Guys I know what the circuit should be. I'm not a novice any more. I was back then.

Thanks for the diagrams but thats not what the circuit is and I cant change it now because its sealed in a model. I just need to power whats there without the fear of killing the lights.

In that case you can use the circuit I gave you for reference. It has paralel and serial connections.

1. On every serial connection there has to be a current limitation, since you can not do that, the only option you have is to limit the current for all LEDs, like suggested before. That can be done with a current regulator or if possible a resistor right before the device (capsulated box). For this to work, you need "Power supply/All LEDs current = Resistor needed".

 

There is a resistor on each bank of 3 LEDs I just don't know what value it is. I will do some wizardry with my multimeter later and see what I can determine and work out a way forward.

Cheers!

 

Three 240 ohm resistors in parallel will limit the LED string current to 15 mA.

When three 240 Ohm resistors are in parallel... 240 / 3 = 80 Ohms.
If you want 240 Ohms, the calculations say three 720 Ohm resistors in parallel will give you 240 Ohms.
However, 720 is an uncommon value, so put four 1K Ohm resistors in parallel. That will give you 250 Ohms.
If you are using .25 Watt resistors, the four 1K's in parallel will have a power rating of around 1 watt. Plenty for heat dissipation.

 

If you want a group of four resistors to equal 240 ohms, use a series/parallel combination. In that way, four 240 ohm resistors will combine to equal exactly 240 ohms.

However ...

An equivalent resistance of 240 ohms would decrease the LED current to only 5 mA. That is below the requirement in post #3. Three 240 ohm resistors in parallel was suggested because it yields 15 mA LED current as requested. Other options are either two 160 ohm resistors in parallel or a single 82 ohm resistor, but the circuit already is built with three resistors in parallel so neither of these apply to this question.

ak

 

I disagree, sir.
According to Ohm's Law... and you can breadboard this if you like...

with resistors in series:
Rt = R1 + R2 + R3... etc

with two or more LIKE value resistors in parallel:
Rt = R / n where R is the value of the resistor and n is the number of resistors in parallel.
Example: four 1K resistors in parallel.
1,000 / 4 = 250 ohms
... using the formula below for unlike values also works but takes longer to do.
1 / (1/1,000) + (1/1,000) + (1/1,000) + (1/1,000)
1 / .001 + .001 + .001 + .001
1 / .004
Rt = 250 ohms

with two UNLIKE values in parallel:
Rt = R1 x R2 / R1 + R2
Example: R1 = 470 ohms, R2 = 330 ohms
470 x 330 / 470 + 330
155,100 / 800
Rt = 193.875 ohms

with three or more UNLIKE values in parallel:
Rt = 1/(1/R1)+(1/R2)+(1/R3)... etc
Example: R1 = 100 ohms, R2 = 220 ohms, R3 = 470 ohms
1 / (1/100) + (1/220) + (1/470)
1 / .01 + .0045 + .0021
1 / .0166
Rt = 60.24 ohms

 

what is the target resistance you are trying to reach?

 

I disagree, sir.
According to Ohm's Law... and you can breadboard this if you like...

with resistors in series:
Rt = R1 + R2 + R3... etc

with two or more LIKE value resistors in parallel:
Rt = R / n where R is the value of the resistor and n is the number of resistors in parallel.
Example: four 1K resistors in parallel.
1,000 / 4 = 250 ohms
... using the formula below for unlike values also works but takes longer to do.
1 / (1/1,000) + (1/1,000) + (1/1,000) + (1/1,000)
1 / .001 + .001 + .001 + .001
1 / .004
Rt = 250 ohms

with two UNLIKE values in parallel:
Rt = R1 x R2 / R1 + R2
Example: R1 = 470 ohms, R2 = 330 ohms
470 x 330 / 470 + 330
155,100 / 800
Rt = 193.875 ohms

with three or more UNLIKE values in parallel:
Rt = 1/(1/R1)+(1/R2)+(1/R3)... etc
Example: R1 = 100 ohms, R2 = 220 ohms, R3 = 470 ohms
1 / (1/100) + (1/220) + (1/470)
1 / .01 + .0045 + .0021
1 / .0166
Rt = 60.24 ohms

What are you disputing @AnalogKid on? Your math all looks good, but you're not refuting anything he said in the process. Everything he said was correct too.

There are a lot of ways to combine resistors and get close to certain target values. What he was pointing was that if you really want 240 out of four resistors, there's no reason to settle for four 1k resistors yielding 250, when there's a very simple way to get exactly 240. What part of that do you disagree with?

 

Of which all is conjecture because my model has what it has to make it run on 12 v. It ran before and I can't change whats in there. I just need to work out what the best course of action to remove the risk of killing leds or resistor when I power it up for long periods.

 

Just limit the total current to 90 mA as has already been sggested. Do the by using a variable supply and an ammeter to determine the voltage required to get 90mA.

Bob

 

I disagree, sir.

Which part?

According to Ohm's Law...

I'm familiar with Ohm's Law. You have used it correctly.

ak

 

So I have a project that I started 10 years ago involving a circuit of LEDs sealed inside a model. My electronics knowledge was poor back then and when I wired them up I connected the 18 white LEDs in 6 series banks of 3 LEDs which was to run on 12V. These banks were then connected in parallel. Horrible I know

These banks were then protected with a bank of 3 resistors soldered in in parallel to each other to give the right resistance and that soldered in series with the banks of LEDs. These are not high power resistors.

The problem is now that I know more about electronics (I've dug out the project to finish it) I am very very concerned that the resistors will not be able to take the power required for the circuit. I estimate the circuit will draw probably about 1.8W maybe?

My issue is I have 2 wires going into the model to power the circuit and was wondering if anyone could suggest a method of keeping the power down in the circuit be external means to stop the resistors inside melting themselves.

I cannot get inside the model to change the circuit and the circuit will be controlled by an Arduino board via Mosfets or similar.

Help!!

Perhaps you can implement PWM in the Arduino to limit the current.

eT

 

if you adjust the voltage to a level that provides the equivalent of 20 milliamps to each string then at worst case the resistors will be carrying 20 milliamps each. So to reach the 250 milliwatt rating you would need to be dropping over ten volts across the resistors, if you use 12 volts and have 3 leds at 3.6 volts each , that is 10.2 volts on the LEDs, leaving 1.8 volts dropping on the resistors. 1.8 volts times 20 milliamps is 36 milliwatts. So you have no problems at 12 volts.