Lec4 Diode Applications
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This document discusses half wave and full wave rectification in power supplies. It explains that half wave rectification only uses half of the input waveform, resulting in a lower average output voltage than full wave rectification, which uses both halves of the input waveform. The peak inverse voltage that the diode must withstand is equal to the peak input voltage. Full wave rectification using a bridge circuit provides higher output voltage than a single-diode full wave circuit, and its diodes only need to withstand the peak input voltage minus 0.7 volts.
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Lec4 Diode Applications
- 3. 3 CHAPTER 2
- 6. 6 HALF WAVE RECTIFICATION Average voltage (Vavg) • Vavg is the area under the curve Vp = Peak voltage Effect of barrier potential on o/p Vp(out) = Vp(in) – 0.7 v Peak inverse voltage (PIV) • Diode must be capable of withstanding this amount of repetitive reverse voltage PIV = Vp(in) VAVG = 𝑉𝑝 π
- 9. 9 HALF WAVE RECTIFICATION Transformer coupling Vsec = n Vpri Vp(out) = Vp(sec) – 0.7 V PIV = Vp(sec)
- 12. 12 FULL WAVE RECTIFICATION Vavg = 2𝑉𝑝 π Vout = 𝑉𝑠𝑒𝑐 2 − 0.7 PIV = Vp(sec) – 0.7
- 14. 14 BRIDGE FULL WAVE RECTIFICATION Vp(out) = Vp(sec) -1.4 PIV = Vp(sec) – 0.7
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