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Help with voltage/resistor Answered

Hello.  I bought some Christmas LED replacement bulbs but they have no specs with them.  I decided to wire one up.   I figured I would go with 1 . 5 volts at 7 1/2 mA.   Using a 9 volt battery, I have a 1K ohm resistor.   9 - 1.5 = 7/ .0075. 

My problem is, when I put my meter on it, the one side of the resistor is showing 9 volts, correct, but the other side is showing 3.5 volts.   Shouldn't it only be reading 1.5 volts?  What is wrong? 

*I've had the LED lit up for over 25 minutes so far and it hasnt burned out yet.

Thank You.

EDIT:   I just added  a 10K resistor, so now I have 11K (in series), yet I am still getting a reading of 3.5 volts.   I'm missing something here. lol   I'm also getting a reading of about 9mA with 1K and 11K ohms.   I'm lost.


is it a white or blue led?  those take around that much to light up, so that'd make sense

do you go to penn state?  I live near there, just asking :P

I just pulled one out.  It was green.   Since they are all replacement bulbs for christmas lights, I figured they would all be the same voltage and current. 

I left it on for an hour before unhooking it and it didnt seem to fade.  

But why didnt I get less current/voltage when I upped the resistance?  I started with 1K, then added a 10K but the readings stayed the same. 

greens are generally 2.1 volts, but maybe just because they might have a forward voltage of 3.5 volts.  When you up the resistance the voltage will always stay the same cause the LED will be continuosly dropping the same voltage.

If you lower the resistance, the voltage will stay the same until the LED burns out form over current.

By upping the resistance the LED just gets dimmer because there is less current, but no matter what the voltage will stay the same.

But why did the current not change when I upped the resistance?  I was reading 9 mA with a 1K resistor, and also still reading 9mA  with 11K.  

Oh, and I dont go to penn state, but I live 10 miles from it.

that does not sound right at all :P

Are you sure you are measuring it correctly?  Cause without the LED the max current that could be flowing would be a bit less than 1 mA, it's like impossible for it to be 9mA.

I measured it from ground to the battery side of the resistor and got 9 volts, which was normal.  I measured it after the resistor and got the 3.5 volts.  I measured it on the + side of the led and got 3.5 volts.  

I added a 10K resistor, measured the same way and got 3.5 volts. 

well the voltage sounds right, but the current doesn't

My fuse blew in my meter last night, so perhaps just before it blew I was getting bad readings.  I'll know for sure tomorrow after I get a new fuse.

I consider the current (I in Ampere) only for the performance of the LED.
Forget the voltage drop in the LED (and the resistance if it is just one LED).
The current I results from U (Voltage) devided by R (resistance in Ohm).
So we have 9/1000 = 0.009A = 9mA (milliAmpere) which is a good value for
an ordinary LED.
LowPower LEDs suffice with much less current like 9/11k = 0,82mA.
Between this two samples, there should be a significant difference in the
brightness of the LED, when using one and the same with 9mA and 0,82mA.
If possible, I would take a good original LED to make a comparison.
So my approach in your sample would be to change the resistor to achive
what I would think is the correct brightness. From that on I have to
find out the wiring of the ...whatever is the candidate to be replaced.
You did not detail what for you need the new LEDs, it matters if/how many
are put in series for the further progress...
Don't blow your meter by trying to measure the current directly of a source
(e.g. on an main outlet).
An Amperemeter always needs to be put in series with a load, the range switched
to an sufficiently high position to cover the current to be expected. If not
known I need to start with the highest range. The latter is valid also for
voltage measurements.
In case the fuse of your meter seems not to be blown consider a second,
higher range but super fast fuse inside the meter.
And it is upmost impossible that the condition of the fuse reflects on a
measurement in the way you describe. The fuse is blown or good. It is
not eaten up like a battery.
Please continue your story, with some more details if possible, if there
are any. I like continuing stories.
sorry, my English is as it is. ciao Hubi


8 years ago

The "voltage drop" of an LED is pretty much a constant, like any diode.

It won't read like a second resistor in the circuit, I.E., it's not going to act like a voltage divider...