Electronic systems such as HDTVs, audio amplifiers, and computers require a

**DC voltage**to operate effectively.Power-line voltage is

**alternating (AC)**and typically too high, so it must be converted.

**Voltage Reduction****Objective:**Lower the high AC line voltage to a suitable level.

**DC Conversion****Objective:**Convert the reduced AC voltage to a constant DC output.

**Power Supply Unit (PSU)**The section of the electronic system responsible for generating the DC voltage.

Includes various internal circuits for different functions.

**Rectifiers****Function:**Allow current to flow in only one direction.**Purpose:**Convert AC voltage to DC voltage.

**Filters****Function:**Smooth out fluctuations in the DC output.**Purpose:**Ensure a steady and consistent DC voltage.

**Voltage Regulators****Function:**Maintain a constant DC output voltage despite variations in input or load conditions.

**Other Circuits****Clippers:**Modify the shape of the voltage waveform.**Clampers:**Shift the voltage level of the waveform.**Voltage Multipliers:**Increase the voltage to a higher level.

Explore the essential elements of power supplies, including rectifiers, filters, voltage regulators, clippers, clampers, and voltage multipliers.

**Chapter Outline**

The Half-Wave and Full-Wave Rectifier

The Choke and Capacitor Filter

Peak Inverse Voltage and Surge Current

Other Power-Supply Topics

Clippers and Limiters

Clampers

Voltage Multipliers

Figure shows a half-wave rectifier circuit with an AC source.

The AC source produces a sinusoidal voltage.

**Positive Half-Cycle:**Forward-biases the diode.

As the switch is closed, the positive half-cycle appears across the load resistor \(\Rightarrow\) unidirectional load current.

**Negative Half-Cycle:**Reverse-biases the diode.

The ideal diode acts as an open switch.

No voltage appears across the load resistor.

Input voltage waveform is a sine wave with an instantaneous value of \(v_{in}\) and a peak value of \(V_p(\text{in})\).

A pure sinusoid has an average value of zero over one cycle.

Each instantaneous voltage has an equal and opposite voltage half a cycle later.

DC voltmeter reading: Zero (indicates average value).

Fig. (b): Diode conducts during positive half-cycles and does not conduct during negative half-cycles.

Fig. (c): The circuit clips off the negative half-cycles, producing a half-wave signal.

The half-wave voltage produces a unidirectional load current.

If the diode were reversed:

It would become forward-biased during the negative half-cycle.

Output pulses would be negative (Fig. (d)).

Fig. (c): The half-wave signal is a pulsating DC voltage.

Characteristics:

Increases to a maximum.

Decreases to zero.

Remains at zero during the negative half-cycle.

This is not the constant DC voltage needed for electronic equipment.

To obtain a constant voltage (like a battery), filtering of the half-wave signal is required (discussed later).

For troubleshooting, use the ideal diode to analyze a half-wave rectifier.

Peak output voltage equals peak input voltage: \[\text{Ideal half-wave}:~V_p(\text{out}) = V_p(\text{in})\]

DC Value of a Half-Wave Signal

**Definition:**DC value = Average value of the signal.**Formula:**

\[\text{DC Value (Half-Wave)} = \frac{V_p}{\pi}\]**Approximation:**

\(\frac{1}{\pi} \approx 0.318\)

\[V_{DC} \approx 0.318 \, V_p\]**Example:**

If \(V_p = 100 \text{ V}\), then

\[V_{DC} = 31.8 \text{ V}\]

RMS Value of a Half-Wave Signal

**RMS Formula:**

\[V_{rms} = 1.57 \, V_{avg}\]**Average Value:**

\[V_{avg} = V_{DC} = 0.318 \, V_p\]**Alternative Formula:**

\[V_{rms} = \frac{V_p}{\sqrt{2}} = 0.707V_p\]**Definition:**

RMS value represents the DC value that produces the same heating effect as the AC waveform.

Output Frequency

**Same Frequency:**

\[f_{out} = f_{in}\]Each input cycle produces one output cycle.

Second Approximation

**Diode Forward Voltage:**

Diode conducts when \(V_{in} > 0.7 \text{ V}\).**Adjusted Voltage:**

\[V_{p(out)} = V_{p(in)} - 0.7 \text{ V}\]**Example:**

If \(V_{p(in)} = 5 \text{ V}\), then

\[V_{p(out)} \approx 4.3 \text{ V}\]

Source voltage \(\Rightarrow 10 \mathrm{~V}_{\mathrm{rms}}\Rightarrow\) calculate the peak ac source value .

The rms value of a sine wave equals: \[V_{\text {rms }}=0.707 V_p\]

The peak source voltage : \[V_p=\frac{V_{\text {rms }}}{0.707}=\frac{10 \mathrm{~V}}{0.707}=14.1 \mathrm{~V}\]

With an ideal diode, the peak load voltage is: \[V_{p(\text { out })}=V_{p(\text { in })}=14.1 \mathrm{~V}\]

The dc load voltage is: \[V_{\mathrm{dc}}=\frac{V_p}{\pi}=\frac{14.1 \mathrm{~V}}{\pi}=4.49 \mathrm{~V}\]

With the second approximation, we get a peak load voltage of: \[V_{p(\text { out })}=V_{p(\text { in })}-0.7 \mathrm{~V}=14.1 \mathrm{~V}-0.7 \mathrm{~V}=13.4 \mathrm{~V}\]

The dc load voltage: \[V_{\mathrm{dc}}=\frac{V_p}{\pi}=\frac{13.4 \mathrm{~V}}{\pi}=4.27 \mathrm{~V}\]

**United States Nominal Line Voltage:**\(120~\mathrm{V_{rms}}\) at 60 HzActual voltage varies from \(105-125~ V_{rms}\).

**Purpose of Transformer:**

Steps down line voltage to safer levels for electronics.Prevents damage to circuits like diodes and transistors.

**Transformer Basics:**Primary winding: Applies line voltage.

Secondary winding: Produces stepped-down voltage.

**Turns Ratio Formula:**\[V_2 = \frac{N_2}{N_1} V_1\]

**Phasing Dots:**Dotted ends have the same phase.

Positive half-cycle on primary = positive half-cycle on secondary.

If dots were on opposite ends, secondary voltage would be 180° out of phase.

**Half-Wave Rectification:**Positive half-cycle: Diode forward biased.

Negative half-cycle: Diode reverse biased.

Result: Half-wave load voltage.

**Step-Up/Step-Down:**Step-up: Secondary voltage \(>\) Primary voltage.

Step-down: Secondary voltage \(<\) Primary voltage.

\[\begin{aligned} V_2 & =\frac{120 \mathrm{~V}}{5}=24 \mathrm{~V}\\ V_p & =\frac{24 \mathrm{~V}}{0.707}=34 \mathrm{~V}\\ V_{\mathrm{dc}} & =\frac{V_p}{\pi}=\frac{34 \mathrm{~V}}{\pi}=10.8 \mathrm{~V}\\ V_{p(\text { out })} & =34 \mathrm{~V}-0.7 \mathrm{~V}=33.3 \mathrm{~V}\\ V_{\mathrm{dc}} & =\frac{V_p}{\pi}=\frac{33.3 \mathrm{~V}}{\pi}=10.6 \mathrm{~V} \end{aligned}\]