Hi there, im now doing my final year project. One of my task is using igbt to cut off the voltage. The problem i facing is whenever there's voltage or non at the igbt gate, the Vce still cut off to half of the input. It suppose to be Vce same with the input voltage when there's non voltage at gate, and drop when theres voltage at gate. Can anyone who expert in igbt help me pls. I already burnt 7 igbt so far which cost me lots of money. Im using irg4bc40w. Thanks in advance.

My schematic( igbt driver) - http://db.tt/qdsdFkyA​

 

I'm guessing you're not an English major.

One thing that might help is swapping the positions of Q1 and X2. Then the voltage on the gate won't have to overcome the voltage across X2 as it powers on.

 

his link is legit? When I click it, there's a security warning, so I dare not enter.

 

Here's the image as posted.

 

are all 3 of your grounds connected as you show in the drawing?

 

Dear wayneh, sorry for my bad english.. hopefully you can understand what i mean. Back to the circuit, so thats mean the voltage at emitter is uncorrect due to bulb. Because everytime i supply the 240v Dc, the voltage at emitter drop to 120 v when there is or non voltage at gate..

 

Dear Strantor, sorry bout the picture. Only The ground at the optocoupler and tip29 are connected.

 

Well there you go. The IGBT cannot turn on until it sees more voltage on the gate than on the emitter pin. As drawn, if those grounds are not connected, there's no ∆Vge.

 

Means that i need to connect all the ground? And igbt need Vge to function too? Sorry. I really know nothing about igbt..

 

Like wayneh said you are using the IGBT as a "high side switch" in the drawing. Put 'X2' between the collector and supply, to make it into a "low side switch".

Also you only have 5V for your gate drive voltage, this is barely above the "threshold voltage" of the gate. "Threshold voltage" is the point that the gate is turning off. The data sheet lists threshold voltage as 3 to 6 volts, so the IGBT is not on. The data sheet says 15V as the turn on voltage for the gate, so another voltage is needed in your circuit.

 

So thats mean i need to use the op-amp to double up my input voltage right? Thanks for helping me shortbus. i will try the circuit later.

 

An op-amp cannot output a voltage higher than it is supplied, so I'm not sure what you are suggesting. You need a higher voltage source from somewhere.

Also be careful connecting grounds. They're not always isolated from each other, and can actually allow a current to flow from one to another. That's usually a bad thing.

 

I mean like this..
http://db.tt/YC1zJfoP

 

That would work, but if you have 10v available, just use it to power your original circuit, and you won't need the op-amp.

You may or may not still have a ground problem.

 

But, i need to make my Vg variable. In the same time to be 6-15 vDC. Because the microcontroller only supply 0-5vDC. Do you know how i can make my 0-15 vDc in the same time variable.?

 

Maybe you should explain what you are trying to do, so we can help better.

Ideally an IGBT of mosfet is better of used as a switch. Why would you want or need the gate voltage variable? If you are trying for a lamp dimmer and need to use a IGBT, PWM is how to do it. The PWM circuit will be used to pulse the gate on and off, at 15V. You don't need vary the gate voltage, that will only make heat in the IGBT. Using a lower gate voltage makes the IGBT into a large resistor, not a good idea.