Ohms Law Expert Help

[Mark R.]

Having a mental block ....

Example - I have two white LEDs in parallel (3.4 volt / 20ma rated) running on 12 volts using a 1000 ohm resistor and I like the brightness.

If I were to put a resistor on each LED, what value do I need to maintain the same brightness level ?

Would a 1000 ohm resistor on each be the same brightness, or do I need a slightly higher value ? I forget how a parallel load calculates into this.

Mark.

[cacole]

Resistors in series ADD their resistance; in parallel they DIVIDE, so in order to keep the same brightness with parallel LEDs you need to double the resistance value if they are also in parallel.

Where you have complicated the matter is having two LEDs in parallel through one series resistor.  In this case, the LEDs are dividing the voltage between them, but with equal current flow through each, which is actually a series/parallel circuit.

 

[Mark R.]

Maybe I'm not explaining it correctly ....

Right now, I have one resistor (1000 ohms) feeding two leds (the LEDs are wired in parallel). I want to separate the two parallel LEDs and place a single resistor on each LED, yet retain the same brightness.

Mark.

[CSX Robert]

The LEDs are not dividing the voltage, they are dividing the current.  The current being consumed is (12-3.4)/1000 = 0.0086 amps, so they should be at 0.0043 amps each.  To calculate the resistor to drive one of the LEDs at 0.0043 amps: (12-3.4)/0.0043 = 2000 ohms.

[Mark R.]

CSX Robert

The LEDs are not dividing the voltage, they are dividing the current.  The current being consumed is (12-3.4)/1000 = 0.0086 amps, so they should be at 0.0043 amps each.  To calculate the resistor to drive one of the LEDs at 0.0043 amps: (12-3.4)/0.0043 = 2000 ohms.

 

 

Thanks - That was my knee-jerk thought, but the more I thought about it, the more I questioned my thinking !  

So, for the record, and the math .... two LEDs in parallel draw double the current (20ma X 2) but the voltage stays the same (3.4 volts). Two LEDs in series, the current stays the same (20ma), but the voltage draw is doubled (6.8 volts). (?)

Guess I'm getting to the point of needing to keep cheat notes ....

Mark.

[CSX Robert]

Correct, except it is called voltage drop, not draw.

[rrinker]

 Yes, this is the 'other side' of the equations, it's not Ohm's Law any more, these are Kirchoff's Laws

Voltage in parallel is the same, voltage in series adds

current in series stays the same, current in parallel adds

 

If you apply this concept to the circuit of a 12V source feeding a 3.4V LED with a 1K resistor in series, you can see how Ohm's Law then determins how much current will flow through the resistor and the LED (the same) and how much voltage will be dropped by the LED and how much by the resistor (3.4V in the LED because it's a propery of the LED, and 12-3.4 in the resistor)

You know know that the resistor is dropping 8.6V. You know it is a 1K resistor, so using Ohm's Law you cna find the current, E/R, 8.6/1000, 8.6mA. Applying Kirchoff's Laws, current in series are the same, so if 8.6mA is going through the resistor, then 8.6mA will also be flowing through the LED. ANd since this is below the maximum allowed for the LED, we are done and can use a 1K resistor safely.

 Now, if you put two LEDs in parallel, with just one resistor, assuming they have the exact same voltage drop, they will each get only half the current, in this case 4.3mA. And here you can start to see why you shold use 1 resistor per LED and not try to skimp on a 2 cent part. You may be empirically familiar with light bulbs in series vs parallel, especially Christmas lights that have the little bypass shunts in them so that despite being wired in series, if one burnsout, the rest stay on. That comes at a cost. Assume 100 lights in series, each rated at 1.2V, plugged in to 120VAC outlet. Perfect, each bulb gets 1.2V - voltage in series adds, remember.

Now, say 10 of them burn out, but because they have those shunts, the rest stay on. 90 bulbs. Awith 120VAC feeding them. Now each bulb gets 1.33 volts - which will just reduce the life of each one. Until more burn out - increasing the death rate of the remaining ones. Like an avalanche.

OK, back to our LEDs. Say you put 10 in parallel and use just 1 resistor. It would need to be a 100 ohm resistor so that each LED gets 8.6mA. Something happens and 5 of the LEDs become disconnected. Now you have twice the current, 17.2mA - in this case still ok, most white LEDs are good to 20-25mA. But play with the numbers, you can easily create a scenario for the LEDs just liek the lights, drop enough out of the circuit and the current through the remainign ones will exceed the LED rating, making more fail, raising the current still more.. avalanche!

Resistors are like 2 cents, if you get them from some palce other than Radio Shack. Maybe as much as 5 cents. Just use 1 resistor per LED and avoid conflict. Each combo of a resistor and LED can be taken as a signle entity, connect multiples of this to a 12V power supply and eachLED/resistor pair gets 12 volts (voltage in parallel) and the total current will be the sum of each pair's current draw. So 10 pairs of 1K resistors and a 3.4V LED in parallel will need 12V at 86mA.

                   --Randy

 

[crhostler61]

I learned basic electronics and many of it's laws about...um...er...35-40 years ago and honestly forgot far more than I remember. So found a couple of calculator web sites that have saved me from further rehearsal for altzheimers.

http://www.ohmslawcalculator.com/ohms_law_calculator.

http://www.calculatoredge.com/

 

Mark H

[BroadwayLion]

LIONS LAW:

Put a single LED in series with a 1000 Ohm resistor, and apply 12 v dc.

If not light up, reverse polarity.

If too bright add resistance.

If not brigt enough reduce resistance.

Like really, you are going to keep a drawer full of resistors between 100 Ohn and 2000 Ohm?

LION buys 1K Ohm resistors by the THOUSAND (1.2c ea)  and had proclaimed this to be correct.

ROAR