Hi,
Ive read many articles that state that the voltage drop across a diode is 0.7v.
However when i make this simple circuit with 1N4148 Diodes the voltage drop is 7V!
For some reason i cant measure the voltage drop across resistor and led!
Sorry to not include crucial info.
No led does not glow as it must only be receiving pitiful voltage, As far as i can tell i have connected everything right. If i bypass the diode and connect v+ to resistor, led glows.
Have tried the diode both ways, The resistor heats up violently if i change the direction.
The 1N4148 diode and diodes of a similair technology, have forward voltage drops 'around' 0.7Vfwd.
If you slowly increased the forward voltage across the diode, [with a limiting resistor], you find the diode starts to conduct at about 0.6Vfwd.
As you increase the applied voltage, the current will increase thru the diode and the Vfwd go to 0.7V, if you continue to increase the applied voltage, the Vfwd increase to 1.0v. If didn't limit the current, the diode would be destroyed.
The diode has whats referred to as a 'knee' on its applied voltage/conduction curve, were the Vfwd increases with current.
Also the Vfwd is dependant upon its body temperature. check the datasheet.
So don't take for granted you will get 0.7Vfwd across the diode.
By your latest diagram, if you are reading, with a multimeter, 9volts across the diode (positive probe on battery side, neg probe at cathode [striped] end of diode) that means no current is flowing through your circuit.
Possible causes:
Battery weak or dead
LED backwards
Diode backwards
An open connection in your circuit
Resistor open (probably not)
Check, then double-check everything. The circuit as you have drawn it should work.
By your latest diagram, if you are reading, with a multimeter, 9volts across the diode (positive probe on battery side, neg probe at cathode [striped] end of diode) that means no current is flowing through your circuit.
Possible causes:
Battery weak or dead
LED backwards
Diode backwards
An open connection in your circuit
Resistor open (probably not)
Check, then double-check everything. The circuit as you have drawn it should work.
even if the battery is weak, 7V is still enough to power up the entire circuit isnt it?
another thing is that, he said that he read 7V from the multimeter accross the diode. If the diode is not conducting (due to faulty diode, resistor open, faulty LED,and whatever), then isnt the reading should be 9V instead of 7? even if you consider the leakage, it would not be up to 2V right?