Example: quiz answers

New Lecture 20 - Walter Scott, Jr. College of Engineering

1 Lecture 20Bi-directional Switching Problem Solutions andOverview of insulated gate bipolar TransistorSwitchesA. Illustrative Bi-directional Current ProblemSolution : Erickson s Problem. Erickson s : Current quadrants of operation for real solidstate switchesa. Two quadrant illustrative problem 1. Voltage bi-directional switchexample3. Four quadrant switch and exampleB. Introduction to the IGBT Switch DeviceStructure and I-V Characteristics1. Overview of the IGBT Device Goals2. IGBT Device Cross-sections3. SOA regions and long tail turn-off time ofIGBT devices4. Illustrative IGBT Energy loss Problem Solution : Problem. (1) Erickson s problem : Current bi-directional switchingFor Chapter 2 of Erickson in problems #4 & #5 you solved for VOFFand ION of switches leading to the transistor circuit. Follow thesame procedure as your old homework.+-vRLCVg1221iLVgI1vL+-v+-LRCICS W1 on/SW2 offVgI1vL+-v+-LRCICI2SW1 off/SW2 onTime DTsVL = Vg - Vc1I = I - VRVg1221 Time D TsVL = - VgcI = - VRVolt-sec balance on L: <VL> = 0, D = 1 - DD(V-V)-D V = 0 DV-D+DV = 0gggg 3DV = 2 Dvg - VgV = V(2D-1)Dg output voltage in steady stateCharge balance on the capacitor: <ic> = 0D (I-VR) -DVR = 0 DI - DVR - VR + DVR = 011 1g2I = VDR = V(2D-1)DRD > Gives bipolar currentD < when onVg-+-+dabc-+-++-VLL

and the MOSFET is the insulated gate bipolar transistor (IGBT). This device was invented to capture the two separate advantages of each the bipolar and the MOSFET transistor in one device.

Tags:

  Gate, Transistor, Insulated, Bipolar, Insulated gate bipolar transistor

Information

Domain:

Source:

Link to this page:

Please notify us if you found a problem with this document:

Other abuse

Transcription of New Lecture 20 - Walter Scott, Jr. College of Engineering

1 1 Lecture 20Bi-directional Switching Problem Solutions andOverview of insulated gate bipolar TransistorSwitchesA. Illustrative Bi-directional Current ProblemSolution : Erickson s Problem. Erickson s : Current quadrants of operation for real solidstate switchesa. Two quadrant illustrative problem 1. Voltage bi-directional switchexample3. Four quadrant switch and exampleB. Introduction to the IGBT Switch DeviceStructure and I-V Characteristics1. Overview of the IGBT Device Goals2. IGBT Device Cross-sections3. SOA regions and long tail turn-off time ofIGBT devices4. Illustrative IGBT Energy loss Problem Solution : Problem. (1) Erickson s problem : Current bi-directional switchingFor Chapter 2 of Erickson in problems #4 & #5 you solved for VOFFand ION of switches leading to the transistor circuit. Follow thesame procedure as your old homework.+-vRLCVg1221iLVgI1vL+-v+-LRCICS W1 on/SW2 offVgI1vL+-v+-LRCICI2SW1 off/SW2 onTime DTsVL = Vg - Vc1I = I - VRVg1221 Time D TsVL = - VgcI = - VRVolt-sec balance on L: <VL> = 0, D = 1 - DD(V-V)-D V = 0 DV-D+DV = 0gggg 3DV = 2 Dvg - VgV = V(2D-1)Dg output voltage in steady stateCharge balance on the capacitor: <ic> = 0D (I-VR) -DVR = 0 DI - DVR - VR + DVR = 011 1g2I = VDR = V(2D-1)DRD > Gives bipolar currentD < when onVg-+-+dabc-+-++-VLLILCR+-VICSW a Va = Vg - V Ia = ILSW b Vb = V - Vg Ib = ILSW c VC = - Vg Ic = ILSW d Vd = Vg Id = IL(a) Vg-V-ILIL(b) V-Vg-ILIL(c) -Vg-ILIL(d) Vg-ILILKEY:Show why Vtr(OFF) is unipolarIL = IR is bipolar4A bipolar implementation of the bi-directional current switch isshown below.

