Simple question: What resistor to buy?

[ajd344]

I want to buy 10 LEDs and put them all together in a circuit. Each LED needs 1.7VOLTS and 20mA. I have 10, so that would be 17 volts. I want to power it off 2 9Volts, aka 18VOLTS. So, I went (18-17)/.02 = 50. So I need a 50-ohm resister, right? Where would I buy one? I found a 50OHM, but it is rated for 5WATTs. Is that OK? Also, would a 51OHM work? If so, should I get 1/4 or 1/2 watt?

THANKS!

 

[FusionITR]

V=IR (Ohms Law)
(9V - 1.7V)/20mA = 365

You can use ~ 350ohm resistors.

Power dissapation in the resistor
P=IV
P=(20mA)(9V - 1.7V) = 0.146W

1/4 W resistor should be fine.

 

[ajd344]

V=IR (Ohms Law)
(9V - 1.7V)/20mA = 365

You can use ~ 350ohm resistors.

Power dissapation in the resistor
P=IV
P=(20mA)(9V - 1.7V) = 0.146W

1/4 W resistor should be fine.

Thanks, but I have 10 leds going at 18Volts. Isnt it different? Would I have 10 350 ohm resisters in there? BTW, I couldnt find a 350ohm resistor on Jameco.

 

[Hero999]

No, use higher value resistors rather than lower - always round up never down. I would go for the next value up in the e12 series which is 390hm: or 470hm: (a more common e6 value).

 

[ajd344]

Alright, I found a 390ohm'er. Thanks. My other question is: to use 10 LEDs all together, and have input source of 18V, what would I use?

 

[Hero999]

You can't simply parallel the LEDs and use one resistor.

Connect the LEDs in series, you might be able to get away with connecting them in series but 1V really isn't enough room for accounting for the change in power supply voltage. I'd use two chains of five LEDs, each with a series resistor.

Use the equation below to work out the resistor values.

[latex]R= \frac{V_{IN}-V{LED}}{I_{LED}}[/latex]

 

[ajd344]

Well, I get 350OHMs when I do that (Total.Voltage.In-Voltage.Of.All.LEDs)/.02 . So, do I connect 10 of them with one 350 OHM resister?

If it was 2 sets of 5, I guess it would be 25OHMs. So, I connect 5 of them with one 25 ohm resister? Thanks!

 

[FusionITR]

Thanks, but I have 10 leds going at 18Volts. Isnt it different? Would I have 10 350 ohm resisters in there? BTW, I couldnt find a 350ohm resistor on Jameco.

If your powering LEDs (or any device) you dont add the voltage drops, you can power them on a single supply rail (like a single 9V battery).

See attachment.

 

[FusionITR]

Well, I get 350OHMs when I do that (Total.Voltage.In-Voltage.Of.All.LEDs)/.02 . So, do I connect 10 of them with one 350 OHM resister?

If it was 2 sets of 5, I guess it would be 25OHMs. So, I connect 5 of them with one 25 ohm resister? Thanks!

See my attachment.

 

[ajd344]

Cool, that makes sense. How would I determine what resisters to use on each one? I have 10 LEDs (1.7V,20mA) and an input of 18V. Thanks!

 

[Richard Principal]

you said

. My other question is: to use 10 LEDs all together, and have input source of 18V, what would I use?

you did not say if they are going in series or parallel

if they all go in series they take up 17 Volts the resistor needs to drop 1 Volt
at 20 mA = 50 Ohm best prefered value in E12 range is 56 Ohms at 20 milli Watts.

If they all go in parallel the resistor needs to drop 16.3 Volt at 200 mA = 81.5 Ohms or 82 Ohm at 3.26 Watts. trouble if they all go in parallel some LED might be brighter than some others

 

[ajd344]

Well, I was thinking the way Fusion ITR said, with the supply rail. Would that be 50 OHM then for each one? Thanks!

 

[Richard Principal]

Well, I was thinking the way Fusion ITR said, with the supply rail. Would that be 50 OHM then for each one? Thanks!

no all 10 LED in series and 56 Ohm in series as well (like a Christmas tree lights)
if you get one more LED you might be able to do away with resistor.

 

[FusionITR]

Its wasteful to power LEDs in series because of the higher voltage needed.

Would you rather buy another 9V battery for $3 or a pack of 350ohm resistors for 10 cents?

Not to mention you cant turn them on and off individually. It's all or nothing.

 

[FusionITR]

no all 10 LED in series and 56 Ohm in series as well (like a Christmas tree lights)
if you get one more LED you might be able to do away with resistor.

You wouldnt want to do away with the resistor because you will need something to limit the current. 0.1V different in the diode group could mean an extra 100mA of current going through it (diode voltage-current relationships are not linear).

 

[audioguru]

It is wasteful to power individual 1.7V LEDs from a 9V battery because 7V is not used for the LEDs. Connect 2 LEDs in series and in series with a 220 ohm resistor for 25.5mA with a new 9V battery and 11.8mA when the battery voltage has dropped to 6V. Make 5 groups like this.

The LEDs probably have a "typical" voltage of 1.7V but it could be 2V max. Then 3 in series wouldn't light when the battery voltage drops to 6V and they would be very dim for most of the time.

 

[ajd344]

OK, thanks for your help! Actually, my friend's father is an electrical engineer and I went over to his house and he helped me out with the plans. It makes sense now! Thanks a lot!

 

[Richard Principal]

You wouldnt want to do away with the resistor because you will need something to limit the current. 0.1V different in the diode group could mean an extra 100mA of current going through it (diode voltage-current relationships are not linear).

you dont need any resistor if you put say 10 1.7 Volt LED in series and run it off 17 Volts you would not have to limit the cureent,

it would have to be the best way to stop power wasteing if you had the right power for the LED.

 

[audioguru]

you dont need any resistor if you put say 10 1.7 Volt LED in series and run it off 17 Volts you would not have to limit the current.

The voltage rating for the LEDs "might" be 1.7V, but it isn't guaranteed. They might actually be 1.6V then without a current-limiting resistor their current would be way too high and they will blow up. They could actually be 1.8V or even 2.0V then they won't light unless some extra voltage is provided.

He doesn't have 17V. He is thinking of using two 9V batteries, that produce about 19V when new and drop to 12V when finished.

 

[em2006]

Using a constant current generator is a good solution.