Recall-Lecture 6 Zener effect and Zener diode Avalanche Effect When a Zener diode is reverse-biased, it acts at the breakdown region, when it is forward biased, it acts like a normal PN junction diode Avalanche Effect Gain kinetic energy – hit another atom –produce electron and hole pair

Voltage Regulator using Zener Diode 3. The remainder of VPS drops across Ri 2. The load resistor sees a constant voltage regardless of the current 1. The Zener diode holds the voltage constant regardless of the current

Voltage Regulator

For proper function the circuit must satisfied the following conditions. The power dissipation in the Zener diode is less than the rated value When the power supply is a minimum, VPS(min), there must be minimum current in the Zener diode IZ(min), hence the load current is a maximum, IL(max), When the power supply is a maximum, VPS(max), the current in the diode is a maximum, IZ(max), hence the load current is a minimum, IL(min) AND Or, we can write

Considering designing this circuit by substituting IZ(min) = 0 Considering designing this circuit by substituting IZ(min) = 0.1 IZ(max), now the last Equation becomes: Maximum power dispassion in the Zener diode is EXAMPLE 1 Consider voltage regulator is used to power the cell phone at 2.5 V from the lithium ion battery, which voltage may vary between 3 and 3.6 V. The current in the phone will vary 0 (off) to 100 mA(when talking). Calculate the value of Ri and the Zener diode power dissipation

Solution: The stabilized voltage VL = 2.5 V, so the Zener diode voltage must be VZ = 2.5 V. The maximum Zener diode current is The maximum power dispassion in the Zener diode is The value of the current limiting resistance is

Example 2 Range of VPS : 10V– 14V RL = 20 – 100  VZ = 5.6V Find value of Ri and calculate the maximum power rating of the diode

Rectifier

Rectifier Circuits A DC power supply is required to bias all electronic circuits. A diode rectifier forms the first stage of a dc power supply. Diagram of an Electronic Power Supply Rectification is the process of converting an alternating (ac) voltage into one that is limited to one polarity. Rectification is classified as half-wave or full-wave rectifier.

Rectifier Parameters 𝑉 𝑃 𝑉 𝑆 = 𝑁 1 𝑁 2 Relationship between the number of turns of a step-down transformer and the input/output voltages 𝑉 𝑃 𝑉 𝑆 = 𝑁 1 𝑁 2 The peak inverse voltage (PIV) of the diode is the peak value of the voltage that a diode can withstand when it is reversed biased Duty Cycle: The fraction of the wave cycle over which the diode is conducting.

vs< V, diode off, open circuit, no current flow, Vo = 0V vs > V, diode conducts, current flows, vo = vs – V V Equation of VO and current when diode is conducting

Consider a sine wave where vs = vm sin t and vm is the peak value vs < V, diode off, open circuit, no current flow, vo = 0V vs > V, diode conducts, current flows and vo = vs – V Consider a sine wave where vs = vm sin t and vm is the peak value vm Notice that the peak voltage of Vo is lower V vs > V

Example Consider the rectifier circuit in the figure below. Let R = 1 k, and the diode has the properties of V = 0.6 V and rf = 20 . Assume vs= 10 sin t (V) Determine the peak value of the diode current Sketch vO versus time, t. Label the peak value of vO. vs

Solution vO vs > V

FULL WAVE RECTIFIER Center-Tapped Bridge

Full-Wave Rectification – circuit with center-tapped transformer Positive cycle, D2 off, D1 conducts; vo– vs + V = 0 vo = vs - V Negative cycle, D1 off, D2 conducts; vo– vs + V = 0 vo = vs - V Since a rectified output voltage occurs during both positive and negative cycles of the input signal, this circuit is called a full-wave rectifier. Also notice that the polarity of the output voltage for both cycles is the same

vs = vm sin t vm V -V Notice again that the peak voltage of Vo is lower since vo = vs - V vs < V, diode off, open circuit, no current flow, vo = 0V

Full-Wave Rectification –Bridge Rectifier Positive cycle, D1 and D2 conducts, D3 and D4 off; V + vo + V – vs = 0 vo = vs - 2V Negative cycle, D3 and D4 conducts, D1 and D2 off V + vo + V – vs = 0 vo = vs - 2V Also notice that the polarity of the output voltage for both cycles is the same

A full-wave center-tapped rectifier circuit is shown in the figure below. Assume that for each diode, the cut-in voltage, V = 0.6V and the diode forward resistance, rf is 15. The load resistor, R = 95 . Determine: peak output voltage, vo across the load, R Sketch the output voltage, vo and label its peak value.   ( sine wave )

SOLUTION peak output voltage, Vo vs (peak) = 125 / 25 = 5V V +ID(15) + ID (95) - vs (peak) = 0 ID = (5 – 0.6) / 110 = 0.04 A vo (peak) = 95 x 0.04 = 3.8V   ii. 3.8V V -V