# Lighting Project

#### Butterworth

Joined May 6, 2009
135
Hi Everyone, I am working on a very small LED lighting project that I am trying to gain as much luminance as possible with the least amount of components.

The application is for a large metal mask that is to lit up from behind. The mark is about 16" wide x 18" tall.

I am planning to use the proposed pcb layout to accomplish my task. The SMD's used are white 1.2mm LED's that put out 160mca @ 10mA, they work off of 3.5-4V.

Using ohms law, plugging in 3.5V @ 10mA = 350Ω @ 35mW (per LED).

Using 2 leds, 7V @ 20mA = 350Ω @ 140mW (total). I assume twice the W rating for my resistors, and since I am only @ 140mW, the closest rating to 1/4W is 1/2W, this is what I will be using.

I ask that this math be checked as I am a first timer for LED circuits. The attached layout is my proposed PCB for the project.

If anyone has insight to a brighter / compact / low power update to my current design, by all means let me know.

Thanks in advance everyone.

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Joined Dec 26, 2010
2,148
No, the resistor does not need to be 350Ω. The fact that this value is given by dividing the LED voltage by the current is not relevant.

In addition, the resistor power rating is not necessarily equal to the LED power. This would only be so if the resistor voltage was exactly equal to the total of the two LED voltages, which with two 3.5V devices could only be so if you had about a 14V supply.

Note that the sum of emfs and voltage drops around any closed loop is zero (except in very special cases which really do not concern us here).
Therefore, if you have a 9V supply, and two LEDs in series which each have Vf = 3.5V,the voltage across the resistor is 9V-2*3.5V = 2V.

The resistor required is therefore 2V/10mA = 200Ω. If you can only get E12 values, you night select 220Ω. If the LEDs are safe with a little more current (many, but not all can take 20mA), you might use a slightly lower value like 180Ω, to allow for the fact that the battery voltage drops with use, and that the forward voltage could be a bit over 3.5V.

The resistor will be dropping 2V at about 10mA, so it will dissipate about 20mW. Even if it is a bit more, a 125mW (1/8 W) resistor would do. Naturally a 1/4 W or even 1/2 W resistor would also be fine if that was all you had.

EDIT: I see you do not have the LEDs in series: I had assumed this because you mentioned 7V. Details for the single LED case to follow.

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#### nickelflipper

Joined Jun 2, 2010
280
I saw these small surface mount leds by Osram referenced on another site. It's hard to imagine anything coming close in terms of efficiency and brightness in this small of a package.

Joined Dec 26, 2010
2,148
Using the two LEDs in series with one resistor uses less current, but the LEDs will dim fairly fast as the battery voltage falls.

Using two branches each of a single LED in series with a resistor across the battery will use more current, but will allow the battery to be used down to a lower voltage before the LEDs dim too far. I am not sure which would last the longer!

Your reference to 7V confused me: with the LEDs not in series, their voltages do not add. Anyway, with 9V Vs and 3.5V Vf and 10mA required we have R = (9V-3.5V)/0.01A = 550Ω.

You might use 560Ω, or again if the LEDs can handle a bit more than 10mA consider maybe 470Ω (≈12mA at 9V). The resistor dissipation for 560Ω is (9V-3.5V)*10mA = 55mW, or 64mW for 470Ω. Again, a 1/8 W resistor would do

#### Butterworth

Joined May 6, 2009
135
No, the resistor does not need to be 350Ω. The fact that this value is given by dividing the LED voltage by the current is not relevant.

In addition, the resistor power rating is not necessarily equal to the LED power. This would only be so if the resistor voltage was exactly equal to the total of the two LED voltages, which with two 3.5V devices could only be so if you had about a 14V supply.

Note that the sum of emfs and voltage drops around any closed loop is zero (except in very special cases which really do not concern us here).
Therefore, if you have a 9V supply, and two LEDs in series which each have Vf = 3.5V,the voltage across the resistor is 9V-2*3.5V = 2V.

The resistor required is therefore 2V/10mA = 200Ω. If you can only get E12 values, you night select 220Ω. If the LEDs are safe with a little more current (many, but not all can take 20mA), you might use a slightly lower value like 180Ω, to allow for the fact that the battery voltage drops with use, and that the forward voltage could be a bit over 3.5V.

The resistor will be dropping 2V at about 10mA, so it will dissipate about 20mW. Even if it is a bit more, a 125mW (1/8 W) resistor would do. Naturally a 1/4 W or even 1/2 W resistor would also be fine if that was all you had.

