means 'forward' which is the mormal operating condition, if it said (r) then it would be reverse and is an indication to designers what characteristics the LED has when you reverse the polarity. (basically minimal current until you get to a 'breakdown' voltage (usually aboyt 4 volts or thereabouts) where you may damage the LED.
I(f) is the suggested operating current (20mA here) at which point the Volts across it will be V(f) which is 3.2 Volts. To find the resistor needed for this use Ohms law. You have a 6 volt battery, 3.2 volts across the LED so 6 - 3.2 = 2.8 Volts.
R=V/R so you want 140 Ohms.
An LED will try to draw LOTS of current once above say 3.5 Volts (your LED) until it overheats and fries.
As your battery may not be 'perfect' and has it's own internal resistance, this excessive current probably accounts for the indicated drop. Pots also have some resistance at each 'end' and are not zero Ohms.
Matt S
Old 26th December 2010
  #4
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Thanks for the replies, but I fear I still don't understand where the remaining 1.3Volt have gone too.

It appears that when the potentiometer is fully engaged, ie. the LED is very bright, that the LED draws 3.56Volt, the potentiometer draws 0.02Volt, but what about the remaining Volt? The thing I don't get is if I measure the Volt output DIRECTLY from the battery in this configuration, it seems that there is only 3,59Volt available, but how can that be, because if I turn the potentiometer counterclockwise again, I can measure 4,9Volt DIRECTLY at the outputs of the battery. Am I missing the obvious?!?
Old 26th December 2010
  #5
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mister sunshine's Avatar
 

Voltage builds behind the resistance.

Your DMM should have a high enough input resistance that the additional resistance provided by the potentiometer is such a small percentage that the accuracy of the measurement is not shifted to badly.

My guess is that the potentiometer you are using has a very high value.

Measuring voltage on a small battery is a challenge to all DMMs because the circumstance you have just discovered is common. You can measure the voltage but the result may not be meaningful.

Did you view EEVblog I linked for you in another thread? You can ignore the product reviews and just focus on learning about the universal challenges all DMM must be designed for.

Have Fun.

best regards,
mike
Old 26th December 2010
  #6
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Ok, I have added a little video to the original post, showing exactly what I am confused about, perhaps this helps to portray my question, hopefully leading to a conclusive answer.

@mike_mccue Thanks, I did indeed take a glance at that blog, but with all due respect, I don't think my $120 multimeter is to be compared with what this guy is talking about. He seems to be targeting $9.99 DMM's, my multimeter is actually ISO certified, along with a calibration certificate, last calibrated in July 2010.
Old 26th December 2010
  #7
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Quote:
Originally Posted by datune View Post
Thanks for the replies, but I fear I still don't understand where the remaining 1.3Volt have gone to.
The voltage hasn't "gone" anywhere.

As Matt pointed out, with the pot fuil up, the load on the battery is excessive because there is no current limiting resistor, so the battery voltage is significantly reduced. You can compare this to a car on an uphill slope, with the accelerator in a certain position. In neutral (not loaded by the transmission) the motor will turn at a certain RPM. When the motor is engaged to the transmission and drive train and delivers energy to the wheels, it will run slower.

If you measure the battery voltage with the LED off and then on you will confirm this. The battery voltage will equal the voltage across the two terminals of the pot, plus the voltage across the LED, regardless of the position of the pot setting, but the battery voltage will change depending on how much it is loaded.
Old 26th December 2010
  #8
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The is a resistance inherent inside the battery, so when you draw a lot of current from the battery it's terminal voltage drops due to internal IxR losses (ohms law).

You can damage the pot and the LED by adjusting it to such a low resistance that the current is too large. At the very end of the range there is a metal wiper touching a metal terminal inside the pot, but when you vary just off the full pot extreme, you can burn up the resistive element inside the pot. Likewise the LED can get very bright and overheat from too much current, not exactly the same mechanism but it could burn out just like a light bulb turned up too bright.

JR
Old 26th December 2010
  #9
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Hi
The batteries (presumably not quite new alkaline or similar) have an 'internal' resistance and this is where the volts are 'going' Think of the battery as a 'perfect' 6 volt source that does NOT vary with a resistor, say 10 or more Ohms in series, INSIDE THE BATTERY CASING. OK you can work this resistance out but I am off out for the night!.
IF you used a freshly charged lead acid battery (like a car battery) the INTERNAL resistance is VERY low (0.001 Ohm or threabouts) and you would have blown your LED and possibly wires completely.
Yout meter is not lying, it is indicating that Ohms law works but you have not calculated in the battery's internal resistance.
Matt S
Old 26th December 2010
  #10
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datune's Avatar
Thanks for all the replies!

