You should know this curve and be able to 'interpret' this characteristic that means explain how it shows that:

• The current through a diode effectively only flows in one direction only.
• It's resistance is very low when connected in forward bias as long as it has a potential difference of more than 0.6 volts (this varies but is usually about 0.6 to 0.7 volts) across it.
• The diode has a very high resistance when it is connected in 'reverse bias' - the opposite direction - therefore only a tiny current flows when this is the case.

 

You should note that:

• At 0V no current flows.
• At +0.6V the forward current rises sharply.
• At -ve voltage there is a tiny current.

When connected into a circuit in forward bias the diode is simply like a conductor wire - it has such a low resistance that it hardly affects current flow..

The p.d. across the diode in a circuit is about 0.6V (it's operating voltage - sometimes the question will state that it is 0.65V or 0.7V). So when analysing circuits you have to remember this. Sometimes the examiner will give you a graph to read the operating voltage from.

When connected in reverse bias the diode acts like an open switch in the circuit (it has a very high resistance) so all of the components on that strand will have a negligible current flowing through them - bulbs will effectively be 'off' because so little current will flow that they will not light up.

LOJ March 2003