Electronics I - Diode Circuits

Transcription

1 Chapter 5 Electronics I - Diode Circuits p n A K Fall 2017 talarico@gonzaga.edu 1

2 Diode Circuits Applications: Rectifiers Limiting Circuits (a.k.a. clippers) Detectors Level Shifters (a.k.a. clampers) Regulators Voltage doublers Switches 2

3 Warm-up examples Example #1: diode and resistor in series source: Razavi (or V in < V γ ) (or V in < V γ ) V out V γ V γ V in Side note: 1. When the diode is forward biased the current though the diode is V in /R: we cannot make make V in get so large that V in /R > I F,peak otherwise the diode melts 2. When the diode is reverse biased the voltage across the diode is Vin: we cannot make Vin get so small that V in > V R,peak otherwise the diode breaks V R,peak is a.k.a. PIV (Peak Inverse Voltage) (e) The input/output characteristics with ideal and constant-voltage models yields two different break points. Applying an inappropriate diode s model can be misleading! talarico@gonzaga.edu 3

4 Warm-up examples Example #2: diode and resistor in series (half-wave rectifier) source: Razavi V in > 0 (or V in > V γ ) 1 V in < 0 (or V in < V γ ) V γ V in talarico@gonzaga.edu 4

5 Warm-up examples Example #3: diode implementation of OR gate source: Razavi VA (V) VB (V) Vout (V) D1 D OFF OFF OFF ON ON OFF ON ON Let s try a few cases: V A = 5V and V B = 4V Let's guess D 1 is ON V out (A side) = V A V γ = 4.3V Let's guess D 2 is ON V out (B side) = V B V γ = 3.7V BAD GUESS!! V out (A-side) must bethe same as V out (B-side) otherwise we violatekvl!! D 2 is OFF V A = 3V and V B = 0V Let's guess D 1 is ON V out = V A V γ = 2.3V Let's guess D 2 is OFF CONSISTENCY METHOD: It is sometime difficult to correctly predict the region of operation of each diode by inspection. In such cases, we may simply make an educated guess proceed with the analysis, and eventually determine if the final result agrees or conflicts with the original guess. V A = 0.6V and V B = 0V Let's guess D 1 is OFF V out = 0V Let's guess D 2 is OFF talarico@gonzaga.edu 5

6 Warm-up examples Example #4: source: in = V D,on V out = V D,on When the diode is ON we can model the circuit as follow: From the model of the circuit is easy to see that if V in < V D,on we cannot have current flowing through the diode => the diode must be OFF V out = V in If we prefer to use the ideal diode model all we have to do is to assume V D,on =0 rather than V D,on =0.7V. We do not need a lot of deep thinking to get the associated I/O characteristic V out = V in V D,on R 1 + R 2 R 2 +V D,on = slope R 2 R 1 + R 2 V in +V D,on R 1 R 1 + R in = V D,on V out = V D,on talarico@gonzaga.edu 6

7 Warm-up examples Example #5: source: Razavi For V in < 0 the diode is definitely OFF When the diode is OFF we can model the circuit as a resistive divider: V out = R 2 R 1 + R 2 V in When the diode goes ON => V out = V D,on => => so the turn on point is V D,on = R 2 R 1 + R 2 V in V in = V D,on R 1 + R 2 R 2 talarico@gonzaga.edu 7

8 Warm-up examples Example #6: source: Hambley I D1 4kΩ I R V R 6kΩ I D2 1. Let s assume both diodes are ON V R = 3V I R = 3 / 6k = 0.5mA I D1 = 10 3 =1.75mA 4k KCL : I D1 + I D2 = I R I D2 = I R I D1 = 1.25mA 2. D 1 is ON and D 2 is OFF The result is not consistent: current cannot flow from K to A 4kΩ 6kΩ I D1 = I R = 10 4k + 6k =1mA V R = I R R =1m 6k = 6V V D2 = 3V 6V = 3V talarico@gonzaga.edu 8

9 Warm-up examples Example #7: source: Razavi + V D,on For V in < 0 the diode is definitely ON. When the diode is ON: V out = V in +V D,on (straight line with slope 1 and crossing x axis at V D,on ) When the diode is OFF, the circuit can be modeled as a voltage divider: V out = R 1 R 1 + R 2 V in (straight line passing through the origin and with slope R1/(R1+R2) The break point between ON and OFF is when V out = V in + V D,on V in +V D,on = R 2 R 1 + R 2 V in V D,on = R 2 R 1 + R 2 V in V in = V D,on R 1 + R 2 R 2 talarico@gonzaga.edu 9

