3.6 Limiting and Clamping Circuits

Transcription

1 3/10/2008 secton_3_6_lmtng_and_clampng_crcuts 1/1 3.6 Lmtng and Clampng Crcuts Readng Assgnment: pp (.e., neglect secton 3.6.2) Another applcaton of juncton dodes Q: What s a lmter? A: A 2-port devce that restrcts (.e., lmts) the voltage across a devce to some specfed regon. HO: ode Lmters Q: A: HO: Steps for Analyzng Lmter Crcuts Example: A ode Lmter Jm Stles The Unv. of Kansas ept. of EECS

2 3/10/2008 secton_3_6_lmtng_and_clampng_crcuts 1/1 3.6 Lmtng and Clampng Crcuts Readng Assgnment: pp (.e., neglect secton 3.6.2) Another applcaton of juncton dodes Lmters Q: What s a lmter? A: A 2-port devce that restrcts (.e., lmts) the voltage across a devce to some specfed regon. HO: ode Lmters Q: A: HO: Steps for Analyzng Lmter Crcuts Example: A ode Lmter Jm Stles The Unv. of Kansas ept. of EECS

3 3/10/2008 secton_3_6_lmtng_and_clampng_crcuts 1/1 3.6 Lmtng and Clampng Crcuts Readng Assgnment: pp (.e., neglect secton 3.6.2) Another applcaton of juncton dodes Lmters Q: What s a lmter? A: A 2-port devce that restrcts (.e., lmts) the voltage across a devce to some specfed regon. A lmter s a protecton devce! HO: ode Lmters Q: A: HO: Steps for Analyzng Lmter Crcuts Example: A ode Lmter Jm Stles The Unv. of Kansas ept. of EECS

4 3/10/2008 secton_3_6_lmtng_and_clampng_crcuts 1/1 3.6 Lmtng and Clampng Crcuts Readng Assgnment: pp (.e., neglect secton 3.6.2) Another applcaton of juncton dodes Lmters Q: What s a lmter? A: A 2-port devce that restrcts (.e., lmts) the voltage across a devce to some specfed regon. A lmter s a protecton devce! HO: ode Lmters Q: So how do we determne the transfer functon of a lmter? A: HO: Steps for Analyzng Lmter Crcuts Example: A ode Lmter Jm Stles The Unv. of Kansas ept. of EECS

5 3/10/2008 ode Lmters 1/4 ode Lmters Often, a voltage source (ether C or AC) s used to supply an electronc devce that s very expensve and/or very senstve. n ths case, we may choose nsert a dode lmter between the source and the devce ths lmter wll provde over-voltage protecton! To see how, we should frst consder a typcal transfer functon for a juncton dode lmter: + v ( t ) Juncton - ode v ( t) Lmter + O Senstve evce v O L + K v L K L+ K L - Jm Stles The Unv. of Kansas ept. of EECS

6 3/10/2008 ode Lmters 2/4 Note that ths transfer functon ndcates that the output voltage v O can never be more than a maxmum voltage L +, nor less than a mnmum voltage L -. * Thus, the devce places some lmts on the value of the output voltage: L < v < L for any v O + * The lmts L - and L + provde a safe operatng value for v O, the voltage across our senstve electronc devce. * Presumably, f no lmter were present, we mght fnd that vo > L + or v O < L, resultng n damage to the devce! * Note althoughl+ > L, the values of L - and L + may be both postve, both negatve, or even zero. For example, a lmter wth L - =0 (L + >0) would prevent the voltage from ever becomng negatve (postve). We fnd that for many devces, the wrong voltage polarty can be destructve! To llustrate, let s consder an example nput voltage v (t), and the resultng output voltage when passed through a lmter wth values L - =0 and L + =20 V (K=1)..E.: 0 f v < 0 vo = v f 0 < v < f v > 20 Jm Stles The Unv. of Kansas ept. of EECS

7 3/10/2008 ode Lmters 3/4 v L + =20 v (t) v O (t) L - =0 t Note there are a couple of hccups n the nput voltage that take the voltage value outsde the safety range of the senstve devce. However, the lmter does n fact lmt these excursons, such that the voltage across the senstve devce always remans between 0 and 20 Volts. Q: Why would these hccups occur? A: There are many possble reasons, ncludng: 1. A power surge (e.g., lghtnng strke) 2. Statc dscharge 3. Swtchng transents (e.g., at power up or down). Jm Stles The Unv. of Kansas ept. of EECS

8 3/10/2008 ode Lmters 4/4 Perhaps the most prevalent reason, however, s operator error. Someone connects the wrong source to the senstve devce! Thus, lmters are often used on expensve/senstve devces to make them fool-proof. Your book has many examples of lmter crcuts, ncludng: Jm Stles The Unv. of Kansas ept. of EECS

9 3/10/2008 Steps for Analyzng Lmter Crcuts 1/4 Steps for Analyzng Lmter Crcuts The juncton dodes n most lmter crcuts can/wll be n forward bas, or reverse bas, or breakdown modes! Thus, the dstncton between a Zener dode and a normal juncton dode s essentally meanngless. But, ths presents us wth a bg problem what dode model do we use to analyze a lmter? Recall that none of the dode models that we studed wll provde accurate estmates for all three juncton dode modes! The soluton we wll use s to change the dode model we mplement, as we consder each of the possble juncton dode modes. Specfcally: Juncton ode Mode Forward Bas Reverse Bas Breakdown Juncton ode Model CV model wth deal dode f.b. deal dode model wth deal dode r.b Zener CV model wth deal dode f.b. Jm Stles The Unv. of Kansas ept. of EECS

