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Topic: Forward voltage vs Current (Read 19225 times)previous topic - next topic

GaryP #15
Apr 03, 2011, 12:41 pm

Forward voltage is the voltage drop across an LED.

So, it is just a voltage drop? Why wouldn't just use that term?
Waste of time...

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 #16
Apr 03, 2011, 12:46 pm
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So, it is just a voltage drop? Why wouldn't just use that term?

Why don't we just use electron flow instead of conventional?  Because engineers are silly that way apparently. Forward Voltage is the term used in every datasheet I've seen.  It confused the hell out of me for the longest time too.  If it had been called Voltage Drop it would have made more sense.  And I would have figured out how circuits work a lot faster if I wasn't being confused by which way the electricity was flowing. :-)

floresta #17
Apr 03, 2011, 04:55 pm
Here's the important part first:
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What I was asking is if the forward voltage and the current you put through the LED are related, if I could lower the current I put through them from the 50mA used in the datasheet, and then always be certain these LEDs will work in a 5v circuit.
Your problem is right here.  The 50mA value is the 'Absolute Maximum' rating.  This is a value that you never want to exceed, not the value that you design for.  Typically you operate at somewhere around half the absolute maximum rating.  At 25mA your LED should have a (nominal) forward voltage of around 3.75 volts so your resistor should be around 50 ohms.

Here's the nit-picking part:

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Forward voltage is the voltage drop across an LED.  In other words, if you put an LED in a circuit with a 5v source, and its forward voltage is 2.4v, then you will read 5v-2.4v, ie 2.6v at the anode of the LED.  (I think!)
This is ambiguous because we don't know where the resistor is and it is wrong regardless of how you connect the circuit.  You really measure voltage between two points, not 'at' a point.  When you express the voltage 'at' a certain point then the other point is assumed and the assumption is typically some common point, frequently referred to as 'ground'.  So in this case if you have the resistor connected to the + side of the battery and the LED between the resistor and the - side of the battery (ground) then the voltage 'at' the anode is 2.4v.  If you have the LED connected to the + side of the battery and the resistor between the LED and the - side of the battery (ground) then the voltage 'at' the anode is 5v (unless you have the LED in backwards).

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You would then do your Ohm's law calculation on the remaining voltage, and the amount of current you want to put through the LED, and you get:
Although "through the LED" is technically correct you really should use "through the resistor" since that's where you are using Ohm's law.

You are concentrating on the voltage aspects because of the fact that your particular LED has a forward voltage drop that is greater than your available voltage when it is operating at high currents.  You have correctly surmised that you must settle for a lower current in order to operate such an LED from a 5v supply.  Once you have determined an appropriate current (from the graph) then you should concentrate on that current.  Don't worry about the power or anything else.

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So if I get a bunch of LEDs with a forward voltage of 5.2v, they won't work
.But they will work if you lower the current.

Don

floresta #18
Apr 03, 2011, 04:57 pmLast Edit: Apr 03, 2011, 04:59 pm by floresta Reason: 1
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Forward Voltage is the term used in every datasheet I've seen.  It confused the hell out of me for the longest time too.  If it had been called Voltage Drop it would have made more sense.
Because there is also an important parameter called the 'Reverse Voltage' rating.  It is on the next line of your data sheet, right below the 'Forward Voltage' rating.   That is why you must use the term forward voltage for the parameter that you are talking about.

Don

floresta #19
Apr 03, 2011, 05:03 pm
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So, it is just a voltage drop? Why wouldn't just use that term?
Waste of time...
Because, since the LED is not a resistor, the voltage across the LED is different when it is 'forward biased' than when it is 'reverse biased'.  This is really where the terms forward and reverse come from.  Are you beginning to understand why an engineering education takes so long?

Don

GaryP #20
Apr 03, 2011, 05:11 pm

Quote
So, it is just a voltage drop? Why wouldn't just use that term?
Waste of time...
Because, since the LED is not a resistor, the voltage across the LED is different when it is 'forward biased' than when it is 'reverse biased'.  This is really where the terms forward and reverse come from.  Are you beginning to understand why an engineering education takes so long?

Don

Yes, I really do, and I would never be ready to go that road!!!
XD

What I always consern, is the max. current with the Vcc I'm using, and for some reason, most of the times nothing fries. Lucky me!?

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 #21
Apr 03, 2011, 06:19 pm
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Your problem is right here.  The 50mA value is the 'Absolute Maximum' rating.  This is a value that you never want to exceed, not the value that you design for.  Typically you operate at somewhere around half the absolute maximum rating.

I'm aware of the need to run at below the Absolute Maximum rating.  The reason I'm quoting the 50mA over and over is because that's also the listest test condition for the LED.  I'm not sure why the manufacturer did that other than to pad their stats if they don't expect the LED to be run at 50mA.  Most seem to list current levels around 66% lower than the Absolute Maximum for their test conditions.

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At 25mA your LED should have a (nominal) forward voltage of around 3.75 volts so your resistor should be around 50 ohms.
You have correctly surmised that you must settle for a lower current in order to operate such an LED from a 5v supply.
But they will work if you lower the current.

Thank you.  I've asked about that several times, but this is the first straight answer I've gotten.

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Forward voltage is the voltage drop across an LED.  In other words, if you put an LED in a circuit with a 5v source, and its forward voltage is 2.4v, then you will read 5v-2.4v, ie 2.6v at the anode of the LED.  (I think!)

