Diodes Two - Engineering Electrical Circuits - Lecture Slides, Slides of Electrical Circuit Analysis

Download Diodes Two - Engineering Electrical Circuits - Lecture Slides and more Slides Electrical Circuit Analysis in PDF only on Docsity! Diodes-2 Docsity.com Learning Goals • Understand the Basic Physics of Semiconductor PN Junctions which form most Diode Devices • Sketch the IV Characteristics of Typical PN Junction Diodes • Use the Graphical LOAD-LINE method to determine the “Operating Point” of Nonlinear (includes Diodes) Circuits Docsity.com Diode Models • Consider an Electrical Diode →  We can MODEL the V-I Behavior of this Device in Several ways V I REAL Behavior IDEAL Model OFFSET Model LINEAR Model Docsity.com Ideal Model (Ideal Rectifier) • Analyze Ckts containing Ideal Diodes 1. Assume (or Guess) a “state” for each diode. Ideal Diodes have Two states 1. ON → a SHORT Ckt when Fwd Biased 2. OFF →an OPEN Ckt if Reverse Biased 2. Check the Assumed Opens & Shorts – Should have Current thru the SHORTS – Should have ∆V across the OPENS D io de O N Diode OFF Docsity.com Ideal Model (Ideal Rectifier) 3. Check to see if guesses for i-flow, ∆V, and BIAS-State are consistent with the Ideal- Diode Model 4. If i-flow, ∆V, and bias-V are consistent with the ideal model, then We’re DONE. – If we arrive at even a SINGLE Inconsistency, then START OVER at step-1 D io de O N Diode OFF Docsity.com Example  Ideal Diode • Using ID2 = 1 mA • Thus • Now must Check that both Diodes are indeed conducting • From the analysis • Thus the current thru both Diodes is positive which is consistent with the assumption  1dI↓ ↓2dI A [ ]( ) mA 1 k 9.9 V 100 1 −Ω −− =dI mA 0101.01 +=dI mA 1µA 10 21 +=+= dd II Docsity.com Example  Ideal Diode • Since both Diodes conduct the Top of Vo is connected to GND thru D2 & D1 • Another way to think about this is that since VD2 = 0 and VD1 = 0 (by Short Assumption) Find Vo = GND+VD2+VD1 = GND + 0 + 0 = 0 • Thus the Answer 1dI↓ ↓2dI A V 0µA 101 =+= od VI Docsity.com Example  Ideal Diode • Find For Ckt Below find: – Use the Ideal Diode Model – Note the different values on R1 & R2 • Swapped od VI &1 1dI↓ ↓2dI B Docsity.com Example  Ideal Diode • Must Iterate • Assume – D1 → OFF D2 → ON • In this Case D1 is an OPEN → ID1 =0 • Current ID2 must flow thru BOTH Resistors • Then Find by Ohm [ ]( ) ( ) mA 005.1k 9.910 V 1010 2 =Ω+ −− =dI 1dI↓ ↓2dI B Docsity.com Example  Ideal Diode • Must Check that D1 is REVERSE Biased as it is assumed OFF • By KVL & Ohm • Thus D1 is INDEED Reverse-Biased, Thus the Ckt operation is Consistent with our Assumption  1dI↓ ↓2dI B mV 50 V 05.0 mA005.1kΩ10V10 +=+= ⋅+−= B B V V Docsity.com Example  Ideal Diode • Calculate Vo by noting that: • D2 is ON → VD2 = 0 D1 is OFF → Current can only flow thru D2 • In this case Vo = VB • By the Previous Calculation, Find 1dI↓ ↓2dI B mV 50A 01 +== od VI Docsity.com Point Slope Line Eqn • Using the 2nd Point • Can easily convert to y = mx+b • Multiply by m, move −5 to other side of = 2 4 6 8 10 12 14 16 18 20 4 6 8 10 12 14 16 18 x y ( ) ( )19 4 35 −−=− xy (3,17) (19,5) ( ) ( ) 4 77 4 3 4 20 4 57 4 3 5 4 57 4 3 4 57 4 35 19 4 35 +−= ++−= ++−= +−=− −−=− xy xy xy xy xy Docsity.com Slopes on vi Curve • With Reference to the Point-Slope eqn v takes over for x, and i takes over for y • The Slope on a vi Curve is a conductance • If the curve is NONlinear then the local conductance is the first Derivative • Recall the Op-Pt