Shockley diode

Working

Unlike other semiconductor diodes, the Shockley diode has more than one p–n junction. The construction includes four sections of semiconductors placed alternately between the anode and cathode in the pattern of PNPN. Though it has multiple junctions, it is termed a diode for being a two-terminal device.

The Shockley diode remains in an OFF state, with a very high resistance, until a voltage greater than the trigger voltage is applied across its terminals. When the voltage exceeds the trigger value, the resistance drops to an extremely low value and the device switches ON. The constituent transistors help in maintaining the ON and OFF states. Since the construction resembles a pair of interconnected bipolar transistors, one PNP and other NPN, neither transistor can turn ON until the other is turned ON due to the absence of any current through the base-emitter junction. Once sufficient voltage is applied and one of the transistors breaks down, it starts conducting and allows base current to flow through the other transistor, resulting in saturation of both the transistors, keeping both in ON state.

On reducing the voltage to a sufficiently low level, the current flowing becomes insufficient to maintain the transistor bias. Due to insufficient current, one of the transistors will cut off, interrupting the base current to the other transistor, hence sealing both transistors in the OFF state.