2 Vgdac-CR+-VbFor HW#4 you do Erickson Problem another two quadrantexample with bipolar current.(2)Limited quadrants of operation forreal solid state switchesa. Voltage bi-directional switch exampleswitchon-statecurrentswitchoff-st atevoltageIdeal voltage-bi-directional 2-quadrant switch (diodeblocks voltage)off (transistorblocks voltage)vi(b)(a)i10+-vCRealizing a voltage-bi-directional2-quadrant SPST switch:(a) implementation using atransistor and series diode, (b)idealized switch voltage bi-directional switch finds use in various circuits suchas the buck-boost DC to ac 3 converter below. aVgiL++--vab(t)vab(t) b cvac(t)DC-3 ac buck-boost inverter.(3).Four quadrant real switches can conducteither polarity of current, and can blockeither polarity of +-vi1i10+-v0i1+-vThese switch topologies find use in 3 mains to 3 voltages @other frequencies, in short 3 f converters.

3 These could be usedfor speed control of synchronous AC phase60 Hz @ VThree phasevoltage atanotherfrequency f6 Below all v and i are ac therefore 4-quadrant switches are requiredas ac power could flow in either acoutput3-phase acinputA 3 ac-3 ac matrix converter, which requires nine SPST four-quadrant IGBT SWITCH1. Overview of IGBT Device GoalsAnother solid state switch choice besides the bipolar transistorand the MOSFET is the insulated gate bipolar transistor (IGBT). This device was invented to capture the two separate advantagesof each the bipolar and the MOSFET transistor in one device. Specifically, the IGFET has the high input impedance of theMOSFET input circuit with the high current and low impedance ofthe BJT. We will outline below the basic properties of the IGBT asa switch to achieve higher output switch current, lower input gatecurrent and the ability to block both polarity voltages across the circuit shown below which captures the intent of thedevice Darlington connection for low IG drivepower switchingThe Darlington combination of an input FET and an output bipolarallows for:1.

4 Convenient low current gate control on the FET driven bystandard analog Higher output current capability from the bipolar integrated form cross-section is shown below with a PNPN structure that also gives rise to an inadvertant parasiticthyristor(SCR). Cross-sectionsA brief sketch of emitter, collector and MOSFET locations .8A little clearer device cross-section that betterillustrates the parasitic SCR is shown next employ the dual transistor model of the SCR adding thedevice resistance in the emitter leg as shown below on page 9 tobetter explain dynamic performance of the multi-device model for IGBTRBA voidance of thyristor latch-up and avoiding inadvertent turn-on of the bipolar transistor are both important for proper IGBT Also, unlike a pure MOSFET switch the IGBT can block voltagesof either polarity. The safe operating area of the IGBT is shownbelow for both the forward and reverse and long tail turn-off of IGBTFor this capability we pay a price in slow turn-off as off pointCollector current (A)Fig.

5 IGBT long tail turn-off blocking capability and any problems that arise are bestdiscussed after we look at the device cross-section below to seethe various regions. Note the THREE JUNCTIONS from the top ofthe device down. Especially note the J1 junction at the bottom thatdistinguishes the IGBT. Also note that the extended drain at thebottom of the cross-section and the two source regions at the topof the cross-section are separated spatially. This is accomplishedby a large N- region of low doping as shown, so we can achieve aLARGE stand-off voltage. Unfortunately, this also introduces thepossibility of a parasitic SCR. A N+ buffer layer speeds up theIGBT turn-off. Below is the cross-section for your static or DC model for the IGBT is a diode in series with apower MOSFET to better explain: forward voltage behavior, lowgate current and reverse blocking ability as shown Static model for IGBTT radeoffs are made in traditional IGBT manufacture between thestatic model Von and the turn-off speed.