EDIT: I see you do not have the LEDs in series: I had assumed this because you mentioned 7V. Details for the single LED case to follow.
OK, so I had the calculations mixed up. I used your example to come up with the following (based on another LED I found):
6V-(1x4V) = 2V
2V/20mA = 100Ω

Single LED specs: 10mm White 20,000mcd / 4.0V @ 20mA

I plan to use only this single LED, it should throw a lot more light than the SMD model. The 6V supply is based on convenience of the battery holder (4x1.5V AA). I could use a 9V but the mAh rating isn't as much.

I saw these small surface mount leds by Osram referenced on another site. It's hard to imagine anything coming close in terms of efficiency and brightness in this small of a package.
I will consider this, thank you!

Using the two LEDs in series with one resistor uses less current, but the LEDs will dim fairly fast as the battery voltage falls.

Using two branches each of a single LED in series with a resistor across the battery will use more current, but will allow the battery to be used down to a lower voltage before the LEDs dim too far. I am not sure which would last the longer!

Your reference to 7V confused me: with the LEDs not in series, their voltages do not add. Anyway, with 9V Vs and 3.5V Vf and 10mA required we have R = (9V-3.5V)/0.01A = 550Ω.

You might use 560Ω, or again if the LEDs can handle a bit more than 10mA consider maybe 470Ω (≈12mA at 9V). The resistor dissipation for 560Ω is (9V-3.5V)*10mA = 55mW, or 64mW for 470Ω. Again, a 1/8 W resistor would do
Sorry about the confusion, the 7V was the Vf from the 2 LED's. I had misinterpreted the equation. (I'm still learning)

I don't get what you mean about "Using two branches each of a single LED in series with a resistor across the battery"
See my example attached of what I think you mean.

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Joined Dec 26, 2010
2,148
OK, so I had the calculations mixed up. I used your example to come up with the following (based on another LED I found):
6V-(1x4V) = 2V
2V/20mA = 100Ω

Single LED specs: 10mm White 20,000mcd / 4.0V @ 20mA

I plan to use only this single LED, it should throw a lot more light than the SMD model. The 6V supply is based on convenience of the battery holder (4x1.5V AA). I could use a 9V but the mAh rating isn't as much.

I will consider this, thank you!

Sorry about the confusion, the 7V was the Vf from the 2 LED's. I had misinterpreted the equation. (I'm still learning)

I don't get what you mean about "Using two branches each of a single LED in series with a resistor across the battery"
See my example attached of what I think you mean.
No, that diagram does show two LEDs in series, with a single resistor, so 7V would be relevant. Note also that your switch will not work as shown, you need to connect the middle terminal somewhere.

You really need to get the meanings of series and parallel sorted out in your mind or it will be hard to discuss this meaningfully.

Series connections are along a single path. The configuration I was referring to is what seems to be on your PCB drawing. I will have another look at that and try to draw a diagram for you.

#### Butterworth

Joined May 6, 2009
135
No, that diagram does show two LEDs in series, with a single resistor, so 7V would be relevant. Note also that your switch will not work as shown, you need to connect the middle terminal somewhere.

You really need to get the meanings of series and parallel sorted out in your mind or it will be hard to discuss this meaningfully.

Series connections are along a single path. The configuration I was referring to is what seems to be on your PCB drawing. I will have another look at that and try to draw a diagram for you.
I understand the differences of parallel & series, I was just trying to understand your wording to this application. Please show me a diagram of your example?

Joined Dec 26, 2010
2,148
Here are some diagrams from a simulation which may help your understanding.

Note that with the single branch case the battery current and the LED currents are identical. With two branches the battery current is twice the LED current.

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Joined Dec 26, 2010
2,148
I agree that four cells in a holder will be better if the LEDs are not in series pairs. Less wasteful on voltage than 9V, and as you say more capacity, at least if the cells are bigger than the AAAA equivalents normally used in small 9V batteries.

#### Butterworth

Joined May 6, 2009
135
Here are some diagrams from a simulation which may help your understanding.

Note that with the single branch case the battery current and the LED currents are identical. With two branches the battery current is twice the LED current.
OK that makes sense to me. Thank you for the graphical comparison. If I plan to use the single LED in series with the resistor, will it come out the same as the first graph?

I agree that four cells in a holder will be better if the LEDs are not in series pairs. Less wasteful on voltage than 9V, and as you say more capacity, at least if the cells are bigger than the AAAA equivalents normally used in small 9V batteries.
My plan was to use those rechargable NiMH batteries with 800mAh or more in them, which would give a lot of intermitent usage, I would think.

Joined Dec 26, 2010
2,148
That sounds OK: with 4 NiMH cells you probably won't get quite 6V even when newly charged, and it will be around 5V or a little less for much of its life, but that's not such a bad thing. You might like to lower the resistor a bit to allow for that, provided you are sure that alkaline batteries will never be used instead - the latter could give about 6.4V brand-new (admittedly, not for long).

