The diodes used in a half-wave rectifier circuit have the following parameters:
[tex]$V_{T}=0 V, \quad r_{f}=0 \Omega, \quad R_{R}=\infty, \quad V_{Z}=\text { very large }$
The secondary winding of the transformer has a resistance [tex]$R _{ S }=0 \Omega$[/tex]. A load of [tex]$370 \Omega$[/tex] is connected across its output. If the voltage across the secondary is [tex]$v_s=12 \sin w t$[/tex] volts, determine Use the above information to solve question 32 to 35
32. PIV rating of the diodes
(a) 10.0 V
(b) 13.0 V
(c) [tex]$12 . V$[/tex]
(d) 11.5 V
(e) None of the above.
Answer (2)
The Peak Inverse Voltage (PIV) is the maximum reverse voltage a diode experiences in a rectifier circuit.
Identify the peak voltage V m from the secondary voltage v s = 12 sin ( ω t ) , which is 12 V.
The PIV rating of the diode is equal to the peak voltage.
The PIV rating of the diodes is 12 V .
Explanation
Understanding PIV In a half-wave rectifier circuit, the Peak Inverse Voltage (PIV) is the maximum voltage that the diode experiences when it is reverse biased. This occurs during the negative half-cycle of the input AC voltage. The PIV rating of the diode must be greater than or equal to the peak value of the reverse voltage to prevent the diode from breaking down.
Identifying Peak Voltage The secondary voltage is given by v s = 12 sin ( ω t ) volts. The peak voltage V m is the amplitude of the sinusoidal voltage, which is 12 V.
Determining PIV Rating Therefore, the PIV rating of the diode should be equal to the peak voltage, which is 12 V.
Final Answer The PIV rating of the diodes is 12 V.
Examples
In designing power supplies, understanding PIV is crucial. For instance, if you're building a power supply that converts AC voltage from a wall outlet (120V RMS) to DC voltage for charging a laptop, you need to select diodes with an appropriate PIV rating. If the peak voltage after the transformer is, say, 20V, the diodes must have a PIV rating greater than or equal to 20V to prevent them from failing under reverse bias. This ensures the power supply operates safely and reliably, protecting the connected devices from voltage spikes.
The PIV rating of the diodes in a half-wave rectifier circuit is determined by the peak voltage from the secondary winding of the transformer. Given that the output voltage is v s = 12 sin ( ω t ) volts, the peak voltage is 12 V. Therefore, the PIV rating of the diodes is 12 V.
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