6 You can get both smallonly with difficulty. Consider two IGBT IGBTSlow IGBTVon = V @ 100 AVon = V @ 100 AW(switch) = 4mJ/switchW(switch) = 12mJ/switchWhen VGE = 10V is applied to the IGBT the turn-on time of500 ns is similar to BJT but slower than MOSFET s. To turn theIGBT off VGE is set to zero allowing the CGS voltage to dischargeto zero cutting off bipolar MOSFET conduction. The parallelbipolar transistor current however falls very slowly (1-20 s) ascarriers are eliminated in the base only by slow Po=2kW13In summary, the IGBT device replaces the bipolar transistor orMOSFET primarily because with it we can achieve higher V and Iswitching characteristics by using an of a base we have aMOS gate as input. However, turn-off characteristics of an IGBT are very slow(typically 1-20 s), causing additional switching , we note in passing that MODERN IGBT device designemploys vertical trenches cut into the bulk silicon and toachieve An increased density of IGBT cells per unit areabecause of more confined vertical current flow and lesscurrent spreading laterally Improved latch-up protection fron the parasitic SCR bybetter confining vertical current flowThe modern trench isolated IGBT structure is shown on thefollowing page.

7 We will also note on page 14 two tricks to reducethe long turn-off time of conventional IGBT devices: a heavilydoped buffer layer to recombine excess charge and post devicefabrication irradiation of the device by MeV electron beams tocreate recombination centers in the bulk silicon and thereby14increase the recombination time for carriers. The N-buffer layer lies on the bottom of the device structure The MeV electron irradiation is not shown, but it achievesincreased carrier recombination times in the bulk siliconWe are now done with our brief overview of IGBT devices. Thisreview would be a good starting point for aterm paper. Please feel free to do a termpaper on modern IGBT Design orpreformance154. Illustrative Problem for IGBT Energy lostduring on page 15 are VCE across IGBT and Ic out of the IGBT. Note the Ic possesses a long tail in time compared to the rapidchange VCE during turn-off.

8 Also the IGBT has characteristic highcurrent spikes, due to the internal diode, during turn-on. Bothtransients cause increased switching losses. Next we outline thesolution to Erickson Chapter 4 Pbm. which quantitativelyillustrates all of the above (t)vCE(t) iC(t) iC(t)400V40A200V20At sFor a typical IGBT, two very different switching transitions occur inTs: an on and an off transition as shown above with unique ic - tcharacteristics for each.(1)Given the measured ON transition below with currentpeaking find the energy to switch and the power DTs periodends when VCE returnsto to draw a piece-wise linearapproximation to get thepower (sw. loss) = WON*fswAs fsw so do dynamiclossesThere are 4 time regions involved during IGBT turn-on. Eachcontributes a triangle of VA*time (energy) as shown (VA) * t = off transition with current tailingWith a piece-wise linearapproximation wedivide IGBT current turnoff into 3 regionsPower dissipated alsohas three regionscontributingWon + Woff = mJTotal switch lossPloss = Wsw*fswb) For operation in abuck converter insteady state withVon= for IGBT andVon(diode)= findthe DC switch look at DC losses for the situation where VD = , Von = But first we determine the D ( transistor on time) and D(diode on time)Volt-sec balance on L:-DVg + DVon + D VD + V = 0 D =.

9 505, D = .49518 Then the two powers followTransistor:Pon = (IL = 10A)(.505)(Von = ) = WDiode:PD = (IL = 10A)(.495)(VD = ) = WNext we get the buck converter total switching power loss versusswitching frequency and see operating efficiency ) Pout = 2000 W = 10A * 200 V but Ploss = PDC + fsw*WswPDC is fixed at Wfsw(kHz) (%) the particular choice ofswitching when:P(sw. loss) = P(dc loss)This means a frequency given by: fcrit = Ptot/Wsw = = fcrit kHz which is far below anormal switching frequency, which isusually 50 - 500 kHz. Such a lowfrequency would mean large size L andC are (loss) varies with fswWsw = JPsw = *fs@1 KHz, Psw = W@10 KHz, Psw = W = Pout/(Pout + Ploss) = 2000/(2000 + + )19 Remember that in Erickson Chapter 4you must do problems 2,4,5,and 6 aswell as questions asked in the lecturenotes for HW#4


Related search queries