3.4 Operation in the Reverse Breakdown Region Zener Diodes


 Duane Shaw
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1 3/3/2008 secton_3_4_zener_odes 1/4 3.4 Operaton n the everse Breakdown egon Zener odes eadng Assgnment: pp A Zener ode The 3 techncal dfferences between a juncton dode and a Zener dode: The practcal dfference between a Zener dode and normal juncton dodes: Manufacturer assumes dode wll be operated n breakdown regon. Therefore: Jm Stles The Unv. of Kansas ept. of EECS
2 3/3/2008 secton_3_4_zener_odes 2/ HO: Zener ode Notaton A. Zener ode Models Q: How do we analyze zener dodes crcuts? A: Same as juncton dode crcuts Bg problem > Bg soluton > HO: Zener ode Models Example: Fun wth Zener odes Jm Stles The Unv. of Kansas ept. of EECS
3 3/3/2008 secton_3_4_zener_odes 3/4 B. Voltage egulaton Say that we have a 20 V supply but need to place 10 V across some load: 20 V 10 V The soluton seems easy! 20 V = 10 V Ths, n fact s a very bad soluton HO: The Shunt egulator Jm Stles The Unv. of Kansas ept. of EECS
4 3/3/2008 secton_3_4_zener_odes 4/4 Two prmary measures of voltage regulator effectveness are lne regulaton and load regulaton. HO: ne egulaton HO: oad egulaton Example: The Shunt egulator Another mportant aspect of voltage regulaton s power effcency! egulator Power and Effcency One last pont; voltage regulaton can (and s) acheved by other means. Voltage egulators Jm Stles The Unv. of Kansas ept. of EECS
5 2/20/2008 Zener ode Notaton 1/4 Zener ode Notaton To dstngush a zener dode from conventonal juncton dodes, we use a modfed dode symbol: Anode Cathode Generally speakng, a zener dode wll be operatng n ether breakdown or reverse bas mode. For both these two operatng regons, the cathode voltage wll be greater than the anode voltage,.e.,: v < 0 (for r.b. and bd) kewse, the dode current (although often tny) wll flow from cathode to anode for these two modes: < 0 (for r.b. and bd) Q: Ykes! Won t the the numercal values of both and v be negatve for a zener dode (assumng only rb and b.d. modes). A: Wth the standard dode notaton, ths s true. Thus, to avod negatve values n our crcut computatons, we are gong to change the defntons of dode current and voltage! Jm Stles The Unversty of Kansas ept. of EECS
6 2/20/2008 Zener ode Notaton 2/4 z. v v z Conventonal dode notaton _ Zener dode notaton _ * In other words, for a Zener dode, we denote current flowng from cathode to anode as postve. * kewse, we denote dode voltage as the potental at the cathode wth respect to the potental at the anode. Note that each of the above two statements are precsely opposte to the conventonal juncton dode notaton that we have used thus far: v = v and = Z Z or Z V ZK 0.7V v 0.7V V ZK v Z Two ways of expressng the same juncton dode curve. Jm Stles The Unversty of Kansas ept. of EECS
7 2/20/2008 Zener ode Notaton 3/4 The Z versus v Z curve for a zener dode s therefore: Z bd 0.7V V ZK v Z r. b. f. b. Thus, n forward bas (as unlkely as ths s): Z v I exp nv Z = s T or approxmately: v 07V. and < 0 Z Z Jm Stles The Unversty of Kansas ept. of EECS
8 2/20/2008 Zener ode Notaton 4/4 kewse, n reverse bas: I and 0 < v < V Z s Z ZK And fnally, for breakdown: > 0 and v V Z Z ZK Jm Stles The Unversty of Kansas ept. of EECS
9 2/20/2008 Zener ode Models 1/6 Zener ode Models The conventonal dode models we studed earler were based on juncton dode behavor n the forward and reverse bas regons they dd not match the juncton dode behavor n breakdown! V ZK 0.7V v CV Model However, we assume that Zener dodes most often operate n breakdown we need new dode models! Specfcally, we need models that match juncton/zener dode behavor n the reverse bas and breakdown regons. New Zener Model Juncton ode Breakdown Z v Z V ZK Jm Stles The Unv. of Kansas ept. of EECS
