Structure. Symbol. physical structure. directions for current and voltage. circuit symbol

Transcription

1 iodes 1/18

2 Structure. Symbol physical structure circuit symbol directions for current and voltage The arrow in the diode s symbol points in the direction of forward current flow 2 / 5

3 Current voltage characteristic The current flowing through the diode is controlled by the voltage drop across the diode itself nonlinear semiconductor device i S v nv ( e T 1) Shockley diode equation S - saturation current (~ na - pa) KT V T thermal voltage q K - Boltzmann s constant q elementary charge (electric charge carried by a single electron) T absolute temperature measured in K degrees n=2 discrete diodes n=1 integrated diodes V T O 20 C

4 Operating regions i S e v nv T i [ma] (on) (off) V0.5 0 Th reverse bias v < 0V forward bias v > 0V (off) v < v Th ; i = 0 (on) v > v Th ; i > 0 v [V] V Th 0.6V

5 llustration is a rectifier diode, 1N400x with S =14nA, n=2 Assuming a voltage drop across the diode v 0.7V the current through the diode results as: i e 16.8mA

6 Operating (quiescent) point llustration for 1N400x with S =14nA, n=2 Q( V ; ) 100 i [ma] 80 3 Q3(0.78V; 70.8mA) Q2(0.7V; mA) 2 Q1(0.3V; 0mA) v [V] V V 2 3

7 Temperature dependence i S e v nv T S, V T - depend directly on the temperature At a constant current the voltage across the diode decreases by approximately 2 mv for every 1 o C increase in temperature. Negative tempco TC 2mV/ Ο C v ( T2 ) v ( T1 ) TC ( T2 T1 ) cst At a constant voltage across the diode the current increases with the temperature

8 etermining the operating point Circuit with a dc voltage source and a resistor =? V =? S e V nv T iode equation V V R S e V nv T V RV Circuit equation (load line equation) Transcendental equation Two solving methods: 1. Graphical method 2. Numerical method (successive approximation)

9 Graphical method iode equation: S e V nv T Load line equation: V RV

10 Numerical analysis - simplified Assume the voltage drop across the diode V = 0.7V and compute the current using the load line (circuit) equation Circuit equation: V RV V 0. 7V V V R

11 llustration V = 9V, R = 0.5K a) What is the operating (quiescent) point of the diode? V 0.6V (on) Assume V 0.7V across the conducting diode V V R mA Q(0.7V, 16.6mA)

12 Numerical analysis - iteratively 1. Consider an initial value of diode voltage, eg. V (0) =0.7V and compute the current (0) using the load line equation. (V (0), (0) ) initial solution 2. With (0) compute the diode voltage from diode equation, than the current (1) from load line equation (V (1), (1) ) solution after first iteration We finalize one iteration. f a more accurate solution is necessary further iteration should be performed. For quick, first order analysis of the circuit, usually the initial solution is considered!

13 llustration Consider V =3V, R=0.5K, is 1N400x with S =14nA and n=2. What is the operating (quiescent) point of the diode? Quick, first order analysis: V 0.6V (on) Assume V 0.7V in conduction V V R mA Q(0.7V, 4.6mA)

14 0.7V (0) V 0.635V 14nA 4.6mA ln ln (0) (1) S T nv V 4.73mA (1) (1) R V V R V V S T nv V ln (0) ma 0.637V 14nA 4.73mA ln ln (1) (2) S T V n V 4.726mA (2) (2) R V V etailed analysis: 4.726mA) Q(0.637V,

15 Constant-voltage-drop model f v < 0.7V (off) f v tends to be > 0.7V (on) v < 0.7V i = 0 v = 0.7V i > 0

16 R two-port networks analysis VTC voltage transfer characteristic 1. Consider all possible situations resulting from the combination of the diode states (on, off) 2. For each situation : i. draw the equivalent circuit ii. find v O iii. determine the range of v for that particular situation 3. raw VTC.

