The half wave rectifier is obtained by using a single diode to let only half of the sine wave go through.

The source can be any signal, but the most frequent would be a sine signal, for instance coming out of a transformer.If we neglect the drop due to the diode, the potential difference on the load will be half of the sine wave.

Now comes a big question: what DC potential difference would be needed to get the same energy dissipated through a purely resistive load? If we call

*u(t)*the varying, periodic signal and

*U*the potential difference needed to generate the same energy, we get

*T*is the signal's period. Remembering that

*I=U/R*(Ohm's law), we get that

*R*is constant in time, it can be taken out of the integral, and we end with

Let's go through a few examples.

###
Sine wave of frequency *f* and peak *U*_{p}.

_{p}

Such a signal can be described by the function

Where ω is the

###
Rectangular wave of amplitude

The duty cycle ω is the fraction of the period during which the signal will be at

And thus

And finally ...

###
Half wave rectified sine of amplitude

*pulsation*and is###
Rectangular wave of amplitude *U*_{p} and duty cycle ω

_{p}

The duty cycle ω is the fraction of the period during which the signal will be at

*U*. Otherwise, the signal is 0._{p}And thus

And finally ...

###
Half wave rectified sine of amplitude *U*_{p} and frequency *f*.

_{p}

This is almost the same as for the full sine wave, except that instead of taking the integral from

*0*to*T*, the integral goes only to*T/2*. This gives the result
Before moving to a different topic, let's establish a property we will use later.

Let's consider the signal

And

We will prove that

Let's go!

The part labeled

The part labeled

Substituting back, we conclude that

This result will help us later.

{To be continued!}

Let's consider the signal

*u(t)*with period*T*defined byAnd

We will prove that

Let's go!

The part labeled

*A*develops like thisThe part labeled

*B*is easy to develop and givesSubstituting back, we conclude that

This result will help us later.

{To be continued!}

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