10 2/20/2008 Zener ode Models 2/6 We wll study two mportant zener dode models, each wth famlar names! 1. The Constant Voltage rop (CV) Zener Model 2. The PeceWse near (PW) Zener Model The Zener CV Model et s see, we know that a Zener ode n reverse bas can be descrbed as: I 0 and v < V Z s Z ZK Whereas a Zener n breakdown s approxmately stated as: > 0 and v V Z Z ZK Q: Can we construct a model whch behaves n a smlar manner?? A: Yes! The Zener CV model behaves precsely n ths way! eplace: z v z _ wth: z Ideal dode V ZK _ v z _ Note the drecton of deal dode! Zener CV Model Jm Stles The Unv. of Kansas ept. of EECS
11 2/20/2008 Zener ode Models 3/6 Z = v d V ZK _ 0 < 0 vz < VZK _ Analyzng ths Zener CV model, we fnd that f the model voltage v Z s less than V ZK (.e., v Z < V ZK ), then the deal dode wll be n reverse bas, and thus the model current Z wll equal zero. In other words: = 0 and v < V Z Z ZK Just lke a Zener dode n reverse bas! kewse, we fnd that f the model current s postve ( Z >0), then the deal dode must be forward based, and thus the model voltage must be v Z =V ZK. In other words: > 0 and v = V Z Z ZK Just lke a Zener dode n breakdown! z > v d V ZK _ 0 = 0 v Z ZK _ = V Problem: The voltage across a zener dode n breakdown s NOT EXACTY equal to V ZK for all z > 0. The CV s an approxmaton. Jm Stles The Unv. of Kansas ept. of EECS
12 2/20/2008 Zener ode Models 4/6 In realty, v Z ncreases a very small (tny) amount as Z ncreases. Z CV model eal zener dode characterstc V ZK v Z Thus, the CV model causes a small error, usually acceptable but for some cases not! For these cases, we requre a better model: The Zener (PW) PeceWse near model. The Zener Pecewse near Model eplace: Z V Z _ wth: z Ideal dode V Z0 _ r Z v z _ Note the drecton of deal dode! Zener PW Model Jm Stles The Unv. of Kansas ept. of EECS
13 2/20/2008 Zener ode Models 5/6 Z = v d Please Note: V Z0 _ 0 < 0 r z * The PW model ncludes a very small seres resstor, such that the voltage across the model v z ncreases slghtly wth ncreasng z. * Ths small resstance r Z s called the dynamc resstance. * The voltage source V Z0 s not equal to the zener breakdown voltage V ZK, however, t s typcally very close! v Z Z 0 _ < V Analyzng ths Zener PW model, we fnd that f the model voltage v Z s less than V Z0 (.e., vz < VZ 0), then the deal dode wll be n reverse bas, and the model current Z wll equal zero. In other words: = 0 and v < V V Z Z Z 0 ZK Just lke a Zener dode n reverse bas! kewse, we fnd that f the model current s postve ( Z > 0), then the deal dode must be forward based, and thus: Z > 0 and vz = VZ 0 Z rz Note that the model voltage v Z wll be nearv ZK, but wll ncrease slghtly as the model current ncreases. Just lke a Zener dode n breakdown! z > v d V Z0 _ 0 = 0 r z v V Z = Z 0 z z _ r Jm Stles The Unv. of Kansas ept. of EECS
14 2/20/2008 Zener ode Models 6/6 Comparson between CV and PW models Z CV model Zener ode PW model V Z0 V ZK v Z Q: How do we construct ths PW model (.e., fnd VZ 0 and r z )? A: Pck two ponts on the zener dode curve (v 1, 1 ) and (v 2, 2 ), and then select r z and V Z0 so that the PW model lne ntersects them. Z 2 1/r z 1 v Z.e., v2 v1 rz = 2 1 V Z0 v 1 v 2 and V = v r or V = v r z0 1 1 z z0 2 2 z Jm Stles The Unv. of Kansas ept. of EECS
15 2/14/2007 Example Fun wth Zener ode Models 1/6 Example: Fun wth Zener ode Models Consder ths crcut, whch ncludes a zener dode: 5.0 V 5.0 ma 17.0 V v Z Z 4 K 10.0 V V ZK = 20.0V 1K et s see f we can determne the voltage across and current through the zener dode! Frst, we must replace the zener dode wth an approprate model. Assumng that the zener wll ether be n breakdown or reverse bas, a good choce would be the zener CV model. Carefully replacng the zener dode wth ths model, we fnd that we are left wth an IEA dode crcut: Jm Stles The Unv. of Kansas ept. of EECS
