Connecting a LED without resistor.(HELP)

[justaskliba]

I saw a way but I don't remember it could anyone help It included only wires and a LED and a raspberry pi with a breadboard of course. A picture would help.

[mahjongg]

Its possible, but not recommended!

Normally you should ALWAYS use a current limitation resistor.
Ive seen some people use the trick of using the 1,8K pullup resistor of the two I2C GPIO's as the current limiting resistor, but its a very nasty trick, not recommended at all.

The trick is that if you open the GPIO as an output to GND OR as an input you can directly connect a LED.
Problem is that if you accidentally program the GPIO as an output to 3V3, you will blowup the LED AND the PI!

[justaskliba]

Well thanks!

[PeterO]

These 5V LEDs aren't as bright on 3V3 but they do work.

http://www.rapidonline.com/Electronic-C ... -12V-60113

PeterO

[mahjongg]

These 5V LEDs aren't as bright on 3V3 but they do work.

http://www.rapidonline.com/Electronic-C ... -12V-60113

PeterO

Yes because they have the resistor built right in! Normal LEDs have not, and thus these LEDs are only confusing the matter.

[W. H. Heydt]

The thing to remember is that (a) a LED is a diode, and is forward biased to make it light, and (b) a diode connected from a +v to ground is a short circuit (less the voltage drop across the junction). Would you even consider shorting a connection through a diode? No? Then don't do it through a LED, either.

[PeterO]

These 5V LEDs aren't as bright on 3V3 but they do work.

http://www.rapidonline.com/Electronic-C ... -12V-60113

PeterO

Yes because they have the resistor built right in! Normal LEDs have not, and thus these LEDs are only confusing the matter.

Yes thanks for that, but I did actually read the link that I posted so I had already read that "The Kingbright 5mm LEDs are a superb, high quality range of components incorporating an inbuilt series resistor enabling the LED to be directly connected to 5V or 12V supply lines."

How are they confusing ? Since the OP asked " It included only wires and a LED and a raspberry pi " , and since this is exactly what these do , I would have thougth they were helpful, but I'll remember next time not to post helpful replies !

PeterO

[mikerr]

If the reason you want to avoid resistors is because you have multiple LEDS, you can have a single resistor serving many LEDs like this:

[mahjongg]

These 5V LEDs aren't as bright on 3V3 but they do work.

http://www.rapidonline.com/Electronic-C ... -12V-60113

PeterO

Yes because they have the resistor built right in! Normal LEDs have not, and thus these LEDs are only confusing the matter.

How are they confusing ? Since the OP asked " It included only wires and a LED and a raspberry pi " , and since this is exactly what these do , I would have thougth they were helpful, but I'll remember next time not to post helpful replies !

PeterO

Its unhelpful claiming there are LED's that don't need the resistor as it might confuse naive users into thinking that the average LED they have might also not need one, with disastrous consequences. LED's with built in resistors are rare and relatively expensive, and ones designed for 3V3 are non existent (yeah I know that 5V ones might "work").

LED's need a series resistor, lets teach users that!

Yes, if you are lazy you can buy a "5V LED", but I won't recommend it.

[sprinkmeier]

If the reason you want to avoid resistors is because you have multiple LEDS, you can have a single resistor serving many LEDs like this:

The more LEDs you turn on the more current through the resistor, the more voltage it drops, the less voltage for the LEDs, the dimmer the LEDs.
Also, if the LEDs are different colour then they'll have different turn-on voltages meaning the closer-to-the-red-end-of-the-rainbow ones may be the only ones that light up.

[drgeoff]

The more LEDs you turn on the more current through the resistor, the more voltage it drops, the less voltage for the LEDs, the dimmer the LEDs.

Close but no cigar.

Turning on more LEDs gives only slight increase in current through the resistor. The percentage increase tends to zero as the supply voltage is increased.

By the same token the voltage increase across the resistor is mimimal.

Yes the LEDs become dimmer but not because the voltage across them is less. It is because the almost contant current through the resistor is shared between them.

[boyoh]

The more LEDs you turn on the more current through the resistor, the more voltage it drops, the less voltage for the LEDs, the dimmer the LEDs.

Close but no cigar.

Turning on more LEDs gives only slight increase in current through the resistor. The percentage increase tends to zero as the supply voltage is increased.

By the same token the voltage increase across the resistor is mimimal.

Yes the LEDs become dimmer but not because the voltage across them is less. It is because the almost contant current through the resistor is shared between them.

I might be missing something here, How is the supply voltage
Increased with more than one led switching on. They are all
comming off the same 3.3v+ Pi power supply GPIO out/puts.

[rpdom]

I'm not understanding those posts completely, but I can understand

Yes the LEDs become dimmer but not because the voltage across them is less. It is because the almost constant current through the resistor is shared between them.

The LEDs, being diodes, will have a fixed forward voltage (Vled). Therefore the resistor will have a fixed voltage across it of (Vsupply - Vled), no matter how many LEDs. If V is a constant and R is a constant, then I must be a constant too. That means that current (I) will be shared over however many LEDs there are. The more LEDs the less current through each one and the dimmer they will be.

