EECS 314 Fall 2007 HW 09 Problem 1 Student's name ___________________________ (Last name, first name, IN INK) Discussion section # __________ The big picture This problem introduces several important aspects of rectifier circuits: the actual voltages across the load resistance (they are lower than the peak voltage of the sinusoidal AC source due to the voltage drops across the diodes: in the bridge circuit, two diodes are in series with the load), the polarity of output voltage, and the careful choice of the ground node. Refer to the file Rectifier circuits and to the lecture notes already posted on the CTools web site. Problem Part 1 (10 points) Consider the bridge rectifier circuit shown on the first diagram. Assume the offset diode model with VDO = 0.7 V and the following circuit parameters: VIN = (5 V)cos(260t) RLOAD = 20 1. Sketch the output voltage VOUT across the load resistor; calculate its maximal and minimal values (in volts) 2. Calculate the maximal and minimal current ILOAD through the load resistor (in mA) 3. Calculate the interval of time (in msec), over which the output voltage remains equal to zero during each period of the input sinusoidal voltage at 60 Hz. Part 2 (5 points) Repeat for the second circuit: calculate 4. Sketch the output voltage VOUT across the load resistor; calculate its maximal and minimal values (in volts) 5. Calculate the maximal and minimal current ILOAD through the load resistor (in mA) 6. Calculate the interval of time (in msec), over which the output voltage remains equal to zero during each 2007 Alexander Ganago Page 1 of 2 EECS 314 Fall 2007 HW 09 Problem 1 Student's name ___________________________ (Last name, first name, IN INK) period of the input sinusoidal voltage at 60 Hz. Discussion section # __________ Part 3 (10 points) Consider the circuit shown on the third diagram, which is very similar to that on the second diagram above, but the student who built it decided to ground the transformer's output coil in addition to grounding the load resistor. Analyze and explain how the circuit operates over one period of the input sinusoidal voltage: for simplicity, consider the two moments of time, labeled T1 and T2 on the sketch below. For each moment of time T1 and T2, do the following: 7. Explain which diodes (A, B, C, E) are conducting 8. Explain whether the current flows through the load resistor 9. Calculate the the output voltage VOUT across the load resistor Write your conclusion on the operation of the rectifier circuit with two ground nodes shown here: 10. Explain whether adding the second ground node is effective 11. Explain whether adding the second ground node is safe 12. Add any other comments on the operation of this circuit if you wish 2007 Alexander Ganago Page 2 of 2 EECS 314 Fall 2007 HW 09 Problem 2 Student's name ___________________________ (Last name, first name, IN INK) Discussion section # __________ For The big picture, refer to the 3-page document Power Supply and Voltage Regulation attached to this set of problems. Problem Part 1 (18 points = 3 points for each correct sketch) Consider a Full-Wave Rectifier with a transformer and a bridge circuit with four diodes as in Problem 1 of this set; assume offset diode model with VDO = 0.7 V. Consider the 60-Hz sinusoidal input and the peak amplitude of the sinusoidal voltage across the secondary coil VIN, PEAK = 5 V. Sketch the following waveforms (three periods of each): 1. Input voltage (indicate the time in msec) 2. Output voltage measured across the resistor, without any capacitor in the circuit; indicate the peak output voltage in volts 3. Output current through the load resistor, in the circuit without any capacitor 4. Output voltage measured across the load resistor, to which a capacitor is connected in parallel (for clarity, show the ripple voltage of about of the peak output voltage) 5. The current through the load resistor, in the same circuit 6. The current through the capacitor (positive when the capacitor is charging, and negative when the capacitor is discharging), in the same circuit. Clearly label the intervals of time (do not calculate the moments of time; a qualitative picture is enough) during which the diode conducts and charges the capacitor, and during which the diode does NOT conduct and the capacitor is discharged. Part 2 (4 points) Calculate what capacitance C is needed to ensure that the ripple voltage in the output of the Full-Wave Rectifier does not exceed 1% of the peak voltage, with the load resistance equal to 1 k. Determine the ripple voltage (as % of the peak voltage), which will be obtained in the circuit with a Half-Wave rectifier, the same load resistance and capacitance. Part 3 (3 points) Repeat Part 2 calculations for the load resistance equal to 50 . Show your work for each part of the problem: use separate sheets. 