Hi everybody,

I don't have enough knowledge in eletronics to calculate the resistance I need in my resistors to power a LED (3.3 VDC - 20 mA) on a 24 VDC - 1.6 A circuit.

I just need someone to explain to me how I calculate the resistance I need in my resistors.

Feel free to ask for more details !

 

Hi everybody,

I don't have enough knowledge in eletronics to calculate the resistance I need in my resistors to power a LED (3.3 VDC - 20 mA) on a 24 VDC - 1.6 A circuit.

I just need someone to explain to me how I calculate the resistance I need in my resistors.

Feel free to ask for more details !

1. Look up the LED forward voltage (Vf) dropped across the LED with the specified forward current (If) through the LED.
2. Subtract Vf from the supply voltage.
3. Divide the difference in voltage by If, and the quotient will be the resistance required.

In your example: Vsupply = 24VDC, Vf = 3.3VDC, and If = 0.03A, so...

\( R = \frac {Vsupply - Vf}{If} = \frac {24V - 3.3V}{0.03A} = 690 \Omega,\)

and the resistor will dissipate:

\( P = I^2 R = 0.03A^2 \times 690 \Omega = 0.621 \text { watts.}\)

750 is the closest E24 value which will limit If to less than but close to 30 mA, so a safe bet for the ballast would would be a 750 ohm +/- 5%, 1 watt resistor.

 

In your example: Vsupply = 24VDC, Vf = 3.3VDC, and If = 0.03A, so...

He stated it was .02A.

 

1kΩ should do it. If the LED is too bright, try 2kΩ, etc.

 

Here are two ways to choose.
1. Using LM317 and setup the output voltage as 3.3V.
2. Choosing 80% of 20mA is equal to 16mA,
R=(24V - 3.3V)/16mA=20.7V/16mA=1.294 K
Wr= 20.7V*16mA=0.33W, choosing the w value as 5 times of calculation value is 1.65W, you can choose 1.2K/2W.

 

He stated it was .02A.

Yup, he did, and now that he knows how to fish he can ignore my misteaks.

 

24v ÷ 0.02 (20 mA) = 1200Ω

OR

(24V - 3.3) ÷ 0.02 = 1035Ω

IF the LED you're using has a 3.3 volt forward rating then you subtract that from the supply. In your case, 24 volts. The fact that the supply is capable of delivering 1.6 amps ONLY means that it is strong enough to run your LED. If your supply were only capable of delivering 0.010 amps then it wouldn't work because you want (or need) 0.020 amps. So knowing your supply is big enough you don't need to concern yourself with factoring the supply amperage.

The amperage moving through the entire circuit is what you want. In your case it's 20 mA. OK, that's fine. Nothing wrong with that. A bit bright but if it suits your purpose - well enough. You can take two approaches (as shown above). If you want to be EXACT then you subtract the forward voltage of the LED from the supply voltage. Then you divide voltage by amperage to come up with a resistance. But since your voltage supply is greatly more than you need you can ignore (in most cases) the forward voltage drop across the LED. As you can see from my first example I ignored the forward voltage and came up with 1.2KΩ resistance. In the second example I took for granted that the forward voltage of your LED was 3.3 volts. I don't KNOW that to be the fact. None of us can know that except you. This is where you need to know what you're working with. But assuming it's 3.3 volts forward then you subtract that from the supply and then divide what's left by the desired amperage.

If you used the 1.2KΩ resistance and have a 3.3 volt drop your final amperage would be 17 mA, pretty close to the 20 mA you asked for. I doubt you'll be able to tell any difference in brightness between the two (17 and 20 mA).

To calculate the total amperage going through the circuit you divide the voltage by the resistance. Again, don't forget to subtract the forward voltage. If it's 3.3 volts then your amperage would be based off of 20.7 volts and a resistance of 1.2KΩ

20.7 V ÷ 1.2K = 17.25 mA

If you MUST have 20 mA (don't see why you would) then use a 1KΩ resistance. That'll give you 20.7 mA

[edit]
Something I neglected to mention is wattage. EM Fields didn't forget the wattage part of it, and that's important. To calculate wattage multiply voltage by amperage. Don't subtract any voltage drops, use the full value, the 24 volts. 24 ÷ 0.020 = 0.48 watts. Nearly half a watt (0.5 watts). IF you use a half watt resistor you're running close to the limit of the capacity of the resistor, so always use twice the wattage rating you need. So if you need to handle 480 mW (0.48 watts) then use 960 mW (or 0.960 watts) And since you can't get resistors in those kinds of ratings you'd use a 1 watt (1.000 watt) resistor.