10 Warm-up examples Example #8: source: Razavi + V D,on For V in < 0 the diode is definitely ON. When the diode is ON: V out = V in +V D,on (straight line with slope 1 and crossing x axis at V D,on ) When the diode is OFF, the circuit can be modeled as a voltage divider: V out = R 1 R 1 + R 2 V in (straight line passing through the origin and with slope R1/(R1+R2) The break point between ON and OFF is when V out = V in + V D,on V in +V D,on = R 2 R 1 + R 2 V in V D,on = R 2 R 1 + R 2 V in V in = V D,on R 1 + R 2 R 2 talarico@gonzaga.edu 10

11 Half-wave rectifiers source: Neamen Rule of thumb it is good practice to select a diode with V BR 1.5 PIV and I F,max 1.5xI D,max v I v S = N 1 N 2 When the diode is ON the max current flowing through the diode is: I D,max = V V S γ R V S R When the diode is OFF the PIV across the diode is: V S V γ V γ v O source: Sedra & Smith PIV = V S talarico@gonzaga.edu 11

12 Half wave rectifier as a signal strength indicator source: Razavi V out (t) V p V out (t) = V p sinωt for 0 t T/2 0 for T/2 t T V out,avg = 1 T T 0 V (t)dt out = 1 T T /2 0 V p sinωt dt = 1 T V p [ ω cosωt T /2 ] 0 = V p π V "#$,&'( = V * 2 2 talarico@gonzaga.edu 12

13 Half wave rectifier as a battery charger source: Neamen V - > V 0,1"' V 6 ωt if V B < V B,nominal the battery get recharged (diode is ON from t1 to t2) otherwise the battery is left alone (the diode is OFF all period T) talarico@gonzaga.edu 13

14 Precision half-wave rectifier a PIV = V SS I F,max = V O,max R L = V I,max R L source: Sedra & Smith If v I > 0 the diode is ON. With the diode ON the circuit becomes a follower. The transfer function is almost ideal: it doesn t suffer from having one or two diode drops If v I < 0 the diode is OFF with the diode OFF the load is at ground For the o.a. to start to operate and turn-on the diode, v I has to exceed only a negligibly small voltage equal to V γ /A d talarico@gonzaga.edu 14

15 Full-wave rectifiers source: Sedra & Smith v I = 2v S source: Neamen V γ V S v O I D,max = V S V γ R V S R D1 ON D2 ON D1 ON D2 ON D1 ON PIV = V S V γ ( V S ) = 2V S V γ talarico@gonzaga.edu 15

16 Diode-bridge full wave rectifier a.k.a. Grätz bridge v I = D 3 D 4 (a) (b) when v S is positive, D 1 and D 2 are turned ON when v S is negative, D 3 and D 4 are turned ON In either case current flows through R in the same direction (a) V X V S 2V γ v O (b) source: Neamen source: Sedra & Smith v D4 = v s v D1 v D3 = v D1 +v O v O = v s 2V γ I D,max = V S 2V γ R PIV = V S V γ V S R talarico@gonzaga.edu 16

17 Clippers (a.k.a. Limiters) The idea behind clippers is quite simple. We have already built one in the past V in All we have to do to shift the clipping threshold to a different value is to add a battery Positive-cycle limiting circuit V in source: Razavi talarico@gonzaga.edu 17

18 Negative-cycle clipping V B V B V B source: Razavi talarico@gonzaga.edu 18

19 Positive and negative cycle clipping source: Razavi 19

20 A very common clipper s application Protection circuitry: keep the signals below certain thresholds source: Harris & Weste CMOS IC V DD Diode Limiters V in input pin D 1 V X D 2 R V X V B1 =V DD V B2 =0 source: Razavi talarico@gonzaga.edu 20

21 Unconventional clippers D V B V γ (V B V γ ) source: Millman talarico@gonzaga.edu 21

22 Clippers with the battery in series Assuming ideal diode model source: Neamen 22

23 Non-idealities in limiting circuits source: Razavi The clipping region is not exactly flat since as Vin increases, the currents through diodes change, and so does the voltage drop. 23

