IGBT Gate Driver Calculation

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# IGBT Gate Driver Calculation - PowerPoint PPT Presentation

Gate Driver Requirement. IGBT Gate Driver Calculation. What is the most important requirement for an IGBT driver ?. Gate Peak current. Which gate driver is suitable for the module SKM 200 GB 128D ?.

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Presentation Transcript

### What is the most important requirement for an IGBT driver ?

Gate Peak current

reverse recovery current Diode should be -

1.5 x I diode by 80 degree case

130A x 1.5 = 195A

Design parameters:

fsw = 10 kHz

Rg = ?

Gate resistor in range of “test – gate resistor”

Conditions for a safety operation

195A – max reverse recovery current

Rg = 7 Ohm

Two gate resistors are possible for turn on and turn off

Ron = 7 Ohm

Roff = 10 Ohm

How to find the right gate resistor ?
Trench Technology needs a smaller Gate charge
• Driver has to provide a smaller Gate charge
• SPT Technology needs more Gate charge compared to Trench Technology
• Driver has to provide a higher Gate charge
Difference between Trench- and SPT Technology

The suitable gate driver must provide the required

• Gate charge (QG) – power supply of the driver must provide the average power
• Average current (IoutAV) – power supply
• Gate pulse current (Ig.pulse) – most important
• at the applied switching frequency (fsw)

Demands for the gate driver

Gate charge (QG) can be determined from fig. 6 of the SEMITRANS data sheet

The typical turn-on and turn-off voltage of the gate driver is

VGG+ = +15V

VGG- = -8V

15

 QG = 1390nC

-8

1390

Determination of Gate Charge

Calculation of average current:

• IoutAV = P / U V = +Vg + [-Vg]
• with P = E * fsw = QG * V * fsw
•  IoutAV = QG * fsw

= 1390nC * 10kHz = 13.9mA

Absolute value

Calculation of the average current

Gate charge
• The power supply or the transformer must provide the energy (Semikron is using pulse transformer for the power supply, we must consider the transformed average power from the transformer)
• Average current
• Is related to the transformer
Power supply requirements
• E.g. RG.on = RG.off = 7
• Ig.puls≈ V / RG + Rint= 23V / 7 + 1 = 2.9 A

Calculation of the peak gate current

P total – Gate resistor
• Ppulse Gate resistor = I out AV x V

The problem occurs when the user forgets about the peak power rating

of the gate resistor.

The peak power rating of many "ordinary" SMD resistors is quite small.

There are SMD resistors available with higher peak power

ratings. For example, if you take an SKD driver apart, you will see

that the gate resistors are in a different SMD package to all the other

resistors (except one or two other places that also need high peak power). The

problem was less obvious with through hole components simply because the

resistors were physically bigger.

The Philips resistor data book has a good section on peak power ratings.

Pulse power rating of the gate resistor

The absolute maximum ratings of the suitable gate driver must be equal or higher than the applied and calculated values

• Gate charge QG = 1390nC
• Average current IoutAV = 13,9mA
• Peak gate current Ig.pulse = 2.9 A
• Switching frequency fsw = 10kHz
• Collector Emitter voltage VCE = 1200V
• Number of driver channels: 2 (GB module)
• dual driver

Choice of the suitable gate driver

According to the applied and calculated values, the driver e. g. SKHI 22A is able to drive SKM200GB128D

Calculated and

applied values:

• Ig.pulse = 2.9 A@ Rg = 7 + R int
• IoutAV = 13.9mA
• fsw = 10kHz
• VCE = 1200V
• QG = 1390nC

Comparison with the parameters in the driver data sheet

Simple

• Expandable
• Short time to market
• Two versions
• SKYPER™ (standard version)

Driver core for IGBT modules

SKYPER

• Driver board
• SEMIX 3 IGBT half bridge
• with spring contacts

Assembly on SEMiXTM 3 – Modular IPM

take 3 for 6-packs

solder directly in your main board

modular IPM using SEMiX®

SKYPER™ – more than a solution

### Selection of the right IGBT driver

vGE,T1(t)

vGE,T2(t)

VGG+

T1

D1

VGE, Io

VGE(th)

0

t

vCE,T1(t)

iC,T1(t)

VCC

T2

D2

IO

iv,T2

0

t

vCE,T2(t) =vF,D2(t)

iF,D2(t),iC,T2(t)

VCC

IO

• Why changes VGE,T2 when T1 switches on?

0

t

Cross conduction behavior

• This leads to a displacement current iV

IGBT - Parasitic capacitances

vCE,T2(t)

VCC

vGE,T2(t)

VGG+

VGE(th)

0

t

0

iC,T2(t)

iv,T2(t)

t

0

t

iC,T2

CGC,T2

iv,T2

vCE,T2

RGE,T2

vGE,T2

• Diode D2 switches off and takes over the voltage
• T2 “sees” the voltage over D2 as vCE,T2
• With the changed voltage potential, the internal capacitances change their charge
• The displacement current iv,T2 flows via CGC,T2, RGE,T2 and the driver
• iv,T2 causes a voltage drop in RGE,T2 which is added to VGE,T2
• If vGE,T2 > VGE(th) then T2 turns on (Therefore SK recommends: VGG- = -5…-8…-15 V)

Switching: Detailed for T2

Z18

PCB design because no cable close to the IGBT

Gate clamping ---- how ?
Over voltage
• 1200V ----- is chip level ---- consider internal stray inductance
• +/- 20V----- gate emitter voltage ---- consider switching behavior of freewheeling diode
• Over current
• Power dissipation of IGBT (short circuit current x time)
• Chip temperature level
Problem 4 ---------------------------- Short circuit
Example:
• Dead time = 3 us logic level
• Turn on delay 1 us
• Turn off delay 2.5 us
• Td – toff delay + ton delay = real dead time
• Real dead time: 3us – (2.5us+1us) = 1.5 us