16 2/14/2007 Example Fun wth Zener ode Models 2/6 5.0 V 5.0 ma 17.0 V v 4 K 10.0 V 20V 1K Snce ths s an IEA dode crcut, we know how to analyze t! Q: But wat! The deal dode n ths crcut s part of a zener dode model. on t we need to thus modfy our deal dode crcut analyss procedure n some way? In order to account for the zener dode behavor, shouldn t we alter what we assume, or what we enforce, or what we check? A: NO! There are no zener dodes n the crcut above! We must analyze ths deal dode crcut n precsely the same way as we have always analyzed deal dode crcuts (.e., secton 3.1). Jm Stles The Unv. of Kansas ept. of EECS
17 2/14/2007 Example Fun wth Zener ode Models 3/6 ASSUME: Ideal dode s forward based. ENFOCE: v = 0 ANAYZE: 5.0 V 5.0 ma 17.0 V v = 0 4 K 10.0 V 20V 1 2 1K From KV: 2 ( ) 17 v 20 1 = = = 3.0 ma 1 kewse from KV: 1 ( ) 17 v 20 4 = = = 3.25 ma 4 Now from KC: Jm Stles The Unv. of Kansas ept. of EECS
18 2/14/2007 Example Fun wth Zener ode Models 4/6 = = = ma CHECK: = ma < 0 X Ykes! We must change our deal dode assumpton and try agan. ASSUME: Ideal dode s reverse based. ENFOCE: = 0 ANAYZE: 5.0 V 5.0 ma 17.0 V v 4 K 10.0 V = 0 20V 1 1K 2 From KC: 2 = 1 5 From KV: Jm Stles The Unv. of Kansas ept. of EECS
19 2/14/2007 Example Fun wth Zener ode Models 5/6 ( ) ( ) = ( ) ( )( ) = = = 1 ma 4 1 Now, agan usng KV: 1 ( ) 17 v 20 4 = 10 v ( )( ) = = 11.0 V CHECK: v = 11.0V < 0 Q: Our assumpton s good! Snce our analyss s complete, can we move on to somethng else? A: Not so fast! emember, we are attemptng to fnd the voltage across, and current through, the zener dode. To (approxmately) determne these values, we fnd the voltage across, and current through, the zener dode model. Jm Stles The Unv. of Kansas ept. of EECS
20 2/14/2007 Example Fun wth Zener ode Models 6/6 So, and v = v V Z ZK = = 90. V Z = = 0 We re done! Q: Wat! on t we have to somehow CHECK these values? A: NO! We assumed nothng about the zener dode, we enforced nothng about the zener dode, and thus there s nothng to explctly check n regards to the zener dode solutons. However lke all engneerng analyss we should perform a santy check to see f our answer makes physcal sense. So, let me ask you the queston Q:oes ths answer make physcal sense? A: Jm Stles The Unv. of Kansas ept. of EECS
21 2/20/2008 The Shunt egulator 1/5 The Shunt egulator V S V 0 =V ZK Z The shunt regulator s a voltage regulator. That s, a devce that keeps the voltage across some load resstor ( ) constant. Q: Why would ths voltage not be a constant? A: Two reasons: (1) the source voltage V s may vary and change wth tme. (2) The load may also vary and change wth tme. In other words, the current delvered to the load may change. What can we do to keep load voltage V O constant? Employ a Zener dode n a shunt regulator crcut! Jm Stles The Unv. of Kansas ept. of EECS
22 2/20/2008 The Shunt egulator 2/5 et s analyze the shunt regulator crcut n terms of Zener breakdown voltage V ZK, source voltage V S, and load resstor. eplacng the Zener dode wth a Zener CV model, we ASSUME the deal dode s forward based, and thus ENFOCE v = 0. ANAYZE: V S v d = 0 V ZK d v z = V O From KV: v = V = v V = V Z O ZK ZK From KC: = where from Ohm s aw: V = S V ZK Jm Stles The Unv. of Kansas ept. of EECS
23 2/20/2008 The Shunt egulator 3/5 and also : Therefore: CHECK: = VZK = V V V = V S ZK ZK S = ZK ( ) V Note we fnd that deal dode s forward based f: V ( ) V S ZK = > 0 or therefore: ( ) V V S ZK > 0 V S ZK ( ) V V S ZK > > V Hence, the Zener dode may not be n breakdown (.e., the deal dode may not be f.b.) f V S or are too small, or shunt resstor s too large! Jm Stles The Unv. of Kansas ept. of EECS