[PeterO]

Its unhelpful claiming there are LED's that don't need the resistor

Yeah, lets assume they are all too stupid to understand what they are reading and dumb everything down to the point of uselessness....
PeterO

[boyoh]

There is one calculation I have not seen
mentioned, is the diversity factor, that not
all the led's will be on at the same time,
But we must assume that at some time they
will all be on at the same time,

[drgeoff]

There is one calculation I have not seen
mentioned, is the diversity factor, that not
all the led's will be on at the same time,
But we must assume that at some time they
will all be on at the same time,

The first mention of more than one LED being on was by sprinkmeier - "The more LEDs you turn on...."

Where has anyone confined the discussion to the case of all the LEDs being on?

[drgeoff]

The LEDs, being diodes, will have a fixed forward voltage (Vled). Therefore the resistor will have a fixed voltage across it of (Vsupply - Vled), no matter how many LEDs. If V is a constant and R is a constant, then I must be a constant too.

No, that is a good engineering approximation but is not strictly correct. A diode does not have a fixed forward voltage. Although it is approximately 0.6 volts for a conducting silicon PN junction, it does vary with the current. As more LEDs are turned on the current through each one decreases and the voltage across the LEDs decreases slightly. The total current increases slighly each time another LED is turned on. (But not by enough to get worried about!)

[drgeoff]

The more LEDs you turn on the more current through the resistor, the more voltage it drops, the less voltage for the LEDs, the dimmer the LEDs.

Close but no cigar.

Turning on more LEDs gives only slight increase in current through the resistor. The percentage increase tends to zero as the supply voltage is increased.

By the same token the voltage increase across the resistor is mimimal.

Yes the LEDs become dimmer but not because the voltage across them is less. It is because the almost contant current through the resistor is shared between them.

I might be missing something here, How is the supply voltage
Increased with more than one led switching on. They are all
comming off the same 3.3v+ Pi power supply GPIO out/puts.

I didn't write that the supply voltage increased with more than one LED switching on. The circuit given by mikerr with a single resistor does not specify a supply voltage. I was saying that the effect becomes less significant when this arrangement is used with higher voltages.

[boyoh]

I admire a man who will admit when he is wrong
But not a man who will will wriggle out of a statement,
he made by using words.

[drgeoff]

I admire a man who will admit when he is wrong
But not a man who will will wriggle out of a statement,
he made by using words.

I have not wriggled out of anything. The supply voltage increasing with more LEDs being switched on was a figment of your imagination.

As I explained above, the voltage across the LEDs and across the resistor does change very slightly depending on the number of LEDs that are on. That changes the current in the resistor. The change is less and less when the circuit is used with higher and higher voltages.

[boyoh]

I admire a man who will admit when he is wrong
But not a man who will will wriggle out of a statement,
he made by using words.

I have not wriggled out of anything. The supply voltage increasing with more LEDs being switched on was a figment of your imagination.

As I explained above, the voltage across the LEDs and across the resistor does change very slightly depending on the number of LEDs that are on. That changes the current in the resistor. The change is less and less when the circuit is used with higher and higher voltages.

( Turning on more LEDs) gives only slight increase in current through the resistor . The percentage increase tends to zero (as the( supply voltage is increased.) drgoff Did you write this post

[drgeoff]

Either you genuinely do not understand or you are trolling. I'll give you the benefit of the doubt and explain again.

Let the supply voltage be S
Let the nominal current through the resistor be I
Let the voltage drop of a LED passing current I be L
Assume the slope of the volts versus current of a LED be linear and of value D. (That assumption of linearity is false but does not invalidate the discussion.)

With one LED operating at current I the voltage across the resistor is S-L.

The resistor value is therefore (S-L)/I

With two LEDs operating the voltage across the resistor is close to S-L+D/(2I). (Not exact because the current through the resistor is no longer I so the LED drop is not precisely L+D/(2I).)

With three LEDs operating the voltage across the resistor is close to S-L+2D/(3I).

With N LEDs operating the voltage across the resistor is close to S-L+(N-1)D/(NI).

The change of resistor voltage from one LED on to N LEDs on is (N-1)D/(NI).

Dividing by the resistor value gives the change in resistor current as (N-1)DI/((NI(S-L)).

This reduces to (N-1)D/((N(S-L))

from which, noticing that (S-L) is in the divisor, we conclude that:

1. When S is not much more than, such that (S-L) approaches zero, the change of current increases sharply.

2. If we operate the circuit with a high S the change is much less. In the limiting case of infinite S, the change is zero.

That is the meaning of "Turning on more LEDs gives only slight increase in current through the resistor. The percentage increase tends to zero as the supply voltage is increased."

[boyoh]

Either you genuinely do not understand or you are trolling. I'll give you the benefit of the doubt and explain again.

Let the supply voltage be S
Let the nominal current through the resistor be I
Let the voltage drop of a LED passing current I be L
Assume the slope of the volts versus current of a LED be linear and of value D. (That assumption of linearity is false but does not invalidate the discussion.)

