A diode ONLY conducts in one direction (well their is the avalache case, but thats another story). An LED is just a DIODE that has been "tweaked" to emit light as well as block voltage

So a diode is a perfect silicon component to only allow current to flow in one direction (hence its use in a rectifier).


There is another proporty of a Diode. the forward voltage. When a diode conducts there is a voltage across the diode (it is exponential but for quick cct design assume it is constant).

So when the diode conducts it has 0.7V across it and it is constant (or there abouts).

An LED has a higher volt drop across it, 1.2V is a good rule of thumb as well.


IF you had a battery of 0.5V and put a diode across it, it would not conduct becuase there is not enough voltage to get over its forward-voltage requirement.


So you have a LED in series with a standard DIODE. The total volt-drop across it is 0.7+1.2 ~ 1.9V. So 1.9V needs to be present to allow that branch to conduct (and for the RED-LED to emit light).

In parallel with this Red-LED+DIODE branch is a GREEN-LED+test leads.

With a faulty fuse (ie open cct) the RED-LED+DIODE will conduct.
With a working fuse (ie it is a short-cct) the GREEN-LED will conduct, this will clamp that parallel branch at 1.2V, this is lower then the 1.9V needed to make the RED-LED branch conduct, thus the RED-LED will not light up and only the GREEN.


In thinking abt it you might want to put a momentary switch between the battery and the burn-resistor.


Different LED's/Diodes will have different current capability so be careful.
What LED's you choose (and what current capability) will determine what value of resistor is needed. That resistor has to drop the rest of the battery voltage (say a 9V battery and 1.2V is dropped via the GREEN-LED, then the resistor has to drop 9-1.2 = 7.8V)

another rule of thumb (that I use) is 10mA is a good current for a standard LED. more current the brighter they go

__________________
Nothing is impossible.
Once a problem is realised, the rest is just details

 

a genius

thanks, i will put a momentary switch there aswell

 

so, in theorey, if i put anything that uses a current in place of the diode, it will still work.... am i correct?

Mike

 

What diode, they are all diodes?

if you mean the non-LED, not really you need something there that has a signincant voltage across it (ie a diode with 0.7V) to increase the forward voltage for the path.

IF you put a resistor in its place then there will still be a slicht leackage through that path and the RED-LED will come on slightly - but then we start getting intot he exponential characteristics of diodes

__________________
Nothing is impossible.
Once a problem is realised, the rest is just details

 

yes, i meant the non-LED, ok, thanks for the help. it is realy usefull

 

sorry to bother you all again ops: but i have 3 diodes that were lying around, i don't know the values of them but I have 1 silicone one (verry small) and 2 germanium diodes (slightly bigger) they are different to what i have seen them like, they are clear with 2 black stripes on one side (tha ones i have seen are black with a yellow or gold stripe)..... would any of these work? btw our teacher tried the black ones (dont know values as they were just in a draw he said) and that didnt work.....

Thanks,
Mike

 

doesnt matter bout the last post, i got it working, thanks alot Styx for giving me the diagram and helping me understand everything more clearly it workes...... well it worked until the red LED stopped working (going to maplins tomorrow so i will get another one then to replace it)
I am a bit behind because i started again, i didnt leave any space in between each row on the stripboard, i tried to solder everything next to each other, because i am not a skilled solderer, it didnt work... well it worked up to the 2nd to last row and then it turned into a mess with solder shorting things out (i.e the current wasnt going through the resistor ect... luckily i didnt connect it up to see if it works :lol: )