ECE32 Electronics Lecture 2: Basic Circuits with iodes Payman Zarkesh-Ha Office: ECE Bldg. 230B Office hours: Tuesday 2:00-3:00PM or by appointment E-mail: payman@ece.unm.edu Slide: Review of Last Lecture Semiconductor technology trend and Moor s law Benefits of transistor scaling: More functionality in the same foot print Faster device evices with less switching energy Less cost/function Challenges of transistor scaling: evice size reaching quantum level Power dissipation and heat removal concerns nterconnect worsen by scaling Manufacturing yield issues Basic Logic Gates emorgan s Law Review of Basic Circuit Theory ynamic Power Analysis for igital Circuits Slide: 2
Today s Lecture Circuits with Nonlinear evices (iode) iode Basic Characteristics iode Approximations iode Application Circuits (Rectifiers) Slide: 3 acuum Tubes: The First Nonlinear Element Fleming iode 904 Courtesy of Prof. Ed. Graham Slide: 4 2
Evolution of Electronic evices X 00,000,000 Capacity increase BM Memory Bank 950 s (~KB) Today s Sanisk Memory Stick 204 (28GB) Slide: 5 Basic iode - Characteristics + - (A) Forward bias ( > 0) Reverse bias ( < 0) () = ( ) th e S Where s and th are constants (we will later show how to compute s and th based on device physics models) Slide: 6 3
4 ECE32 - Lecture 2 Slide: 7 University of New Mexico Example: Basic iode Characteristics At room temperature (300K), th is about 26m. Assume that s is pa and the diode current is 20mA. ) What is the diode region of operation? 2) What is the diode voltage,? 3) f the diode current is 0A, what is the diode voltage? 4) f the diode voltage is -0, what is the diode current? ECE32 - Lecture 2 Slide: 8 University of New Mexico Reverse bias: Zero bias: Forward bias: [ ] S S T S = 0 exp [ ] 0 exp = S T S = T S T S exp exp iode Zero, Reverse, and Forward Bias
Some iode Circuit Examples Slide: 9 iode in a Circuit: Exact Solution Assume that th is about 26mv and s is pa. Find and (Q-point). KΩ 5 Answer: Q-point (4.42mA, 0.58) Slide: 0 5
Load Line Analysis Graph the - relationships for the non-linear element (e.g. diode) and for the rest of the circuit (Thevenin oltage and Resistor) The operating point of the circuit is found from the intersection of these two curves. R Th Th + + Th /R Th operating point Th The - characteristic of all of the circuit except the non-linear element is called the load line Slide: deal iode Approximation f diode is forward-biased, voltage across diode is zero. f diode is reverse-biased, current through diode is zero. v =0 for i >0 and i =0 for v < 0 Thus diode is assumed to be either on or off. Analysis is conducted in following steps: Guess diode s region of operation from circuit. Analyze circuit using diode model appropriate for assumed operation region. Check results to check consistency with assumptions. Slide: 2 6
Example: deal iode Approximation Find and (Q-point). KΩ 5 deal diode approximation: Q-point (5.00mA, 0.0 ) Exact answer: Q-point (4.42mA, 0.58) Slide: 3 Constant oltage iode Model f <,on : The diode operates as an open circuit. f,on : The diode operates as a constant voltage source with value,on. Slide: 4 7
Example: Constant oltage iode Model Assume that,on =0.7. Find and (Q-point). KΩ 5 Constant voltage diode model: Q-point (4.30mA, 0.70 ) Exact answer: Q-point (4.42mA, 0.58) Slide: 5 How to Analyze Circuits with iodes A diode has only two states: forward biased: > 0, = 0 (or 0.7 ) reverse biased: =0, < 0 (or 0.7 ) Procedure:. Guess the state(s) of the diode(s) 2. Check to see if KCL and KL are obeyed. 3. f KCL and KL are not obeyed, refine your guess 4. Repeat steps -3 until KCL and KL are obeyed. Slide: 6 8
Two iode Circuit Analysis Analysis: deal diode model is chosen. Since 5 source is forcing positive current through and 2 and -0 source is forcing positive current through 2, assume both diodes are on. Since voltage at node is zero due to short circuit of ideal diode, (5 0) = 0kΩ = 2.5 ma = 0 ( 0) 5kΩ 2 = = + =.5 2= 0.5mA 2 ma Q-points are (-0.5 ma, 0 ) and (2.0 ma, 0 ) But, <0 is not allowed by diode, so try again. Slide: 7 Two iode Circuit Analysis (Contd.) 5 0k 5k 2 ( 0) = 0 25 = =.67mA 5kΩ = 5 0k = 5 6.7=.67 Analysis: Since current in 2 but that in is invalid, the second guess is off and 2 on. Slide: 8 9
Rectifier: iode Practical Application Basic rectifiers convert an AC voltage to a pulsating C voltage. A filter then eliminates pulsating components of the waveform to produce a nearly constant C voltage output. Rectifier circuits are used in virtually all electronic devices to convert the 20-60Hz AC power line source to the C voltages required for operation of the electronic device. n rectifier circuits, the diode state changes with time and a given piecewise linear model is valid only for a certain time interval. Slide: 9 Half-Wave Rectifier with Resistive Load For positive half-cycle of input, source forces positive current through diode, diode is on, v o = v s. uring negative half cycle, negative current can t exist in diode, diode is off, current in resistor is zero and v o =0. An ideal diode model is assumed here. Slide: 20 0
Half-Wave Rectifier Circuit (contd.) Using C model, during on state of diode v o =( P sinωt)- on. Output voltage is zero when diode is off. Often a step-up or step-down transformer is used to convert 20-60 Hz voltage available from power line to desired ac voltage level as shown. Time-varying components in circuit output will be removed using filter capacitor. Slide: 2 Half-Wave Rectifier with Capacitive Load As input voltage rises, diode is on and capacitor (initially discharged) charges up to input voltage minus the diode voltage drop. At peak of input, diode current tries to reverse, diode cuts off, capacitor has no discharge path and retains constant voltage providing constant output voltage dc = P - on. With no load, filtering is easy. Slide: 22
Half-Wave Rectifier with RC Load As input voltage rises during first quarter cycle, diode is on and capacitor (initially discharged) charges up to peak value of input voltage. At peak of input, diode current tries to reverse, diode cuts off, capacitor discharges exponentially through R. ischarge continues till input voltage exceeds output voltage which occurs near peak of next cycle. Process then repeats once every cycle. This circuit can be used to generate negative output voltage if the top plate of capacitor is grounded instead of bottom plate. n this case, dc = -( P - on ) Slide: 23 Half-Wave Rectifier with RC Load Current charging up capacitor Slide: 24 2
Full-Wave Rectifier with RC Load Full-wave rectifiers cut capacitor discharge time in half and require half the filter capacitance to achieve given ripple voltage. All other specifications are the same as for half-wave rectifiers. Reversing polarity of diodes gives a fullwave rectifier with negative output voltage. Slide: 25 Full-Wave Bridge Rectifier with RC Load Requirement for a center-tapped transformer in the full-wave rectifier is eliminated through use of 2 extra diodes. The four diodes in the bridge are available in a single 4-terminal package. Slide: 26 3