24 2/20/2008 The Shunt egulator 4/5 Summarzng, we fnd that f: then: V S > 1. The Zener dode s n breakdown. 2. The load voltage V O = V ZK. V ZK 3. The load current s = V ZK. 4. The current through the shunt resstor s = V V. ( ) S ZK 5. The current through the Zener dode s = > 0. We fnd then, that f the source voltage V S ncreases, the current through shunt resstor wll lkewse ncrease. However, ths extra current wll result n an equal ncrease n the Zener dode current Z thus the load current (and therefore load voltage V O ) wll reman unchanged! Z V S Extra current goes n here! V 0 =V ZK Z Jm Stles The Unv. of Kansas ept. of EECS
25 2/20/2008 The Shunt egulator 5/5 Smlarly, f the load current ncreases (.e., decreases), then the Zener current Z wll decrease by an equal amount. As a result, the current through shunt resstor (and therefore the load voltage V O ) wll reman unchanged! V S Z Extra current comes from here! V 0 =V ZK Q: You mean that V O stays perfectly constant, regardless of source voltage V S or load current?? A: Well, V O remans approxmately constant, but t wll change a tny amount when V S or changes. To determne precsely how much the load voltage V O changes, we wll need to use a more precse Zener dode model (.e., the Zener PW)! Jm Stles The Unv. of Kansas ept. of EECS
26 2/20/2008 ne egulaton 1/4 ne egulaton Snce the Zener dode n a shunt regulator has some small (but nonzero) dynamc resstance r Z, we fnd that the load voltage V O wll have a small dependence on source voltage V S. In other words, f the source voltage V S ncreases (decreases), the load voltage V O wll lkewse ncrease (decrease) by some very small amount. Q: Why would the source voltage V S ever change? A: There are many reasons why V S wll not be a perfect constant wth tme. Among them are: 1. Thermal nose 2. Temperature drft 3. Coupled 60 Hz sgnals (or dgtal clock sgnals) As a result, t s more approprate to represent the total v t ), consstng of source voltage as a tmevaryng sgnal ( ( ) both a C component (V S ) and a smallsgnal component ( Δ vs ( t) ): vs ( t ) = VS Δ vs ( t ) v S S V S t Jm Stles The Unv. of Kansas ept. of EECS
27 2/20/2008 ne egulaton 2/4 As a result of the smallsgnal source voltage, the total load voltage s lkewse tmevaryng, wth both a C (V O ) and smallsgnal ( Δ v ) component: o v ( t) = V Δ v ( t) O O o So, we know that the C source V S produces the C load voltage V O, whereas the smallsgnal source voltage Δ v results n the smallsgnal load voltage Δ vo. s Δ v s V O Δ v o V S Q: Just how are Δ vs and Δ vo related? I mean, f Δ vs equals, say, 500 mv, what wll value of Δ be? vo A: etermnng ths answer s easy! We smply need to perform a smallsgnal analyss. In other words, we frst replace the Zener dode wth ts Zener PW model. Jm Stles The Unv. of Kansas ept. of EECS
28 2/20/2008 ne egulaton 3/4 Δ v s V S V Z0 _ r z V O Δ v o We then turn off all the C sources (ncludng V ZO ) and analyze the remanng smallsgnal crcut! Δ v s Δ v o r z From voltage dvson, we fnd: v v r Z Δ o = Δ s rz However, recall that the value of a Zener dynamc resstance r Z s very small. Thus, we can assume that r Z >>, and therefore rz rz, leadng to: Jm Stles The Unv. of Kansas ept. of EECS
29 2/20/2008 ne egulaton 4/4 earrangng, we fnd: r Z Δ vo = Δvs rz r Z Δvs rz Δvo rz = Δ v r s Z lne regulaton Ths equaton descrbes an mportant performance parameter for shunt regulators. We call ths parameter the lne regulaton. * ne regulaton allows us to determne the amount that the load voltage changes ( Δ vo ) when the source voltage changes ( Δ vs ). * For example, f lne regulaton s 0.002, we fnd that the load voltage wll ncrease 1 mv when the source voltage ncreases 500mV (.e., Δ v = Δ v = 0 002(0.5). = V ). o s * Ideally, lne regulaton s zero. Snce dynamc resstance r Z s typcally very small (.e., r Z ), we fnd that the lne regulaton of most shunt regulators s lkewse small (ths s a good thng!). Jm Stles The Unv. of Kansas ept. of EECS
