Ohm’s law help
Sep 6, 2009 at 12:17 PM Thread Starter

#### johnwmclean

##### Aka: capone, bignurse.
I need a little help figuring out the value of a resistor I need to run in series from my psu to my latched Bulgin switch.

The psu (twisted pair LCBPS) is 15V dc and the LED in the Bulgin switch is 12V dc. I tried a couple of online calculation sites, I still don’t get it, any help would be most appreciated.

Data sheet for the switch is http://www.farnell.com/datasheets/60853.pdf
MP0045/1E2

Cheers
John

Sep 6, 2009 at 12:40 PM

#### tomb

##### Member of the Trade: Beezar.com
Quote:

 Originally Posted by johnwmclean /img/forum/go_quote.gif I need a little help figuring out the value of a resistor I need to run in series from my psu to my latched Bulgin switch. The psu (twisted pair LCBPS) is 15V dc and the LED in the Bulgin switch is 12V dc. I tried a couple of online calculation sites, I still don’t get it, any help would be most appreciated. Cheers John

That's because you need the current to know how to solve the three-variable equation represented by Ohm's law. You can't solve for one variable when there are two.

Most LED's are rated for a maximum of 20ma. So, if we plug in 0.02 with your voltages, you'd have: 15V-12V = 0.02 x R, R = 3V/0.02 = 150 ohms.

Personally, I try to size LED resistors so that they're no more than half that current rating - there are other variables that either make this calc too much trouble or are unaccountable. (Technically, you'd include some calculated resistance of the LED in the total series resistance and then correct accordingly. Also, there may be variances on your voltage that are not presently known.) If you exceed the LED's current limit by much for an extended period - they'll burn out. Yet, brightness is diminished by very little, even at half current rating.

So, if we say 300 ohms, that would result in I = 3V/300 = 0.01A (10ma). Note that you should also check the Power equation to insure you have the proper rating for the resistor. P = (I^2) * R = 0.0001 * 300 = 0.03W. So, a 1/8th W resistor would be plenty. It's also interesting to see that if we used the original values, we'd end up with 0.06W, which is close to one-half of 0.125W (1/8th W). We typically size resistors for twice the power rating for a good safety factor, so it would be right on the edge of bumping up to a 1/4W resistor.

Hope that helps.

P.S. I didn't see anything on the switch's data sheet that would be of much help, unfortunately.

Sep 6, 2009 at 1:08 PM

#### digger945

Sep 6, 2009 at 1:45 PM

#### amb

##### Member of the Trade: AMB Laboratories
johnwmclean, the Bulgin switch you have already has an internal current-limit resistor, so all you need to do is to feed it 12V without additional current limiting. Since you have 15V, you just need to insert a 3V zener diode in series (reverse-biased) to drop it to 12V for the LED. A 1N5225B or BZX55C3V0 would work nicely.

Sep 6, 2009 at 9:22 PM

#### johnwmclean

##### Aka: capone, bignurse.
Thankyou so much guys, very much appreciated.

Ti if a Zener diode is reversed biased, can be placed over either + or - when in series, is there a preference?
Lastly just to confirm does the anode side of the Zener diode face the psu?

Sep 6, 2009 at 9:54 PM

#### Ferrari

When a zener diode is used in reverse-biased mode, the cathode of a zener diode should be connected to the positive supply (higher potential).
By doing this way, no current can flow through the diode => the voltage is forced to drop across the diode.
If a 3V zener diode is used, the voltage will drop 3V across the diode.
(Btw, I'm not Ti)

Sep 6, 2009 at 10:09 PM

#### amb

##### Member of the Trade: AMB Laboratories
Quote:

 Originally Posted by johnwmclean /img/forum/go_quote.gif Ti if a Zener diode is reversed biased, can be placed over either + or - when in series, is there a preference? Lastly just to confirm does the anode side of the Zener diode face the psu?

Either of the circuits below will work.

Sep 6, 2009 at 11:01 PM

#### johnwmclean

##### Aka: capone, bignurse.
I love this community! Thank you!

Sep 7, 2009 at 12:30 AM

#### DaKi][er

Alternatively, if we take the assumption that it is designed to run at 20mA, as pretty much all LED's are, then (15-12)V / 0.02 A = 150ohm

Sep 7, 2009 at 4:02 AM

#### amb

##### Member of the Trade: AMB Laboratories
Quote:

 Originally Posted by DaKi][er /img/forum/go_quote.gif Alternatively, if we take the assumption that it is designed to run at 20mA, as pretty much all LED's are, then (15-12)V / 0.02 A = 150ohm

Actually, that's a fairly bold (and probably incorrect) assumption. Most of the time (for small LEDs), 20mA is a maximum current rating, not what you'd want to run it continuously at. Many LEDs will be too bright with that much current. I routinely run discrete LEDs at 2-5mA and get very good brightness.

For this Bulgin switch with a "12V" LED, the LED element inside is not 12V. There is an internal current-limit/dropping resistor, whose resistance is determined by Bulgin to produce good brightness when 12V is applied. We do not know what that resistance is, nor do we know what the resultant LED current is (unless you measure it with 12V applied). We also do not know the exact LED element forward voltage. If you simply add a 150 ohm resistor, which is calculated based on 3V drop and 20mA of current, the resultant current will most probably NOT be 20mA. That resistor will be in series with the internal resistor, and without knowing what the value is, we'd just be shooting in the dark.

While most LEDs tolerate a fairly wide range of operating currents, a more prudent approach is not to second-guess. The better solution is thus to use the zener diode. It will drop 3V, provde 12V to the switch's resistor and LED combination, and operate it the way it was spec'ed.

Sep 7, 2009 at 1:28 PM

#### wink

##### His amps are made out of recycled beer cansand his source from tomatos.
Yay, AMB. A very erudite explanation. It's good to see someone who knows his electronics.

Sep 8, 2009 at 11:24 AM

#### johnwmclean

##### Aka: capone, bignurse.
Quote:

 Originally Posted by amb /img/forum/go_quote.gif Either of the circuits below will work.

I tried this 3.3v zenor diode - and I get 29v current? I used IN4728A what’s going on?

Sep 8, 2009 at 12:07 PM

#### Ferrari

Quote:

 Originally Posted by johnwmclean /img/forum/go_quote.gif I tried this 3.3v zenor diode - and I get 29v current? I used IN4728A what’s going on?

Are you sure you are using 1N4728A zener diode? How do you measure that value of 29 things ???
That is not a parameter for current. Your measurement is very questionable!

Sep 8, 2009 at 12:08 PM

#### amb

##### Member of the Trade: AMB Laboratories
johnwmclean, how are you measuring current? Current should not be in volts.
Also, in this circuit the difference between V+ and V- is assumed to be 15V, so how do you get 29V? Did you connect it across +15V and -15V (which makes it 30V)? Also, make sure you have the zener oriented correctly. If installed wrong, it would drop less than 1V rather than 3.3V.

Sep 8, 2009 at 12:09 PM

#### wink

##### His amps are made out of recycled beer cansand his source from tomatos.
29V current,,??? are you sure that's not 29 milliamps.
How did you get 29v..??
what was your meter set to read, milliamps or volts - and where did you have the meter leads connected?