24 Zener diode source: Sedra & Smith V ZT = zener voltage i D V ZT = BV Zener diode symbol Often is convenient to model the breakdown region with a piece-wise linear model: i D constant voltage I-V model of breakdown region V D,on = IBV v D K stands for knee V ZT V Z0 v D The lower the value of r Z the more constant the zener s voltage remain Q I ZT 1 = ΔI Z r Z ΔV The model is valid for I Z > I ZK and V Z > V Z0 : V Z = V Z 0 + r Z I Z talarico@gonzaga.edu 24

25 Zener diode A zener diode is a diode specifically manufactured to be be used in breakdown region. The zener s I-V curve in breakdown region is very steep (more than usual) Diode breakdown is normally not destructive, provided the power dissipated in the diode is limited to a safe level The fact that the diode I/V characteristic in breakdown is almost a vertical line (just like a battery) enables it to be used in voltage regulation (more to come soon!) There are two mechanism causing the behavior we have in breakdown region ( despite the mechanism the end result is the same) Avalanche: occurs when the minority carriers swept by the electric field in depletion region have enough kinetic energy to be able to break covalent bonds in atoms with which they collide Zener: occurs when the electric field in the depletion region increases to the point that it can tear out a bound electron from its covalent bond talarico@gonzaga.edu 25

26 Zener diode: data sheet example On Semiconductor: Zener Voltage Regulator with V Z,nom =2.4V talarico@gonzaga.edu 26

27 Zener diode: data sheet example ΘV Z V Z,nom = 2.4V talarico@gonzaga.edu 27

28 Zener diode: data sheet example If the current exceed a certain limit the power dissipated P D =V D I D rises the junction temperature too much (> 150 C in our case) and the device may get damaged T J = T A + P D R ΘJA A device may get damaged also in the case the junction temperature becomes too small (< 65 C in our case) The max power rating of the diode (P D,max = 300 mw) goes down of 2.4 mw/ C for temperatures above 25 C talarico@gonzaga.edu 28

29 Clipping with Zener diodes Basic idea: Replacing batteries with Zener diodes source: Neamen 29

30 More clipping with Zener diodes source: Sedra & Smith For large positive v I the diode D Z1 is forward biased and D Z2 is biased in zener region (v I > V z ) For large negative v I the diode D Z1 is biased in zener region and D Z2 is forward biased (v I < V Z1 0.7) In the range (V Z1 +0.7) < v I < V Z one of the diodes is in forward region and the other one in reverse region (therefore v O =v I ) talarico@gonzaga.edu 30

31 source: Razavi Another application of clippers: soft limiters cell phone far from base station cell phone near a base station source: Hambley talarico@gonzaga.edu 31

32 Detectors Peak Detector source: Sedra & Smith v O v I t talarico@gonzaga.edu 32

33 Dissecting the peak detector a little more source: Razavi Note: the voltage across the diode (V D1 ) is just like Vin, only shifted down talarico@gonzaga.edu 33

34 Detectors: AM demodulator AM Demodulator source: Neamen Modulated input signal Detector circuit RC >> T C period carrier Demodulated output signal talarico@gonzaga.edu 34

35 Clampers (a.k.a. level shifters) Clampers shift the entire signal applied at the input by a DC level. In steady state, the output signal is an exact replica of the input waveform, but the output signal is shifted by a DC value Common application: Suppose there is a stage (e.g. an amplifier) that does not operate properly with the DC level provided at its input, the issue can be solved by putting a level shifter in front of the stage talarico@gonzaga.edu 35

36 Positive Peaks Clamper v O = v I v C source: Neamen Assuming r f 0 Ω and V γ =0 V This is the same circuit of the peak detector, but now we take the output across the diode! When the diode goes OFF there is no path to ground so the capacitor cannot discharge: v C stays constant at V M (so after T/4: v O = v I V M ) D1 ON D1 OFF The positive peaks of the output voltage are clamped at 0 V Negative DC level C gets charged Level shifter with peak at -2V (at T/4 v C V M ) M talarico@gonzaga.edu 36

37 Positive peaks clamper The positive peaks of the output voltage are clamped at 0 V Negative DC Level Shifter D1 OFF C1 blocks Vin= 6V DC D1 ON C1 charges very quickly to V C1 =4V D1 OFF C1 remains charged at V C1 =4V D1 ON C1 already charged at V C1 =4V V out = V in V C1 V out = 0 V out = 4 4 = = 0 V out = 6 4 = = 10 V out = 4 4 = = 0 talarico@gonzaga.edu 37

