# Output DC voltage and Frequency of Half-wave, Full-wave and Bridge Rectifier

An AC signal is converted to DC signal using an electronics circuit which is called a rectifier. There are mainly three types of rectifiers called half wave rectifier, full wave rectifier and bridge rectifier. These are build using diodes. The DC voltage and frequency of the output signal from a rectifier circuits are determined by the type of rectifier being used. Half-wave rectifiers provide lower DC voltage output but higher output signal frequency than the full-wave rectifiers and bridge rectifiers. The full wave and bridge rectifier have the same output signal DC voltage and output signal frequency.

In this tutorial it is shown how the average value of DC voltage and frequency of these three types of rectifier are measured and calculated. The DC value or the average DC voltage is the voltage one measures using a DC voltmeter. Here we will use the 1N4001 diode as an example for the rectifiers.

### Half Wave Rectifier

Half-wave rectifiers use a single diode to convert alternating current (AC) to direct current (DC). The output is a continuous waveform with a pulsating DC voltage, but with a peak voltage twice that of the AC input voltage. The frequency of the output is equal to the frequency of the AC input.

The following shows a half wave rectifier circuit with 1N4001 diode and 1kOhm load resistor driven by 10Vrms and 60Hz sine wave signal.

The input and output waveform are shown below.

Average DC value or dc load voltage of a half wave signal is given by the following relation.

$$V_{dc} = \frac{V_p}{\pi}$$

The peak voltage $$V_p$$ is related to the $$V_{rms}$$ by the following equation.

$$V_{rms} = \frac{V_p}{\sqrt{2}}$$

In the above example, $$V_{rms}=10V$$ so the peak voltage $$V_p$$ is given by,

$$V_p = \sqrt{2} V_{rms}$$

and so,  $$V_p = \sqrt{2}\times 10V = 14.14V$$

For quick approximation we can say there is no voltage drop across the diode, that is an ideal diode and for this case we can the peak voltage at the output is,

$$V_{p(out)} = V_{p(in)} = 14.14V$$

and the dc voltage is given by,

$$V_{dc} = \frac{V_{p(out)}}{\pi}=\frac{14.14V}{\pi}=4.5V$$

For accurate calculation we have to take into account the voltage drop across the diode. The following relates input and output voltage across the diode.

$$V_{p(in)} - V_{p(out)} = V_D$$

where $$V_D$$ is the voltage across the diode which we will take as 0.7V.

So,   $$V_{p(in)} - V_{p(out)} = 0.7V$$

or,  $$V_{p(out)} = V_{p(in)} - 0.7V$$

or,  $$V_{p(out)} = 14.14V - 0.7V = 13.44V$$

and the dc load voltage is,

$$V_{dc} = \frac{13.44V}{\pi}=4.28V$$

The frequency of the signal at the diode output is the same as the input signal in a half wave rectifier.

$$f_{out}=f_{in}=60Hz$$

The following Fourier graph shows the frequency of the input and the output signal.

As can be seen both the input and output signal have frequency of 60Hz, so no change in frequency in a half rectifier.

For half wave rectifier, the dc voltage value at the output and the frequency is shown in the diagram below.

#### Full Wave Rectifier

Full-wave rectifiers use two diodes allowing the current flowing through them to alternate directions. This creates a DC voltage output that is almost twice that of the AC input voltage, with a frequency equal to half the frequency of the AC input.

Below is a circuit diagram where the input is an AC signal of 120Vrms and frequency of 60Hz.

Here a center tapped transformer is used with turn ratio 10. That is if the number of turns in the primary is Np then the number of turns in the secondary winding is Ns=10xNp. We can also get the primary and secondary inductance using the online transformer calculator. Here if primary inductance Lp=100mH then the secondary winding inductance Ls=1mH.

The input and output waveform of the above full wave rectifier is shown below.

Here we will calculate the dc voltage at the output of the rectifier and also the frequency of the output signal of the diode.

