-DIY- Powering a 3 volt LED from a 4.5V supply

[T800mc]

Hello,

I'm wanting to power 1 x 3v led from a 4.5v power supply and 2 x 1.5v leds from another 4.5v supply.
Can anyone advise what resistor I have to install on each one. Using an online calculator I'm getting 82 Ohm on a calculator. is that correct? Is there anything else I need to take into account?

Thanks

 

[Lucien Nunes]

The actual chip in an LED requires the current to be set by the resistor, not the voltage. You feed in a known current, and the voltage drop that occurs across it is whatever it is, usually close to the stated value. This is due to the exponential relationship between current and voltage in a semiconductor.

To calculate the resistor value, you first need to decide what current is to flow, which depends on the desired brightness, whether power saving is important e.g. in a battery application, the efficiency of the LED, and of course must be within the LED's rated maximum. A typical current for an indicator LED is 20mA, a high-efficiency one might work fine on 1mA.

Then, subtract the forward voltage (Vf) from the supply voltage (Vs) and divide by the current (If).

If Vs=4.5V, Vf=3V, If=20mA
R = (Vs - Vf) / If
= (4.5 - 3) / 0.02
=75Ω

 

[T800mc]

The actual chip in an LED requires the current to be set by the resistor, not the voltage. You feed in a known current, and the voltage drop that occurs across it is whatever it is, usually close to the stated value. This is due to the exponential relationship between current and voltage in a semiconductor.

To calculate the resistor value, you first need to decide what current is to flow, which depends on the desired brightness, whether power saving is important e.g. in a battery application, the efficiency of the LED, and of course must be within the LED's rated maximum. A typical current for an indicator LED is 20mA, a high-efficiency one might work fine on 1mA.

Then, subtract the forward voltage (Vf) from the supply voltage (Vs) and divide by the current (If).

If Vs=4.5V, Vf=3V, If=20mA
R = (Vs - Vf) / If
= (4.5 - 3) / 0.02
=75Ω

Thanks for the input. That lines up with the calculator I used. It also said "Nearest higher rated resistor" as 82 Ohms. Is there a recommendation for having a higher rated one?

 

[marconi]

Re #1: Is there anything else I need to take into account?

Once you have calculated the required resistance value, the next thing to work out is the power rating of the resistor. Because a current is flowing through the resistance there will be Ohmic heating inside it - so it will warm up and this heat must be lost to the surroundings. If the resistor warmed up too much its value would increase and may be damaged too.

The power dissipated P is calculated by P = Isquared x R where I is the current through the resistor and R is its value.

In your case P = 0.02 x 0.02 x 75 = 0.03Watts

You should select a 0.25Watt rated resistor because it is greater than 0.03W and one of the standard power ratings for axial resistors.

 

[Lucien Nunes]

By 'higher rating' we usually mean power rating, not resistance, which is normally referred to as the 'value.' The higher the resistance, the lower the current. You'll hardly notice the difference here, but in any case you will probably want to assess the brightness and adjust accordingly. 20mA is just a starting point for vanilla-flavour LEDs. If you choose a high efficiency type you might be OK on 2mA, in which case 750 ohms is the value you need.

 

[T800mc]

Re #1: Is there anything else I need to take into account?

Once you have calculated the required resistance value, the next thing to work out is the power rating of the resistor. Because a current is flowing through the resistance there will be Ohmic heating inside it - so it will warm up and this heat must be lost to the surroundings. If the resistor warmed up too much its value would increase and may be damaged too.

The power dissipated P is calculated by P = Isquared x R where I is the current through the resistor and R is its value.

In your case P = 0.02 x 0.02 x 75 = 0.03Watts

You should select a 0.25Watt rated resistor because it is greater than 0.03W and one of the standard power ratings for axial resistors.

Thank you for that.
[automerge]1597131115[/automerge]

By 'higher rating' we usually mean power rating, not resistance, which is normally referred to as the 'value.' The higher the resistance, the lower the current. You'll hardly notice the difference here, but in any case you will probably want to assess the brightness and adjust accordingly. 20mA is just a starting point for vanilla-flavour LEDs. If you choose a high efficiency type you might be OK on 2mA, in which case 750 ohms is the value you need.