38 Positive peaks clamper with Battery source: Neamen The positive peaks of the output voltage are clamped at V B steady state (it takes T/4 to reach it) Superposition of + steady state (it takes T to reach it) talarico@gonzaga.edu 38

39 Positive peaks clamper with battery If we take the circuit we just analyzed, and reverse the polarity of the battery we clamp the positive peaks of the signal to a negative voltage value. This is no surprise: we still clamp the positive peaks to V B (but now V B happens to be negative) The source: circuit Hambley is designed to have RC >> T NOTE: when v in is at +5V the diode is ON and the cap is charged to V C =10V talarico@gonzaga.edu 39

40 Negative peaks clamper source: Razavi v C1 initially the cap is uncharged: v C1 (0)=0 v out = v in v C1 = v in The diode turns ON and the cap. charges to V P The diode turns OFF for good: v out = v in ( V P ) = v in +V P The negative peaks of the output voltage are clamped at 0 V Positive DC level t 1 t 2 t 4 t Level shifter with peak at +2V P talarico@gonzaga.edu 40

41 Negative peaks clamper Positive DC Level Shifter (it is better to use PWL rather than PULSE) The negative peaks of the output voltage are clamped at 0 V D1 ON D1 OFF D1 ON C1 charges very quickly to V C1 = 6V C1 remains charged at V C1 = 6V C1 already charged at V C1 = 6V V out = V in V C1 V out = 6 ( 6)= = 0 V out = 4 ( 6)= =10 V out = 6 ( 6)= = 0 talarico@gonzaga.edu 41

42 Positive DC level shifter: effect of load In practice the clamper will be driving a load. we need to make sure that R 1 C 1 >> T/2, otherwise when D 1 is OFF the cap. C 1 loses too much charge on the load Example showing the effect of having R 1 C 1 too small (R 1 C 1 = T/2) talarico@gonzaga.edu 42

43 Negative peaks clamper with battery Example: This circuit clamps the negative peaks of an AC signal to +6V source: Hambley v in = V P sin ωt V B V B = 6 if we assume: V γ 0V or V B = 6.7 if we assume: V γ 0.7V talarico@gonzaga.edu 43

44 Negative peaks clamper with battery v in = V P sin ωt V B ideal diode V γ 0V V B = 6V V B = 1V V p = 2V, f in =1KHz talarico@gonzaga.edu 44

45 What about replacing batteries with Zeners? It kind of works, but we need to keep in mind that (differently from what happened with the limiters) here the zener must work in zener region at all time. So it must be biased in zener region at all time!! source: Hambley when the v in is at 2V the diode is ON and the cap charges at 2 5= 3V 4.3V V γ =0.7V Circuit that clamps the negative peaks to 5V If we take off the 15V bias voltage and return R directly to ground the diode never turns ON and the circuit doesn t work talarico@gonzaga.edu 45

46 Replacing batteries with Zeners Example of circuit for clamping positive peaks source: Hambley 5.3V Circuit that clamps the positive peaks to +6V 46

47 Example: another clamper source: Millman In steady state the cap is charged to V m V R talarico@gonzaga.edu 47

48 Example: another clamper f in =1/20ns Circuit s elements: C=1nF R=100KΩ Ideal diode V R =11V V R =2V talarico@gonzaga.edu 48

49 Alternative ways of clamping Inside CMOS ICs DC level shifting is usually achieved using current sources (i.e. MOS transistors) and cascade of diodes (or diode connected MOS transistors) Assumption: the current pulled by the next stage is negligible (or at least constant), so that the current through the diode establish a drop of V D,on across the diode source: Razavi shift down circuit shift up circuit talarico@gonzaga.edu 49

50 Alternative ways of clamping Inside CMOS ICs, another common way of a achieving DC level shifting is by using a Common Drain stage source: B. Murmann & R. Dutton talarico@gonzaga.edu 50

51 Application: DC Power supply Let s take a look at how to build a DC power supply (AC-DC power converter) smooth the rectified waves source: Sedra & Smith detect the peak (smoothing cap or filter cap) Figure 4.22 Block diagram of a dc power supply. talarico@gonzaga.edu 51

