how to use device safely by reading specification

Thread Starter

anukalp

Joined Jul 28, 2018
146
I have so many questions for the reading datasheet of device/ component. if I have a datasheet of LED then How to find safe voltage and current required for a device like LED.

I have tried to find the safe voltage and current for LED in the attached document but I am a little bit confused

How do you know the following points
  1. What voltage and current required to operate LED?
  2. What would be the maximum value of current, where LED can damage ?
  3. What would be the minimum value of current, where LED can damage?
  4. What would be the minimum value of voltage, where LED can damage?
 

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Travm

Joined Aug 16, 2016
282
Hi, LED's are current devices, so they don't exactly have "safe voltages"
You limit the current to the LED, using a resistor typically. There are other more complicated options, but at this early learning stage, keep it simple.

From your data sheet, Continuous Forward Current (If) absolute maximum is 30 mA. So this tells you, anything more than 30mA will destroy this LED. Remember, its absolute maximum, use something less than this.
Forward Voltage is 3v, this is how much voltage drop the LED will induce in the circuit.
Now you use Ohms law to calculate the required resistance in your circuit to keep the LED forward current below 30mA (i'd suggest 15mA). To do this, you need to specify the voltage your circuit will run at. Your LED doesnt care about the voltage, as long as you keep the current under 30mA.
lets say you use 12v;
12v-3v = 9v This is V - Vf
we need a resistor that will drop 9v @ 15mA, ohms law produces
9/0.015A = 600ohms
Use a 600ohm or higher resistor.

To specifically answer your numbered questions;
1. Any voltage higher than the forward voltage can be used. Current cannot exceed 30mA (Continuous Forward Current).
2. This is the "Absolute Maximum Rating" Continous Forward Current, 30 mA.
3. There is no minimum current. There will be a minimum current where it wont light. This isnt typically on the datasheet, because who wants to use an LED that wont light?
4. There is no minimum voltage that would damage it. Anything less than 3v and you will not be able to light the LED, possibly even 3.6v. To ensure it works, I wouldnt try to use anything less than 5v. But you wont damage it with less voltage.
 
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dl324

Joined Mar 30, 2015
9,606
What voltage and current required to operate LED?
You can operate the LED at any intensity you desire as long as you observe the maximum current restrictions.

This graph gives intensity as a function of current:
upload_2019-3-9_10-32-26.png

What would be the maximum value of current, where LED can damage ?
Most won't operate a device at it's maximum ratings, so you could have a DC current of up to 30mA with Ta <= 25C. You need to derate power dissipation above 25C.
upload_2019-3-9_10-33-55.png

What would be the minimum value of current, where LED can damage?
It's in the table above.
What would be the minimum value of voltage, where LED can damage?
That information isn't given, you have to extrapolate from this graph:
upload_2019-3-9_10-38-29.png

You can avoid damaging the LED by choosing an operating point in it's safe operating range.

Just because an LED is characterized at 20mA, you don't have to operate at 20mA.

You need to decide what brightness you desire and choose the corresponding current. You can use a resistor to limit current, or use a current source.
 

Raymond Genovese

Joined Mar 5, 2016
1,658
Your LED doesnt care about the voltage, as long as you keep the current under 30mA.
How would Pd, power dissipation be considered: Power Dissipation Pd 100 mW
with regard to V, I and safe operation?

So:
if I have 5v and a 330 resistor, it is operating at 15 mA and ~75 mW and it is ok?
If I have 15v and a 990 ohm resistor, it is operating at 15 mA and ~225 mW and it is not ok?

right or ?
 

Travm

Joined Aug 16, 2016
282
How would Pd, power dissipation be considered: Power Dissipation Pd 100 mW
with regard to V, I and safe operation?

So:
if I have 5v and a 330 resistor, it is operating at 15 mA and ~75 mW and it is ok?
If I have 15v and a 990 ohm resistor, it is operating at 15 mA and ~225 mW and it is not ok?

right or ?
That is the dissipation of the resistor. You need to be sure you use a resistor large enough to handle the power. Typically 1/4W resistors are fine, but as you point out at 15v, 990 ohm, you are dangerously close to 1/4w. A 1/2W resistor would be a better choice in that case. The LED will dissipate very little power.
 