Resistors typically come in 1/8 watt, 1/4 watt, 1/2 watt, 1 watt, (I've seen 3 watts), 5, 10 and bigger wattage ratings. Wattage is the amount of heat the resistor has to dissipate. So if you have a high wattage to dissipate you also need to ventilate. Don't put a 1 watt resistor directly on the board if you're running close to 1 watt. Stand it off the board with a quarter inch air gap between the bottom of the resistor and the board (approximately).

Hope this helps.

 

Here are two ways to choose.
1. Using LM317 and setup the output voltage as 3.3V.
2. Choosing 80% of 20mA is equal to 16mA,
R=(24V - 3.3V)/16mA=20.7V/16mA=1.294 K
Wr= 20.7V*16mA=0.33W, choosing the w value as 5 times of calculation value is 1.65W, you can choose 1.2K/2W.

There's a negative advantage to using an LM317 since the wasted power will be a little higher if an LM317 is used instead of a single ballast resistor, not to mention the extra $$ for the heat sink, PCB real estate, assembly, and the LM317 itself.

For a 3.3 volt, 20mA LED running from a 24 volt supply, the series ballast resistance looks like:

\( Rs = \frac{Vs - Vf(led)}{If(led)} = \frac{24V-3.3V}{0.02A} = 1035 \Omega\)

1100 ohms is a good E24 value, and assuming Vf will stay close to 3.3 volts with a small deviation in If, we'll have:

\( If = \frac{Vs-Vf(led)}{Rs} = \frac {24V - 3.3V}{1100\Omega} = 0.019 \text { ampere,}\)

and the power dissipated by the ballast will then be:

\( P = If(led)^2 \times Rs = 0.019A^2 \times 1100\Omega \approx 0.4 \text { watts.}\)

A standard 1100 ohm +/- 5% 1 watt resistor should work nicely, is readily available, and is inexpensive.

 

There's a negative advantage to using an LM317 since the wasted power will be a little higher if an LM317 is used instead of a single ballast resistor, not to mention the extra $$ for the heat sink, PCB real estate, assembly, and the LM317 itself.

For a 3.3 volt, 20mA LED running from a 24 volt supply, the series ballast resistance looks like:

\( Rs = \frac{Vs - Vf(led)}{If(led)} = \frac{24V-3.3V}{0.02A} = 1035 \Omega\)

1100 ohms is a good E24 value, and assuming Vf will stay close to 3.3 volts with a small deviation in If, we'll have:

\( If = \frac{Vs-Vf(led)}{Rs} = \frac {24V - 3.3V}{1100\Omega} = 0.019 \text { ampere,}\)

and the power dissipated by the ballast will then be:

\( P = If(led)^2 \times Rs = 0.019A^2 \times 1100\Omega \approx 0.4 \text { watts.}\)

A standard 1100 ohm +/- 5% 1 watt resistor should work nicely, is readily available, and is inexpensive.

View attachment 129385

When we using the led, we should be considering the lumens depreciation of led, using rating current for 100% maybe does not damage the led, but the brightness will be reducing little by little, so choosing around 80% of rating current or less is better.

1.65W for a LM317 is not a big problem, LM317 can be easy to afford it, I offered two methods to let the TS to choose, if he can do the experiments then he will know which one is better for him.

 

When we using the led, we should be considering the lumens depreciation of led, using rating current for 100% maybe does not damage the led, but the brightness will be reducing little by little, so choosing around 80% of rating current or less is better.

Why are you trying to pick a fight with me?
As far as I can tell, all the OP was asking for was a way to determine the value of the ballast resistor, which is what I taught him, while you seem to be intent on lording over the "discussion" with off-topic and unasked-for irrelevance.

1.65W for a LM317 is not a big problem, LM317 can be easy to afford it,

It really isn't a question of being "easy to afford", it's a question of being conservative and not being profligate with precious resources.

I offered two methods to let the TS to choose, if he can do the experiments then he will know which one is better for him.