52 Rectifier + Filter Capacitor + Load The following circuit (peak rectifier or peak detector) provides a DC voltage equal to the peak of the input sine wave v S (t) V m source: Sedra & Smith So at a first glance it would seem a reasonable solution to use it as a DC power supply to drive a load. i C (t) However, once we connect the load if we look at the circuit a little harder we realize it presents some issues source: Hambley talarico@gonzaga.edu 52

53 Rectifier + Filter Capacitor + Load The voltage across the load is not constant. There is a ripple. A ripple more than 5% 10% of V m (or V m V γ with the constant voltage model) is a problem! exponential decay with time constant τ=r L C v D (t) = v S (t) v O (t) source: Hambley We have two design decisions: - pick C (how much ripple we want to tolerate) - pick D (how much current the diode must withstand in forward region, and what is the PIV in reverse region) i L R L = v O (t)/r L source: Neamen Waveforms for half-rectifier with smoothing capacitor talarico@gonzaga.edu 53

54 Ripple for different capacitor values source: Razavi V D = V in V out PIV 2 V M The amplitude of the ripple is given by the decaying exponential For V out to have small ripple we need large C talarico@gonzaga.edu 54

55 i L R L when diode is OFF v O = V m e t RLC Ripple and I D,max Assuming R L C >> T: V m v S PIV V m R L C V R T V R V m R L T C 0 dv 9 dt ; $<9 slope of exponential decay at t=0 = d V 'e >$/(A B C) dt slope by using simple geometry E $<9 The current supplied to RL is almost constant and is bounded by V m /RL I D,max = C dv out dt t= Δt + V m R L The diode s current is max at the beginning of the conduction interval and it goes down as the diode tends to turns off This is also when the current through the cap. is max (this is because the slope of Vout is max) = I C,max We need to find I C,max! talarico@gonzaga.edu 55

56 Ripple and I D,max I C,max = C dv out dt t= Δt = C d ( dt V m cosωt) t= Δt Looking at the geometry of V out we see that: = CV m ω[ sin( ωδt)] = CV m ω sinωδt The diode conducts current only a small portion of the period (Δt/T << 1) therefore ωδt is a small angle and sin(ωδt) ωδt I C,max ωcv m ( ωδt) v out (t) V m cos( ωδt) = V m cosωδt = V m V r cosωδt =1 V r Taylor for small angles cosωδt 1 1 ( 2 ωδt ) 2 V m ( 2 ωδt ) 2 1 V r ωδt V m 2V Finally r I V C,max ωcv m ωδt m ( ) 2π T CV m 2V r V m Δt T 1 2π talarico@gonzaga.edu 56 2V r V m % of time the diode is ON

57 Ripple and I D,max I D,max = 2π T CV m 2V r V m + V m R L = V m R L 1+ 2π CR L T 2V r V m V m 1+ 2π 2V m R L V r V m R L C V r T slope of exponentialdecay at t=0 Δt Δt 0 T I D,avg 1 T V m 1+ 2π 2V m R L V r Δt 2 = 1 2 V m Δt 1+ 2π 2V R L T m V r V m 1 R L 4π 2V r +1 V m area triangle Δt T 1 2π 2V r V m talarico@gonzaga.edu 57

58 Can we further reduce the ripple? Yes it is. Instead of using a simple diode rectifier we can use a bridge source: Razavi Since C discharges only for ½ period, the ripple voltage is decreased by a factor of 2 Also each diode is approximately subjected to only one V m reverse bias drop (versus the 2V m we had with the half-wave rectifier). B V D,on V AB = V D,on +V out V m V out talarico@gonzaga.edu 58

59 Bridge Rectifier + Filter Capacitor + Load % of time the diode is ON V m R L C V r T / 2 V r 1 2 V m R L T C Δt T 1 π 2V & V ' slope of exponential decay at t=0 T replaced by T/2 V m I D,max V m 1+ π R L 2V m V r source: Sedra and Smith I D,avg V m 1 R L π V r +1 2V m talarico@gonzaga.edu 59

60 Voltage Regulator source: Razavi V ss Variations in V ss may be ripple due to the rectifier but also fluctuations in the AC line The ripple created by the rectifier can be unacceptable to sensitive loads. Therefore, a regulator is required to obtain a more stable output. source: Hambley talarico@gonzaga.edu 60

61 Voltage Regulator source: Razavi V ss R 1 Regulator As long as r d << R 1, the use of a Zener diode provides a relatively constant output despite input variations V ss ΔV ss ΔV out ΔV out = r d r d + R 1 ΔV ss Example: r d =5Ω, R 1 =1K changes in V ss are attenuated by about 200 times at the output talarico@gonzaga.edu 61

