## Electronics Reference

# Semiconductor Diodes

I_{D} - Diode current

I_{S} - The reverse saturation current

V_{D} - The applied forward-bias voltage across the diode*n* - Ideality factor which is a function of the operating conditions, physical construction, and has a range between 1 and 2 depending on a wide variety of factors

V

_{T}- Thermal voltage

*k* - Boltzmann’s constant =1.38X10^{-23} (J/K)

T_{K} - is the absolute temperature in Kelvins = 273 + the temperature in °C *q* - is the magnitude of electronic charge 1.6X10

^{-19}°C

When a DC voltage is applied to a circuit that includes a diode, its characteristic curve will have an operating point that will not change with time. The diode’s DC resistance is not sensitive to the shape of the curve and can be simply determined through the ratio of the voltage and current of the diode at the point of interest. As you’ll notice with the equation, the DC resistance of a diode will decrease if there’s an increase with the diode current or voltage.

If an AC voltage is applied to a diode, it becomes sensitive to the shape of the characteristic curve. The AC voltage applied will move the instantaneous operating point up and down in a region of the characteristic curve. Instead of sketching tangent lines, this equation simplifies the process of determining the AC or dynamic resistance of a diode.

Power dissipation of a diode where V_{D} and I_{D} are the voltage and current of the diode, respectively, at a particular point of operation.

The ideal diode model is the least accurate representation of how a diode operates. The diode is simply represented by a switch. When it is forward-biased, the diode ideally operates like a closed switch. Therefore, there’s no voltage drop across the diode. When reverse-biased, it ideally operates like an open switch. The forward current is simply determined through the ratio of the external bias voltage and the series resistor.

In practical diode model, the diode is represented by a voltage source (V_{F}), which is equal to the barrier potential of the diode, and a switch in series. So basically, the barrier potential of the diode is now included in the calculation of the forward current. To calculate the diode’s forward current, simply subtract the barrier potential to the external bias voltage and divide the result with the resistance of the series resistor.

The complete diode model is the most accurate representation. When the diode is forward-biased, the diode is represented by a closed switch in series with the barrier potential (V_{F}) and dynamic resistance (r’_{d}) of the diode. When reverse-biased, the diode acts like an open switch in parallel with a very large internal reverse resistance (r’_{R}). The forward current is calculated by subtracting the barrier potential to the external bias voltage and divide the result with the resistance of the series resistor plus the diode’s dynamic resistance.

Peak Output Voltage | |

Peak Inverse Voltage | |

Output Average Value |

The ability of diodes to only conduct current in one direction makes them ideal to be used as rectifiers. Rectifiers are devices that can convert an AC voltage into a DC voltage and they are always part of DC power supplies that operate from an AC voltage source. A half-wave rectifier only uses a single diode to convert an AC voltage into a pulsating DC voltage. As you can see in the diagram, only the positive half-cycle of the AC voltage can be converted by the have-wave rectifier because the negative cycle will reverse-bias the diode. The net result is that the load will only get the positive half-cycle of the AC voltage input.

To get the peak output voltage, V_{P(OUT)},_{} across the load, just subtract the peak value of the input, V_{P(IN)}, with the barrier potential or forward voltage, V_{F}, of the diode.

Peak inverse voltage (PIV) is the voltage that a rectifier diode must be able to withstand during the reverse-bias operation. In half-wave rectification, the PIV is equal to the peak value of the input AC voltage, V_{P(IN)}.

If you will measure the voltage across the load (R_{L}) using a DC voltmeter, what the voltmeter will get is the average value of the half-wave rectified output voltage (V_{AVG}). You can calculate V_{AVG} by dividing the peak output voltage, V_{P(OUT)}, with a pi (π) value or multiplying it with 0.318.

Output Voltage | |

Peak Inverse Voltage | |

Output Average Value |

While half-wave rectifier can only rectify the positive half-cycle of the AC voltage, full-wave rectifier can rectify both the positive and negative cycles of the input AC voltage. For center-tapped full-wave rectifier, it uses a transformer with a center-tapped type secondary output and two diodes as you can see in the diagram. During the positive half-cycle, D1 rectifies the input as it is forward-biased while D2 is open since it is reverse-biased. When the input voltage switches to the negative half-cycle, D1 becomes reverse-biased and D2 rectifies the input. Since the rectified output current during the positive and negative cycles goes in the same direction through the load, the result across the load would be a full-wave pulsating DC voltage.

The output voltage, V_{P(OUT)}, across the load for a center-tapped full-wave rectifier is equal to half of the total secondary voltage minus the forward voltage (V_{F}) of the diode.

The peak inverse voltage (PIV) across either diode in a center-tapped full-wave rectifier is twice the output voltage, V_{P(OUT)}, plus the forward voltage (V_{F}) of just one diode.

For a full-wave rectifier, the average value of the output voltage is approximately 63.7% of the output voltage. We can also say that it is twice of the half-wave rectifier average value.

Output Voltage | |

Peak Inverse Voltage | |

Output Average Value |

Bridge full-wave rectifier uses four diodes to convert AC voltage into DC. During the positive half-cycle, D1 and D2 are forward-biased and allows the current to pass through the load while D3 and D4 are reverse-biased. When the input shifts into the negative half-cycle, D1 and D2 are reverse-biased. This time D3 and D4 are forward-biased and conducts current in the same direction during the positive half-cycle. As a result, a full-wave rectified output voltage, V_{P(OUT)}, appears across the load (R_{L}).

Since there are two diodes conducting for each half-cycle in a bridge rectifier, the output voltage, V_{P(OUT)}, across the load is subtracted by the voltage drop across the two diodes.

The peak inverse voltage (PIV) for each diode in a bridge rectifier is equal to output voltage, V_{P(OUT)}, across the load plus the forward voltage (V_{F}) of the forward-biased diode inside the loop.

Same with the center-tapped full-wave rectifier, the average value of the output voltage for a bridge rectifier is approximately 63.7% of the output voltage.