With most electricity generated to have alternating current, it is useful in some cases for this to be converted into direct current. This process is called rectification.
At its heart is the diode, an electrical component that only conducts electricity in one direction.
Key Concepts
A diode is an electronical component that conducts current in one direction (forward bias).
An ideal diode would have zero resistance in one direction and infinite) resistance in the other.
Half wave
The simplest form of rectification is the insertion of a diode in the series circuit. All current in the forward direction is transmitted; no current flows in reverse bias.
Notice that the diode symbol acts as an arrow to show the direction in which conventional current will flow.
This circuit has the downside that current is only flowing for half of the time.
Full wave
Full wave rectification uses two diode pathways (in opposite directions) to enable the current to flow at all times. Each diode path runs through the same resistor (or desired electrical component).
See if you can trace a finger around the following circuit diagram, first starting upwards from the power supply and then downwards.
While an improvement on half wave rectification, the output is still sinusoidal in magnitude.
An understanding of rectification circuit diagrams can also be used to determine unknown values. The following circuit is known as a Wheatstone Bridge; the current directions are arbitrary.
\(R_1\) and \(R_3\) are known fixed resistors. \(R_2\) is adjusted until \(V_G\) is zero, at which point there is no net current from D to B. \(R_\text{x}\) can be determined:
\(R_1+R_2=R_3+R_\text{x}\)
Use quizzes to practise application of theory.
START QUIZ!
The order of magnitude of the resistivity of silicon in Ωm is:
Silicon is a semiconductor. Its resistivity is the most moderate value, in between a conductor (10-8) and insulator (109).
The graph shows the \(IV\) characteristics for an LED.
\(V_1\) is approximately:
The energy required to produce photons of visible light is in the order of 1 eV - therefore 1 V is needed to accelerate an electron across the diode.
Conduction in a P-type semiconductor is due to the movement of:
Electrons are minority charge carriers.
One could argue that hole current is also due to electron movement but the correct answer is the best presented.
The image represents an LED and a 5 V AC source.
Which graph represents the potential difference across the LED?
The LED requires about 1.5 V forward bias before it conducts.
The image shows an LED connected to a 5 V battery and a 1 kΩ resistor.
The potential difference across the LED is:
The diode is in reverse bias so no current flows. The potential difference across the resistor \(V_\text{R}=IR=0\text{ V}\)
\(\Rightarrow V_\text{LED}=5\text{ V}\)
The image shows an LED connected to a 5 V battery and a 1 kΩ resistor.
The potential difference across the LED is:
The diode is in forward bias. This means that the potential difference across it is that required to overcome the depletion layer - a value that you should be prepared to estimate.
The image shows two diodes connected to a 5 V battery and a 1 kΩ resistor.
The potential difference across diode A is:
A is in reverse bias so has very large resistance; all potential is dropped across A.
The colour scheme in the simulation gives us a clue: grey is 0 V and green is 5 V.
The image shows a regular diode and an LED connected to a 5 V AC source and a 1 kΩ resistor.
Which graph represents the potential difference across the resistor?
The LED requires a larger forward bias potential difference than a regular diode. Notice the asymmetry on the graph.
Which of the following circuits represents a full wave rectifier?
The current must pass through the resistor in the same direction for both halves of the cycle - so C is the correct response.
In A, notice that the two lower diodes are in the same direction.
In B, the resistor flow direction would change for each half of the cycle.
How much of Diodes and rectification have you understood?
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