Analytically, there is really only one right way.

 

Why are you trying to pick a fight with me?
As far as I can tell, all the OP was asking for was a way to determine the value of the ballast resistor, which is what I taught him, while you seem to be intent on lording over the "discussion" with off-topic and unasked-for irrelevance.

It really isn't a question of being "easy to afford", it's a question of being conservative and not being profligate with precious resources.

Analytically, there is really only one right way.

Assuming that If using your designed that the brightness will be lost half for one year later, and if using my designed that the brightness will be similar for one year later and then which design is better, is that a fight?

Who told you that if the TS didn't ask then we can't tell him some more since the TS mentioned that he is not familiar with ee, the rights of selection is by the TS not you, since when the rules setup by you?

Please google something like what's the relationship between lumens depreciation (evanescence), luminous decay and led.

 

Assuming that If using your designed that the brightness will be lost half for one year later, and if using my designed that the brightness will be similar for one year later and then which design is better, is that a fight?

Since the OP's question wasn't "Which way is better?" and your intent seems to be to trivialize the OP's query in order to replace it with your own agenda, it starts to look, to me, like you're spoiling for confrontation.

Who told you that if the TS didn't ask then we can't tell him some more since the TS mentioned that he is not familiar with ee, the rights of selection is by the TS not you,

True enough, and I respected the OP's rights of selection by responding to his query without interjecting irrelevant fluff. You, on the other hand, took it upon yourself to look past his rights and to post largely off-topic material for your own self-aggrandisement. Or so it seems...

since when the rules setup by you?

Heh! Throwing down the gauntlet? This isn't my site, so I don't set up the rules, but I do like to defend them rigorously when I see them violated in flagrante delicto.

Please google something like what's the relationship between lumens depreciation (evanescence), luminous decay and led.

Please note that the subject of this thread is "24 VDC - 1.6 A to 3.3 VDC - 20 mA Using Resistors", which makes your request off-topic, so I'll not follow it up here.

However, if you'd like to start a new thread devoted to discussing the minutiae of LED evanescence I'd be happy to hop on board.

 

Yup, he did, and now that he knows how to fish he can ignore my misteaks.

I tought someone to fish once. The dynamite sticks in the canoe scared him bad enough he never asked me for help with learning anything again!

 

Geez fella's, I think you scared him away. I think I heard rumor he's going to become a plumber now.

If you wanna go off topic; what about a current mirror? Two simple transistors and a single resistor. Voltage can vary but the current will remain constant. (relatively speaking)

http://leachlegacy.ece.gatech.edu/ece3050/notes/bjt/bjtmirr.pdf

Just In Case the TS is interested.

 

Geez fella's, I think you scared him away. I think I heard rumor he's going to become a plumber now.

If you wanna go off topic; what about a current mirror? Two simple transistors and a single resistor. Voltage can vary but the current will remain constant. (relatively speaking)

http://leachlegacy.ece.gatech.edu/ece3050/notes/bjt/bjtmirr.pdf

Just In Case the TS is interested.

Seems like these super simple circuit concepts always get turning into a 'who can make it as overly complicated as possible for no real reason' contest every time. My vote would have been for a 1 - 1.2 K 1/2 watt resistor and call it good.

Rather surprised a micro SMPS buck converter circuit was not mentioned so the OP could theoretically save a 1/10th of a watt.

 

In the final analysis, TS should try it and see if he likes it. I never run a 20mA LED at 20mA.

I would start with 10kΩ ½W resistor and work my way down. A good sweet spot is probably at 2k2Ω ½W.

 

Thanks for all the fast answers everybody, looks like I need 1kΩ to 2kΩ resistors, close to what I tought.

So, I tried the resistance that most of you guys told me and I'm probably doing something wrong, because the light just doesn't work.

Here's my setup :

 

Can you make sure the led is ok, this is just a very simply circuit, you can check all the wiring and measure the voltage of battery and three resistors and led.

 

So, I tried the resistance that most of you guys told me and I'm probably doing something wrong, because the light just doesn't work.

Is the LED wired backwards? They only work in one direction.

 

Is the LED wired backwards? They only work in one direction.

I don't think so. The rounded side (anode) is wired to DC+ and the flat side (cathode) is wired to DC-.