Raymond Genovese

Joined Mar 5, 2016
1,658
That is the dissipation of the resistor. You need to be sure you use a resistor large enough to handle the power. Typically 1/4W resistors are fine, but as you point out at 15v, 990 ohm, you are dangerously close to 1/4w. A 1/2W resistor would be a better choice in that case. The LED will dissipate very little power.
Ahh ok. How do I calculate the dissipation measure relevant to the LED?

[I realize that I am not the TS, but it seems like a relevant question]
 

WBahn

Joined Mar 31, 2012
25,107
To specifically answer your numbered questions;
1. Any voltage higher than the reverse voltage can be used. Current cannot exceed 30mA (Continuous Forward Current).
Uh... no. The reverse voltage is something very different. It is the maximum reverse-biased voltage (hence the name) that can be applied to the diode and have assurance that it will not breakdown. If you are operating the LED in such a way that it can be reverse-biased, you have to ensure that the reverse bias is LESS than the reverse-voltage listed (5V in this case).

For forward operation, you can use whatever power supply voltage you want, as long as it is at least as much as the forward voltage of the diode (3.6 V in this case). The higher the voltage, the more accurate and stable you can make the current using a simple current-limiting resistor.
 

Travm

Joined Aug 16, 2016
282
Uh... no. The reverse voltage is something very different. It is the maximum reverse-biased voltage (hence the name) that can be applied to the diode and have assurance that it will not breakdown. If you are operating the LED in such a way that it can be reverse-biased, you have to ensure that the reverse bias is LESS than the reverse-voltage listed (5V in this case).

For forward operation, you can use whatever power supply voltage you want, as long as it is at least as much as the forward voltage of the diode (3.6 V in this case). The higher the voltage, the more accurate and stable you can make the current using a simple current-limiting resistor.
Good Catch, thanks.
 

WBahn

Joined Mar 31, 2012
25,107
How would Pd, power dissipation be considered: Power Dissipation Pd 100 mW
with regard to V, I and safe operation?

So:
if I have 5v and a 330 resistor, it is operating at 15 mA and ~75 mW and it is ok?
If I have 15v and a 990 ohm resistor, it is operating at 15 mA and ~225 mW and it is not ok?

right or ?
The device specs for the LED can only talk about power dissipation limits in the LED, not the entire circuit.

Since the max forward voltage is 3.6 V (at 20 mA), a first cut at the max current you can run the diode at for any length of time (since max power dissipation is normally a thermal issue, which requires a bit of time to build up), you would have about 28 mA. The voltage will be a bit higher at a higher current, but probably not appreciably more. This is likely where the 30 mA absolute max continuous forward current spec comes from, since the typical voltage is less than 3.6 V, specs are given in round numbers, and there's always some headroom allowed.

If the diode is going to be allowed to zener, that requires a different calculation since the zener voltage is different than the forward voltage.
 
The device specs for the LED can only talk about power dissipation limits in the LED, not the entire circuit.

Since the max forward voltage is 3.6 V (at 20 mA), a first cut at the max current you can run the diode at for any length of time (since max power dissipation is normally a thermal issue, which requires a bit of time to build up), you would have about 28 mA. The voltage will be a bit higher at a higher current, but probably not appreciably more. This is likely where the 30 mA absolute max continuous forward current spec comes from, since the typical voltage is less than 3.6 V, specs are given in round numbers, and there's always some headroom allowed.

If the diode is going to be allowed to zener, that requires a different calculation since the zener voltage is different than the forward voltage.
I follow that and would add that they (the data sheet) gives a non-continuous higher current value:
Peak Forward Current(Duty /10 @ 1KHZ)IFP 100 mA

But my question is, how do you calculate the power dissipation for the LED to know whether you have come close to the max? For me, in practice, it has never been an issue because I am 3.3V/5V with a current-limiting resistor to somewhere around 5-20 mA. But the spec and the question is making me wonder.
 

Travm

Joined Aug 16, 2016
282
I follow that and would add that they (the data sheet) gives a non-continuous higher current value:
Peak Forward Current(Duty /10 @ 1KHZ)IFP 100 mA

But my question is, how do you calculate the power dissipation for the LED to know whether you have come close to the max? For me, in practice, it has never been an issue because I am 3.3V/5V with a current-limiting resistor to somewhere around 5-20 mA. But the spec and the question is making me wonder.
This may be an incomplete thought, but with respect to LED's my understanding has always been, keep the current under the max, and you will never exceed the max power dissipation. So basically having the Pd max on the datasheet along with max current, is like stating the same thing twice. If you need to calculate the power dissipation its Vf x I. Vf is given on the datasheet, I is calculated based on your circuit.