#### mcgyvr

Joined Oct 15, 2009
5,394
Did you actually proto this first to make sure it was even enough light for your needs? Just trying to save you from finding out too late and having to junk the PCB's you just payed good money for.

Also any plans for optics for these LED's? Or sight shields,etc.. I'd hate that you caused vision problems because someone could see through parts of the mask and was just staring into the LED.. Many LED's are powerful enough now to cause serious vision issues.

#### Butterworth

Joined May 6, 2009
135
Did you actually proto this first to make sure it was even enough light for your needs? Just trying to save you from finding out too late and having to junk the PCB's you just payed good money for.

Also any plans for optics for these LED's? Or sight shields,etc.. I'd hate that you caused vision problems because someone could see through parts of the mask and was just staring into the LED.. Many LED's are powerful enough now to cause serious vision issues.
Well I was thinking about using an old flashlight reflector to bounce the light off of, almost how you see a dentist's light. If I can find an old flashlight that can fit this enormous led, (it measures almost 1/2" across) I think it will do nicely. It does throw a lot of light, but it needs to be controlled to meet my needs. I am also using a breadboard and a spare proto-board I had laying around. I am trying out the single LED with 12V (8AA cells) and 18V (2-9V cells). The difference was quite amazing once I passed the 9V mark, the LED jumped in brightness substantially.

That sounds OK: with 4 NiMH cells you probably won't get quite 6V even when newly charged, and it will be around 5V or a little less for much of its life, but that's not such a bad thing. You might like to lower the resistor a bit to allow for that, provided you are sure that alkaline batteries will never be used instead - the latter could give about 6.4V brand-new (admittedly, not for long).
I did find out that the 4 newly charged AA's gave about 6.6V, but for how long right? If I use the 1.2V rechargables, I should hope to get about 9.6V out of them. Though this isn't as high as I would like to use given the statement above about the significant increase over 9V, but at least it will give the flexibility of using either 1.5V or 1.2V AA's.

On a side note, will making a constant current design help with the luminence of the LED? or will it just improve the overall stability of the lighting?

*EDIT* This might be too simple of a solution to be true, but why can't I use a Zener Diode to limit the voltage to exactly 4V for the LED? Will it work? or will is not work due to the LED being current driven?

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#### Audioguru

Joined Dec 20, 2007
11,249
A white LED with a very bright rating of 20000mcd was mentioned.
Using the reflector from a flashlight was also mentioned.
But a very bright LED has a very narrow beam (almost like a laser) and a reflector will not work.

#### Audioguru

Joined Dec 20, 2007
11,249
*EDIT* This might be too simple of a solution to be true, but why can't I use a Zener Diode to limit the voltage to exactly 4V for the LED? Will it work? or will is not work due to the LED being current driven?
You cannot buy a "4.0V LED". LEDs have a range of forward voltage. Maybe 3.2V to 4.5V. the forward voltage also changes with temperature change.

An LED is lighted with a limited current (NOT a voltage). The current limiter can be a series resistor or a complicated circuit. The current limiter uses voltage so extra voltage must be supplied to the LED plus current limiter.

#### Butterworth

Joined May 6, 2009
135
A white LED with a very bright rating of 20000mcd was mentioned.
Using the reflector from a flashlight was also mentioned.
But a very bright LED has a very narrow beam (almost like a laser) and a reflector will not work.
My idea was to try and aim the LED at the reflector, similar to the light your dentist may use. The LED is then bounced off of the reflector giving a wider beam, the intention was to create more lighting as an ambient rather than a focused beam, but I don't know how wide of an angle I can get out of a pre-fabricated reflector... Any ideas on what to use?

You cannot buy a "4.0V LED". LEDs have a range of forward voltage. Maybe 3.2V to 4.5V. the forward voltage also changes with temperature change.

An LED is lighted with a limited current (NOT a voltage). The current limiter can be a series resistor or a complicated circuit. The current limiter uses voltage so extra voltage must be supplied to the LED plus current limiter.
I agree, the Vf was stated @ 4V, so unless I read the spec wrong, I do not know what the range is, this is why I was inquiring about the 4V only.

Joined Dec 26, 2010
2,148
Oh dear, this old chestnut again: how many times will it come back? It's not your fault of course, it seems to be a very widespread difficulty. Perhaps there should be a "sticky" about it, or maybe there already is. I don't know, but here goes. I will borrow some text from something I posted the other day, when I was talking about one LED in series with two resistors:

...For an LED, and most other diodes for that matter, Vf does not vary much as the forward current changes over quite a wide range. Typically we can assume it to have a constant value, except in quite accurate work, or where the current will be quite a long way from the normal operating level. The Vf is different for different designs of LED, in particular it is higher for blue or white devices, less for longer wavelengths. It also does vary a bit between devices, and generally it falls with increasing temperature.

