Basic Voltage drop question using LED

[onpiopah]

Hi I am just starting out so please bear with me as I realise this is probably a very basic problem I have. I am currently learning about Voltage and have made a simple circuits with a 6volt battery, resistor and LED.
• The voltage drop across the battery is 6volts.
• Across the resistor is 4.5v and across the Led 1.5 v which all adds to 6volts. This makes perfect sense to me.
• If I take out the LED the drop across the resistor is 6v and the battery 6v – again it makes sense.
• But If I just use the LED the drop across the battery is just over 2 v and across the LED is just over 2v.
• And If I try the experiment in software such as “Circuit Wizard” the voltage drops over both remains at 6v – can someone explain what is happening – I thought it should always be 6v?
thanks for your reply.

[piratepaul]

Circiut wizard is right... there is an internal resistance of the battery too but I dont think it matters too much.
I am suprized the LED is not fried... may be the battery is flat.

[Pauldf]

2v is the forward voltage of the diode, if you go above this then as Piratepaul says things can go pop!
You calculate the resistance taking into account the volt drop you need and the current required, mr ohms does the rest! Bog standard LEDs need around 20mA to work, 4v/0.02A = 200 ohms.

[cilek]

Hmmm....interesting. I am also surprised that your LED didn't burn out. I agree with both piratepaul and Pauldf about the LED. I think the most common advice to give out is that use zener diode in every electronic circuit you design so that the chances of damaging any other components will be minimized.

printed circuits assembly