Hello, I'm working with a group for school and we are trying to use this gate driver to drive an IGBT. The signal into this driver is coming from an Arduino Mega which is to signal the igbt to turn on and off. We were told that we needed this gate driver between the Arduino and the IGBT gate. I am having trouble understanding where each of the components would connect to this driver. I attached the data sheet for the gate driver. Again, what i'm looking for most is to understand what the different pins represent so I can better understand how to include this device in our circuit.
Any help is greatly appreciated.

Please and Thank you!!,
Adam

 

What IGBT Gate are you trying to drive? Also do you require the Fault detection output and Active Miller Clamp functionality of this driver? if not then TLP352 would be a better option for you.

But using this IC in it's most simple configuration you will simply connect pin 7 to your arduinos GPIO output, and pin 8 to your arduinos ground.

Pin 13 to your higher voltage source, 11 to the gate of your IGBT.

9, 10, 12, 14, 16 to ground

and 1 micro farad capacitors between 9 and 13 as well as 16 and 13.



Effectively we have disabled many features of this IC and left in place a very simple circuit consisting of an optocoupler, and an undervolt protection circuit feeding a cmos AND gate that when it is true closes a IGBT allowing VCC2 to flow to vOut.

You could in all reality even simplify down to a simple optocoupler rated for the voltage you require to drive your IGBT depending on what your requirements are. However the TLP5214 can also be used to enable fault detection that will allow your arduino to know if you lost voltage on VCC2, triggered over current protection, or desat was triggered.

Let us know your exact requirements and we can help a little more with selecting the best IC and configuration for your needs.

 

Thank you for your timely and detailed response! This really helps me out. The IGBT we are using is MPN IXYS IXGH100N30C3. We do not require the fault detection or the active Miller clamp functionality. I will look into the TLP352 as you have requested.
One more question I have is that it was recommended to me that we include a resistor between the GPIO and the driver to limit the current. I believe for the sake of the IGBT but also the returning current into the Arduino. Does this sound like it makes sense? At least from what I can see, the limit for the turn on of the driver current I_F is between 7.5-10 mA. And I believe the Arduino limit is 40 mA(need to double check). Am I interpreting that part of the data sheet right?(in regards to I_F).

Thank you again for your help!!

 

For the most part you will nearly always have a resistor between your micro controller and an ic input or output noth for the sake of the transistors in the Ic and the microcontroller itselft.

Always check the spec sheets for both the controller and the ic for Max current. Remember your Arduino will have two rated currents one for total current draw and a maximum per pin.

 

Well, I shall add Toshiba to my list of manufacturers who produce really bad datasheets. All sorts of test circuit diagrams, and not a single "typical application" circuit. Maybe an applications note exists.

With regard to driving the input:
The input is an infrared emitting diode (IRED). The forward voltage is specified as between 1.4 and 1.7 V. The recommended input current range is 7.5 to 10 mA.
[EDIT - the Figures mentioned below are in the processor datasheet]
As with any LED driven from a voltage source, you need a resistor to limit the current. Figure 32-23 plots the voltage at an I/O versus the current flowing into the pin (that is, the pin is "sinking" "conventional current" from some positive voltage to "ground"). Figure 32-25 plots the voltage at the pin if the pin is "sourcing" conventional current from the processor's positive supply pin through a load that is connected to ground. (Conventional current is considered to flow from positive to negative.) In our target load current range, with a 5 V supply, the pin will produce about 4.75 V when sourcing or about 0.25 V when sinking - it is quite symmetric, which is typical of most modern CMOS outputs.
You can drive the IRED either by connecting the cathode to ground and sourcing current into the anode with a HIGH. or by connecting the anode to +5 V and sinking current to ground with a LOW.
So our resistor needs to be
R = (5 V - 1.4 V - 0.25 V) / 9 mA that is, supply voltage minus IRED forward voltage minus voltage between the I/O pin and either VDD or 0V, with LED current of 9 mA, picked to be in the recommended range, so
R = 3.35 V / 9 mA = 372 ohms
360 ohms and 390 ohms are the closest common values. I'll leave it to you to check the expected current with each. The resistor can go anywhere in the series circuit that is convenient.

In critical and difficult applications, the graphs aren't sufficient to determine what to do because they show "typical" performance. Somewhere in the tabular data there should be "worst case" values. You often need to compare the worst case limits with the typical values and then try to extrapolate that to the graphs. For your circuit, typical is likely entirely satisfactory - especially since we've picked a IRED current that is between the min and max recommended values.

 

Well, I shall add Toshiba to my list of manufacturers who produce really bad datasheets. All sorts of test circuit diagrams, and not a single "typical application" circuit. Maybe an applications note exists.

With regard to driving the input:
The input is an infrared emitting diode (IRED). The forward voltage is specified as between 1.4 and 1.7 V. The recommended input current range is 7.5 to 10 mA.
[EDIT - the Figures mentioned below are in the processor datasheet]
As with any LED driven from a voltage source, you need a resistor to limit the current. Figure 32-23 plots the voltage at an I/O versus the current flowing into the pin (that is, the pin is "sinking" "conventional current" from some positive voltage to "ground"). Figure 32-25 plots the voltage at the pin if the pin is "sourcing" conventional current from the processor's positive supply pin through a load that is connected to ground. (Conventional current is considered to flow from positive to negative.) In our target load current range, with a 5 V supply, the pin will produce about 4.75 V when sourcing or about 0.25 V when sinking - it is quite symmetric, which is typical of most modern CMOS outputs.
You can drive the IRED either by connecting the cathode to ground and sourcing current into the anode with a HIGH. or by connecting the anode to +5 V and sinking current to ground with a LOW.
So our resistor needs to be
R = (5 V - 1.4 V - 0.25 V) / 9 mA that is, supply voltage minus IRED forward voltage minus voltage between the I/O pin and either VDD or 0V, with LED current of 9 mA, picked to be in the recommended range, so
R = 3.35 V / 9 mA = 372 ohms
360 ohms and 390 ohms are the closest common values. I'll leave it to you to check the expected current with each. The resistor can go anywhere in the series circuit that is convenient.

In critical and difficult applications, the graphs aren't sufficient to determine what to do because they show "typical" performance. Somewhere in the tabular data there should be "worst case" values. You often need to compare the worst case limits with the typical values and then try to extrapolate that to the graphs. For your circuit, typical is likely entirely satisfactory - especially since we've picked a IRED current that is between the min and max recommended values.

They do have a use guide for this IC posted on their site. I did not link it however, because the OP stated he wouldn't be using the clamp, or fault detection features of the IC and thought that guide would do nothing but lead to further confusion by posting for the OP as it doesn't show application without using these features. I'd still suggest to the OP to simply go with the simpler version of this IC as it's cheaper and he won't be using any of the added functionality.

To be honest this is my first time browsing a Toshiba datasheet, about 98% of my IC's i stock are Motorola, Texas Instruments, Fairchild, or ON Semiconductors.