Trouble with driving IGBTs with OptoIsolators

Currently I am trying to drive an IGBT with two OptoIsolators.

They are setup in series to as follows:

Vin-Collector, Emitter[Gate]Collector, Emitter-Gnd
3.3v-Anode, Cathode[Signal]Anode, Cathode-Gnd

Well, when using the simple Blink example sketch the Signal pin alternates normally and drives the Isolators
just fine. But when using any other pins, there is no alternating signal. Right now I use two leds to indicate the altering signal, but again for some reason, this only works for pin 13.

Are there any ideas, or workarounds by chance?

Vin-Collector, Emitter[Gate]Collector, Emitter-Gnd
3.3v-Anode, Cathode[Signal]Anode, Cathode-Gnd

I can't follow that can you post a diagram.

this only works for pin 13.

When you try the other pins do you set the pinMode of the pin you use to be an output?

Grumpy_Mike:
When you try the other pins do you set the pinMode of the pin you use to be an output?

That was the problem, thanks!

As for the diagram:

A B

1 2 1 2
|Opto| |Opto|
3 4 3 4

Pins A2, and B1 are tied along with A4, and B3.
A1 goes to HighVoltageIn, and B2 goes to ground.
A3 goes to Arduino 3.3v Supply, B4 goes to Gnd

Where the pins are tied on the High side goto the gate of the IGBT. (A2 + B1)
And for the Low side ones, they goto the pin for the control signal.

In theory, when the pin is HIGH the OptoB will fire on. And when the pin is LOW OptoA will fire.
This will cause an inverting output to either pull up the gate of the IGBT or pull it down to turn it off.

The next thing I need to do is manage to flip this around because when the arduino is first powered up I notice that the pins are LOW, which is a bad thing when i will be having 400+ vdc on the IGBT. I need the OptoB to be on when the pin is LOW so that the IGBT stays off until the pin goes HIGH.

Again, thanks for the help.

Update: Tied A3+B4, and put 3.3v on B3, and Gnd on A4. This fixed the pinLOW=IGBTon, and pinHIGH=IGBToff

Optos are basically LEDs, you need a current limiting resistor somewhere to protect your opto and your arduino's output.

Grumpy_Mike:
Optos are basically LEDs, you need a current limiting resistor somewhere to protect your opto and your arduino's output.

I have one on the gnd from the led.

Sorry I still can't work out what you are doing. The terminal letters do not convey anything.

I have one on the gnd from the led.

If you have two LEDs you need two resistors.

Grumpy_Mike:
If you have two LEDs you need two resistors.

It sounds to me that the OP is only turning one of the opto isolators on at a time, since one is used to turn the IGBT on and the other is used to turn it off. In which case, they can share a resistor to ground.

dc42:

Grumpy_Mike:
If you have two LEDs you need two resistors.

It sounds to me that the OP is only turning one of the opto isolators on at a time, since one is used to turn the IGBT on and the other is used to turn it off. In which case, they can share a resistor to ground.

Grumpy_Mike is right. The resistor atm is in between A4 and 3.3v. Which is good while the signal pin is LOW. but when it goes HIGH, there is no current limiting for OptoB. But, if we use a single resistor from the signal pin to A3+B4, then there will be protection either way. I think two resistors, ie; one between Vin and A4, and one between B3 and ground will be safer. imo

A full diagram including the IGBT would be useful - what voltages are you talking about? What IGBT?
What load is the IGBT expected to handle?

MarkT:
A full diagram including the IGBT would be useful - what voltages are you talking about? What IGBT?
What load is the IGBT expected to handle?

Thanks....
but that is not a diagram is it?
It is a physical layout and as such is very difficult to follow.

I spent some time trying to draw out a real schematic of what you had and I must say I am a bit confused about your design. You seem to have two opto isolators driving a FET in a push / pull mode. Why is that? All you need with a FET is to put the gate high to turn it on and have a pull down resistor so that when the top opto's transistor is off then the FET is off.
There seems to be a lot of unnecessary complication unless I am missing something.
Have you a real schematic?