is also the Q-Pt Gm V Im vi vi = =⇒ ∆ ∆ = iemens S Volt Amps g dv dim dv dim Op Opvi Op Opvi == =⇒= , , SiemensVolt Amps g dv dimm Q QviOpvi === ,, Docsity.com Slopes on vi Curve • Finally recall that conductance & resistance are Inverses • Example: Find the RESISTANCE of the device associated with the VI curve that follows 2 4 6 8 10 12 14 16 18 20 4 6 8 10 12 14 16 18 V (volts) I ( am ps ) Linear VI Curve R Gmvi 1 == ( ) ( ) Siemens 4 3 V16 A12 Volts319 Amps517 === = − − ⇒ ∆ ∆ = Gm V Im vi vi Docsity.com PieceWise Linear Zener • m for Segment A • Us Pt-Slp eqn with (0.6V,0mA) for Pt-1 • Segment- B is easy ( ) ( ) Ω == − − ⇒ ∆ ∆ == 10 1mS 100 V0.61.6 mA01001 A A A m V I R m ( ) mA60mS100 :OR V6.0mS1000 −⋅= −=− DD DD vi vi 0=Di Docsity.com PieceWise Linear Zener • m for Segment C • Us Pt-Slp eqn with (−6V,0mA) for Pt-1 • Thus the PieceWise Model for the Zener [ ]( ) [ ]( ) Ω === −−− −− ⇒ ∆ ∆ == 12 1mS 33.83 1.2V mA100 V2.76 mA10001 C C C m V I R m [ ]( ) mA500mS3.83 :OR V6mS3.830 +⋅= −−=− DD DD vi vi      −≤+⋅ <<− ≥−⋅ = V6ifmA5003.3mS8 V6.0V6if0 V6.0ifmA60mS100 DD D DD D vv v vv i Docsity.com Example PieceWise Linear Model • Alternatively in terms of Resistances • ADVICE: remember the Pt- Slope Line-Eqn        −≤+ Ω <<− ≥− Ω = V6ifmA500 12 V6.0V6if0 V6.0ifmA60 10 D D D D D D vv v vv i Docsity.com Smoothed HalfWave Rectifier • Adding a Cap to the Circuit creates a Smoothing effect • In this case the Diode Conducts ONLY when vs>vC and vC=vL • This produces vL(t) and iL(t) curves • Note that iL(t) is approx. constant ( ) LCD LC iii dtdvCi += ⋅= Docsity.com Smoothed HalfWave Rectifier • The change in Voltage across the Cap is called “Ripple” • Often times the load has a Ripple “Limit” from which we determine Cap size • From the iL(t) curve on the previous slide note: – Cap Discharges for Almost the ENTIRE Cycle time, T (diode Off) – The Load Current is approx. constant, IL • Recall from EARLY in the Class Ripple TIQ ⋅= ×= :lysymbolical Or, Time Current Charge Docsity.com Smoothed HalfWave Rectifier • Also from Cap Physics (chp3) • In the Smoother Ckt the Cap charges during the “Ripple” portion of the curve • Equating the Charge & Discharge “Q’s find • Note that both these equations are Approximate, but they are still useful for initial Ckt Design • Solving the equations for the Cap Value needed for a given Vr capcap VCQ ∆⋅=∆ TIVCQ Lrcap ⋅=⋅=∆ Charge Discharge r L V TIC ⋅= Docsity.com Full Wave Rectifier • The half-wave ckt will take an AC-Voltage and convert it to DC, but the rectified signal has gaps in it. • The gaps can be eliminated thru the use of a Full-Wave rectifier ckt • The Diodes are – Face-to-Face (right) – Butt-to-Butt (left) • This rectified output has NO Gaps Docsity.com Full Wave Rectifier Operation D1 Supplies V to Load D4 Supplies V to Load Docsity.com Full Wave Rectifier Smoothing • The Ripple on the FULL wave Ckt is about 50% of that for the half-wave ckt • Since the Cap DIScharges only a half- period compared to the half-wave ckt, the size of the “smoothing” cap is then also halved: r L V TIC 2 ⋅ = Docsity.com Small Signal Analsyis • Now let y→iD, and x→ vD • Use a DC power Supply to set the operating point on the diode curve as shown at right – This could be done using LoadLine methods • From Calculus • Next Take derivative about the Q-Pt ( ) Siemensor A/V of units DDDD vidvdi ∆∆≅ Docsity.com Small Signal Analysis • About Q-Pt • Now if we have a math model for the