[automerge]1597131289[/automerge]

By 'higher rating' we usually mean power rating, not resistance, which is normally referred to as the 'value.' The higher the resistance, the lower the current. You'll hardly notice the difference here, but in any case you will probably want to assess the brightness and adjust accordingly. 20mA is just a starting point for vanilla-flavour LEDs. If you choose a high efficiency type you might be OK on 2mA, in which case 750 ohms is the value you need.

By 'higher rating' we usually mean power rating, not resistance, which is normally referred to as the 'value.' The higher the resistance, the lower the current. You'll hardly notice the difference here, but in any case you will probably want to assess the brightness and adjust accordingly. 20mA is just a starting point for vanilla-flavour LEDs. If you choose a high efficiency type you might be OK on 2mA, in which case 750 ohms is the value you need.

I honestly don't know what to do with this information. It's beyond me.

This is the LED:

Vishay3 V Green LED 3mm Through Hole, TLHG4900 | RS

uk.rs-online.com

This is the power supply:

http://www.soapstudio.com/index.php?route=product/product&path=477&product_id=645

It has 4.5v 0.03a output.

 

[marconi]

resistors | Rapid Online - https://www.rapidonline.com/Catalogue/Search?Query=resistors&Size=20&Attributes={%22Case%20Type%22:[%22Axial%22]}

Here is a link to a component supplier from which you can buy suitable resistors. You can use 'the select attributes' to search for your desired resistor.

You can buy leds too and obtain their datasheets to find out forward voltage (-changes with colour - why? - look up Fermi levels) and forward current - eg:

Kingbright | Rapid Online - https://www.rapidonline.com/brands/kingbright?ra_source=tier-page

 

[marconi]

https://www.vishay.com/docs/83009/tlhg4900.pdf

Studying page 2 of the reference at link, the tlhg4900 green led has a typical forward voltage of 2.4V and a forward current of typically 10mA.

From a 4.5V supply, the required series voltage dropper resistance to drop 2.1 V at 10mA is (4.5 - 2.4)/0.01 =

2.1/0.01 = 210 Ohms. Nearest value is 220 Ohms and at least 250mW

Buy some of these:

Royal Ohm CFR01SJ0221A10 220R 5% 1W Axial Carbon Film Resistor - https://www.rapidonline.com/royal-ohm-cfr01sj0221a10-220r-5-1w-axial-carbon-film-resistor-73-0065

and wire in series with the led and then connect to power supply. Your power supply delivers up to 30mA per output so in theory you could power three of these LEDs.from one output; the device has eight 4.5V 30mA outputs so better to spread your LED supplies over these outputs. In theory 24 such LEDs could be illuminated.

Do it like this noting that the longer lead of the LED connects to the + side of the battery. (In your case the 9V battery is one of the SOAPBOX PSU outputs and you will use a 220 Ohms resistor instead of the 470 one shown below):

 

[T800mc]

https://www.vishay.com/docs/83009/tlhg4900.pdf

Studying page 2 of the reference at link, the tlhg4900 green led has a typical forward voltage of 2.4V and a forward current of typically 10mA.

From a 4.5V supply, the required series voltage dropper resistance to drop 2.1 V at 10mA is (4.5 - 2.4)/0.01 =

2.1/0.01 = 210 Ohms. Nearest value is 220 Ohms and at least 250mW

Buy some of these:

Royal Ohm CFR01SJ0221A10 220R 5% 1W Axial Carbon Film Resistor - https://www.rapidonline.com/royal-ohm-cfr01sj0221a10-220r-5-1w-axial-carbon-film-resistor-73-0065

and wire in series with the led and then connect to power supply. Your power supply delivers up to 30mA per output so in theory you could power three of these LEDs.from one output; the device has eight 4.5V 30mA outputs so better to spread your LED supplies over these outputs. In theory 24 such LEDs could be illuminated.

Do it like this noting that the longer lead of the LED connects to the + side of the battery. (In your case the 9V battery is one of the SOAPBOX PSU outputs and you will use a 220 Ohms resistor instead of the 470 one shown below):

View attachment 60053

Okay great, thanks. I will power that LED from one output and use another output for the other two.
For the two other LEDs, I don't have much information other than they are normally powered by 2 x LR44 batteries for which I work out as 3v total so one battery per LED, can you advise what resistors to fit that cable?