2007 Alexander Ganago Power supply and voltage regulation Power Supply and Voltage Regulation A rectifier produces output voltage of constant polarity but NOT constant amplitude. Filters are needed to reduce the ripples in the output voltage. The simplest filter is a capacitor in parallel with the load resistor. 2007 A. Ganago Power supply &amp; voltage regulation 1 Half-Wave Rectifier V VIN RL VRL constant polarity, but not constant amplitude VRL has The diode conducts if VIN &gt; VDO , according to the offset model; here VDO =0.7 V 2007 A. Ganago Power supply &amp; voltage regulation 2 Half-Wave Rectifier's Output Is NOT Constant DC Voltage 5V 0V -5V VIN Time A power supply includes a Low-Pass filter to suppress AC and keep DC in the output voltage. The simplest filter is a capacitor in parallel with the load resistor VIN C R VOUT VRL 4.3 V Time 0V VRL, MAX = VIN, MAX -VDO = 5 V - 0.7 V= 4.3 V 2007 A. Ganago Power supply &amp; voltage regulation 3 What does the capacitor do? 2007 A. Ganago Power supply &amp; voltage regulation 4 First, consider only the capacitor without any load resistor VIN C VC R Without a load resistor, the capacitor charges during the first quarter-period and never discharges thereafter VIN 5V Time 0V -5V 4.3 V 0V VC (without R) Time 6 The diode conducts if VIN - VDO &gt; VC and charges the capacitor to VIN, MAX -VDO 2007 A. Ganago Power supply &amp; voltage regulation 5 2007 A. Ganago Power supply &amp; voltage regulation 2007 A. Ganago Page 1 of 3 1 Power supply and voltage regulation What happens if the load resistor is added? IC VIN C VC IR R VR = VC With a load resistor, the capacitor charges when VIN &gt; VC (the diode conducts) then discharges when VIN &lt; VC (the diode V does not conduct) 5V 0V -5V 4.3 V 0V VC = VR Time Power supply &amp; voltage regulation 8 IN Time The diode conducts if VIN - VDO &gt; VC and charges the capacitor to VC, MAX = VR, MAX = VIN, MAX -VDO 2007 A. Ganago Power supply &amp; voltage regulation 7 2007 A. Ganago The capacitor is charged IC &gt; 0 and discharged IC &lt; 0 every period 4.3 V 0V IC 0 2007 A. Ganago Power supply &amp; voltage regulation When IC &gt; 0, the diode conducts; and when IC &lt; 0 , the diode does NOT conduct 4.3 V 0V Time IC 0 Time VOUT = VC = VR Time Time The smaller ripples you demand, the higher diode peak current you need 9 2007 A. Ganago Power supply &amp; voltage regulation 10 The output voltage is nearly constant (with ripples) 4.3 V 0V VRIPPLES Half-Wave Rectifier: VRIPPLES &quot; VOUT, PEAK # T = VOUT, R #C PEAK # 1 f # R #C VOUT = VC = VR Time Full-Wave Rectifier: VRIPPLES ! VRIPPLES &quot; VOUT,PEAK # T 1 = VOUT,PEAK # 2 # R #C 2 # f # R #C T 1 &quot; VPEAK # = VPEAK # R #C f # R #C ! The ripples are smaller (by ~ half) in a Full-Wave Rectifier circuit because the discharge time is shorter (by ~ half) T is the period of the input sine wave T is the period of input sine wave ! 2007 A. Ganago Power supply &amp; voltage regulation 11 2007 A. Ganago Power supply &amp; voltage regulation 12 2007 A. Ganago Page 2 of 3 2 Power supply and voltage regulation To get rid of the ripples, use <a href="/keyword/voltage-regulator/" >voltage regulator</a> s: The simplest <a href="/keyword/voltage-regulator/" >voltage regulator</a> is a reversebiased Zener diode R VIN 2007 A. Ganago I -VZ V VD0 If the Zener diode dissipates I too much power, it can be destroyed. Resistor R is -VZ V needed to limit the current IZ VD0 through the Zener diode R IZ, MAX RL VOUT = VZ 14 C Power supply &amp; voltage regulation RL VOUT = VZ 13 VIN 2007 A. Ganago C Power supply &amp; voltage regulation Zener diodes regulate (maintain constant) the output voltage: they maintain it at VZ and eliminate ripples. The above circuit with a Zener diode is a simple <a href="/keyword/voltage-regulator/" >voltage regulator</a> . 2007 A. Ganago Power supply &amp; voltage regulation 15 Zener diodes are available with VZ ranging from about 2 V to nearly 200 V, for various applications See for example, the data sheet http://www.fairchildsemi.com/pf/BZ/BZX79C5V6.html 2007 A. Ganago Power supply &amp; voltage regulation 16 I -VZ V VD0 Many other voltage regulating circuits are available. 