1. 10/1/2004EE 42 fall 2004 lecture 141 Lecture #14 Example circuits, Zener diodes, dependent sources, basic amplifiers Reading: 4.10, 5.1, 5.8 Next: transistors (chapter 6 and 14)

2. 10/1/2004EE 42 fall 2004 lecture 142 Topics Today: Examples, circuit applications: Diode circuits, Zener diode Use of dependent sources Basic Amplifier Models

3. 10/1/2004EE 42 fall 2004 lecture 143 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

4. 10/1/2004EE 42 fall 2004 lecture 144 Example 1: Ideal Diode Model Find I D and V D 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. +-+- IDID VDVD +_+_ 2 V 1 kW V I Reverse bias Forward bias I V +_+_

5. 10/1/2004EE 42 fall 2004 lecture 145 Guessing the Diode Mode: Graphing Look at the diode circuit as a Thevenin equivalent linear circuit attached to a diode. V L = V D I L = -I D Graph the diode I-V curve and the linear circuit I-V curve on the same graph, both in terms of I D and V D. This means draw the diode I-V curve normally, and draw the linear I-V curve flipped vertically (I L = -I D ). See where the two intersect—this gives you I D and V D. Linear circuit ILIL +VL-+VL- IDID VDVD +_+_

6. 10/1/2004EE 42 fall 2004 lecture 146 Example 1: Ideal Diode Model Forward biased V D = 0 V I D = 2 mA VDVD IDID 2 V 2 mA

7. 10/1/2004EE 42 fall 2004 lecture 147 Guessing the Diode Mode: “Common Sense” Notice the polarity of the 2 V falling over the resistor and diode The 2 V is in same direction as V D 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). +-+- IDID VDVD +_+_ 2 V 1 kW

8. 10/1/2004EE 42 fall 2004 lecture 148 Example 1: Ideal Diode Model Guess reverse bias: Since no current is flowing, V D = 2 V (by KVL) This is impossible for reverse bias (must have negative V D )  So the diode must be forward biased V D = 0 V I D = 2 V / 1 kW = 2mA Same as what we got graphically. +-+- IDID VDVD +_+_ 2 V 1 k  +-+- IDID VDVD +_+_ 2 V 1 k 

9. 10/1/2004EE 42 fall 2004 lecture 149 Example 2: Large-Signal Diode Model The large-signal diode model takes into account voltage needed to forward bias, (V F = 0.7 for silicon) to find I D and V D. 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 => I D = 0 A, V D = 0.5 V +-+- IDID VDVD +_+_ 0.5 V 1 k 

10. 10/1/2004EE 42 fall 2004 lecture 1410 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:

11. 10/1/2004EE 42 fall 2004 lecture 1411 Zener diode as a simple regulator The Zener diode shown here will keep the regulated voltage equal to its reverse breakdown voltage. ~ Constant voltage power supply to load R

12. 10/1/2004EE 42 fall 2004 lecture 1412 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.

13. 10/1/2004EE 42 fall 2004 lecture 1413 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

14. 10/1/2004EE 42 fall 2004 lecture 1414 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. 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. + - c d V cd + - V = A v x V cd Here the voltage V is proportional to the voltage across the element c-d. OR +-+-

15. 10/1/2004EE 42 fall 2004 lecture 1415 WHY DEPENDENT SOURCES? EXAMPLE: MODEL FOR AN AMPLIFIER V 0 depends only on input (V +  V - ) +  A V+V+ VV V0V0 Differential Amplifier AMPLIFIER SYMBOL output 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. EXAMPLE: A =20 Then if input (V + -V - ) = 10mV; V o = 200mV. input

16. 10/1/2004EE 42 fall 2004 lecture 1416 EXAMPLE OF THE USE OF DEPENDENT SOURCE IN THE MODEL FOR AN AMPLIFIER V 0 depends only on input (V +  V - ) +  A V+V+ VV V0V0 Differential Amplifier AMPLIFIER SYMBOL +  +  V0V0 AV 1 +  V1V1 RiRi Circuit Model in linear region AMPLIFIER MODEL 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.

17. 10/1/2004EE 42 fall 2004 lecture 1417 NODAL ANALYSIS WITH DEPENDENT SOURCES Example circuit: Voltage controlled voltage source in a branch Write down node equations for nodes a, b, and c. (Note that the voltage at the bottom of R 2 is “known” so current flowing down from node a is (V a  A v V c )/R 2.) R5R5 R4R4 V AA +  I SS R3R3 R1R1 V a V b +  R6R6 V c R2R2 CONCLUSION: Standard nodal analysis works AvVcAvVc

18. 10/1/2004EE 42 fall 2004 lecture 1418 NODAL ANALYSIS WITH DEPENDENT SOURCES Finding Thévenin Equivalent Circuits with Dependent Sources Present Method 1: Use V oc and I sc as usual to find V T and R T (and I N as well) Method 2: To find R T 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.

19. 10/1/2004EE 42 fall 2004 lecture 1419 NODAL ANALYSIS WITH DEPENDENT SOURCES Example : Find Thévenin equivalent of stuff in red box. With method 2 we first find open circuit voltage (V T ) and then we “measure” input resistance with source I SS turned off. Verify the solution: I SS R 3 V a +  A v V cs R 6 V c R 2

20. 10/1/2004EE 42 fall 2004 lecture 1420 EXAMPLE: AMPLIFIER ANALYSIS USING THE AMPLIFIER MODEL WITH R i = infinity: A: Find Thévenin equivalent resistance of the input. B: Find the Ratio of the output voltage to the input voltage (“Voltage Gain”) Method: We substitute the amplifier model for the amplifier, and perform standard nodal analysis You find : R IN and V O /V IN +  +  V0V0 AV 1 +  V1V1 RiRi Circuit Model in linear region AMPLIFIER MODEL +  A V-V- V+V+ V0V0 RFRF RSRS V IN

21. 10/1/2004EE 42 fall 2004 lecture 1421 EXAMPLE: AMPLIFIER ANALYSIS USING THE AMPLIFIER MODEL WITH R i = infinity: How to begin: Just redraw carefully! Method: We substitute the amplifier model for the amplifier, and perform standard nodal analysis Verify the solution: R IN = V O /V IN = +  AV 1 - + V1V1 V-V- V+V+ V0V0 RFRF RSRS V IN +  A V-V- V+V+ V0V0 RFRF RSRS