10 3/10/2008 Steps for Analyzng Lmter Crcuts 2/4 Step 1: Assume that the lmter dode s forward based, so replace wth a CV model, where the deal dode s forward based: or A C C A A V Now, usng ths model, determne: C 1. The output voltage v O n terms of nput voltage v. 2. The deal dode current n terms of nput voltage v. Fnally, we solve the nequalty > 0 for v, thus determnng when (.e., for what values of v ) ths assumpton, and thus the derved expresson for output voltage v O, s true. Step 2: Assume that the lmter dode s n breakdown, so replace A C or A C Jm Stles The Unv. of Kansas ept. of EECS

11 3/10/2008 Steps for Analyzng Lmter Crcuts 3/4 wth a Zener CV model, where the deal dode s forward based: A V ZK + C Now, usng ths model, determne: 1. The output voltage v O n terms of nput voltage v. 2. The deal dode current n terms of nput voltage v. Fnally, we solve the nequalty > 0 for v, thus determnng when (.e., for what values of v ) ths assumpton, and thus the derved expresson for output voltage v O, s true. Step 3: Assume that the lmter dode s reverse based, so replace A C or A C wth an deal ode model, where the deal dode s reversed based: A + v C Jm Stles The Unv. of Kansas ept. of EECS

12 3/10/2008 Steps for Analyzng Lmter Crcuts 4/4 Now, usng ths model, determne the output voltage v O n terms of nput voltage v. Q: What about v? on t we need to lkewse determne ts value, and then determne when v < 0? A: Actually, no. f the juncton dode s not forward based and t s not n breakdown, then t must be reverse based! As obvous as ths statement s, we can use t determne when the juncton dode s reverse based t s when the juncton dode s not n forward bas and when t s not n reverse bas. For example, say that we fnd that the juncton dode s forward based when: v > 20 V, and that the juncton dode s n breakdown when: v < 15 V. We can thus conclude that the juncton dode s reverse based when: 15V < < 20 V v Step 4: We take the result of the prevous 3 steps and form a contnuous, pecewse lnear transfer functon (make sure t s contnuous, and that t s a functon!). Jm Stles The Unv. of Kansas ept. of EECS

13 3/10/2008 Example A ode Lmter 1/7 Example: A ode Lmter Consder the followng juncton dode crcut: +5V 1K V ZK =10V open crcut v v O 1K Ths crcut s a juncton dode lmter! Perhaps that would be clearer f we redrew ths crcut as: 1K v V - V ZK =10V 1K + v O - Ths s the same crcut as above! Jm Stles The Unv. of Kansas ept. of EECS

14 3/10/2008 Example A ode Lmter 2/7 Now, let s determne the transfer functon of ths lmter. To do ths, we must follow the 4 steps detaled n the prevous handout! Step1: Assume juncton dode s forward based Replace the juncton dode wth a CV model. ASSUME the deal dode s forward based, ENFORCE v = 0. v 1K 1 +5V 1K - 0.7V + 2 v O We fnd that the output voltage s smply: v = = 57V. O whle the deal dode current s more dffcult to determne. From KCL: where from Ohm s Law: = Jm Stles The Unv. of Kansas ept. of EECS

15 3/10/2008 Example A ode Lmter 3/7 v = = v and: = = Thus, the deal dode current s: = = v = v Now, for our assumpton to be correct, ths current must be postve (.e., > 0 ). Thus, we solve ths nequalty to determne when our assumpton s true: So, from ths step we fnd: v > 0 v > V v = 57V. when v > 114V. O Step2: Assume the juncton dode s n breakdown Replace the juncton dode wth a Zener CV model. ASSUME the deal dode s forward based, ENFORCE v = 0. Jm Stles The Unv. of Kansas ept. of EECS

16 3/10/2008 Example A ode Lmter 4/7 +5V v 1K V - v O 1 1K 2 We fnd that the output voltage s smply: v = 5 10 = 5 0V. O whle the deal dode current s more dffcult to determne. From KCL: where from Ohm s Law: and: = V 1 = = v = = 50V. 1 Thus, the deal dode current s: Jm Stles The Unv. of Kansas ept. of EECS

17 3/10/2008 Example A ode Lmter 5/7 = = v = v Now, for our assumpton to be correct, ths current must be postve (.e., > 0 ). Thus, we solve ths nequalty to determne when our assumpton s true: So, from ths step we fnd: v > 0 v > V v < V v = 50V. when v < 100V. O Step 3: Assume the juncton dode s reverse based Replace the juncton dode wth the deal ode model. ASSUME the deal dode s reverse based, ENFORCE = 0. v 1K +5V + v v O 1K A voltage dvder! Jm Stles The Unv. of Kansas ept. of EECS

18 3/10/2008 Example A ode Lmter 6/7 Thus the output voltage s: v O ( 1) v = 1+ 1 v = 2 Ths output voltage s true when the juncton dode s nether forward based nor n breakdown. Thus, usng the results from the frst two steps, we can nfer that t s true when: < v < Step 4: etermne the contnuous transfer functon Combnng the results of the prevous 3 steps, we get the followng contnuous, pece-wse lnear transfer functon: 57V. f v > 114V. vo = v 2 f 100. < v < 114V. 50.V f v < 100V. Jm Stles The Unv. of Kansas ept. of EECS

19 3/10/2008 Example A ode Lmter 7/7 v O v Note that at v = 10 : and at v = : v 10 v O = = = 50V. 2 2 v v O = = = 57V. 2 2 Thus, ths functon s contnuous! Jm Stles The Unv. of Kansas ept. of EECS