EE362L, FallIsolated Firing Circuit for HBridge Inverteree362l季隔离全桥逆变触发电路.ppt

EE462L, Spring 2013Isolated Firing Circuit forH-Bridge Inverter(partially pre-Fall 2009 approach)1Isolation is needed because we have three separate MOSFET source nodes, and these three nodes are ground references for the respective firing circuitsOne logic signal toggles A+,A–One logic signal toggles B+,B–Vdc(source of power delivered to load)LoadA+B+A–B–Local ground reference for A+firing circuitLocal ground reference for B+firing circuitLocal ground reference for B−firing circuitLocal ground reference for A−firing circuitSSSS!2 8 5 Comp 1 4 270kΩ VtriVcont–Vcont270kΩ1kΩ1.5kΩ1.5kΩV(A+,A–)–12Vfrom DC-DC chip+12Vfrom DC-DC chipCommon (0V) from DC-DC chip+12V–12VComparator Gives V(A+,A–) wrt. Common (0V)Vcont > VtriVcont < Vtri+24V0VVcont > VtriVcont < VtriUse V(A+,A–) wrt. –12VOutput of the Comparator ChipSince the comparator compares signals that can be either positive or negative, the comparator must be powered by ±V supply3 8 5 Comp 1 4 270kΩ VtriVcont–Vcont270kΩ1kΩ1.5kΩ1.5kΩV(B+,B–)–12Vfrom DC-DC chip+12Vfrom DC-DC chipCommon (0V) from DC-DC chip+12V–12VComparator Gives V(B+,B–) wrt. Common (0V)–Vcont > Vtri– Vcont < Vtri+24V0V– Vcont > Vtri– Vcont < VtriUse V(B+,B–) wrt. –12VOutput of the Comparator ChipSince the comparator compares signals that can be either positive or negative, the comparator must be powered by ±V supply4V(A+,A−) control signalV(B+,B−) control signalReference (is −12V from DC-DC chip)The control signals at the open-circuited output of the PWM control circuit are +24V, or 0VWhen V(A+,A−) is 24V, MOSFET A+ is on, MOSFET A− is offWhen V(A+,A−) is 0V, MOSFET A+ is off, MOSFET A− is onMOSFETs B+ and B− work the same way with V(B+,B−)50V+24V0V+24V0V+24V0V+24VLook for symmetry of pulse centersLook for symmetry of pulse centers6 One firing circuit for each MOSFET, with each firing circuit mounted on a separate protoboard. Protoboards A– and B– can share a power supply and ground . A+ and B+ must each use separate power supplies and grounds. Do not connect any of these grounds to the ground of the control circuit. O+ O– (see Figure 2 for connections) Powered by +12V that is isolated from the PWM control circuit 10kΩ 0.1µF 10Ω 1.2kΩ MOSFET G D S 100kΩ 5 4 Opto 8 1 5 4 Driver 8 1 Outline of protoboard A+ and B+ use inverting drivers (1426’s). A– and B– use non -inverting drivers (1427’s) . The optocouplers provide an additional inversion. green green green blue for A+,B+, violet for A–,B– blue blue red blue Grounds (isolated from control circuit) Wait until next week Switching diode 14mAOptocoupler is current-controlled. Gate current turns on the transistor, which pulls down the collector voltage. Isolating barrierOnce the MOSFET is connected, this asymmetrical circuit will add blanking by making the turn-on slower than the turn-off. (blanking is the opposite of overlap)Overlap is the time that A+ and A− are simultaneously “on,” which should be avoided. Hence, some blanking (time between one turning off and the other turning on) is desirable.714mA from control circuit10kΩ+12V+ Vdriver = 0V–Isolatingbarrier14mA to Opto Input Yields ≈ 0V to Input of Driver Chip, so Inverting Driver Chip Turns MOSFET ONTo driver1.2mA (will pull down Vdriver to zero)Spec. sheet current transfer ratio   0.2 to 0.3 (times 14mA)!810kΩ+12V+Vdriver = 12V–0mAIsolatingbarrier0mA to Opto Input Yields 12V to Input of Driver Chip, so Inverting Driver Chip Turns MOSFET OFFTo driver0mA from control circuit!9We use the control signals to send 14ma through optocouplers on each of the four firing circuit boardsA+ and A− are daisy chainedB+ and B− are daisy chained(for complementary outputs)So, each 14mA control signal passes through two optocouplers in series10•24V control signals from the comparators, less 3.2V drop across two series optocoupler LEDs, and with 14mA, requires about 1.5kΩ of resistance in series with the daisy-chained optocouplersWith 14mA, the LED of each optocoupler has about 1.6V drop•If applied half the time, 24V across a 1.5kΩ resistor would produce about 0.2W. So, it is a good idea to size up to ½W resistors.11Thus, you use ½W series resistors between the comparator chip and the output terminals12Layout of inverter control circuit and isolated firing circuits A+ A− B+ B− No MOSFETs connected yet (i.e., the drivers are open-circuited)13•Keep the 0.1µF capacitors across the drivers to prevent driver failure•Use the same pattern for B+ and B–•One DC converter chip feeds A+•Another DC converter chip feeds B+•Wall wart feeds A− and B−Zoom-in view of A+ and A– isolated firing circuits14 wall wart input chip output − + − + Side view of A+ and A– isolated firing circuit and single 12V isolated DC-DC converter chip that powers A+Socket each single DC-DC converter chip, using one half of an 8-pin SIP socket.Carefully break an 8-pin SIP socket in half. Do this by clamping on one-half with your long-nose pliers, and then bending the other half down with your fingers. It should be a clean break.15 V(A+,A–) Opto A+ output Save screen snapshot #1 0.5V3.2VInput and Output Voltages of Optocoupler Vcont = 0 (i.e., ma = 0) in this Snapshot12V0VOpto Input (the 1.5kΩ resistor drops the voltage from 24V to 3.2V)As expected, the opto output is invertedThis phototransistor turn off delay will limit your PWM operating frequencyPhototransistor turning onPhototransistor turning offLook for Symmetry Among all Four Circuits!Different time-constants to avoid shoot-through (i.e. to provide a “dead-time”)16Look for Nearly Perfect Alignment Between V(A+,A−) Signal to Optocoupler, and Output of A+ Inverting Driver Chip V(A+,A–) A+ driver output In phase17Look for Nearly Perfect Out of Phase Alignment Between V(A+,A−) Signal to Optocoupler, and Output of A− Non-Inverting Driver Chip V(A+,A–) A– driver output Out of phase18Now the present circuit based on PCBs:19Key new component: IRS2184420IRS21844High outputLow outputActual pinout21IRS21844Blanking time and isolation already integrated in a single IC22。