# Embarrassing question. What resistance to make this lamp/led brighter?

#### Videodrome

Joined Nov 12, 2009
64
i am using these dialight 610-2221-120f 6v lamps from a carboot sale I picked up and have already burned through a pair in getting them to be brighter with what I thought was the proper resistance. Powering via 9vdc battery for now, need to be used for a 9-12vdc circuit). The specs on them can be seen here.

https://www.mouser.com/datasheet/2/109/diac_s_a0008883141_1-2265592.pdf

for now they are just too dim for my needs and hope I can brighten them up a bit but could anyone be so kind as to substantiate what value resistance I would need to brighten up? thanks for any help on my embarrassing question here.

#### ElectricSpidey

Joined Dec 2, 2017
2,144
You are going to have to swap out the LED with brighter ones and use the correct resistors/drivers for the new LEDs.

#### SamR

Joined Mar 19, 2019
4,132
You have to remember that LEDs are current devices, not voltage. The data sheet says 15mA and I have to assume that is MAX since it is not designated otherwise. So your I = V/R applies with I being less than 15mA.

#### dl324

Joined Mar 30, 2015
14,332
Powering via 9vdc battery for now, need to be used for a 9-12vdc circuit). The specs on them can be seen here.
Why are you operating an LED rated for 6V at 9V? That's a good way to kill things.

Assuming they're spec'ed to operate at 6V because they have an integral current limiting resistor... If we assume an LED forward voltage of 2V, Ohm's law tells us that the series resistance would be about 270 ohms. If you wanted to operate at 12V and maintain the same current, you'd need an additional resistance of 390 ohms.

$$R = \frac{V}{I} = \frac{6V-2V}{15mA} = 267\Omega$$

To maintain the current at 15mA with a 12V supply:
$$R_{tot} = \frac{12V-2V}{15mA} = 667\Omega$$

If you want a constant current with 9-12V, use a current source.

for now they are just too dim
Use a different indicator.

#### ElectricSpidey

Joined Dec 2, 2017
2,144
Looking at the mcd ratings for those LEDs I would say they are not going to get much brighter before they burn out.

#### sparky 1

Joined Nov 3, 2018
706
If I understand correctly you are working with a 9.5V battery when the battery gets down to 7.5V the light is too dim.
You would like an efficient way to maintain same brightness that you get when the battery is new ?
Possibly constant current would work also leds will also run on AC however there may be an easier way.
Keeping it simple, a small 5 or 6V linear voltage regulator rated at 200mA would deal with battery voltage drop.
Once the voltage is fixed at say 5V the resistor would maintain the same current. No question should be embarrassing.
An led calculator that even draws the circuit:
https://ledcalculator.net

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#### MisterBill2

Joined Jan 23, 2018
11,624
The data sheet states that these are 6 volt devices and so the external resistor will need to be R= V/I=6V/0.015Amp. If you are using two of them, simply put them in series and they should each have the correct six volts to operate as intended. No external resistor needed, just use two in series.
But these devices in the data listing are indicator lights, not suitable for providing useful illumination.

#### ThePanMan

Joined Mar 13, 2020
359
I'm going to make a leap of faith and assume the lights have their own internal resistors limiting them to 15mA. Depending on the color (or forward voltage) of the LED, the internal resistor would have been selected to provide 15mA at 6V. That's 180mW. If you up the voltage to 9V then, assuming some internal values for the 15mA on a 2Vf LED, there should be a 267Ω resistor internally. Under that assumption, (9V - 2Vf)/267Ω = 26mA (on a 9V battery). That's a lot hotter than 15mA. Still, 9V x 0.026A = 236mW. You're very close to failure. Especially if these LED's are the common 5mm types with average operating ranges of amperages of 5 to 20mA. Some can be run at 30mA, but they won't last long running that brightly. Mostly because of the heat. The hotter they get the faster they degrade and begin to cascade into failure. If you pumped one with 12 volts - I bet you got a flash out of it. Maybe it lit for a whole second before it was too dim to see, then in an instant later it was completely dead.

You don't ADD resistance to increase the brightness, you DECREASE resistance. And without opening those lamps up you're not going to be able to change the resistance. The only other way to increase brightness is to increase the voltage. At 9V you're probably real close to failure.