Grumpy_Mike:
Thanks....
but that is not a diagram is it?
It is a physical layout and as such is very difficult to follow.

I spent some time trying to draw out a real schematic of what you had and I must say I am a bit confused about your design. You seem to have two opto isolators driving a FET in a push / pull mode. Why is that? All you need with a FET is to put the gate high to turn it on and have a pull down resistor so that when the top opto's transistor is off then the FET is off.
There seems to be a lot of unnecessary complication unless I am missing something.
Have you a real schematic?

I dont have a schematic, and the software i am using does not have optoisolators.

Push-pull mode, yes. I figured that if I used that second opto i could pull the gate down faster. the 10k resistor
is a fall back. You are right, there is a bit unnecessary bits to the design.

With the design, the concept is basically an IGBT H-bridge. Didnt intend for this to be complicated.

I'm not an expert on IGBTs, but my understanding is that in order to turn them off quickly, charge needs to be pulled out of the gate - preferably by driving the gate negative. I guess that's why the OP used an optoisolator to pull the gate down. However, opto isolators are themselves rather slow, especially to turn off.

the software i am using does not have optoisolators.

That is no problem the software doesn't need to know.

I dont have a schematic,

I here this sort of thing a lot from beginners. Tell me how on earth you can make anything if you don't have a schematic so you can see what you are making?
Maybe it's me but I would not even think about trying to do that.

Grumpy_Mike:

the software i am using does not have optoisolators.

That is no problem the software doesn't need to know.

I dont have a schematic,

I here this sort of thing a lot from beginners. Tell me how on earth you can make anything if you don't have a schematic so you can see what you are making?
Maybe it's me but I would not even think about trying to do that.

How could one even do the board layout without a schematic drawing to start with? Color me confused also.

Lefty

I think he probably had the board layout supplied with the board, but not the schematic. Depending on experience, the board layout is handy for troubleshooting as one can see tha actual connections. A schematic can be difficult because the layout of the board and the layout of the schematic can be quite different. Many is the time I have had to draw a board layout so I could figure out where the signals were actually going.

Actually, I think the IGBT could be driven directly from the Arduino. Worst case would be a couple resistors, one from the Arduino and one to ground. I am using a 4 of these in a washing machine (commonly called SSRs) but they are rated 5 to 20V control signal.

You are trying to make an H-bridge with IGBTs and using the same supply to drive the gates on the high-side switches
as on the low-side switches - there is no chance of this working.

You need a floating power supply for each of the high side switches that follows the high-side emitter voltage. The
way to do this is with a high-side-low-side driver chip. These are available upto 600V or so, and for higher voltages
you'll need to engineer your own high-voltage charge-pumps...

If the voltage isn't beyond 600V I'd suggest looking at chips like IRS2112, FAN7380, FAN7388 ...
For lower voltages there are more choices of chip. 80V and below the HIP3081 does everything you
need in one.

Ideally for high power use you would want to drive gates negative in an IGBT to reduce turn-off time - this
matters more with high voltage and high current. Higher voltage IGBT's tend to want higher gate drive
voltages (+15/-5 rather than +12/0)

Grumpy_Mike:

I dont have a schematic,

I here this sort of thing a lot from beginners. Tell me how on earth you can make anything if you don't have a schematic so you can see what you are making?
Maybe it's me but I would not even think about trying to do that.

lol, its in my head.

Why didn't you say. I will put my mind reading hat on.

Dam the batterie is flat.

This is all setup on a breadboard btw. It seems the arduino cannot supply enough current to switch the IGBTs properly. I decided to add another 12V battery so i have +12 and -12 on the high side of the optos. the problem is that the arduino can only push-pull one opto when its tied to the gate of an igbt. my next step is to get some of these TC4422's to drive them. http://www.ebay.com/itm/170985238551

I need the design to have a low side for the arduino+other stuff (3.3v|5v). a mid side for higher power devices(such as driving the gate of the IGBTs and lighting,etc(12v|15v|24v), and a high side for heavy loads.(12v|24v to 800v)