PowerPoint PresentationTopicsNotes on Use of ModelsExample 1: Ideal Diode ModelGuessing the Diode Mode: GraphingSlide 6Guessing the Diode Mode: “Common Sense”Slide 8Example 2: Large-Signal Diode ModelZener diodesZener diode as a simple regulatorResistor sizingRipple calculationSlide 14Slide 15Slide 16Slide 17Slide 18Slide 19Slide 20Slide 2110/1/2004 EE 42 fall 2004 lecture 14 1Lecture #14 Example circuits, Zener diodes, dependent sources, basic amplifiersReading: 4.10, 5.1, 5.8 Next: transistors (chapter 6 and 14)10/1/2004 EE 42 fall 2004 lecture 14 2TopicsToday:Examples, circuit applications:•Diode circuits, Zener diode•Use of dependent sources•Basic Amplifier Models10/1/2004 EE 42 fall 2004 lecture 14 3Notes on Use of Models•Most of the diode models are piecewise defined:– One function for reverse bias– Another for forward bias•You will need to:– “Guess” which diode or diodes are reverse (or forward) biased– Solve for V, I according to your guess– If result is impossible, guess again•Rarely, both guesses may lead to impossibility.– Then you must use a more detailed model10/1/2004 EE 42 fall 2004 lecture 14 4Example 1: Ideal Diode ModelFind ID and VD using the ideal diode model.•Is the diode reverse biasedor forward biased?•Make a guess, substitutecorresponding circuitfor diode.•“Reality check”answer to see if we need to re-guess.+-IDVD+_2 V1 kWVIReverse biasForward biasIV+_10/1/2004 EE 42 fall 2004 lecture 14 5Guessing the Diode Mode: Graphing•Look at the diode circuit as a Thevenin equivalent linear circuit attached to a diode.VL = VDIL = -ID•Graph the diode I-V curve and the linear circuit I-V curve on the same graph, both in terms of ID and VD.•This means draw the diode I-V curve normally, and draw the linear I-V curve flipped vertically (IL = -ID).•See where the two intersect—this gives you ID and VD.Linear circuitIL+VL-IDVD+_10/1/2004 EE 42 fall 2004 lecture 14 6Example 1: Ideal Diode Model•Forward biased•VD = 0 V•ID = 2 mAVDID2 V2 mA10/1/2004 EE 42 fall 2004 lecture 14 7Guessing the Diode Mode: “Common Sense”•Notice the polarity of the 2 Vfalling over the resistor and diode•The 2 V is in same direction as VD•Diode is probably forward biased•It’s generally easier to guess reverse bias first since it is easy to check.•No matter what piecewise model we use, reverse bias is always open circuit.•So when you don’t know what to do, put in open circuit for the diode, and see if it violates reverse bias conditions (zero current, negative voltage).+-IDVD+_2 V1 kW10/1/2004 EE 42 fall 2004 lecture 14 8Example 1: Ideal Diode Model•Guess reverse bias:•Since no currentis flowing,VD = 2 V (by KVL)•This is impossible for reverse bias (must have negative VD) So the diode must be forward biased•VD = 0 V•ID = 2 V / 1 kW = 2mA•Same as what we got graphically.+-IDVD+_2 V1 k+-IDVD+_2 V1 k10/1/2004 EE 42 fall 2004 lecture 14 9Example 2: Large-Signal Diode Model•The large-signal diode model takes into account voltage needed to forward bias, (VF = 0.7 for silicon) to find ID and VD.•To be in forward biasmode, the diode needs0.7 V. •The source only provides 0.5 V. •The resistor cannot add to the voltage since the diode could only allow current to flow clockwise. •Reverse bias => open circuit => ID = 0 A, VD = 0.5 V+-IDVD+_0.5 V1 k10/1/2004 EE 42 fall 2004 lecture 14 10Zener diodes•A Zener diode is the name commonly used for a diode which is designed for use in reverse breakdown•Since the diode breaks down sharply, at accurate voltage, it can be used as a voltage reference•The symbol for a Zener diode:10/1/2004 EE 42 fall 2004 lecture 14 11Zener diode as a simple regulator•The Zener diode shown here will keep the regulated voltage equal to its reverse breakdown voltage.~Constant voltagepower supplyto loadR10/1/2004 EE 42 fall 2004 lecture 14 12Resistor sizing•How big should R be in the regulator shown?•If the load draws a current I, then the resistor must carry that current when the unregulated voltage is at the lowest point, without letting the regulated voltage drop.•Lets say the load draws 10 milliamps, the regulated voltage is 2 volts, and the minimum unregulated voltage (low point of ripple) is 2.5 volts)The resistor must be •R=(2.5v-2v)/10 milliamps=500 ohms.10/1/2004 EE 42 fall 2004 lecture 14 13Ripple calculation•How much ripple will be observed on the unregulated supply?•The maximum current will be: •And we can estimate the amount of charge lost:•So the ripple will be RVVIregulatedpeakdunregulatepeak)(tIQ CQV10/1/2004 EE 42 fall 2004 lecture 14 14Dependent Voltage and Current Sources• A linear dependent source is a voltage or current source that depends linearly on some other circuit current or voltage. • We can have voltage or current sources depending on voltages or currents elsewhere in the circuit. Sometimes a diamond-shaped symbol is used for dependent sources, just as a reminder that it’s a dependent source.Circuit analysis is performed just as with independent sources.+ -cdVcd+ -V = Av x VcdHere the voltage V is proportional to the voltage across the element c-d .OR+-10/1/2004 EE 42 fall 2004 lecture 14 15 WHY DEPENDENT SOURCES? EXAMPLE: MODEL FOR AN AMPLIFIERV0 depends only on input (V+ V-) )VV(AV0 +AV+VV0Differential AmplifierAMPLIFIER SYMBOLoutputAn actual amplifier has dozens (to hundreds) of devices (transistors) in it. But the dependent source allows us to model it with a very simple element.EXAMPLE: A =20 Then if input (V+-V-) = 10mV; Vo = 200mV.input10/1/2004 EE 42 fall 2004 lecture 14 16 EXAMPLE OF THE USE OF DEPENDENT SOURCE IN THE MODEL FOR AN AMPLIFIERV0 depends only on input (V+ V-) )VV(AV0 +AV+VV0Differential AmplifierAMPLIFIER SYMBOL++V0AV1+V1RiCircuit Model in linear regionAMPLIFIER MODELSee the utility of this: this model, when used correctly mimics the behavior of an amplifier but omits the complication of the many many transistors and other components.10/1/2004 EE 42 fall 2004 lecture 14 17NODAL ANALYSIS WITH DEPENDENT SOURCESExample circuit: Voltage controlled voltage source in a branchWrite down node equations for nodes a, b, and c.(Note that the voltage at the bottom of R2 is “known” so current flowing down from node a is (Va AvVc)/R2.)R5R4 VAA+
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