2007 A. Ganago Power supply &amp; voltage regulation 17 2007 A. Ganago Power supply &amp; voltage regulation 18 2007 A. Ganago Page 3 of 3 3 EECS 314 Fall 2007 HW 09 Problem 3 Student's name ___________________________ (Last name, first name, IN INK) Discussion section # __________ The Big Picture: Inductive kick An abrupt change of inductor current (for example, an interruption of current when the switch is suddenly open) causes an abrupt change of inductor voltage, known as the inductor's kickback (which can damage the switch and can be dangerous for the human operator). This effect is well-known and discussed on many web sites, such as: http://www.wisc-online.com/objects/index_tj.asp?objID=ACE5803 http://www.coilgun.info/theory/inductivekickback.htm Note that an inductor has many loops of wire around a magnetic core, and an electric motor has many loops of wire around a magnetic core thus motors act as inductive loads, in particular, motors exhibit the kickback. Consider the circuit shown on the diagram. Here the battery is shown as a combination of the voltage source VB and resistance RB; the DC motor is shown as a series combination of the inductance L and resistance RM, and RS is the shunt resistance. Assume DC steady-state condition at t = 0- before the switch was opened. The inductor acts as a short circuit, thus the current through each of the resistors is determined by current division. Due to the continuity demands, the inductor current remains the same at the moment when the switch was opened t = 0. t = 0+, the inductor is pushing the same current through both RM and RS as used to flow through the inductor and RM at t = 0-, thus the voltage drop across RS , which equals the voltage drop across the motor, is solely Right after the switch was opened, at determined by the resistances, not by the inductance (any inductor pushes the same current!). Note the challenge of choosing the shunt resistance RS. If RS is large, it consumes small power at DC steady-state, when the motor is running, but develops a large kickback of voltage when the motor is stopped. On the contrary, if RS is reduced, in order to reduce the voltage spike, it consumes more power at DC steady-state, due to the current division. See the problem on the following page. 2007 Alexander Ganago Page 1 of 2 EECS 314 Fall 2007 HW 09 Problem 3 Student's name ___________________________ (Last name, first name, IN INK) Discussion section # __________ Problem Assume VB = 100 V, RM = 0.5 , and RB = 0 (an ideal battery); DC steadystate conditions at t = 0- before the switch is open. If needed, assume L = 1 H. Consider the role of the shunt resistor RS: if it is large, the inductive kick gets high and dangerous for the operator and the circuit; if RS is small, it consumes too much power (due to current division). Part 1 (10 points) = 50 and calculate the magnitude |VM(t = 0+)| in the given circuit. Also, calculate the power PS consumed by RS under DC steady-state conditions (as % of that absorbed by RM). Your answers: |VM(t Assume RS = 0+)| = ________ ; PS = ___________ Part 2 (5 points) Calculate the shunt resistance RS such that |VM(t the power = 0+)| = 150 V, and calculate PSN consumed by this new RS under DC steady-state conditions (as % of that absorbed by RM). Your answer: PSN = ___________ Part 3 (10 points) Do not assume any numeric value for VB. Given that the magnitude of the motor voltage |VM(t at time = 0+)| = 2 VB , calculate the power absorbed by the shunt resistor RS under DC steady-state conditions = 0- , as the percentage of power absorbed by the motor resistance RM. Show your work for each part. 2007 Alexander Ganago Page 2 of 2 EECS 314 Fall 2007 HW 09 Problem 4 Student's name ___________________________ (Last name, first name, IN INK) Discussion section # __________ The big picture In Problem 3 of this set, you analyzed the inductive kick in a DC motor circuit: on this diagram, the motor is shown as a series combination of the inductor L and resistor RM . When the switch is opened at t = 0+, the inductor current (which must be continuous!) causes a voltage drop in the shunt resistor RS which builds up the voltage VM across the motor. Notice that use of resistors (linear devices) leads to difficult choices: on the one hand, in order to minimize the voltage rise (when the motor is stopped), we need a very small shunt resistance; on the other hand, we also want to minimize the waste of power (when the motor is running) thus we need a very large shunt resistor. Tough choice ... In this problem, you will see how we can use a semiconductor diode (non-linear device) to achieve both goals (small inductive kick and low power waste). In the circuit shown on the second diagram, the shunt resistor is replaced with a semiconductor diode, which is reverse-biased (not conducting) at DC steady state when the switch is closed and the motor is running. If the DC voltage VB is positive, then the current it pushes through the diode would be in the direction opposite to the arrow of the diode symbol: the diode does not conduct and thus does not absorb any power. When the switch is open and the DC source disconnected, the inductor pushes the current through the diode in the direction shown by the arrow of the diode symbol thus the diode is forward-biased: it conducts and the voltage drop across it does not exceed the offset voltage VD0 . Note that a forward-biased diode can burn out if the current through it exceeds the manufacturer's specifications (safe limit). Problem = 100 V, RB = 0.5 , VD0 = 0.7 V, RM = 0.5 . Determine the voltage VM across the motor and the inductor voltage VL at t = 0+ If needed, assume L = 1 H. Calculate the maximal current IMAX through the diode. 