vi curve, and we inject ON TOP of VDQ a small signal, ∆vD find • The derivative is the diode small-signal Conductance at Q Qnear D D QD D v i dv di ∆ ∆ ≅ ( ) DdD QD D D vgvdv dii ∆⋅=∆         ≅∆ Docsity.com Small Signal Analysis • In the large signal Case: R = 1/G • By analogy In the small signal case: • r = 1/g • Also since small signal analysis is associated with small amounts that change with time… • Define the Diode’s DYNAMIC, small-signal Conductance and Resistance QD D QD D d d QD D d di dv dv di g r dv dig =         == = −1 1 Docsity.com Notation: Large, Small, Total • VDQ and IDQ are the LARGE Signal operating point (Q-Pt) DC quantities – These are STEADY-STATE values • vD and iD are the TOTAL and INSTANTANEQOUS quantities – These values are not necessarily steady-state. To emphasize this we can write vD(t) and iD(t) Docsity.com Notation: Large, Small, Total • vd and id are the SMALL, AC quantities – These values are not necessarily steady-state. To emphasize this we can write vd(t) and id(t) • An Example for Diode Current notation Docsity.com Effect of Q-Pt Location • From Analysis DQ T d ddd I nVr rvi = = and ppdi ,2 ppdi 1 ppdv 1 ppdv 2 Docsity.com Large and Small Signal Ckts • Recall from Chps 3 and 5 for Caps: – OPENS to DC – SHORTS to fast AC • Thus if C1 is LARGE it COUPLES vin(t) with the rest of the ckt • Similarly, Large C2 couples to the Load • To Find the Q-point DEcouple vin and vo to arrive at the DC circuit ( )CjZC ω1= Docsity.com Large and Small Signal Ckts • Finding the Large signal Model was easy; the Caps acts as an OPENS • The Small Signal Ckt needs more work – Any DC V-Supply is a SHORT to GND – The Diode is replaced by rd (or gd) – The Caps are Shorts • Thus the Small Signal ckt for the above Docsity.com Example: Small Signal Gain • Find the Small Signal Amplification (Gain), Av, of the previous circuit • Using the Small Signal Circuit • Note that RC, rd, and RL are in Parallel • And vo(t) appears across this parallel combination • The equivalent ckt LdCp RRRR 1111 ++= ( ) − + tvo Docsity.com Bruce Mayer, PE Registered Electrical & Mechanical Engineer BMayer@ChabotCollege.edu Engineering 43 Appendix Docsity.com ( ) − + tvo ( ) − + tvo Docsity.com Small Signal Analysis • In the large signal Case: R = 1/G • By analogy In the small signal case: • r = 1/g • Also since small signal analysis is associated with small amounts that change with time… • Define the Diode’s DYNAMIC Conductance and Resistance QD D QD D d d QD D d di dv dv di g r dv dig =         == = −1 1 Docsity.com ylhen P/ODOE 48 CON, Vo= Vo o. Lox Le ¢ Tz = Sv—-Ve ZA. 3A. Also Lair Vir? Docsity.com a lta By Kee DitLs —te BY #BOVE OFM CANS Ui- Uo 4 Sv-Vo _ Ve LZR SAL ZA Jeo Ww 42d 1. 7ALL BAL ZA /0i —L0V0 + 2Zor—YVo = 6Us JAW t2QVS ZRVO = : ple ov Vo = BHF lV | az) @ y= moth  Bruce Mayer, PE 4 Engineering Instructor Chabot College 25555 Hesperian Blvd i Hayward, CA 94545 eMail: omayer@chabotcollege.edu “ST (2 PPR [ee Docsity.com plo. 8S otQ=st 0.01 ewe (wr) Dp coz (w+) Lp fers BT EE = Av fa arp Con een Cr) Cozlwe) = Ap-3 (3) oO.?e 6ue (2) ™7e C1) 40-3 | oO. Vpz S+0.08 (p-3) Wp = Sst 0.0] 1A 905X272 On gidio ma Acs o ob Uo Jak = 0.0!to dinlwt) do jut — 0. 2Ww4RWt “N > Docsity.com Q-pr THE C-PT IS THE PC oPpatanvar Powr, dae gues CASE TE BC CON RP OverTS or OF Mp tf Lo 7 V= SV > Ip= 374 THos THE Q-PT Vbg = SV Dog = 3A glue — dost _ 0. 0lby tifest) V dho hu fst dt 0. TOO derlat) ma Aes 0°SK = son A @