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Topic: Forward voltage vs Current (Read 19229 times) previous topic - next topic

floresta

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Like Don said, LED's current is not linear when voltage changes. You need to resize the resistor. Just think your LED as a part, that gives you voltage drop and current to work properly.
2.4V, 50mA. Imagine that as a resistor Rled
Rled= 2.4v/50mA= 48 ohm.
R1 will be 2.6v/50mA=52 ohms
The American slang for this would be 'hogwash' which translates to 'nonsense'.  You cannot and should not imagine the LED as a resistor.  It does not exhibit the characteristic that Mr. Ohm defined as 'resistance'.

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Maybe I misunderstand the whole problem
Yes you did.  Nothing was ever said about changing the supply voltage.  If he could have used a higher voltage he wouldn't have his (perceived) problem.

Don

scswift

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In this case the voltmeter is connected across the LED and the LED is forward biased.  Therefore it is reading the forward voltage drop of the diode which happens to be 2.4 volts.

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You would know if you think across instead of at.  If you put the voltmeter across the resistor you get the voltage drop of the resistor.  If you put the voltmeter across the LED you get the forward voltage of the LED (assuming it is connected properly).  If you put the voltmeter across the supply you get the supply voltage.  Those are your only three choices in this circuit.


I don't understand.

I understand the concept of voltage across a component.  But I don't understand why the voltmeter would be reporting the voltage drop, rather than the supply voltage minus the drop.  Isn't what flows through the meter everyhting but what flows through the component between the probes?

GaryP

Don, you are right again, but.. but... but I was trying to get past one thing, maybe it was a stupid idea to use resistor for substitute in that case.
I hope I didn't burn anything with that idiotic post!

Yep, forget the word "resistor" but think about it as a part of the circuit, where we know absolutely what is the nessessary current to make it light bright, and we already know what is the voltage potential OVER it at that point.

Damn, I'm dummy! Have mercy!!!! Lesson learned!
:smiley-red:

Cheers,
Kari
The only law for me; Ohms Law: U=R*I       P=U*I
Note to self: "Damn! Why don't you just fix it!!!"

GaryP


Isn't what flows through the meter everyhting but what flows through the component between the probes?


Your volt meter's impedance is very high, so not much actually flows throught it. It just simply measures voltage over two points.
If you have ten LED's in serial, you can measure over each one, and summing them you will get total voltage. Current through the circuit won't change while measuring, it probably don't even notice that somebody's touching it.

Cheers,
Kari
The only law for me; Ohms Law: U=R*I       P=U*I
Note to self: "Damn! Why don't you just fix it!!!"

scswift

#34
Apr 03, 2011, 10:06 pm Last Edit: Apr 03, 2011, 10:09 pm by scswift Reason: 1
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If I were to replace the 5v source with a 9v source, I would not expect to read the same 2.4v.


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Now we are running in place. Are you playing with us, or what is going on?
If you want to have 50mA to go throught the LED, of course you change the resistor for 9 volts, right?


In the usual circumstance, yes, you would.  But here I was proposing an experiment to better understand things.  And in an experiment, you don't want to change two variables at once, because that makes it much harder to interpret the results.

That said, if I understand what has been told to me thus far:

1) If I use the same resistor with a higher voltage, I will put more current through the LED.
2) As I put more current through the led, the LED's forward voltage increases.
3) If I put my meter probes across the LED, I will get the LED's forward voltage.
4) Therefore if I conducted the above experiment, the value I read with the meter would increase, just as I would expect it to if I was reading supply voltage - forward voltage, but it would not increase nearly so much as I would expect in that case, because what I'm actually reading is just forward voltage, and that will have increased only slightly.

floresta

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Isn't what flows through the meter everyhting but what flows through the component between the probes?
No.  As Gary said essentially nothing flows through the voltmeter (an ammeter is a different story).  The meter merely reports what it sees between it's probes.  If the probes are across the resistor it reports the resistor voltage, if the probes are across the supply it reports the supply voltage.  If you have this circuit (-) --> resistor --> -led+ --> (+)  and you want the LED voltage you can connect the probes across the LED and measure the voltage directly or if you insist you can connect one probe to '-' (ground), measure the voltage 'at' the supply (which is the same as measuring the voltage across the supply), then you can measure the voltage 'at' the resistor / LED junction (which is the same as measuring the voltage across the resistor) and subtract the second measurement from the first.  All of this is in accordance with 'Kirchoff's Voltage Law' mentioned in a previous post.

Don

floresta

#36
Apr 03, 2011, 10:59 pm Last Edit: Apr 03, 2011, 11:02 pm by floresta Reason: 1
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1) If I use the same resistor with a higher voltage, I will put more current through the LED.
2) As I put more current through the led, the LED's forward voltage increases.
3) If I put my meter probes across the LED, I will get the LED's forward voltage.
4) Therefore if I conducted the above experiment, the value I read with the meter would increase, just as I would expect it to if I was reading supply voltage - forward voltage, but it would not increase nearly so much as I would expect in that case, because what I'm actually reading is just forward voltage, and that will have increased only slightly.
You are getting closer.  Let me (try to) fix this up.

1) If I use the same resistor with a higher voltage, I will put more current will flow through the LED.
2) As I put more current flows through the led, the LED's forward voltage increases.
3) If I put my meter probes across the LED, I will get the LED's forward voltage.
4) Therefore if I conducted the above experiment, the value I read with the meter would increase, just as I would expect it to if I was reading supply voltage - forward voltage (you are describing the voltage across the resistor), but it would not increase nearly so much as I would expect in that case, because what I'm actually reading is just forward voltage, and that will have increased only slightly. (Both voltages will increase due to the increased supply voltage - Kirchoff again, but you cannot calculate or predict the new values based on the old due to the non-linearity of the LED).

Don

GaryP


1) If I use the same resistor with a higher voltage, I will put more current through the LED.


It is now time for you to start testing, and compare the measurement results with datasheet.

Do you have adjustable powersupply?

Get about 50 ohm, more ohms better to save the LED.
Adjust max. voltage to 5V, if 50 ohm R1.
Start from 1V, 0.5V increments to 5V.
Measure every step from LED and R1.

If your PSU don't show amps, you can calculate it with two units you know for sure.

Draw a curve, calculated amps agains the measured voltage OVER THE LED, that shows you the  non-linear LEDs effect in the circuit.

Then double the R1, increase the PSU voltage 7V and start measuring the voltage over LED, notice the brightness, draw a curve, calculate the current...

Cheers,
Kari
The only law for me; Ohms Law: U=R*I       P=U*I
Note to self: "Damn! Why don't you just fix it!!!"

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