The near-constant but not entirely predictable value of Vf is why LEDS almost always have to be used with series resistors or more elaborate constant-current drives (too little current = too dim, too much current = short life). We may see cheap lamps that manage to run LEDs directly from small batteries, but such arrangements tend not to get the best out of the battery life. The brightness may also be very variable, and worst of all the LEDs may even burn out if there is an unfavourable combination of low Vf and high battery voltage. A series resistance always wastes some power, but this is usually preferable to having an unpredictable current...

The point is that constant-voltage supplies really do not suit LEDs. It is always highly advisable (many would say mandatory) to use resistors or other current-determiners in association with LEDs. This is the source of a lot of trouble for beginners who try to run LEDs directly from voltage supplies, with zero or minimal resistance.

#### nickelflipper

Joined Jun 2, 2010
280
My idea was to try and aim the LED at the reflector, similar to the light your dentist may use. The LED is then bounced off of the reflector giving a wider beam, the intention was to create more lighting as an ambient rather than a focused beam, but I don't know how wide of an angle I can get out of a pre-fabricated reflector... Any ideas on what to use?

I agree, the Vf was stated @ 4V, so unless I read the spec wrong, I do not know what the range is, this is why I was inquiring about the 4V only.
Use a led with a wide viewing angle to start with, say 120 degree. Even then, you will need several of them, one led isn't going to light up a 15"x18" mask, unless the focal point is wayyyyy back.

#### Butterworth

Joined May 6, 2009
135
Oh dear, this old chestnut again: how many times will it come back? It's not your fault of course, it seems to be a very widespread difficulty. Perhaps there should be a "sticky" about it, or maybe there already is. I don't know, but here goes. I will borrow some text from something I posted the other day, when I was talking about one LED in series with two resistors:

...For an LED, and most other diodes for that matter, Vf does not vary much as the forward current changes over quite a wide range. Typically we can assume it to have a constant value, except in quite accurate work, or where the current will be quite a long way from the normal operating level. The Vf is different for different designs of LED, in particular it is higher for blue or white devices, less for longer wavelengths. It also does vary a bit between devices, and generally it falls with increasing temperature.

The near-constant but not entirely predictable value of Vf is why LEDS almost always have to be used with series resistors or more elaborate constant-current drives (too little current = too dim, too much current = short life). We may see cheap lamps that manage to run LEDs directly from small batteries, but such arrangements tend not to get the best out of the battery life. The brightness may also be very variable, and worst of all the LEDs may even burn out if there is an unfavourable combination of low Vf and high battery voltage. A series resistance always wastes some power, but this is usually preferable to having an unpredictable current...

The point is that constant-voltage supplies really do not suit LEDs. It is always highly advisable (many would say mandatory) to use resistors or other current-determiners in association with LEDs. This is the source of a lot of trouble for beginners who try to run LEDs directly from voltage supplies, with zero or minimal resistance.
I understand. I was just curious. Thank you for the explanation & sorry for dredging up the past. lol

Use a led with a wide viewing angle to start with, say 120 degree. Even then, you will need several of them, one led isn't going to light up a 15"x18" mask, unless the focal point is wayyyyy back.
• The LED in question is almost 1/2" in diameter. It does have a rather tight beam, but it also throws a lot of ambient light. I cannot use mains power with a transformer, so battery power is essential, using about 3 of the proposed lights will give the following: 14.4V-(3*4V) = 2.4V | 2.4V/20mA = 120Ω @ 48mW.
• Asusming that the (12x1.2V AA's) are at least 800mAh each, the collective mAh rating of 12x1.2V AA's would be 9,600mAh. If I take the collective 60mA from 3 LED's, that sould give me 160h of use? It seems a bit much to me, I know its simple math but wow.
• Would not a home-made reflector be good? I could make a cheapo panel reflector or spherical-ish shape with carboard & alum foil (or any other reflective substance).

#### Audioguru

Joined Dec 20, 2007
11,249
Your LEDs have a range of voltages and if they are actually 3.2V then their current will be 14.4V-(3*3.2V) = 4.8V/120Ω= 40mA but will be 70mA when the battery is fresh out of the charger and they will burn out very soon.
If their voltage is actually 4.5V then they will be dim at only 7.5mA.

The difference is less if more voltage is allowed across the current-limiting resistor or if a current regulator circuit is used.

The datasheet for a battery cell shows that its mAh is rated when its voltage drops very low but then the LEDs might not produce any light.

You need a diffuser (the opposite of a reflector) to spread out the narrow beam of your LEDs. The light is less bright when spread out.