richg1998 It appears some here need to do some research. I think I see opinions at times. The below link shows quite a lot about resistors, light bulbs and LED's. Get your test equipment out and do some experimenting. http://members.optusnet.com.au/nswmn1/Lights_in_DCC.htm Online LED resistor calculator. Many here do not realize there are many electronic/electrical calculatros online. http://www.hebeiltd.com.cn/?p=zz.led.resistor.calculator Various online calculators. http://www.ifigure.com/engineer/electric/electric.htm Rich
It appears some here need to do some research. I think I see opinions at times. The below link shows quite a lot about resistors, light bulbs and LED's. Get your test equipment out and do some experimenting.
http://members.optusnet.com.au/nswmn1/Lights_in_DCC.htm
Online LED resistor calculator. Many here do not realize there are many electronic/electrical calculatros online.
http://www.hebeiltd.com.cn/?p=zz.led.resistor.calculator
Various online calculators.
http://www.ifigure.com/engineer/electric/electric.htm
Rich
Good Timing.
Click on the first calculator.
In the calculator select source voltage of 3 volts
Then select voltage drop of 3 volts
Then enter 30miliamps for desired LED current.
Read the value of resistor needed. It's zero
That means no resistor is needed.
While the output will sometimes creap up, the calculations are done based on the rated output of the circuit.
Springfield PA
All is good, except... when the junction heats up, it's characteristics change. Inside a closed shell it will get hotter than it does in free air.
Another point is that doing it the conventional way with a series resistor is less likely to have problems, which will not frustrate people. Powering a LED off a 1.5V output is pushing it, unless there is someway to insure that there is current limiting available. White LEDs have a much higher Vf, while being a lot more sensitive to current.
While connecting an ammeter across a cell will not show a million amps, as the internal resistance will limit it, doing it can result in burns or explosion of the cell. The internal resistance is what results in a dead cell in time.
While your demonstration works, it isn't something most of us would recommend as a normal practice.