30 2/20/2008 oad egulaton 1/5 oad egulaton V S v O Z For voltage regulators, we typcally defne a load n terms of ts current, where: vo = Note that snce the load (.e., regulator) voltage v O s a constant (approxmately), specfyng s equvalent to specfyng, and vce versa! Now, snce the Zener dode n a shunt regulator has some small (but nonzero) dynamc resstance r Z, we fnd that the load voltage v O wll also have a very small dependence on load resstance (or equvalently, load current ). In fact, f the load current ncreases (decreases), the load voltage v O wll actually decrease (ncrease) by some small amount. Q: Why would the load current ever change? Jm Stles The Unv. of Kansas ept. of EECS
31 2/20/2008 oad egulaton 2/5 A: You must realze that the load resstor smply models a more useful devce. The load may n fact be an amplfer, or a component of a cell phone, or a crcut board n a dgtal computer. These are all dynamc devces, such that they may requre more current at some tmes than at others (e.g., the computatonal load ncreases, or the cell phone begns to transmt). As a result, t s more approprate to represent the total load t ), consstng of both a current as a tmevaryng sgnal ( ( ) C component (I ) and a smallsgnal component ( ( t) ( t) = I Δ ( t) Δ ): Ths smallsgnal load current of course leads to a load voltage that s lkewse tmevaryng, wth both a C (V O ) and smallsgnal ( Δ v ) component: o v ( t) = V Δ v ( t) O O o So, we know that the C load current I produces the C load voltage V O, whereas the smallsgnal load current Δ ( t) results n the smallsgnal load voltage Δ vo. We can replace the load resstor wth current sources to represent ths load current: Jm Stles The Unv. of Kansas ept. of EECS
32 2/20/2008 oad egulaton 3/5 = I Δ V S V O Δ v o I Δ ( t ) Q: Just how are Δ and vo say, 50 ma, what wll value of Δ related? I mean, f Δ equals, Δ be? vo A: etermnng ths answer s easy! We smply need to perform a smallsgnal analyss. In other words, we frst replace the Zener dode wth ts Zener PW model. = I Δ V S V Z0 _ V O Δ v o I Δ ( t ) r z Jm Stles The Unv. of Kansas ept. of EECS
33 2/20/2008 oad egulaton 4/5 We then turn off all the C sources (ncludng V ZO ) and analyze the remanng smallsgnal crcut! Δ r z Δ v o Δ ( t ) From Ohm s aw, t s evdent that: earrangng, we fnd: o ( Z ) Δ v = Δ r r rz Z = Δ Δv r = = o Z load regulaton rz rz Ohms Δ r Z Ths equaton descrbes an mportant performance parameter for shunt regulators. We call ths parameter the load regulaton. Jm Stles The Unv. of Kansas ept. of EECS
34 2/20/2008 oad egulaton 5/5 * Note load regulaton s expressed n unts of resstance (e.g., Ω). * Note also that load regulaton s a negatve value. Ths means that ncreasng leads to a decreasng v O (and vce versa). * oad regulaton allows us to determne the amount that the load voltage changes ( Δ vo ) when the load current changes ( Δ ). * For example, f load regulaton s KΩ, we fnd that the load voltage wll decrease 25 mv when the load current ncreases 50mA (.e., Δ v = Δ = (50). = V ). o * Ideally, load regulaton s zero. Snce dynamc resstance r Z s typcally very small (.e., r Z ), we fnd that the load regulaton of most shunt regulators s lkewse small (ths s a good thng!). Jm Stles The Unv. of Kansas ept. of EECS
35 2/20/2008 Example The shunt regulator 1/4 Example: The Shunt egulator Consder the shunt regulator, bult usng a zener dode wth V ZK =15.0 V and ncremental resstance r z = 5Ω: Z V S = 25 V v O V ZK = 15 V r z = 5 Ω 1. etermne f the largest possble value of s 20 ma. 2. Usng the value of found n part 1 determne Z f =1.5 K. 3. etermne the change n v O f V S ncreases one volt. 4. etermne the change n v O f ncreases 1 ma. Jm Stles The Unv. of Kansas ept. of EECS
