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Lecture #14 Example circuits, Zener diodes, dependent sources, basic amplifiers

Lecture #14 Example circuits, Zener diodes, dependent sources, basic amplifiers. Reading: 4.10, 5.1, 5.8 Next: transistors (chapter 6 and 14) . Topics. Today: Examples, circuit applications: Diode circuits, Zener diode Use of dependent sources Basic Amplifier Models.

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Lecture #14 Example circuits, Zener diodes, dependent sources, basic amplifiers

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  1. Lecture #14 Example circuits, Zener diodes, dependent sources, basic amplifiers • Reading: 4.10, 5.1, 5.8 • Next: transistors (chapter 6 and 14) EE 42 fall 2004 lecture 14

  2. Topics Today: Examples, circuit applications: • Diode circuits, Zener diode • Use of dependent sources • Basic Amplifier Models EE 42 fall 2004 lecture 14

  3. Notes 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 model EE 42 fall 2004 lecture 14

  4. Example 1: Ideal Diode Model 1 kW Find ID and VD using the ideal diode model. • Is the diode reverse biased or forward biased? • Make a guess, substitute corresponding circuit for diode. • “Reality check” answer to see if we need to re-guess. ID + _ + - 2 V VD I I Forward bias + _ Reverse bias V V EE 42 fall 2004 lecture 14

  5. IL + VL - Linear circuit Guessing the Diode Mode: Graphing • Look at the diode circuit as a Thevenin equivalent linear circuit attached to a diode. VL = VD IL = -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. ID + _ VD EE 42 fall 2004 lecture 14

  6. Example 1: Ideal Diode Model • Forward biased • VD = 0 V • ID = 2 mA ID 2 mA 2 V VD EE 42 fall 2004 lecture 14

  7. Guessing the Diode Mode: “Common Sense” 1 kW • Notice the polarity of the 2 V falling 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). ID + _ + - 2 V VD EE 42 fall 2004 lecture 14

  8. Example 1: Ideal Diode Model 1 kW • Guess reverse bias: • Since no current is 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. ID + _ + - 2 V VD 1 kW ID + _ + - 2 V VD EE 42 fall 2004 lecture 14

  9. Example 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 bias mode, the diode needs 0.7 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 1 kW ID + _ + - 0.5 V VD EE 42 fall 2004 lecture 14

  10. Zener 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: EE 42 fall 2004 lecture 14

  11. Zener diode as a simple regulator • The Zener diode shown here will keep the regulated voltage equal to its reverse breakdown voltage. R ~ Constant voltage power supply to load EE 42 fall 2004 lecture 14

  12. Resistor 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. EE 42 fall 2004 lecture 14

  13. Ripple 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 EE 42 fall 2004 lecture 14

  14. Here the voltage V is proportional to the voltage across the element c-d . c + V = A x V v cd + V cd - - d Dependent 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. + - OR 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. EE 42 fall 2004 lecture 14

  15. input Differential Amplifier V0 V+ + A  V V0 depends only on input (V+  V-) WHY DEPENDENT SOURCES?EXAMPLE: MODEL FOR AN AMPLIFIER AMPLIFIER SYMBOL output EXAMPLE:A =20 Then if input (V+-V-) = 10mV; Vo = 200mV. An 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. EE 42 fall 2004 lecture 14

  16. AMPLIFIER SYMBOL AMPLIFIER MODEL Differential Amplifier Circuit Model in linear region V0 V+ + A AV1 Ri  V +  V0 depends only on input (V+  V-) + + V0 V1   EXAMPLE OF THE USE OF DEPENDENT SOURCE IN THE MODEL FOR AN AMPLIFIER See 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. EE 42 fall 2004 lecture 14

  17. NODAL ANALYSIS WITH DEPENDENT SOURCES Example circuit: Voltage controlled voltage source in a branch R5 R1 R3 V V V a b c R2 + ISS VAA R4 R6 +  AvVc  Write 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.) CONCLUSION: Standard nodal analysis works EE 42 fall 2004 lecture 14

  18. NODAL ANALYSIS WITH DEPENDENT SOURCES Finding Thévenin Equivalent Circuits with Dependent Sources Present Method 1: Use Voc and Isc as usual to find VT and RT (and IN as well) Method 2: To find RT by the “ohmmeter method” turn off only the independent sources; let the dependent sources just do their thing. See examples in text (such as Example 4.3) and in discussion sections. Pay most attention to voltage-dependent voltage sources and voltage-dependent current sources. We will use these only. EE 42 fall 2004 lecture 14

  19. NODAL ANALYSIS WITH DEPENDENT SOURCES Example : Find Thévenin equivalent of stuff in red box. R V V a c 3 R 2 ISS R + 6 A V  v cs With method 2 we first find open circuit voltage (VT) and then we “measure” input resistance with source ISS turned off. Verify the solution: EE 42 fall 2004 lecture 14

  20. Circuit Model in linear region AV1 Ri +  + + V1 V0   EXAMPLE: AMPLIFIER ANALYSIS USING THE AMPLIFIER MODEL WITH Ri = infinity: A: Find Thévenin equivalent resistance of the input. B: Find the Ratio of the output voltage to the input voltage (“Voltage Gain”) RF AMPLIFIER MODEL RS VIN V-  A V0 V+ + Method: We substitute the amplifier model for the amplifier, and perform standard nodal analysis You find : RIN and VO/VIN EE 42 fall 2004 lecture 14

  21. RF RF RS RS VIN V- VIN V- +   A V0 V0 V+ + V+ AV1 - V1 + EXAMPLE: AMPLIFIER ANALYSIS USING THE AMPLIFIER MODEL WITH Ri = infinity: How to begin: Just redraw carefully! Method: We substitute the amplifier model for the amplifier, and perform standard nodal analysis Verify the solution: RIN = VO/VIN = EE 42 fall 2004 lecture 14

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