# Does running LED at higher voltage damages the LED?

I want to run 12v 6 watt LED at 19v by keeping maximum watts constant (current controlled). By doing so will damage the LED?

I.e

12v x 0.5000 A = 6W

19v x 0.3157 A = 6W

update1: I am running these LED with buck based PT4115 home made led driver

• You can run it from a 19V supply, but if it's current controlled then the voltage across the LED will be its nominal forward voltage of 12V. You won't be able to increase that without exceeding the current limit. Jul 11 '16 at 8:51
• What you want is impossible, see Andy's answer. What you could do is get a DCDC buck converter to make 12 V from the 19V, then run the LEDs from that 12 V. The current at the 19 V side will be about 0.3 A despite 0.5 A flowing through the LEDs. That sounds weird to the uninitiated but is perfectly normal for a buck converter. This is an example of a suitable converter: goo.gl/kgxqme Jul 11 '16 at 9:15
• How can you expect while using more voltage will allow you lesser current flow? Jul 11 '16 at 19:53
• @AlwaysConfused A switch mode regulator works like that. If you increase the input voltage, the input current will decrease.
– pipe
Jul 12 '16 at 6:14
• @pipe Exactly. Now I have learned how the buck type cc regulator works Jul 12 '16 at 6:25

The idea sounds good but won't work. LEDs cannot magically adjust their incoming current to suit the prevailing applied voltage unless it has some type of switching regulator built into it.

On the understanding that it doesn't, if you increase the voltage a little bit across a partially conducting LED, it will draw a lot more current and quite possibly enough to near-instantly fry it.

Take the example of a 24 ohm resistor. With 12 V applied it takes 0.5 A but with 19 V applied it takes 0.792 A. There is no power regulation implied in ohms law.

If you were able to simultaneously control voltage and current across a device, all the Current-Voltage Characteristics (including the Ohm law) would be nonsense.

If you regulate in current the voltage will step down, if you regulate in voltage the current will step up, but you will always be on this curve:

If you are using a constant current supply, it will attempt to adjust it's output voltage until the load draws the expected current, within its voltage range. So if your led setup has a forward current of 500 mA at a forward voltage of 12V, the CC supply will try to lower it's Voltage output to 12V. If it can't go that low, it may result in a higher voltage and current than you want, blowing your led.

The other option is using a simple series ballast resistor to soak up the extra voltage, as @EM has answered. But a 5W power resistor is not exactly an ideal solution.

### Update:

As Op has listed the driver they intend to use, from the data sheet introduction:

The PT4115 is a continuous conduction mode inductive step-down converter, designed for driving single or multiple series connected LED efficiently from a voltage source higher than the total LED chain voltage. The device operates from an input supply between 6V and 30V and provides an externally adjustable output current of up to 1.2A. Depending upon the supply voltage and external components, the PT4115can provide more than 30 watts of output power.

The IC is designed for this very function, stepping down a higher input voltage, to a current controlled LED. Just set it to 500mA, and it will adjust the voltage down until the led load pulls 500mA. If that means 12V, it will adjust down to that.

• Which means that the LED driver bucks up voltage till the Diode draws allowable current? Jul 12 '16 at 3:20
• @SaiArun it attempts to, within its current and voltage range. And depending if it is a buck or boost or linear type driver. Jul 12 '16 at 3:54
• thanks . This is the info which I was searching for Jul 12 '16 at 4:39

How can you expect that using more voltage will allow you lesser current flow?

If I agree, a diode does not obey the linear (proportional) relationship of Ohms Law, still, increase of voltage (Volts) increase current (flow) (Ampere) in a non-linear manner as shown in USER @DarioP 's answer.

If you're thinking about a current-limiting resistor in series with that LED-light (as done indeed) , that is, at-the-same-time also decreasing voltage (or in simple-words, electrical-pressure), to that individual LED. Though to the whole series-combined circuit's 2 terminus, the total applied voltage remain (almost) same.

However the question asked, is conceptual and basic.

• I am not running LEDs barely. I am using an adjustable led driver to control input current Jul 12 '16 at 3:15
• This huge image illustrates Ohm's law, and is as such not related to the question. Not sure why you wanted it here, it distracts from the flow without adding anything. See the answer from DarioP which you refer to, on how to use illustrations to add value to an answer.
– pipe
Jul 12 '16 at 6:16
• @pipe I used this diagram is not for describing Ohm's law. I just meant, in a diode also, increasing volt would increase current. (However there could be exceptions such as a tunnel-diode 's negative differential conductance range). However, since an LED follows a graph as by DarioP , and it follows the increase-push increased-flow behavior, so, if the LED gives 0.5A at 12V, then the LED how can give 0.3157A ( < 5A) at 19V ( >5V) and to mean that, I used that diagram. Jul 12 '16 at 14:54
• @SaiArun , Are you sure, in your power-supply device, you can change / adjust the value of current (say 0.3157A you trying to make) (>1 values of current) on the fixed voltage (19V)? You could check one thing. Do you have a multimeter? then you could set the LED at certain tolerable-voltage, and with its parallel, in the voltmeter setup. Now turn the knob for Ampere, and look is there any change in voltage reading. Btw I've seen powerpacks used in molecular biology lab, that contained 2 adjusting knobs (V and A) with associated meters built-in. While adjusting one, the other- meter changed. Jul 12 '16 at 15:15
• here we're talking about "input" voltage to that power-supply device vs "output"-current from that device?? we're not talking about "output" voltage and "output" current?? (after reading datasheet (people.xiph.org/~xiphmont/thinkpad/PT4115E.pdf) and others' comments) Jul 12 '16 at 16:22

If, in fact, the lamp is designed to dissipate 6 watts when there's 12 volts across it, then there's:

$$I = \frac{P}{E} = \frac{6W}{12V} = 0.5 \text{ amperes}$$

through it.

Knowing that, if you want to run the lamp from a 19 volt supply, all that needs to be done is to drop 7 volts through a resistor connected in series between the lamp and the supply. To figure out the value of that resistor we write:

$$R = \frac{V_S-V_{LED}}{I_{LED}} = \frac{19V-12V}{0.5A} = 14 \text{ ohms}$$

That's a standard 5% value, and the resistor will be dissipating $$P = IE = 7V \times 0.5A = 3.5\text{ watts,}$$ so a 14 ohm 5 watt resistor would be a good choice.

• -1 for a good answer? I guess, then, <shrugs> a great one would be worth -10. Jul 14 '16 at 17:09