IF you want to run it on 12V then, again assuming, you would need to increase the total resistance to (12V -2Vf)/20mA=500Ω total. IF we assume there's already 267Ω inside then an external series resistor of 233Ω will get you a little brighter LED that might survive. But in a car - voltages will fluctuate all the way up to 14.5 volts in some circumstances. Your lamp will die in the car. Before I go making any further assumptions - where do you plan on using these indicators?

#### ThePanMan

Joined Mar 13, 2020
359
Where did you go? Haven't heard from the TS since Wednesday. Maybe you guys scared him away.

#### MisterBill2

Joined Jan 23, 2018
11,624
I'm going to make a leap of faith and assume the lights have their own internal resistors limiting them to 15mA. Depending on the color (or forward voltage) of the LED, the internal resistor would have been selected to provide 15mA at 6V. That's 180mW. If you up the voltage to 9V then, assuming some internal values for the 15mA on a 2Vf LED, there should be a 267Ω resistor internally. Under that assumption, (9V - 2Vf)/267Ω = 26mA (on a 9V battery). That's a lot hotter than 15mA. Still, 9V x 0.026A = 236mW. You're very close to failure. Especially if these LED's are the common 5mm types with average operating ranges of amperages of 5 to 20mA. Some can be run at 30mA, but they won't last long running that brightly. Mostly because of the heat. The hotter they get the faster they degrade and begin to cascade into failure. If you pumped one with 12 volts - I bet you got a flash out of it. Maybe it lit for a whole second before it was too dim to see, then in an instant later it was completely dead.

You don't ADD resistance to increase the brightness, you DECREASE resistance. And without opening those lamps up you're not going to be able to change the resistance. The only other way to increase brightness is to increase the voltage. At 9V you're probably real close to failure.

IF you want to run it on 12V then, again assuming, you would need to increase the total resistance to (12V -2Vf)/20mA=500Ω total. IF we assume there's already 267Ω inside then an external series resistor of 233Ω will get you a little brighter LED that might survive. But in a car - voltages will fluctuate all the way up to 14.5 volts in some circumstances. Your lamp will die in the car. Before I go making any further assumptions - where do you plan on using these indicators?
The TS, in post #1, mentioned using them as lights in the boot of their car. (it must be a british car, in the US cars have trunks.)And the last car that I had with a trunk, the light ran on 12 volts and drew about an amp. But it did provide adequate illumination. An indicator light would look nicer but be almost worthless for finding stuff in a trunk on a dark night.

#### ThePanMan

Joined Mar 13, 2020
359
The TS, in post #1, mentioned using them as lights in the boot of their car.
i am using these dialight 610-2221-120f 6v lamps from a carboot sale I picked up and have already burned through a pair in getting them to be brighter with what I thought was the proper resistance. Powering via 9vdc battery for now, need to be used for a 9-12vdc circuit). The specs on them can be seen here.
Bold and underlined my edit to the TS quoted comment.

I'm not sure that's what he's doing. You could be right, but that wasn't the way I took it. Yeah, British calls the trunk "The Boot". No problem with that. But the quote is "Carboot" - one word. Not sure of the TS meaning. Anyway, if it IS a trunk (or boot) and the TS has a 12 volt system he can get one of those LED lamps that replace existing lights with matching sockets. No need to make anything fancy like colored lights.

#### MisterBill2

Joined Jan 23, 2018
11,624
OK, I was not reading quite closely enough. But still the complaint that it was not bright enough. And one response was calculating a resistor for a 2 volt LED when the thing is identified as a 6 volt assembly drawing 15 mA. In that case the resistance is 6V/ 0.015A=400 ohms, so select a 390 ohm resistor, or possibly a 430 ohm one. .
Unfortunately the TS apparently does not see how ohms law applies here. Frustration .