17 Example What is the VTC v O (v )?

18 Example - cont What is the VTC v O (v )? (off) (on) v 0.7V v O 0 i 0 v O i v 0 v v 0.7V v 0.7V v v v 0.7V v O i vo R v 0.7V R v 0.7V

19 Example - cont v O v 0 0.7V v v 0.7V 0.7V Application: Voltage transfer characteristic Voltage rectifier slope = 1

20 Waveforms for a voltage rectifier v O v 0 0.7V v v 0.7V 0.7V slope = 1 v O ( t)?

21 Waveforms for a voltage rectifier v O v 0 0.7V v v 0.7V 0.7V slope = 1 input voltage [V] V t output voltage [V] V t

22 The influence of the threshold voltage and voltage drop across the diode in conduction 0.5 v [V] v O [V] t t t f the input voltage is large enough (>> 0.7V) the threshold voltage can be considered 0V the voltage drop across the conducting diode can be neglected; (on); v O = v -50 t

23 Applications of R two-port networks Half-wave rectifier Step-down transformer

24 Pulses selector

25 Voltage limiters VTC? simple double

26 Maximum multi-port networks v v A A v B 0.7V 1 ( on), 2 ( off ); v O va 0.7V v v B B v A 0.7V ( off ), 2 ( on); v O v 1 B 0.7V v v A B 0.7V 0.7V v O ( off ), 2 ( off ); 1 O max( v A 0.7V; v B v 0 0.7V; 0V)

27 v O = max(v A 0.7V; v B 0.7V; 0) v O = max(v A ; v B ; 0) neglecting 0.7V neglecting 0.7V constant-voltage-drop What is the peak value of the current through each circuit element if R=5kΩ? What is the range of values for R, if the peak forward current through diode is 200mA? 27 / 9

28 Minimum multi-port networks v O = min(v A + 0.7V; v B + 0.7V; V PS ) v O = min(v A, v B, V PS ) neglecting 0.7V v O (t)?

29 R logic circuits analog signal digital signal Logic 0 false low Logic 1 true high 0V logic 0 5V logic 1 CMOS logic family supplied at +5V

30 two-input OR circuit 0V logic 0 5V logic 1 4.3V 4.3V 4.3V operating table truth table

31 three-input AN circuit 0V logic 0 5V logic

32 Full wave rectifier - diode bridge neglecting 0.7V across the conducting diode ~ positive half, v >0 1, 3 (on) 2, 4 (off) negative half, v <0 1, 3 (off) 2, 4 (on) ~ v >0V v <0V

33 Backup Supply V PS max(8.3v;11.3v)

34 Problem v ( t) Vˆ sin t v For the circuit in the figure, R L =50Ω. Assume a) v O (t) and i O (t) b) What are the value of the maximum reverse voltage v R across each diode and the maximum forward current through each diode? c) Repeat a) and b) assuming ˆ 6.4V V ˆ 25V V

35 Power-supply filtering v is the voltage in a secondary winding of a step-down line transformer. t is required to obtain an almost dc voltage (on a load resistor) 1 st step Half-wave (full wave) rectifier How to smooth the output voltage (as close as possible to dc)?

36 Power-supply filtering 10 v Half-wave rectifier with capacitive filter (with load) VˆO v O (without C in the circuit) t VˆO 0-10 v O T t d t c Δv t t

37 VˆO VˆO 0-10 v v O v O (without C in the circuit) T t d t c Δv t t t Between two successive voltage peaks, (off) and C discharges through R with τ = RC v t t RC RC c ( t) V (0) e (1 e ) f τ >>T, the capacitor discharge during t d can be approximated with a linear variation of the output voltage (across the capacitor) C v t; t t d T t c The discharging current is supposed to be constant, to its maximum value v T C Vˆ O v T T RC t C V VO ˆ R 1 f Vˆ O RC

38 Example Vˆ 10.7V f=50hz R L 100 v 1.5V C=? Vˆ ˆ O V 0.7V 10V Vˆ O v 1.5V frc Vˆ O 10 C 1333μF C 1333 μf 1.5VfR We chose an electrolytic capacitor C =1500µF/25V What is the actual value of the output ripple? What should be a new value of C if the output ripple have to be reduce to the half? Solve again in the case of full-wave rectification.