Load Voltage In A Half-wave Rectifier

**Strings (S _{i}P_{j}A_{jk}) = S_{7}P_{5}A_{51} Base Sequence = 12735 String Sequence = 12735 - 5 - 51**

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Load Voltage In A Half-Wave Rectifier

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*AC retification* (change of AC signal to DC signal) is one of the important diode applications.

Circuit 121.8 is a *half-wave* rectifier. Determine the DC value of the rectified waveform if the AC voltage source, v_{s} = 52cosωt V. Assume ideal diode.

**The strings**:
S_{7}P_{5}A_{51} (Physical Change).
**The math**:

Pj Problem of Interest is of type *change* (physical change). The *half-wave rectifier* is a *change* device. The positive half of an AC signal is *changed* to a DC signal.

Diode of a half-wave rectifier conducts only during the positive half cycle of the AC signal

So, v_{D} ≥0 only when AC source voltage is positive.

Diode acts an open circuit during the negative half cycle of AC signal.

So, v_{D} <0 when AC source voltage is negative.
*i _{D}* = v

So, v

Average value of load voltage (DC load voltage) is obtained by integrating load voltage over one period, T and dividing by the period, T.

So,DC load voltage v

Where T = 2π/ω

So, v

So, v

So, DC load voltage v

The *point* **.** is a mathematical abstraction. It has negligible size and a great sense of position. Consequently, it is front and center in abstract existential reasoning.

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