Lecture 1. EE 215 Electronic Devices & Circuits. Semiconductor Devices: Diodes. The Ideal Diode

Transcription

1 Lecture 1 EE 215 Electronic Deices & Circuits Asst Prof Muhammad Anis Chaudhary EE 215 Electronic Deices & Circuits Credit Hours: 3 1 Course Book: Adel S. Sedra and Kenneth C. Smith, Microelectronic Circuits, Sixth Edition Reference Book: B. Razai, Fundamentals of Microelectronics, Second Edition B.G. Streetman and S. Banerjee, Solid State Electronic Deices, Sixth Edition Software Tools LTspice (isit for LTspice setup and Getting Started Guide ) Matlab Our emphasis in this course is on semiconductor deices based circuits. Semiconductor Deices: Diodes Diodes Bipolar Junction Transistors (BJT) Field Effect Transistors (FET) the semiconductor electronic deices can be used singly along with resistors, capacitors and inductors for the design of discrete circuits. Or as components of an integrated circuit (IC) chip Integrated Circuit(IC) An integrated circuit (IC) has multiple electronic deices (diodes or transistors) in a small piece of silicon (or any other semiconductor). e.g. A microprocessor has hundreds of millions of components in a small piece of silicon whose area is on the order of A Diode The Ideal Diode 100mm 2 is a two terminal deice is the most fundamental non linear circuit element 1/19

2 from the i characteristics of the ideal diode, one can see that the diode can operate in two distinct modes either as a Reerse Biased Diode or as a Forward Biased Diode Reerse Biased Diode If a negatie oltage '' is applied to the diode, no current flows through the diode the diode behaes as an open circuit and the diode is said to be cutoff Forward Biased Diode If a positie current i is applied to the ideal diode, zero oltage drop appears across the diode. i.e. the diode acts as a short circuit in the forward direction the diode will pass any current with zero oltage drop. 2/19

3 The external circuit must be designed to limit the current through the conducting (forward biased) diode the reerse oltage across a cut off (reerse biased) diode 3/19

4 The external circuit must be designed to limit the current through the conducting (forward biased) diode (10mA in fig a) the reerse oltage across a cut off (reerse biased) diode (10 in fig b) A Simple Application: The Rectifier A diode can be used as a rectifier i.e. to conert AC to DC A simple circuit consisting of series connection of a diode and a resistor, can act as a rectifier. the diode is assumed to be ideal and the input signal is a sinusoid. 4/19

5 Now during the positie half cycle of the input sinusoid, +e I will cause current i D to flow through the diode i.e. the diode is forward biased = D, the oltage across the diode will be zero Now during the negatie half cycle of the input sinusoid, a e oltage D will appear across the diode hence the diode is reerse biased and will act as an open circuit As a consequence, the output oltage waeform will be of the form Note that I alternated in polarity and has a zero aerage alue. while is unidirectional and has a finite aerage alue (or a DC component) O 5/19

6 Exercise 4.2 For the circuit in Fig. 4.3(a), sketch the waeform of D Solution from the circuit, D can be gien as D = I O 6/19

7 Exercise 4.3 1kΩ In the circuit of Fig. 4.3(a), let I hae a peak alue of 10 and R=. Find the peak alue of and the dc component of. i D O Solution = 10 peak alue of I = P peak alue of i D will occur when I = P and diode is conducting peak alue of i D = I = P P = 10 = 10mA R 1k the aerage alue (or the dc component) of O can be gien as = 1 T O,Aerage (t) t T 0 O T P sin ωt for 0 t from the figure, 2 O (t)=[. 0 for t T = = 1[ (t) t + (t) t]= 1 O,Aerage [ sin ωt t] T/2 T 0 O T T/2 O T 2 T/2 T 0 P 7/19

