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Download Lecture 7 - Voltage Regulator and Half Wave Rectifier
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Recall-Lecture 6 • Diode AC equivalent circuit – small signal analysis – During AC analysis the diode is equivalent to a resistor, rd VDQ + - IDQ DC equivalent rd id AC equivalent DC ANALYSIS DIODE = MODEL 1 ,2 OR 3 CALCULATE DC CURRENT, ID AC ANALYSIS CALCULATE rd DIODE = RESISTOR, rd CALCULATE AC CURRENT, id • Zener effect and Zener diode – 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 Model 1 V = 0 Model 2 V Model 3 V and rf Load Line ID vs VD Forward Biased, DC Analysis At 300K VT = 0.026 V Reverse Biased Group 5 Materials CHAPTER 2 Semiconductor: Group 4 eg. Silicon and Germanium Bandgap Energy Must perform DC Analysis first to get DC diode current, ID PN junction N-type Insulator Conductor Semiconductor AC Analysis Thermal equilibrium, depletion region P-type Group 3 Calculate rd = VT / ID Extrinsic photodiode Intrinsic Other types of diode Solar cells LED Zener Diode Chapter 3 Diode Circuits Voltage Regulator Voltage Regulator - Zener Diode A voltage regulator supplies constant voltage to a load. The breakdown voltage of a Zener diode is nearly constant over a wide range of reverse-bias currents. This make the Zener diode useful in a voltage regulator, or a constantvoltage reference circuit. 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 Example A Zener diode is connected in a voltage regulator circuit. It is given that VPS = 20V, the Zener voltage, VZ = 10V, Ri = 222 and PZ(max) = 400 mW. a. Determine the values of IL, IZ and II if RL = 380 . b. Determine the value of RL that will establish PZ(max) = 400 mW in the diode. ANSWER: Part (a) IL = 26.3 mA IZ = 18.7 mA II = 45 mA ANSWER: Part (b) PZ = IZ VZ IZ = 40 mA IL = 45 -40 = 5 mA RL = 2 k For proper function the circuit must satisfied the following conditions. 1. The power dissipation in the Zener diode is less than the rated value 2. 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), 3. 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) = 0.1 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 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 Relationship between the number of turns of a step-down transformer and the input/output voltages 𝑣𝑃 𝑣𝑆 = 𝑁1 𝑁2 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. Half Wave Rectifier • vs< V, diode off, open circuit, no current flow, Vo = 0V vs > V, diode conducts, current flows, v = vs – V • o V Equation of VO and current when diode is conducting 𝑣𝑂 = 𝑖𝐷 𝑅 = 𝑣𝑆 − 𝑉𝛾 𝑣𝑆 − 𝑉𝛾 𝑖𝐷 = 𝑅 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 v m vs = v m sin t and v m is the peak value 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 i. ii. v = 10 sin t (V) s Determine the peak value of the diode current Sketch vO versus time, t. Label the peak value of vO. v s SOLUTION vO vs > V