The peak voltage $$V_p$$ on the primary side is,

$$V_{pp} = \sqrt{2} V_{rms}$$

Here, $$V_{rms}=120V$$ so the primary peak voltage $$V_{pp}$$ is,

$$V_{pp} = \sqrt{2}\times 120V = 169.7V$$

Next we can calculate the secondary peak voltage($$V_{ps}$$)using the turn ratio(N) of 10.

$$V_{ps} = \frac{V_{pp}}{N}=\frac{169.7V}{10}=16.97V$$

Next we can calculate the voltage at the input of each diodes by using the knowledge that there is a center tap which divides the voltage into equal voltage going into the diodes.

For each diode then we have the input voltage,

$$V_{ps(in)} = \frac{V_{ps(in)}}{2} = \frac{16.97V}{2}=8.48V$$

Then the peak voltage at the diode output is,

$$V_{ps(out)} = V_{ps(in)} - 0.7V$$

or, $$V_{ps(out)} = 8.48V - 0.7V = 7.78V$$

The DC value of full wave rectifier is given by the following equation,

$$V_{dc} = \frac{2 V_{ps(out)}}{\pi}=\frac{2 \times 7.78V}{\pi}=4.95V$$

The dc value of the output signal of a full wave rectifier is twice the peak voltage of signal at the secondary because a full wave rectifier has twice as many pulses as in half wave rectifier.

The frequency of the output signal from a full wave rectifier is given by,

$$f_{out}=2 f_{in}= 2 \times 60Hz = 120Hz$$

For full wave rectifier, the frequency of the output signal is twice than the input signal because there are twice as many pulses as in half wave rectifier case. The Fourier graph below shows the input and output signal frequencies.

So for full wave rectifier the average dc voltage and the frequency and other equation is shown in the following diagram.

#### Bridge Rectifier

Bridge rectifiers contain four diodes arranged in a bridge configuration. They allow for either positive or negative DC voltage output, depending on the AC input signal. The output voltage has a constant DC value, and the frequency of the output is one-fourth of the frequency of the AC input.

A bridge rectifier is shown in the circuit diagram below.

In the above circuit the four diodes forms the circuit which blocks or allow to pass the input ac signal through them. During the positive half cycle the diodes D1 and D2 passes the signal while the didoes D3 and D4 are open(blocks the passage of the signal). Similarly during the negative half cycle the diodes D3 and D4 allows the signal to pass through them while the didoes D1 and D2 are open.

In the above circuit the input AC signal is 120Vrms with frequency of 60Hz. The average value and the output signal frequency is the same as the full wave rectifier which are given below.

The dc voltage of the output signal is,

$$V_{dc} = \frac{2 V_{ps(out)}}{\pi}=\frac{2 \times 7.78V}{\pi}=4.95V$$

This is the same values we have used in the full wave rectifier example.

The input and output waveform is also the same as the full wave rectifier case which is shown below.

And the frequency of the output signal of the bridge rectifier is also the same as for the full wave rectifier which is given below.

$$f_{out}=2 f_{in}= 2 \times 60Hz = 120Hz$$

The frequency graph of the input and output signal of the bridge rectifier is shown below.

### Summary

In all three rectifiers, the current flow is in one direction only, and the output voltage is either constant or pulsating DC. The different output voltages and frequencies depend on the type of rectifier used, with the half-wave rectifier providing comparatively lower DC voltage but highest frequency than the full-wave rectifier and bridge rectifier.

The output DC voltage and frequency of a half-wave rectifier is equal to the peak of the secondary AC signal divided by PI, while the frequency remains the same. The output DC voltage and frequency of a full-wave rectifier is higher, with twice the peak of the secondary AC signal divided by PI and twice the frequency. Finally, the bridge rectifier has four diodes, and produces an output DC voltage of secondary AC signal divided by PI(same as half wave rectifier) and with output signal frequency that is twice the frequency of the input signal(same as half wave rectifier).

The knowledge of the output DC voltage and the output signal frequency from rectifiers are important because diodes are used in this configuration not only in rectifiers but also in other circuit application such as in AM single diode modulator, two diode modulator and in ring diode modulator. As BJT or FET type transistors can be implemented as diodes this knowledge on output signal amplitude and frequency from diode is also important for transistor based circuit analysis.