36 2/20/2008 Example The shunt regulator 2/4 Part 1: From KC we know that = Z. We also know that for the dode to reman n breakdown, the zener current must be postve..e., Z = > 0 Therefore, f can be as large as 20 ma, then must be greater than 20 ma for Z to reman greater than zero..e. > 20mA Q: But, what s?? A: Use the zener CV model to analyze the crcut. V S = 25 V _ v O 15 V Jm Stles The Unv. of Kansas ept. of EECS
37 2/20/2008 Example The shunt regulator 3/4 Therefore from Ohm s aw: VS VZK = = = and thus > 20mA f: 10 < = 0 5. K = 500 Ω 20 Note we want to be as large as possble, as large mproves both lne and load regulaton. Therefore, set = 500 Ω = 0.5 K Part 2: Agan, use the zener CV model, and enforce v = 0: V S = 25 V =0.5K 15 V v O =1.5K Analyzng, from KC: = Jm Stles The Unv. of Kansas ept. of EECS
38 2/20/2008 Example The shunt regulator 4/4 and from Ohm s aw: Vs VZK = = = 20.0 ma 0.5 VZK 15.0 = = = 10.0 ma 1.5 Therefore = = = 10.0 ma ( = 10 > 0 ) And thus we estmate Z = = 10.0 ma Part 3: The shunt regulator lne regulaton s: 5 ne egulaton = rz r = = Therefore f Δ vs = 1 V, then Δ vo = (0.01) Δ vs = 0.01 V z Part 4: The shunt regulator load regulaton s: oad egulaton = r z ( 500) 5 = = 495. Ω r z Therefore f Δ = 1 ma, then Δ v = (4.95) Δ = 4.95 mv o Jm Stles The Unv. of Kansas ept. of EECS
39 3/3/2008 egulator Power and Effcency 1/7 egulator Power and Effcency Consder now the shunt regulator n terms of power. The source V s delvers power P n to the regulator, and then the regulator n turn delvers power P to the load. V S P n V 0 =V ZK P Z Q: So, s the power delvered by the source equal to the power absorbed the load? A: Not hardly! The power delvered by the source s dstrbuted to three devces the load, the zener dode, and the shunt resstor. Jm Stles The Unv. of Kansas ept. of EECS
40 3/3/2008 egulator Power and Effcency 2/7 The power delvered by the source s: P n = V ( Vs VZK) = Vs whle the power absorbed by the load s: s P = V = V ZK V = 2 ZK V ZK Thus, the power absorbed by the shunt resstor and zener dode combned s the dfference of the two (.e., P n P ). Note that the power absorbed by the load ncreases as decreases (.e., the load current ncreases as decreases). ecall that the load resstance can be arbtrarly large, but there s a lower lmt on the value of, enforced by the condton: VS VZK > emember, f the above constrant s not satsfed, the zener wll not breakdown, and the output voltage wll drop below the desred regulated voltage V ZK! Jm Stles The Unv. of Kansas ept. of EECS
41 3/3/2008 egulator Power and Effcency 3/7 We can rewrte ths constrant n terms of : VZK > V V s ZK earrangng the expresson for load power (.e., 2 P V ZK = ): V = P we can lkewse determne an upper bound on the power delvered to the load: 2 ZK and thus: 2 VZK VZK = > P V V s ZK ( ) V V V P < ZK s ZK we can thus conclude that the maxmum amount of power that can be delvered to the load (whle keepng a regulated voltage) s: P max ( ) V V V = ZK s ZK whch occurs when the load s at ts mnmum allowed value: mn VZK = V V s ZK Jm Stles The Unv. of Kansas ept. of EECS
42 3/3/2008 egulator Power and Effcency 4/7 Note, as ncreases (.e., decreases), the load power decreases. As approaches nfnty (an open crcut), the load power becomes zero. Thus, we can state: max 0 P P Every voltage regulator (shunt or otherwse) wll have a maxmum load power ratng P. Ths effectvely s the max output power avalable to the load. Try to lower (ncrease ) such that you exceed ths ratng, and one of two bad thngs may happen: 1) the regulated voltage wll no longer be regulated, and drop below ts nomnal value. 2) the regulator wll melt! Now, contrast load power P wth the nput power P n : P n = V s ( V V ) s ZK Q: Wat! It appears that the nput power s ndependent of the load resstance! oesn t that mean that P n s ndependent of P? Jm Stles The Unv. of Kansas ept. of EECS
