Q6 Flashcards

1
Q

Why do we need gate drive circuits

A
  • Gate drive circuits are an important component of every power electronic system
  • Convey switching state information
  • Gate voltage adaptation
  • Signal amplification
  • Galvanic isolation
  • Circuit protection
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2
Q

Explain “convey switching state information”

A

Means that gate drive circuit receives input signals from control units (microcontroller). The signals have information about desired switching state of power transistor. The gate driver interprets the signals and perform the required actions

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3
Q

Explain “gate voltage adaptation”

A

The gate drive circuit adapts the control signal voltage to the required level to effectively switch ON or OFF the power transistor

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4
Q

Explain “signal amplification”

A

Since control signals come from low-power devices and are not strong enough to drive power transistor, the gate driver amplifies these signals to a level that can switch the power transistor.

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5
Q

Explain “galvanic isolation”

A
  • Gate drive circuit allows to isolate control side of circuit from power side to prevent noise interference and protect control circuitry (i.e. using optocouplers, transformers)
  • Required when control circuit operates at different voltage levels than power circuits
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6
Q

Explain “Circuit protection”

A

Gate drivers often incorporate protection features to protect the power devices and system. i.e. UVLO that prevents transistor from switching if gate voltage is too low, or desaturation detection that protects against overcurrent conditions.

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7
Q

Q6. slide 2 main ideas

A
  • components of optocoupler
  • operation
  • Characteristics of optocoupler
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8
Q

Components of optocoupler

A
  • LED
  • output transistor
  • Schmitt trigger
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9
Q

Operation of optocoupler

A
  • Positive signal from control logic causes the LED to emit light that is collected by the photo transistor.
  • This turns on the transistor, pulling collector to GND and changes state of Schmitt trigger.
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10
Q

Characteristics of using optocoupler

A
  • Electrical isolation
  • Protect low voltage electronics from damages due to faults on high power side.
  • Increases safety if humans are involved on control side (i.e. in case of any fault on HV side, isolation blocks electrical power from reaching user).
  • Functional isolation
  • Noise immunity.
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11
Q

Explain noise immunity

A
  • i.e. in industrial environments there are many sources of noise like high frequency switching transistors
  • noise corrupt control signals and makes systems unreliable
  • isolation can reduce noise interference from affecting control circuits
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12
Q

Explain functional isolation

A
  • In a half bridge S of high side switch is either bus voltage or GND.
  • Turn on G voltage must be higher than bus voltage when switch is ON if there is no isolation.
  • With isolation it is possible to have level shifted control signals referenced to S of high-side switch.
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13
Q

Reasons for using bootrstrap circuit

A
  • because a transistor needs a gate voltage higher than the threshold referenced to potential of S terminal.
  • in a half-bridge topology S of high-side transistor is not fixed but it changes between GND and DC bus voltage.
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14
Q

Q6. slide 4 describe bootstrap circuit and operation when T1 is ON

A
  • Consists of a C and a diode.
  • When T1 is ON capacitor CB charges to the supply voltage through DB and T1 (voltage drop of IC is small so DB is forward biased)
  • Upper and lower drivers are supplied by voltage Vdd
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15
Q

Q6. slide 5 explain operation of bootstrap when T1 is OFF

A
  • Diode is reversed biased (cathode potential is Vdd+Vdc, anode potential is Vdd)
  • Capacitor CB supplies upper driver, with an almost constant voltage Vdd with respect to S potential.
  • Lower driver is supplied by voltage Vdd.
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16
Q

Voltage rating of bootstrap diode

A

Vdc/dc bus voltage

17
Q

T1 can be kept all the time?

A

No, it must be turned on periodically to recharge CB

18
Q

Maximum and minimum voltage of CB

A
  • Max: Vdd-Vf
  • Min: minimum gate-source voltage of power device + voltage drop across gate driver IC
19
Q

Steps to find CB

A
  • Calculate total charge required: Gate charge+(Iic+Ileak)*Ton
  • Find delta Vb: maximum-minimum voltage of CB
  • Calculate CB: Qtotal/deltaVb
20
Q

Describe general operation of CB

A
  • CB charges when low-side transistor is ON
  • Discharges quickly to turn on the high-side switch
  • Discharges slowly to maintain operation of gate driver and high-side transistor
21
Q

Q6. slide 6 describe charging period.

A
  • occurs when low-side transistor is ON, capacitor starts charging and its voltage increases
22
Q

Q6. slide 6 describe high negative dv/dt

A
  • when high-side transistor is turned ON, capacitor voltage rapidly decreases because charge stored in CB is used to turn on the transistor
  • transistor requires a charge equal to total gate Charge Qg.
  • Drop voltage is Qg times CB
23
Q

Q6. slide 6 describe voltage fall

A
  • after high dv/dt capacitor continues discharging at a slower rate, because it only needs to supply small current for operation of gate driver and maintain charge on gate of transistor.
  • leakage current also contributes to decrease in capacitor voltage.
24
Q

what is VDDA bias current

A
  • minimum current required to bias the driver.