8 O,Aerage thus for As P T/2 T 0 and ωt = 2π Another Application: Diode Logic Gates OR Gate when = 1[ sin ωt t] Diodes along with resistors can be used to implement digital logic gates. Truth tables for two input AND gate and a two input OR gate can be gien as A OR Gate A B Y where T/2 T 0 P O,Aerage ω = 2π T O,Aerage O,Aerage = [ sin ωt t]= P [ cos ωt ] = T ω 0 AND Gate A B Y logic 0 corresponds to a otage alue near 0 and logic 1 corresponds to a oltage alue near +5 Both diodes are forward biased (conducting) when A = 0, B = +5, Diode A is cut off, Diode B is conducting when A = +5, B = 0, Diode A is conducting, Diode B is off T/2 P [ cos ωt]t/2 Tω 0 = P [ cos = [( cos ) ( cos 0)] 2π T/2 t] P 2π T 2π 2π T 0 2π T 2 T = P [( cos π)+ cos 0]= P [1 + 1]= 2π P = 10 = = = = O,Aerage P π = B = +5 = logic1 = Y = +5 = Y = +5 = Y = π 2π P π 8/19

9 AND Gate when A = = 0 = logic0 when A B, Both diodes are reerse biased, = 0 = B = 0 = logic0 Both diodes are forward biased (conducting) = = 0 Y when A = 0, B = +5, Diode A is conducting, Diode B is cut off = = 0 Y when A = +5, B = 0, Diode A is off, Diode B is conducting = = 0 Y Y = = 0 = logic0 A = B = +5 = logic1 = +5 when A B, Both diodes are conducting, Y = 0 when, Both diodes are reerse biased (cutoff), Y Similarly 3 input AND, OR functions can be inplemented using 3 diodes. Example 4.2 Assuming the diodes to be ideal, find the alues of I and, in the circuits of Fig /19

10 Note: It is not always clear in a circuit, whether the diode is conducting or not In such a case, make an assumption proceed with the analysis see if the solution is reasonable e.g. if assumption is that the diode is conducting, its current i D must be positie. for circuit (a), lets assume both diodes are conducting = we can replace both diodes with short circuits = B = 0 and = 0 Applying ohms' law across 10k resistor I = 10 0 D2 = 1mA 10k KCL at node formed by and B (0) B ( 10) 0 ( 10) I 10 D2 I5k(Resistor) 5k 5k 5k = I = 2mA I D2 = 1mA As I and I D2 both are positie currents our assumption was correct i.e. both diodes are conducting. + I = = = = = 2mA 10/19

11 for circuit (b), lets assume both diodes are conducting = we can replace both diodes with short circuits = B = 0 and = 0 Applying ohms' law across 5k resistor I = 10 0 D2 = 2mA 5k KCL at node formed by and B (0) B ( 10) 0+10 I D2 I10k(Resistor) 10k 10k = I = 1mA I D2 = 1mA As I D2 is positie but I is negatie = D2 is conducting but D1 is cut off + I = = = = 10 = 1mA 10k Now, D2 is conducting but D1 is cut off = we can replace D2 with short circuit, D1 with open circuit As is cut off D1 = I = /19

12 I D2 is the current flowing through series combination of 5k and 10k resistors Applying ohms' law across (5k+10k) resistor 10 ( 10) I = = 20 D2 = 1.33mA 5k+10k 15k Ohm's law across 5k resistor 10 = I = 1.33mA 5k D2 = 10 = 6.65 or =3.35 I D2 = 1.33mA and = 3.35 = I D2 is +e = D2 Note that B = = 3.35 = = 3.35 D1 thus D1 is reerse biased is conducting Exercise4.4e Find the alues of I and in the circuits shown in Fig. E4.4e. 12/19

13 only +3 diode conducting other two diodes off = +3 I = 3 1k = 3mA this circuit can be used to select one of the 3 DC sources Terminal Characteristics of Junction Diodes Uptill now we hae considered the behaiour of ideal diode lets take a look at the characteristics of real diodes specifically, semiconductor junction diodes made of Silicon i characteristics of a silicon PN junction diode is shown 13/19