62 Voltage Regulation with Zener Diode Example Design a voltage regulator to power a car radio at V out =9V from an automobile battery whose voltage may vary between 11V and 13.6V. The current in the radio will vary between 0 (off) to 100 ma (full volume). source: Neamen V ss = V out V ss,nom=12v, V SS,middle=12.3V Initially, we need to find out the proper input resistance R i. The resistance R i limits the current through the zener diode and drops the excess voltage between V ss and the nominal voltage we want on the load V out,nom = V Z,T = V Z,nom (in other words it sets the diode operating point Q T ) talarico@gonzaga.edu 62

63 Voltage Regulation with Zener Diode Initially, assume ideal diode: Vss,max Vss,min V ZT V Z (V) R i = V SS,nom V Z,nom I Z,nom + I L,nom Q T I ZT I Z (A) I L,nom = V Z,nom R L,nom More thoroughly, for the circuit to work properly, the diode must remain in zener region and the power dissipation of the diode must not exceed its rated value (P D ). In other words: The current in the diode is a minimum I Z,min when the load current is a maximum I L,max, and the source voltage is a minimum V ss,min The current in the diode is a maximum I Z,max, when the load current is a minimum I L,min and the source voltage is a maximum V ss,max Therefore we can impose the two following constraints: R i = V SS,min V Z,nom I Z,min + I L,max and R i = V SS,max V Z,nom I Z,max + I L,min talarico@gonzaga.edu 63

64 Voltage Regulation with Zener Diode R i = V SS,min V Z,nom I Z,min + I L,max R i = V SS,max V Z,nom I Z,max + I L,min Reasonably, we can assume that we know the range of input voltage, the range of output load current, and the Zener voltage. Further, it is reasonable to set the minimum zener current to be I Z,min 0.1 I Z,max More stringent design requirements may require the minimum zener diode current to be 20 or 30 percent of the maximum value. The important point in setting I Z,min is to make sure is far enough from the knee!! By equating the constrains on R i and setting I Z,min 0.1 I Z,max we can write: ( ) I L,min ( V ss,min V Z,nom ) I Z,max = I L,max V ss,max V Z,nom V ss,min 0.9 V Z,nom 0.1 V ss,max The maximum power dissipated in the Zener diode is approximately: K stands for knee I Z (A) V V Z0 ZT V ZK = BV V Z (V) I ZK = IBV I ZT P Z,max I Z,max V Z,nom Therefore: R i = V ss,max V Z,nom I Z,max + I L,min and I Z,min = V ss,min V Z,nom R i I L,max 64

65 Voltage Regulation with Zener Diode and finally make sure P Z,max < P D and I Z,min < I ZK Let s now go back to the example and plug in some numbers: I L,max =100mA V ss = source: Neamen ( ) I L,min ( V ss,min V Z,nom ) I Z,max = I V V L,max ss,max Z,nom = V ss,min 0.9 V Z,nom 0.1 V ss,max 100 (13.6 9) mA P Z,max I L,max V Z,nom = 300mA 9V = 2.7W R i = V ss,max V Z,nom I Z,max + I L,min = m Ω I L,'21 = 0.1 I L,'3P 30mA ( P Ri,max = V ss,max V Z,nom ) 2 ( ) 2 = R i W talarico@gonzaga.edu 65

66 Regulator s figures of merit In reality the zener is not ideal. It has some non zero resistance, therefore if the source voltage or the load current fluctuates, so does the V out =V Z source: Neamen V ss V out Source regulation (a.k.a. line regulation) It is a measure of how much the output voltage changes as the source voltage change (assuming no-load condition R L = ) sourceregulation ΔV out ΔV ss 100% ability to maintain a constant output voltage level on the output despite changes to the input voltage level talarico@gonzaga.edu 66

67 Line regulation example Example: Find the line regulation for the previous example, assuming r Z =2Ω R 1 V out For V ss =13.6V: I Z = V ss V Z R 1 + r Z = mA V out = I Z r Z +V Z = 9.532V For V ss =11V: I Z = V ss V Z R 1 + r Z = mA V out = I Z r Z +V Z = 9.231V source: Neamen ΔV out = 15.6% ΔV in Alternatively by considering just the variations (small signal circuit) ΔV ss ΔV out ΔI Z r Z ΔV out ΔV ss = r Z 2 = R 1 + r Z % source: Razavi talarico@gonzaga.edu 67