#### Tonyr1084

Joined Sep 24, 2015
6,861
OK, I was not reading quite closely enough. But still the complaint that it was not bright enough. And one response was calculating a resistor for a 2 volt LED when the thing is identified as a 6 volt assembly drawing 15 mA. In that case the resistance is 6V/ 0.015A=400 ohms, so select a 390 ohm resistor, or possibly a 430 ohm one. .
Unfortunately the TS apparently does not see how ohms law applies here. Frustration .
Perhaps another read, closer examination would reveal that we don't know what the forward voltage of the LED is. We're making an assumption and working from an assumed Vf. And when calculating the proper resistance for an LED you have to start by subtracting the LED Vf from the source voltage. So when those guys calculated a 6V supply powering an LED that "May Be" 2Vf, 2 from 6 is 4. The resistor selected has to drop four volts at 15mA. Therefore, 4/0.015=266.67Ω. One of the most basic calculations for lighting LED's. The whole point of making an assumption is to demonstrate the proper method of determining what resistance is needed. SourceV minus Vf FIRST. Then divide the result by the desired current to obtain the correct resistor. With higher voltages subtracting the Vf has a smaller impact on the calculations than it does when starting with a low voltage. The same 2Vf LED powered from 24 volts (as an example) will calculate as follows: (24V - 2Vf) / 0.015A = 1466.67Ω. Without subtracting the Vf you get 1.6KΩ. If you don't subtract the Vf and use the 1.6KΩ on the LED you'll get 16.4mA (0.0164A) Less than 1 1/2 mA more. But when you start with a 6V source using a 400Ω resistor, ignoring the Vf you get 15mA (as you stated). BUT when you factor in the 2Vf, 4V/400=10mA. That's a third less amperage.

I'll restate my point this way: Ignoring the Vf on a low voltage circuit has a much bigger impact on brightness than if you ignore the Vf on a much higher voltage circuit.

#### Tonyr1084

Joined Sep 24, 2015
6,861
I'm not seeing any responses from the TS. Looks like we're arguing amongst ourselves over assumed values. Bottom line, if the TS isn't interested in this problem then I don't know why we are. I don't even know what a "Carboot sale" is.
lamps from a carboot sale
What is that? Like a "Swapmeet"? (common in America, probably goes by other names in other parts of the world). Was this something someone sold him out of their trunk (boot)? Who knows what he actually got. May be some companies factory rejects. Quick sale out of the trunk and away you go with no recourse for the buyer.

#### dl324

Joined Mar 30, 2015
14,332
What is that? Like a "Swapmeet"? (common in America, probably goes by other names in other parts of the world). Was this something someone sold him out of their trunk (boot)?
I took it to mean garage sale...

#### Tonyr1084

Joined Sep 24, 2015
6,861

#### MisterBill2

Joined Jan 23, 2018
11,624
WE are not guessing. The data sheet for that light states 6.0 volts and 15 mA, and since that is an integral LED light that is probably what we have. If the TS got a few of them then two in series should be fine for 12 volts, and that 400 ohms is the correct calculated series resistor for 12 volt operation. Reading is a very useful skill, even when it goofs up a small bit.

#### ThePanMan

Joined Mar 13, 2020
359
WE are not guessing. The data sheet for that light states 6.0 volts and 15 mA
OK, so what's the forward voltage of the LED?

First, we don't know what color the LED is.
Second, we don't know what the forward voltage is.
Third, if it's going to be powered by 12 volts - we don't know under what conditions.
Is it automotive? Is it something else? Will the trunk (if that's the application) be opened only when the engine is not running? Or is there a chance the engine might be running at the time? In an automotive application you can be certain that the voltage will be uncertain. You can assume a range, but now we're back to guesswork.
Finally, how would you calculate a 5mm red LED with a Vf of 2 volts to run on 15mA? Now's your chance to shine.

#### MisterBill2

Joined Jan 23, 2018
11,624
OK, so what's the forward voltage of the LED?

First, we don't know what color the LED is.
Second, we don't know what the forward voltage is.
Third, if it's going to be powered by 12 volts - we don't know under what conditions.
Is it automotive? Is it something else? Will the trunk (if that's the application) be opened only when the engine is not running? Or is there a chance the engine might be running at the time? In an automotive application you can be certain that the voltage will be uncertain. You can assume a range, but now we're back to guesswork.
Finally, how would you calculate a 5mm red LED with a Vf of 2 volts to run on 15mA? Now's your chance to shine.
What we know from reading the data sheet is that the pilot lamp assembly, which undoubtedly includes an internal resistor, is specified for 6.0 volts So there is no need to calculate the dropping resistor if we apply 6 volts. It is an assembly, not some random LED stuck in a holder. Designers use them because they save labor and are consistent.

#### ThePanMan

Joined Mar 13, 2020
359
there is no need to calculate the dropping resistor if we apply 6 volts.
The TS wants to connect it to 12 volts. A.K.A. possibly an automobile. In which case, running a 6V LED on 12V will require additional resistance. However, one must take into account the Vf and internal resistance when adding an external resistor. Or am I the only one who took it that way?