With one LED operating at current I the voltage across the resistor is S-L.

The resistor value is therefore (S-L)/I

With two LEDs operating the voltage across the resistor is close to S-L+D/(2I). (Not exact because the current through the resistor is no longer I so the LED drop is not precisely L+D/(2I).)

With three LEDs operating the voltage across the resistor is close to S-L+2D/(3I).

With N LEDs operating the voltage across the resistor is close to S-L+(N-1)D/(NI).

The change of resistor voltage from one LED on to N LEDs on is (N-1)D/(NI).

Dividing by the resistor value gives the change in resistor current as (N-1)DI/((NI(S-L)).

This reduces to (N-1)D/((N(S-L))

from which, noticing that (S-L) is in the divisor, we conclude that:

1. When S is not much more than, such that (S-L) approaches zero, the change of current increases sharply.

2. If we operate the circuit with a high S the change is much less. In the limiting case of infinite S, the change is zero.

That is the meaning of "Turning on more LEDs gives only slight increase in current through the resistor. The percentage increase tends to zero as the supply voltage is increased."

Thanks for the excise in OHMS LAW.
and giving me the benefit of the doubt.
The original post by " Justaskliba"
asking was it possible to connect a Led
to the RaspberryPi ,with out a resistor,
He was advised not to do this, witch was
correct, Knowing that he would be using
a GPIO 3.3v+ out/puts, and the possibility
Of using more LEDs using GPIO 3.3v+
and using one current limiting resistor
This would create a series parallel
Circuit coming from the same 3.3v+
Pi power supply, making the increase
of voltage not possible.

[drgeoff]

The circuit is a generic one and is not restricted to being used only with a RPi. The main point of my first post in this thread was to point out that sprinkmeier's explanation was completely incorrect. I quoted his "The more LEDs you turn on the more current through the resistor, the more voltage it drops, the less voltage for the LEDs, the dimmer the LEDs." As I explained, the increase in current with more LEDs turned on is minimal. That the increase becomes even less when the circuit is used with higher supply voltages was an additional secondary point. You then asked "How is the supply voltage increased with more than one led switching on." but you have never shown how or where I or anyone else claimed that it does. You have not because you cannot. I took the trouble to explain again, hoping it would aid you in realising your misinterpretation of what had been written. You also demonstrated a low level of comprehension when you posted about "diversity factor" in which you were under the mistaken impression that no-one apart from yourself had considered the possibility of only some LEDs being on simultaneously. I quote you: "There is one calculation I have not seen mentioned, is the diversity factor, that not all the led's will be on at the same time".

That you persist in unsubstantiated arguing and have provided no useful input whatsoever to the thread means that I can no longer give you the benefit of the doubt and I will heed three pieces of advice.

“Never wrestle with pigs. You both get dirty and the pig likes it.” George Bernard Shaw:

“Never argue with an idiot. They will only bring you down to their level and beat you with experience.” George Carlin

"Do not feed the trolls". Unknown

I suggest that the time is fast approaching for a moderator to lock this thread, possibly after you have had an opportunity to reply. I rest my case. In the unlikely event that anyone else is really interested I don't have any worries over the the judgement they will come to.

[boyoh]

The circuit is a generic one and is not restricted to being used only with a RPi. The main point of my first post in this thread was to point out that sprinkmeier's explanation was completely incorrect. I quoted his "The more LEDs you turn on the more current through the resistor, the more voltage it drops, the less voltage for the LEDs, the dimmer the LEDs." As I explained, the increase in current with more LEDs turned on is minimal. That the increase becomes even less when the circuit is used with higher supply voltages was an additional secondary point. You then asked "How is the supply voltage increased with more than one led switching on." but you have never shown how or where I or anyone else claimed that it does. You have not because you cannot. I took the trouble to explain again, hoping it would aid you in realising your misinterpretation of what had been written. You also demonstrated a low level of comprehension when you posted about "diversity factor" in which you were under the mistaken impression that no-one apart from yourself had considered the possibility of only some LEDs being on simultaneously. I quote you: "There is one calculation I have not seen mentioned, is the diversity factor, that not all the led's will be on at the same time".

That you persist in unsubstantiated arguing and have provided no useful input whatsoever to the thread means that I can no longer give you the benefit of the doubt and I will heed three pieces of advice.

“Never wrestle with pigs. You both get dirty and the pig likes it.” George Bernard Shaw:

“Never argue with an idiot. They will only bring you down to their level and beat you with experience.” George Carlin

"Do not feed the trolls". Unknown

I suggest that the time is fast approaching for a moderator to lock this thread, possibly after you have had an opportunity to reply. I rest my case. In the unlikely event that anyone else is really interested I don't have any worries over the the judgement they will come to.

Lets call a truce, and put the Sword's away
We are both correct , You are correct with
general application of LEDs and resistor circuits
I was correct with confining the connection of the
LEDs and single resistor to the RaspberryPi.
I'm to old at 84yrs, to keep arguing,
So lets shake Hand's ,and be friends.

Best regards BoyOh ( John )