68 V ss V out Regulator s figures of merit Load regulation It is a measure of the change in output voltage with a change in load current loadregulation V out,noload V out, fullload V out, fullload 100% capability to maintain a constant voltage on the output despite changes in the load (such as a change in resistance value connected across the supply output where: - V out,noload is the load voltage for zero load current - V out,fullload is the load voltage for the maximum rated load current In practice, there are a couple of other ways of defining load regulation. loadregulation V out,noload V out, fullload V out,nomload 100% loadregulation V out,noload V out, fullload I L,noload I L, fullload 100% = ΔV out ΔI L 100%(W) talarico@gonzaga.edu 68

69 Load regulation example Example: Find the load regulation for the usual example. Assume r Z =2Ω V ss V out Note: When measuring the load regulation the source is assumed constant. Since the full load current is reached for V ss = V ss,max for load regulation computations we must assume V ss = V ss,max = const For I L = 0A: For I L =100mA: I Z = V ss,max V Z R 1 + r Z = mA V out = I Z r Z +V Z = 9.532V ( ) I Z = V R1 I L = V V + r I ss,max Z Z Z R 1 R 1 I L I Z = V ss,max V Z I L R m 15.3 = mA V out = I Z r Z +V Z = 9.355V R 1 + r Z V out,noload V out, fullload %= 100% 1.89% V out, fullload talarico@gonzaga.edu 69

70 Load regulation example V ss V out Alternatively by considering just the variations (small signal circuit) ΔI L ΔV ss =0 r Z ΔV out = ( R 1 r Z )ΔI L ΔV out ΔI L = ( R 1 r Z ) = Ω For a ΔI L of 100 ma we have that ΔV out 177 mv (As expected this is the same result we got before ΔV out = = 177 mv) V out,noload V out,fullload talarico@gonzaga.edu 70

71 Evolution of an AC-DC converter source: Razavi Regulator Ideally we want both line regulation and load regulation to be as close as possible to 0% 71

72 Voltage doubler T in = 1ms, V P =2V X clamper peak detector =V in V C1 V out D1 ON V C1 follows V in all the way to V P V X = V in V C1 = 0 From T/4 on D1 is OFF C1 stays charged to VP V X = V in ( V P ) = = V in + V P clamper peak detector If we take the clamper just designed and attach a peak detector at its output we get a voltage doubler talarico@gonzaga.edu 72

73 Voltage doubler: detailed analysis In the reality the charging of C1 and C2 (V C2 = V out ) is not as simple as assumed in the previous slide (slide 52 is a snapshot of the steady state) talarico@gonzaga.edu 73

74 Voltage doubler: detailed analysis 0 t 1 t 6 Each input cycle, the output increases by V p, V p /2, V p /4, etc., eventually settling to 2 V p V out = V P = V P n 1 = V P 2 1 1/ 2 = 2V P n=0 talarico@gonzaga.edu 74

75 Voltage doubler: detailed analysis 75

76 Diodes as Switches Motor In reality the switch is a transistor The transistor goes in smoke If the switch is suddenly opened, a voltage spike develops across the inductor v L = L di L dt Inductors do not like abrupt changes in current 76

77 Diodes as Switches A diode placed across the inductive load, will give the voltage spike a safe path to discharge, looping over-and-over through the inductor and diode until it eventually dies out. talarico@gonzaga.edu 77

78 Special Diodes Schottky-Barrier Diode (SBD) SBD are built using a metal-semiconductor junction current is conducted by majority carriers (electrons). Thus SBD do not exhibit the minority carrier charge storage effect. As a result SBD can be switched from on to off and vice versa much faster The forward voltage drop is lower (0.3V to 0.5V for silicon) Varactors Photodiodes LEDs The wavelength of the light emitted, and thus the color, depends on the band gap energy of the materials forming the p-n junction. (Example. Red LED: λtalarico@gonzaga.edu d =630nm, VF=2.1V IF=50mA, luminous Intensity I V =7500 mcd) 78

79 Voltage doubler modeled with switches 79

80 Voltage doubler modeled with switches no initial charge moving (V C1 =V C2 =V P ) C2 is moving some charge back to C1 (ΔV=V p /4) C2 is moving some charge back to C1 (ΔV=3V p /8) talarico@gonzaga.edu 80