Concepts Shown:

semiconduction, junction, bias, n-type, p-type, electrons, holes, fermi energy level, conduction band


AA battery and housing, 0.5mA light bulb, switch (2-way, 6-point), diode (HEP241 - 7252), mounting board and mounting equipment, wire. Cost less than $20.



Wire the battery to the switch, from the switch to the diode, and from here to the light bulb. Have it wired so that flipping the switch changes the direction of current flow into the diode. Thus having the switch in one direction will cause a forward bias current to enter the diode, the opposite direction will force a reverse bias current into the diode. Thus as will be explained below, the bulb should only light when forward bias current is flowing through the diode. [eq].


  1. Draw up a schematic
  2. Explain that by flipping the switch, you are not shutting the current either off or on, instead you are merely changing the direction of the current flowing through the circuit (in a sense, turning the battery around).
  3. Demonstrate the diode by flipping the switch back and forth and observing the light going from on to off
  4. Pass around the display letting students flip the switch and what's going on when the bias is changed.
  5. Discuss the science.

A diode is a junction device, or a device that uses the connection between n- and p-type semiconductors. One way to explain what is going on is to talk about the junction in terms of Fermi distribution. This is a good approach, because it helps better visualize more complex devices, such as transistors. The diode is made up of two layers of silicon, a layer of n-type next to a layer of p-type silicon. When silicon is doped with electrons, to make n-type silicon, the fermi energy level goes up, and vice versa when doped with holes to make p-type silicon. However when the two are put together to form a junction, the fermi energy levels much line up, this causes the conduction band of p-type silicon to be at a much higher level than that of n-type. [eq]. Start with no current, the conductivity through the junction is low because few electrons reside in the conduction band. However, if a forward bias is places across the junction (current moving from p-type to n-type), there is an exponential increase in the number of carriers with the increased voltage bias. The electrons are pushed from the n-type, across the junction into the p-type, and vice versa. Here, the carriers recombine with opposite carriers and cancel some of the n-type or p-type characteristics of the material, including the different levels at which the fermi energy occurs. So what is happening, is that the conduction band is moving toward the same energy level for both the n- and p-type materials. Thus current flows. However, under a negative bias, the electrons are pulled away from the junction and none of this equalization effect occurs, thus no current flows. [eq].


Tyler vanHouwelingen

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