 

[marconi]

It looks to me that the plug you need to insert into one of the 30mA sockets on the soapbox reactor is of this type:

dc plugs | Rapid Online - https://www.rapidonline.com/Catalogue/Search?Query=dc%20plugs&Size=20&Tier=Low%20Power

If you measure the external diameter and internal diameter you can then order them.

You can even buy pre-wired connectors:

wired low voltage connectors | Rapid Online - https://www.rapidonline.com/Catalogue/Search?Query=wired%20low%20voltage%20connectors&Size=20
[automerge]1597146035[/automerge]
What colour are these two leds and what size roughly?
[automerge]1597146726[/automerge]
If you send me a private message with your address I will post you some resistors gratis.

 

[T800mc]

It looks to me that the plug you need to insert into one of the 30mA sockets on the soapbox reactor is of this type:

dc plugs | Rapid Online - https://www.rapidonline.com/Catalogue/Search?Query=dc%20plugs&Size=20&Tier=Low%20Power

If you measure the external diameter and internal diameter you can then order them.

You can even buy pre-wired connectors:

wired low voltage connectors | Rapid Online - https://www.rapidonline.com/Catalogue/Search?Query=wired%20low%20voltage%20connectors&Size=20
[automerge]1597146035[/automerge]
What colour are these two leds and what size roughly?
[automerge]1597146726[/automerge]
If you send me a private message with your address I will post you some resistors gratis.

I got quite a few leads with it of various type so I can adapt them but handy to know if I want more. And thank you very much for the offer of resistors, I can get them through work luckily.

The two LED's are in the head of this so other than knowing they are powered by 2x LR44 I know nothing else about them:

DC Comics Armored Batman Premium Format(TM) Figure by Sidesh

The Armored Batman Premium Format Figure is now available at Sideshow.com for fans of Batman v Superman: Dawn of Justice and Ben Affleck.

www.sideshow.com

 

[marconi]

Ok. The two leds for Batman's eyes look white to me so forward voltage drop is similar to the green led ie 2.4V with forward current of 10mA. They light up when connected to a 3V battery which suggests the two LEDs are connected in parallel across it - 2.4V < 3V. If the two LEDs were in series the required battery voltage would need to be greater than 2 x 2.4 = 4.8V.

The required voltage drop resistor to illuminate a paralleled pair of these LEDs using the 4.5V supply is (4.5-2.4)/ (2 x 0.01) = 2.1/0.02 = 105 Ohms (nearest value is 100 Ohm) nb (2 x 0.01) because there are two leds in parallel).

You could obtain 100 Ohm by wiring two 220 Ohm resistors in parallel or by buying a 100 Ohm component.

Power rating is 0.02 x 0.02 x 100 = 0.04 W so a 1/4Watt or higher rated resistor.

Please tell us how you fare with this project.

 

[T800mc]

Ok. The two leds for Batman's eyes look white to me so forward voltage drop is similar to the green led ie 2.4V with forward current of 10mA. They light up when connected to a 3V battery which suggests the two LEDs are connected in parallel across it - 2.4V < 3V. If the two LEDs were in series the required battery voltage would need to be greater than 2 x 2.4 = 4.8V.

The required voltage drop resistor to illuminate a paralleled pair of these LEDs using the 4.5V supply is (4.5-2.4)/ (2 x 0.01) = 2.1/0.02 = 105 Ohms (nearest value is 100 Ohm) nb (2 x 0.01) because there are two leds in parallel).

You could obtain 100 Ohm by wiring two 220 Ohm resistors in parallel or by buying a 100 Ohm component.

Power rating is 0.02 x 0.02 x 100 = 0.04 W so a 1/4Watt or higher rated resistor.

Please tell us how you fare with this project.

Okay great! Will do, thanks!

 

[T800mc]

Okay great! Will do, thanks!

I fitted a 100 Ohm resistor today. Voltage is still 4.5 when I test it. What am I doing wrong?

 

[marconi]

Hello. four things first:

1. please draw me circuit diagram of what you have connected together.

2. Show me where you connect your voltmeter to measure 4.5V.

3. Do the LEDs glow?

4. Read out the colours on the resistor from one end to the other.