IGBT driving theory - Page 1
Forum section: electronics / beginners

author: sonnytiger
author_rank: Frequent Contributor
timestamp: April 02, 2013, 01:36:35 am
content: From what I understand of an IGBT, it is similar to a MOSFET in the way the gate works, with it acting like a capacitor and the current through the emitter and collector dependent on the charge contained in this 'capacitor'. Since one coulomb is equal to 1 ampere per second, it would make sense to think that, in the case of the IGBT attached, it would take 95nA to turn the IGBT completely on in one second. and if i wanted to run the IGBT at 100kHZ with the rise time being one hundredth of the period, I would need 0.95A to do that. Is this thinking wrong?

author: amspire
author_rank: Super Contributor
timestamp: April 02, 2013, 01:43:51 am
content: The equation for charging a capacitor is I = C dV/dT where dV/dT is the rate of voltage change in volts per second.So if you need to be able to change a gate from 0V to 10V in 100nS, that is 10v/10-7s = 100,000,000 volts/secondIf the gate capacitance is 1nF, then I = C dV/dT = 10-9x 108= 100mA.In this example, you would need a driver that can deliver 100mA minimum.In practice you often need a bit more current capability due to the Miller capacitance from the Collector/Drain to the Gate/Base.The fast changing of the voltage across the device induces current into the base/gate that opposed the drive current. The amount can be calculated using exactly the same formula. The amount of Collector to Gate capacitance is in the datasheets. The capacitance is small, but the rate of voltage change on the Collector is very big and so this current becomes very significant.

author: sonnytiger
author_rank: Frequent Contributor
timestamp: April 02, 2013, 02:09:12 am
content: Okay that makes sense, but what is the gate charge characteristic useful for?

author: amspire
author_rank: Super Contributor
timestamp: April 02, 2013, 03:01:41 am
content: Quote from: sonnytiger on April 02, 2013, 02:09:12 amIt is just an unavoidable consequence of the mosfet/IGBT/transistor construction. You cannot make an IGBT without gate capacitance, but the designers do make it as low as possible.If you look up the construction of mosfets or IGBTs, the gate looks just like a capacitor - a conductive plate with a silicon dioxide insulalting layer seperating it from another conductive layer. That is exactly what a capacitor looks like. It is the charge on this gate that controls the current in the mosfet channel (IGBTs act like a MOSFET/Transistor pair) and so the capacitance is a vital part of the operation of the device.The reason why the gate capacitance is so big is that to get a device handling big currents, they have to make the total area of the device big, so the gate capacitor is big.

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: April 02, 2013, 03:06:01 am
content: Hi,The gate charge is useful to calculate the dissipation in the gate driver.With a constant current gate drive, the gate voltage will have a linear ramp (like a capacitor), when the gate threshold is reached the voltage will be flat until the device is fully turned on, after this the gate voltage will rise in a linear manner again.You will see a graph like this on the datasheet:Jay_Diddy_B

author: amspire
author_rank: Super Contributor
timestamp: April 02, 2013, 03:16:57 am
content: Sorry, I misunderstood the question.Gate charge is a useful spec when the capacitance can change with voltage. It is not hard to convert it to an equivalent average capacitance.Charge = C * V so a 1nF gate charged to 10V has a gate charge of 10-9x 10 = 10-8Coloumbs.If you have to charge and discharge this gate 100,000 times a second, that means 100,000 x 10-8- 1mA average charging current and 1mA average discharging current. Doesn't sound much, but remember the peak current in this case is still over 100mA.Current = Coulombs/second.So you can use the Gate charge instead of capacitance to calculate the current needed to charge or discharge the gate to the voltage mentioned on the spec sheet. If you are using a different gate voltage, there will be a different gate charge.You still have to allow for the Miller capacitance separately.

author: sonnytiger
author_rank: Frequent Contributor
timestamp: April 02, 2013, 03:58:03 pm
content: okay I understand this, another question is how does the gate charge flow away? just through the way it came? I guess you would have to allow for this charge to pass to ground in the design.

author: Jay_Diddy_B
author_rank: Super Contributor
timestamp: April 02, 2013, 09:31:25 pm
content: Sonnytiger,The Gate is discharged by the driver. The energy is dissipated in the gate series resistor and the driver.Jay_Diddy_B