single conductor diode peak inverse question
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ignore this... gettin confused here....
RMS voltage is 0.7071 of peak, so 7.071V
RMS voltage is 0.7071 of peak, so 7.071V
extension15
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When selecting diodes, two device ratings must be taken into consideration; Peak Reverse Voltage and Maximum Average Current.
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Peak Reverse Voltage or Peak Inverse Voltage is the maximum voltage that a diode can withstand in the reverse direction without breaking down and starting to conduct. If this voltage is exceeded the diode may be destroyed. Diodes must have a Peak Inverse Voltage rating that is higher than the maximum voltage that will be applied to them when reverse biased.
In DC only circuits, diodes should have a Peak Inverse Voltage rating greater than the highest voltage to which diode will be exposed.
In AC circuits, such as power supplies, diodes should have a Peak Inverse Voltage rating greater than 1.4 times the maximum RMS voltage of the transformer secondary.
*
Peak Reverse Voltage or Peak Inverse Voltage is the maximum voltage that a diode can withstand in the reverse direction without breaking down and starting to conduct. If this voltage is exceeded the diode may be destroyed. Diodes must have a Peak Inverse Voltage rating that is higher than the maximum voltage that will be applied to them when reverse biased.
In DC only circuits, diodes should have a Peak Inverse Voltage rating greater than the highest voltage to which diode will be exposed.
In AC circuits, such as power supplies, diodes should have a Peak Inverse Voltage rating greater than 1.4 times the maximum RMS voltage of the transformer secondary.
extension15
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Andy, the question is the Peak Inverse Voltage..
Not the RMS value..
The answer is 20V
B-)
extension15
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extension15
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Look at page 3 of the link (the waveform plot)...
You'll see the PIV (in red), its twice the peak (in blue)
regards
Kev
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The rms value of a peak voltage (in this instance 10V) would be 0.7071V
However, the question asks for the Peak Inverse Voltage, as already stated.
The Godfather
However, the question asks for the Peak Inverse Voltage, as already stated.
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so the first rule is to read the question!
extension15
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OK, I also found this:
Half wave rectifier circuits
As the name implies, half wave AC rectifier circuits only use half of the AC waveform in the rectification process. In other words they allow through one half of the cycle and block the other half. This means that power is supplied, to the output of the rectifier circuit - often the smoothing circuit only over half the cycle and this leaves half a cycle when no power is supplied. Accordingly the voltage across any smoothing capacitor drops over this period as charge is removed from the smoothing capacitor by the load circuit. Accordingly levels of ripple are higher than those experienced with full wave rectification as will be seen below.
The circuits for half wave rectifiers are relatively simple. The rectification process can be achieved using a single diode. It is the simplicity of the circuit that makes the half wave rectifier circuit attractive in many applications. It uses a minimum of components and it is able to provide a voltage quite adequately for many uses.
When choosing diodes for use in AC rectifier circuits, one parameter which is of importance is the reverse voltage rating. This is called the Peak Inverse Voltage, PIV. For a half wave rectifier the PIV for the diode must be at least twice the peak voltage of the AC waveform. The reason for this is that it must be assumed that the smoothing capacitor will hold the peak voltage of the AC waveform. Then as the diode is in the non conducting part of the waveform the AC waveform reaches its peak, the diode rectifier will see this peak on top of the peak voltage held by the capacitor, i.e. twice the peak value of the waveform. It is worth noting that the peak value of a sine wave is 1.414 times the RMS value. Thus the PIV rating for the diode must be 2 x 1.414 times the RMS value of the AC waveform. On top of this it is worth leaving a healthy margin to accommodate any spikes that may appear on the supply line.
Half wave rectifier circuits
As the name implies, half wave AC rectifier circuits only use half of the AC waveform in the rectification process. In other words they allow through one half of the cycle and block the other half. This means that power is supplied, to the output of the rectifier circuit - often the smoothing circuit only over half the cycle and this leaves half a cycle when no power is supplied. Accordingly the voltage across any smoothing capacitor drops over this period as charge is removed from the smoothing capacitor by the load circuit. Accordingly levels of ripple are higher than those experienced with full wave rectification as will be seen below.
The circuits for half wave rectifiers are relatively simple. The rectification process can be achieved using a single diode. It is the simplicity of the circuit that makes the half wave rectifier circuit attractive in many applications. It uses a minimum of components and it is able to provide a voltage quite adequately for many uses.
When choosing diodes for use in AC rectifier circuits, one parameter which is of importance is the reverse voltage rating. This is called the Peak Inverse Voltage, PIV. For a half wave rectifier the PIV for the diode must be at least twice the peak voltage of the AC waveform. The reason for this is that it must be assumed that the smoothing capacitor will hold the peak voltage of the AC waveform. Then as the diode is in the non conducting part of the waveform the AC waveform reaches its peak, the diode rectifier will see this peak on top of the peak voltage held by the capacitor, i.e. twice the peak value of the waveform. It is worth noting that the peak value of a sine wave is 1.414 times the RMS value. Thus the PIV rating for the diode must be 2 x 1.414 times the RMS value of the AC waveform. On top of this it is worth leaving a healthy margin to accommodate any spikes that may appear on the supply line.
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extension15
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The answer is 20V...
Kev
2 x 1.414 x 7.071=20vOK, I also found this:Half wave rectifier circuits
As the name implies, half wave AC rectifier circuits only use half of the AC waveform in the rectification process. In other words they allow through one half of the cycle and block the other half. This means that power is supplied, to the output of the rectifier circuit - often the smoothing circuit only over half the cycle and this leaves half a cycle when no power is supplied. Accordingly the voltage across any smoothing capacitor drops over this period as charge is removed from the smoothing capacitor by the load circuit. Accordingly levels of ripple are higher than those experienced with full wave rectification as will be seen below.
The circuits for half wave rectifiers are relatively simple. The rectification process can be achieved using a single diode. It is the simplicity of the circuit that makes the half wave rectifier circuit attractive in many applications. It uses a minimum of components and it is able to provide a voltage quite adequately for many uses.
When choosing diodes for use in AC rectifier circuits, one parameter which is of importance is the reverse voltage rating. This is called the Peak Inverse Voltage, PIV. For a half wave rectifier the PIV for the diode must be at least twice the peak voltage of the AC waveform. The reason for this is that it must be assumed that the smoothing capacitor will hold the peak voltage of the AC waveform. Then as the diode is in the non conducting part of the waveform the AC waveform reaches its peak, the diode rectifier will see this peak on top of the peak voltage held by the capacitor, i.e. twice the peak value of the waveform. It is worth noting that the peak value of a sine wave is 1.414 times the RMS value. Thus the PIV rating for the diode must be 2 x 1.414 times the RMS value of the AC waveform. On top of this it is worth leaving a healthy margin to accommodate any spikes that may appear on the supply line.
B-)
chrishornby
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also got that question in my 301 exam. lucky guess i put 20v! All exams passed, no moree studying!