Alternatively If you let this LED operate as a Zener, it would be a whole new game. Personally I have no idea why you would want to do that? I dont think i've ever seen an LED datasheet with that info on it, or my selective reading skips it every time.
 
This may be an incomplete thought, but with respect to LED's my understanding has always been, keep the current under the max, and you will never exceed the max power dissipation. So basically having the Pd max on the datasheet along with max current, is like stating the same thing twice. If you need to calculate the power dissipation its Vf x I. Vf is given on the datasheet, I is calculated based on your circuit.
It is bugging me a bit because I am looking for an answer while the thread is ongoing and all I have found so far, is pretty much what you are saying and how I have been doing it...more or less. Still, a definition of terms in a data sheet is not a bridge too far - is it?

For example, here is a lengthy thread on AAC..but does not get to it..
https://forum.allaboutcircuits.com/threads/resistor-led-power-dissipation-calculations-in-depth.119363/

and

https://sciencing.com/calculate-led-power-6455710.html
Calculating LED Power
To calculate an LED's power use, simply multiply the LED's voltage (in volts) by the LED's current (in amperes). The result, measured in watts, is the amount of power your LEDs use. etc..


In a sparkfun tutorial...https://learn.sparkfun.com/tutorials/light-emitting-diodes-leds/all

The power dissipation is the amount of power in milliWatts that the LED can use before taking damage. This should work itself out as long as you keep the LED within its suggested voltage and current ratings.

Maybe they should use that terminology in the datasheet - just look at the other measures and don't worry about this one, it will work itself out. :)

I'm sure there is an answer, I'll keep looking and stop steering the thread off the road.
 
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WBahn

Joined Mar 31, 2012
25,107
I follow that and would add that they (the data sheet) gives a non-continuous higher current value:
Peak Forward Current(Duty /10 @ 1KHZ)IFP 100 mA
The higher current (peak forward current) is spec'ed based on other failure modes. The continuous current limitation is due to physical heating of the die and damage that results, probably from diffusion of the dopants. But if you take the current high enough, then you can get localized effects that happen at much smaller time scales. A common limitation is the bond wires -- if you exceed a certain current even briefly the bond wires can heat up enough (they heat very fast since they have so little thermal mass) to fuse. Notice how the peak forward and peak reverse current limits are the same -- this hints that the limiting factor is not on the die itself, but instead either a package or a package/die interface limitation. I'm guessing bond wire fusing.

In more complicated ICs you can also have limits based on electromigration in which your metalization literally moves as a result of too high a current density in the traces. Depending on the layout, this can happen at currents well below the bond-wire limits and thermal limits.

But my question is, how do you calculate the power dissipation for the LED to know whether you have come close to the max? For me, in practice, it has never been an issue because I am 3.3V/5V with a current-limiting resistor to somewhere around 5-20 mA. But the spec and the question is making me wonder.
It's the voltage multiplied by the rms current over a period of time in which the LED can respond thermally. That time depends on the thermal characteristics, but a rule of thumb that is often used is anything on the order of a second or longer, though the shorter the time frame you use the safer you are.

Also, a common rule of thumb is to operate it at no more than half of the rated power, if possible, but to be very hesitant of going over 85%. That will drastically increase the life expectancy of the device.
 

WBahn

Joined Mar 31, 2012
25,107
It is bugging me a bit because I am looking for an answer while the thread is ongoing and all I have found so far, is pretty much what you are saying and how I have been doing it...more or less. Still, a definition of terms in a data sheet is not a bridge too far - is it?
For this kind of LED, which is expected to be operated in free air with no heat sinking, the max power spec generally matches well with the max continuous current spec.

What really matters is the junction temperature, but to use that you need a lot more thermal information than is normally given for these kinds of devices.

For higher power LEDs that's usually different and there they give you max junction temperature as well as thermal conductivities from the junction to the case so that you can make heat sink calculations.

As with nearly all electronics, you can operate them at much higher power levels provided you can carry the heat away quickly enough. In those cases, you run into other limiting factors first.
 
For this kind of LED, which is expected to be operated in free air with no heat sinking, the max power spec generally matches well with the max continuous current spec.

What really matters is the junction temperature, but to use that you need a lot more thermal information than is normally given for these kinds of devices.