14 the characteristic cure consists of three distinct regions 14/19

15 The forward bias region, determined by The reerse bias region, determined by < 0 The breakdown region, determined by < zk A diode is in forward bias region of operation when the terminal oltage is positie In the forward bias region, i relationship can be gien as n i = ( e T 1) The Forward Bias Region i = ( e T 1) where = Saturation Current is also called scale current. As is directly proportional to the cross sectional area of the diode. Thus doubling of the junction area = doubling of for small signal diodes, is on the order of A alue of is a ery strong function of temperature As a rule of thumb, doubles for eery 5 o C rise in temperature is called ideality factor.. Assume where T is a constant called thermal oltage and is = = T m i = ( e T 1) for T = absolute temperature in kelins= temperature in At room temperature ( o ), for ease of calculations, at room temperature e T, which is normally the case for a forward bias diode i or e T or in logarithmic form taking natural log (base e) on both sides two forms of i relationship for the forward bias region i i e T and T ln( ) > 0 n 1 n 2 n = 1 T = i = e T o C T T kt q = i + = i > i + = i e T e 20 C = T m = (293) m = m 25 m T i ln( ) ln( e T ) i ln( ) ln(e)= T = i e 15/19 T i = T T ln( )

16 if now diode oltage is 1, its current I1 can be gien as I1 = e T...(A) if diode oltage is changed to 2, diide (B) by (A) then its current I2 will be I2 = e T...(B) taking natural log on both sides I2 or 2 1 = T ln( ) I1 I2 2 1 = T ln( ) I1 or in terms of base 10 logarithms using the identity log x log 10 e log 10 e I2 I = = S e T e = T = e T = I1 e 2 1 T = for a decade (factor of 10) change in current, the diode oltage drop changes by only 1 e 2 1 T 2 I2 = ln( )= I1 2 1 T 2 1 T ln x = x = = = 2.3 log x I2 = 2 1 = 2.3 T log( ) I1 log x log e e 2 1 T 2.3 T = 57.5m from the i characteristics in the forward bias region The current is negligibly small for smaller than about 0.5 This alue is called cut in oltage. 16/19

17 Exercise 4.6 Solution Exercise 4.7 Solution Exercise 4.8 Solution Also the current rapidly increases afterwards for a fully conducting diode, the oltage drop lies in a narrow range, approximately 0.6 to 0.8 This obseration leads to a simple model for the diode where it can be assumed that a conducting diode has a 0.7 oltage drop across it. Find the change in diode oltage if the current changes from 0.1mA to 10mA I1 = 0.1mA = 10mA I2 2 1 T I1 or 2 1 I2 = ln = 25m ln = = 115 m A silicon junction diode has =0.7 at i=1 ma. Find the oltage drop at i=0.1 ma and i=10 ma. 1 = 0.7 and I1 = 1 ma 2 1 =? = =? = 10, I2 ma, I3 ma Using the fact that a silicon diode has 0 14 at 2 C and that increases by % o per C rise in termperature, find the alue of at 12 C. at increases by 15% per o C rise in temperature at 125 o C =?,25 = 1 o C 0 14 A,26 = = 1.15 o C,25 o C,25 o C,25 o C,27 o C,26 o C,26 o C,2 C =,125 = = o C (1.15)(125 25),2 C thus at A The Reerse Bias Region = ln = 25m ln 0.1 = I = = 2 = = = ln = 25m ln 10 = I = = = = T I2 3 1 T I3 = A 25 o C 3 = 1 A 5 o 5 o 15 = = = o (1.15)2,25 o C 5 o (1.15)100,25 o C 125 o C = (1.15) = A 17/19

18 As The Breakdown Region The reerse bias region of operation is entered when the diode oltage is made negatie. i = ( e T 1) if is e and a few times larger than e.g. = 4 T e T e 4 = 25m which clearly indicate that exp term is ery small with respect to 1 and can be neglected for the reerse bias region Real diodes hae reerse currents much higher than e.g. for a small signal diode with = 1 reerse current is on the order of 1nA A to 0 15 Still the reerse current is ery small and the diode can be considered as cut off in the reerse bias region of operation. the reerse current also changes with the change in temperature As a rule of thumb, the reerse current doubles for eery 10 C rise in temperature T = 100m = 1 = 1 = = i = ( e T 1) i = ( e T 1) A 18/19

19 The breakdown region is entered when magnitude of the reerse oltage exceeds a certain threshold alue, called the breakdown oltage, ZK Breakdown oltage ZK is specific to the particular diode and is typically in the range 2 ZK 2000 ZK is the oltage at the knee of the i cure The subscript Z stands for zener and K stands for knee. In the breakdown region, the reerse current increases rapidly, while the associated increase in oltage drop is ery small. thus the diode i characteristics in breakdown is almost a ertical line. this region can be used for oltage regulation. 19/19