Diodes

Diodes are semiconductor devices that allow current to flow only in one direction. Diodes look like this:

The diagram is labelled with an anode and a cathode. The voltage drop accross the diode is from anode -> cathode, and the current is conducted in the direction pointed by the really big black arrow.

The type's of diode's we're concerned with are silicon diodes, which have a forward voltage of about 0.7V. This is only an approximation, but is the value to use in calculations.

IV characteristics

Diodes are non-linear components:

• When current is flowing from anode to cathode, the diode is forward-biased, and will conduct current
• When the current is flowing backwards (the wrong way), the diode is reverse-biased.
• At a large negative voltage, the diode will break down, and start to conduct current again
  • Don't let the voltage get this high, you wont like what happens.

Forward Voltage

For the diode to conduct, it must have a minimum voltage accross it, known as the forward voltage. This is also always the total voltage drop accross the diode. For a silicon diode, this is 0.7V, which is why the I-V graph does not go up from zero. The diode can be said to "open" or "switch on" at about this voltage.

• If there is a voltage of 0.2V accross a diode, no current will flow
• If there is a voltage of 0.6V accross a diode, a tiny amount of current may flow
• At >0.7V, the full current will flow with no resistance.

Example 1

Find the current and the voltages accross each component in the circuit below.

By Ohm's law, the current is:

$$I=\frac{V_{in}-V_D}{R_t}=\frac{10-0.7}{100+300}=23.25mA$$

Thefore, the voltages are $$V_{300R}=300\times 23.25=6.98V$$ $$V_{100R}=100\times 23.25=2.32V$$ $$V_D=0.7V$$

Example 2

Find the current through each resistor in the circuit below.

Doing KCL around node $V_x$:

$$\frac{V_x-9}{500}+\frac{V_x}{100}+\frac{V_x-0.7}{50}=0$$ $$V_x=1V$$

The three currents are then:

$$I_{500R}=\frac{9-1}{500}=16mA$$ $$I_{100R}=\frac{1}{100}=10mA$$ $$I_{50R}=\frac{1-0.7}{50}=6mA$$