2007 Alexander Ganago Given: VB EECS 314 Fall 2007 HW 09 Problem 5 Student's name ___________________________ (Last name, first name, IN INK) Discussion section # __________ The big picture Here we introduce the photodiode a non-ohmic semiconductor device used as a light sensor (detector of light). The circuit in this problem takes advantage of the leakage current through a reverse-biased photodiode: IPD increases when the ambient light gets brighter. Since photodiode is a non-ohmic element, the voltage across it cannot be found from Ohm's law; in the circuit below it can be calculated from KVL around the loop that includes the source, the resistor and the photodiode. Recall that the current through the Gate terminal of the MOSFET transistor can be neglected. The current through the Source and Drain terminals can be large; the same current flows through the lamp. If the voltage VGS is below a certain threshold VGS &lt; V0 the MOSFET does not conduct, or acts as an open switch thus the lamp does not glow. If the voltage VGS exceeds the threshold VGS &gt; V0 the conducts, its resistance RDS becomes very small thus current flows through the MOSFET as a closed switch, and the lamp glows. Suppose that the circuit is in the dark so that the leakage current IPD through the photodiode is practically zero. Then we neglect the current through the resistor, and the entire source voltage VS is applied across the diode. Due to the connection, the same voltage On the other hand, if the circuit is illuminated so that IPD VS is applied as VGS , which makes the MOSFET conduct and the lamp glow. 0 there develops a voltage drop across the resistor, and the voltage VGS is decreased. If VGS drops below the For comparison, consider the circuit with two voltage dividers (so called, Wheatstone bridge) and a comparator, familiar from previous HW and shown here for your convenience. Note that the photodiode plays the role of RL - variable resistor whose resistance depends on light threshold, the lamp is switched OFF. Thus MOSFET is used as the switch for the lamp in this circuit. 2007 Alexander Ganago Page 1 of 4 EECS 314 Fall 2007 HW 09 Problem 5 Student's name ___________________________ (Last name, first name, IN INK) Discussion section # __________ intensity, and the lamp plays the role of Actuator. Interestingly, the second voltage divider (resistors R3 and R4) is skipped in the photodiode circuit above: it can be done because the resistance R is carefully chosen so that the MOSFET starts conducting at the desired light intensity. Problem The circuit shown on this diagram automatically turns on a small lamp in the dark. Part 1 (5 points) Assume VS = 10 V, R = 1.2 M, and V0 = 2.5 V. Calculate the current IPD below which the lamp is turned on. Your answer: ________ Part 2 (5 points) Assume VS = 10 V, and V0 = 2.5 V. Calculate the resistance R needed to turn the lamp ON when the current IPD gets below 60 nA. Your answer: ________ Part 3 (15 points) Consider the circuit built to alert the operator that the amount of fluid in the cylindrical tank falls below the lower safe limit H1 or exceeds the upper safe limit H2, as measured with the sensors monitoring the fluid level H. As shown on the side view below, two light-emitting diodes (LED) are installed at two heights H1 and H2, both are turned on all the time. Light from each LED is detected with a photodiode (PD). The mechanical design prevents crosstalk, in other words, light from LED1 does not reach PD2, and light from LED2 does not reach PD1. The big idea is that the fluid attenuates light emitted by the LED. Circuit 2, installed at the maximal safe height H2, was discussed in The Big Picture above. Light from LED2 is detected with PD2, which is connected to the circuit that turns the pilot lamp ON when the light intensity drops below a certain threshold. Thus the pilot lamp activated by circuit 2 turns ON when the fluid level exceeds the upper safe limit H2. Show your work for all parts of the problem on additional pages. 