43 3/3/2008 egulator Power and Effcency 5/7 A: That s correct! The power flowng nto the shunt regulator s constant, regardless of how much power s beng delvered to the load. In fact, even f P =0, the nput power s stll the same value shown above. Q: But where does ths nput power go, f not delvered to the load? A: emember, the nput power not delvered to the load must be absorbed by the shunt resstor and the zener dode. More specfcally, as the load power P decreases, the power absorbed by the zener must ncrease by an dentcal amount! Q: Is ths bad? A: It sure s! Not only must we dsspate the heat that ths power generates n the regulator, the energy absorbed by the shunt resstor and zener dode s essentally wasted. Ths s partcularly a concern f our source voltage V s s from a storage battery. A storage battery holds only so much energy. To maxmze the tme before ts depleted, we need to make sure that we use the energy effectvely and effcently. Jm Stles The Unv. of Kansas ept. of EECS
44 3/3/2008 egulator Power and Effcency 6/7 Heatng up a zener dode s not an effcent use of ths lmted energy! Thus, another mportant parameter n evaluatng regulator performance s ts effcency. Smply stated, regulator effcency ndcates the percentage of nput power that s delvered to the load: regulator effcency e r P P n Ideally, ths effcency value s e r =1, whle the worst possble effcency s e r =0. For a shunt regulator, ths effcency s: e r 2 P VZK = P V ( V V ) n s s ZK Note that ths effcency depends on the load value. As ncreased toward nfnty, the effcency of the shunt regulator wll plummet toward e r =0 (ths s bad!). On the other hand, the best possble effcency occurs when max P = P : Jm Stles The Unv. of Kansas ept. of EECS
45 3/3/2008 egulator Power and Effcency 7/7 e max r P P max n ( ) VZK Vs VZK = V ( V V ) V = V ZK s s s ZK Thus, for the shunt regulator desgn we have studed, the effcency s: ( ) 0 er VZK Vs Q: So, to ncrease regulator effcency, we should make V s as small as possble? A: That would n fact mprove regulator effcency, but beware! educng V s wll lkewse lower the maxmum possble load power P. max Jm Stles The Unv. of Kansas ept. of EECS
46 2/20/2008 Voltage egulators 1/3 Voltage egulators Note that we can vew a shunt regulator as a threetermnal devce, nserted between a voltage source and a load: Input Output V S Common Integrated crcut technology has resulted n the creaton of other three termnal voltage regulator desgns regulators that do not necessarly use zener dodes! V S Input Integrated Crcut Voltage egulator Output Common Jm Stles The Unv. of Kansas ept. of EECS
47 2/20/2008 Voltage egulators 2/3 These ntegrated crcut voltage regulators are small and relatvely nexpensve. In addton, these IC regulators typcally have better load regulaton, lne regulaton, and/or effcency than the zener dode shunt regulator! Q: Wow! The desgners of these IC regulators obvously had a much better electroncs professor than the dope we got stuck wth! Wth what devce dd they replace the zener dode? A: The electronc desgn engneers dd not smply replace a zener dode wth another component. Instead, they replaced the entre shunt regulator desgn wth a complex crcut requrng many transstor components. Jm Stles The Unv. of Kansas ept. of EECS
48 2/20/2008 Voltage egulators 3/3 Integrated crcut technology then allows ths complex crcut to be manufactured n a very small space and at very small cost! Jm Stles The Unv. of Kansas ept. of EECS
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