For higher power LEDs that's usually different and there they give you max junction temperature as well as thermal conductivities from the junction to the case so that you can make heat sink calculations.

As with nearly all electronics, you can operate them at much higher power levels provided you can carry the heat away quickly enough. In those cases, you run into other limiting factors first.
Let me first stipulate that understanding is frequently limited because of knowledge level. So, I am reading what you are saying and much of it is understandable but an explanation for how to calculate the Pd of an LED is not perceived as being forthcoming - again, perhaps because of my knowledge level.

So, I hooked up a resistor measured as 324 Ohm to a power supply, measured at 4.99V. One end of the resistor goes to + and the other end of the resistor goes to one lead of my meter (set to read mA) and the other lead goes to power supply GND. Ok, I read 14.94 mA. That is reasonable and they are not soldered connections, just me touching the leads.

Now I keep the top of the resistor connected to + on the power supply and the bottom to one lead of an LED and I use my meter leads to the other lead of the LED and to power supply GND. The LED glows and I read 5.93 mA.

It seems to me that the power dissipated by the LED, not the resistor, is represented by the difference between 14.94 mA and 5.93 mA. IOW ~ 9 mA is being used by the LED. If I try to turn that into a wattage, so I can compare it to Pd, is it 4.99V X 9.01 mA = 44.96 mW or is it 3.3 (the rated Vf of the LED) X 9.01 = 29.73 mW? Or is it something else?

Edited to add: If I am understanding what is here http://www.alciro.org/alciro/diodo-LED_21/potencia-diodo-led_306_en.htm (and yes I know it is in Spanish [or is that Italian, no matter a diodo is a diodo], but you can see the formula and graphs):

Pd=I (anode) x V(anode-cathode). so, in my example, 3.3 X 9.01 = 29.73 mW gives me the Pd. That is my hope. The thing about ignorance is that it does not always have to be permanent :)
 
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Travm

Joined Aug 16, 2016
282
So, I hooked up a resistor measured as 324 Ohm to a power supply, measured at 4.99V. One end of the resistor goes to + and the other end of the resistor goes to one lead of my meter (set to read mA) and the other lead goes to power supply GND. Ok, I read 14.94 mA. That is reasonable and they are not soldered connections, just me touching the leads.
If i'm following this, all you are getting from that is the current of a 324ohm resistor @4.99v
Now I keep the top of the resistor connected to + on the power supply and the bottom to one lead of an LED and I use my meter leads to the other lead of the LED and to power supply GND. The LED glows and I read 5.93 mA.
Your current is less now, not because your reading the LED current, but because you've added the LED and its "Forward Voltage". assuming 3v for Vf, 4.99-3.3 = 1.69.
1.69v/324ohm=.0052A or 5.2mA, close enough i think. your Vf might be more like 3.1v, or 3.2v actually, it depends on temperature and current, and varies slightly from 1 LED to another, so I've read.
It seems to me that the power dissipated by the LED, not the resistor, is represented by the difference between 14.94 mA and 5.93 mA. IOW ~ 9 mA is being used by the LED. If I try to turn that into a wattage, so I can compare it to Pd, is it 4.99V X 9.01 mA = 44.96 mW or is it 3.3 (the rated Vf of the LED) X 9.01 = 29.73 mW? Or is it something else?
in this simple situation, your LED Pd is 3.3v (your Vf) x .0052A = 0.019W or 19mW
 
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If i'm following this, all you are getting from that is the current of a 324ohm resistor @4.99v

Your current is less now, not because your reading the LED current, but because you've added the LED and its "Forward Voltage". assuming 3v for Vf, 4.99-3.3 = 1.69.
1.69v/324ohm=.0052A or 5.2mA, close enough i think. your Vf might be more like 3.1v, or 3.2v actually, it depends on temperature and current, and varies slightly from 1 LED to another, so I've read.

in this simple situation, your LED Pd is 3.3v (your Vf) x .0059A = 0.019W or 19mW
Where/how did you get .0059A for I(anode)?
 
I just used your reading of 5.9mA. You had an actual circuit measurement above.

The actual circiut current, x the actual Vf (you can measure this too) = Acutal Pd
So, power dissipation of an LED, Pd=Pd=I (anode) x V(anode-cathode) - you agree with that formula?

But, where I am making a mistake is in calculating I(anode). Instead of the subtraction from resistor only vs resistor + LED, I should have taken the current measured with the resistor and the LED in the circuit?

Is that correct?
 
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