2007 Alexander Ganago Page 2 of 4 EECS 314 Fall 2007 HW 09 Problem 5 Student's name ___________________________ (Last name, first name, IN INK) Discussion section # __________ Your challenge is to choose Circuit 1, as explained below. Circuit 1, which uses light from LED1, installed at the minimal safe height H1, should turn the pilot lamp ON when the level of fluid falls below H1. The light of LED1 is detected with photodiode PD1, which is connected to the circuit that turns the pilot lamp ON when the light intensity exceeds the threshold. Briefly explain of the circuit diagrams (A E) shown on the next page can or cannot be used as Circuit 1 for PD1: 3 points for each correct explanation. Use the space below and/or additional pages as needed. Circuit diagram A Circuit diagram B Circuit diagram C Circuit diagram D Circuit diagram E 2007 Alexander Ganago Page 3 of 4 EECS 314 Fall 2007 HW 09 Problem 5 Student's name ___________________________ (Last name, first name, IN INK) Discussion section # __________ Circuit diagrams for HW 09 Problem 5 Part 3 Which circuit diagram should be chosen for PD1, and why? 2007 Alexander Ganago Page 4 of 4 EECS 314 Fall 2007 HW 09 Problem 6 Student's name ___________________________ (Last name, first name, IN INK) Discussion section # __________ The big picture BJT transistors act as current amplifiers. On the circuit diagrams, they are modeled with dependent sources (diamond-shaped). In this circuit, we use a current-controlled current source whose current IB is proportional to the controlling current IB The methods of circuit analysis, which you learned in this course, work fine in circuits with dependent sources. This circuit is easy to solve by either mesh or node analysis. Problem (Show your work on additional pages) Part 1 (10 points) Derive the algebraic equation for the amplifier's output signal as a function of the input signal, resistances and the current gain . Your answer: ____________________________________________ Part 2 (5 points) Assume that the input is a sum of DC offset VIN, DCO and the sinusoidal signal VS, IN = VIN, MAX sin( t) that is: Input signal = VIN, DCO + VIN, MAX sin( t) Use your result of Part 1 to calculate the output signal in the form Output signal = VOUT, DCO + VOUT, MAX sin( t) Note that you do not need phasors for this calculation! Your answer: Output signal = ________________________________________ Determine the DC offset of the output signal VOUT, DCO and the AC gain: AC Gain = (VOUT, MAX ) / (VIN, MAX ) 2007 Alexander Ganago Page 1 of 2 EECS 314 Fall 2007 HW 09 Problem 6 Student's name ___________________________ (Last name, first name, IN INK) Discussion section # __________ Your answers: VOUT, DCO = ____________________________________________ AC Gain = ____________________________________________ Based on the sign of AC gain, determine whether the amplifier is inverting or noninverting. Your answer: ____________________________________________ Part 3 (5 points) From your result in Part 2, calculate the AC gain for the following circuit parameters: RB = 1 k, RC = 15 k, RE = 300 , = 100. Your answer: ____________________________________________ By how many % does the AC gain change if the transistor's varies from 50 to 200 Your answers: Gain =50 ____________________________________________ ? Gain =200 __________________________________________ Gain&quot; =200 # Gain&quot; =50 Gain&quot; =200 $100% = ! Part 4 (5 points) Assume that you have to use the same resistances a transistor with the same current gain RB = 1 k and RE = 300 and = 100 as listed in Part 3. Choose RC, 25 and RC, 50 to ensure the magnitude of AC gain equal to 25 and to 50: AC Gain = 25 and AC Gain = 50 Your answers: RC, 25 = ____________________________________________ RC, 50 = ____________________________________________ 2007 Alexander Ganago Page 2 of 2 EECS 314 Fall 2007 HW 09 Problem 7 Student's name ___________________________ (Last name, first name, IN INK) Discussion section # __________ The big picture This problem introduces digital circuits where the semiconductor diode and MOSFET act either as an open circuit (reverse-biased diode and the MOSFET with VGS ~ 0 V) or as a short circuit (forward-biased diode and the MOSFET with VGS ~ 5 V). For your convenience, basic MOSFET circuit diagrams are shown below. Problem Part 1 For the circuit shown on this diagram, assume the ideal diode model, and sketch the output voltage. Hint: the output voltage is either ~ 0 V or ~ 5 V. Write a brief explanation (with calculations, as needed) of how you found the output voltages at time intervals a and b. 2007 Alexander Ganago Page 1 of 2 EECS 314 Fall 2007 HW 09 Problem 7 Student's name ___________________________ (Last name, first name, IN INK) Discussion section # __________ Part 2 For the circuit shown on this diagram, assume the MOSFET threshold voltage VT ~ 2 V and sketch the output voltage. Hint: the output voltage is either ~ 0 V or ~ 5 V. Write a brief explanation (with calculations, as needed) of how you found the output voltages at time intervals a and b. Part 3 For the circuit shown on this diagram, assume VT ~ 2 V and sketch the output voltage. Hint: the output voltage is either ~ 0 V or ~ 5 V. Write a brief explanation (with calculations, as needed) of how you found the output voltages at time intervals a and b. 2007 Alexander Ganago Page 2 of 2