I have to admit, I'm still trying to wrap my head around this, but I'll get there!
Old 26th December 2010
  #11
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Quote:
Originally Posted by datune View Post
Ok, I have added a little video to the original post, showing exactly what I am confused about, perhaps this helps to portray my question, hopefully leading to a conclusive answer.

@mike_mccue Thanks, I did indeed take a glance at that blog, but with all due respect, I don't think my $120 multimeter is to be compared with what this guy is talking about. He seems to be targeting $9.99 DMM's, my multimeter is actually ISO certified, along with a calibration certificate, last calibrated in July 2010.


Yes, I see that you choose a nice multi meter. The same person discusses multi meters that span the entire range of quality, and FWIW seems to really zero in on the price range you have selected as a best value. He also reviews the Flukes which are $400+.

I found his explanations of the basic operation of multi meters and how there is an ideal vs real life implementation to be very interesting. That stuff is sort of intertwined within the video blog. The guy has made a dozen or more DMM reviews, not just the one I linked to.

Anyways, I'm in over my head with regards to answering your question compared to the help you are getting from the other guys.

I just got done doing some 120vAC wiring today and wished my aging multimeter was faster and less likely to dance around before settling on a value. A new multi meter is on my mind and that's probably why I was attracted to your thread.

wishing you the best,
mike
Old 27th December 2010
  #12
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Thanks Mike, wish you all the best too
Old 27th December 2010
  #13
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Hi
ALL voltage sources have an 'internal resistance'.
This internal resistance is inherent to all devices but it's resistance value can be from thousandths of an Ohm to many MegOhms.
So, tou have a perfect supply in a small box which you cannot access directly as there is the 'internal resistance' resistor in series with the terminals.
If your measuring meter has INFINITE resistance then the value of this intenal resistance will not affect the voltage you read as it must obey Ohms law and as there is no current flow the resistance will not develop a voltage across it.
If the internal resistance were to be say 200 Ohms and your meter had a resistance of 200 Ohms, then with your nominal 6 Volt battery the voltage you read at the battery terminals would be 3 volts.
A typical digital multimeter has an input resistance of 10 Meg Ohms so you can use Ohms law to work out what you would measure with a 6 Volt source with 200 POhm internal resistance being measured with a 10Meg meter. The answer is very slightly under 6 Volts (5.99996) and would be within the error tolerance of the meter unless it were a 61/2 digit type.
Hope this illustration helps.
Matt S
Old 27th December 2010
  #14
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Hi Matt,

thanks for your reply, it will certainly help!
Old 29th December 2010
  #15
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No electronic parts are perfect

Jurgen:

The battery poops out a bit when you rotate the pot to the extreme where it has almost zero resistance. The 1.3 volts is not lost - the battery cant make the higher voltage when it is loaded down when having to push more current into the led. The battery is like an old man - he can walk at a normal pace until he has to carry his wife, then he walks more slowly (speed being analogous to voltage in this example, weight of wife being the load).

BTW, it may help to know that Ohms are specified in a way that is opposite of what you might think, due to a historical mistake that has been maintained to the present day. When the mistake was discovered, there was discussion in Europe to correct the matter and a measurement called mhos was proposed. (mho is ohm spelled backwards). The mho specifies current conductance, so low value of mho means low current conductance - this makes more sense. A vote was taken and the committee decided to keep the Ohm so that books would not have to be rewritten. The mho was approved by the committee for optional use and is used in a few calculations, usually in FET transistors and vacuum tubes.

So, because the resistor is backwards, a small resistor is a large load.

Volts and amperes are not backwards.

There is one more backwards item(s): the battery, all dc power supplies, and any semiconductor like a diode or transistor that has the arrow in it's symbol. Again, historical error. The committe met on this matter also and decided not to fix the matter so books would not have to be rewritten. So actually electrons flow out of the negative terminal of the battery, and into the point of the arrows. The committee concluded that for 99 percent of electronic theory and application this reversal makes no problems. You should not think about this - forget I mentioned it. Only about 1 in 100 electronic engineers know about this, but all battery engineers know.
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