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This is ambiguous because we don't know where the resistor is and it is wrong regardless of how you connect the circuit.  You really measure voltage between two points, not 'at' a point.  When you express the voltage 'at' a certain point then the other point is assumed and the assumption is typically some common point, frequently referred to as 'ground'.  So in this case if you have the resistor connected to the + side of the battery and the LED between the resistor and the - side of the battery (ground) then the voltage 'at' the anode is 2.4v.  If you have the LED connected to the + side of the battery and the resistor between the LED and the - side of the battery (ground) then the voltage 'at' the anode is 5v (unless you have the LED in backwards).

So, you're saying that if I have:
(-) --> resistor --> -led+ --> (+)

And I touch my probes to (-) and the + side of the led, I'll get 5v... whichs makes sense since the + side of the led is also connected directly to (+) and the potential difference between (-) and (+) with nothing else between would be 5v.

But if I have:
(-) --> -led+ --> resistor --> (+)

And I touch my probles to (-) and to the + side of the led, I'll get 2.4v... which I can only assume is because of the resistor being there.  But I'm not sure I understand how that works.

Lastly, to take it a step further, if I have this setup:
(-) --> -led1+ --> -led2+ --> resistor --> (+)

And I touched a probe to (-) and to led1's + terminal, what would I read?
And what would I read if I touched a probe to (-) and led2's + terminal?

Also, why am I not reading 2.6v at those points if my LEDs forward voltage is 2.4v?  I thought you could supply two leds with a 2.4fv off a 5v supply and have 0.2v left over that the resistor needs to dissipate?

tomm #22
Apr 03, 2011, 07:20 pm
Why don't you connect it up and probe it? Much easier to learn that way.

GaryP #23
Apr 03, 2011, 07:32 pm

And I touch my probles to (-) and to the + side of the led, I'll get 2.4v... which I can only assume is because of the resistor being there.  But I'm not sure I understand how that works.

You make it sound very difficult, and you should use GND in your text instead of "-". In the case you are talking about ground potential, of course. If you are talking about components pole, then "-" is fine, right?

It doesn't matter which side of the LED your resistor is, current is the same, and voltage drop over both components remain the same. Or voltage loss,what ever it is in english, please, help me with that term.

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!!!"

retrolefty #24
Apr 03, 2011, 07:45 pm
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It doesn't matter which side of the LED your resistor is, current is the same, and voltage drop over both components remain the same. Or voltage loss,what ever it is in english, please, help me with that term.

It's cover in Kirchhoff's second law. In a series circuit the sum of the individual voltage drops across the components will equal the total applied circuit voltage, or something like that. http://en.wikipedia.org/wiki/Kirchhoff's_circuit_laws

Lefty

scswift #25
Apr 03, 2011, 07:55 pm
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Why don't you connect it up and probe it? Much easier to learn that way.

Not really.  Since I can't know what the LED's forward voltage is independent of the circuit, I wouldn't be able to interpret the results I got.

In my previous example, the LED had a forward voltage of 2.4v. But Floresta is telling me I would read 2.4v if I probed ground and the point between the led the resistor, even though I specified the source was 5v.  I would expect if I was probing those points that I would see the voltage the resistor sees, not the forward voltage of the LED.  If I were to replace the 5v source with a 9v source, I would not expect to read the same 2.4v.

So if I did probe some leds and resistors, if I got 2.2v or something on the multimeter, I still wouldn't know if what I was reading was the forward voltage of the LED, or the voltage the resistor was seeing.  Though I assume it is the latter, and so I cannot understand why Floresta said that I would read 2.4v at that point when 5v-2.4v = 2.6v.

floresta #26
Apr 03, 2011, 08:18 pm
You will be a lot better off when you learn to think of the voltage as being 'across' something in the circuit rather than 'at' some point in the circuit.
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But if I have:
(-) --> -led+ --> resistor --> (+)

And I touch my probles to (-) and to the + side of the led, I'll get 2.4v...
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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... which I can only assume is because of the resistor being there.  But I'm not sure I understand how that works.
If the resistor weren't there you would have a crispy LED.  The resistor limits the current and has a voltage drop across it.  In this case the voltage drop across the resistor (2.6v) and its resistance cause the amount of current to be whatever it takes to get the voltage across the diode to be 2.4v.  It's not a linear circuit and it cannot be explained in simple terms.  Just follow the outline I gave you a few posts back and don't try to figure out why it works that way.

Don

GaryP #27
Apr 03, 2011, 08:19 pm

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It doesn't matter which side of the LED your resistor is, current is the same, and voltage drop over both components remain the same. Or voltage loss,what ever it is in english, please, help me with that term.

It's cover in Kirchhoff's second law. In a series circuit the sum of the individual voltage drops across the components will equal the total applied circuit voltage, or something like that. http://en.wikipedia.org/wiki/Kirchhoff's_circuit_laws

Lefty

Actually, I was only looking for term, and it's potential difference... 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!!!"

floresta #28
Apr 03, 2011, 08:23 pm
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So if I did probe some leds and resistors, if I got 2.2v or something on the multimeter, I still wouldn't know if what I was reading was the forward voltage of the LED, or the voltage the resistor was seeing.
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.

Don

GaryP #29
Apr 03, 2011, 08:33 pm

If I were to replace the 5v source with a 9v source, I would not expect to read the same 2.4v.

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?

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

Then there is no place for guessing when you change the voltage, just get new value for resistor R1
V over R1 is 9v - 2.4v = 6.6V
R1 = 6.6V/50mA=132 ohm

Maybe I misunderstand the whole problem, but I'm trying to figure and help.

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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