High Side Gate Driver Connections.

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james31207

Joined Sep 19, 2018
72
Referring in particular to pages 23, 27 and 33 of the attached datasheet for isolated gate driver UCC 5390 S [from TI] to be used to power the high side gates of IGBTs in an H-Bridge module, is it correct that the 5V input to the VCC1 pin may be provided by an MCU [PVM], as suggested by the diagram on p.23?

Also, when interpreting such diagrams, does the symbol for ground mean that a common ground can be used; for example in the case of the 2 chassis grounds shown on the input side, can this be made common, and if so, how is this done: what is the chassis; is it just the laminate material on which components will be mounted?* Similarly, on the output side, can the other earth grounds be made common for the VCC2 and VEE2 pins, as is suggested in the text? *This is a practical problem: what material to use in order to mount the few necessary components.

Here I'm supposing that the split supply pins [OUTH, OUTL] connect directly to the Gate pin of the IGBT, and the VEE2 pin connects to the Source/Emitter pin of that IGBT; and that 15V will be supplied from an appropriate DC-source.
 

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ronsimpson

Joined Oct 7, 2019
1,227
is it correct that the 5V input to the VCC1 pin may be provided by an MCU [PVM]
Yes
I'm supposing that the split supply pins [OUTH, OUTL] connect directly to the Gate pin of the IGBT
I use a resistor on OUTH and a different resistor on OUTL, then off to the Gate.
VEE2 pin connects to the Source/Emitter pin of that IGBT
Yes. and you need to find a way to power the output .

Typical layout. Normally the "MCU" side is very much separated from the power transistors. See back line running through the IC>
1611065877939.png
 

Thread Starter

james31207

Joined Sep 19, 2018
72
Thanks for very useful information. How about the common grounds? I think so, but am unsure. Possibly this is clear from the layout image you indicated [p.41] -- although I'm slightly confused about what that image actually shows.
 
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Thread Starter

james31207

Joined Sep 19, 2018
72
Yes

I use a resistor on OUTH and a different resistor on OUTL, then off to the Gate.

Yes. and you need to find a way to power the output .

Typical layout. Normally the "MCU" side is very much separated from the power transistors. See back line running through the IC>
View attachment 228196
Regarding the 15V output (which is required by the IGBT gates), I'm still confused about where this comes from? Isn't the purpose of the driver to level shift the 5V input from the MCU/PVM up to 15V?
 
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BobTPH

Joined Jun 5, 2013
3,308
The earth ground symbols on the right side and the chassis ground symbols on the left side indicate separate grounds. That is what is meant by isolated. They should not be connected together.

Both of them can be considered common's rather than earth and chassis ground, but two different commons.

Bob
 

Thread Starter

james31207

Joined Sep 19, 2018
72
Thank you Bob. Do you have any information about the 15V output? That's the thing that's troubling me. For example, on p.27, it says, ''For operation with unipolar supply, the VCC2 supply is connected to 15V with respect to VEE2 for IGBTs....VEE2 is connected to 0V'' [Earth ground]. What exactly does this mean here?

More specifically, does this mean that a fresh supply of 15V must be connected to the VCC2 pin? And is it possible to obtain this from the same 12V DC battery output through the source/drain of the IGBTs?
 
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shortbus

Joined Sep 30, 2009
8,578
Regarding the 15V output (which is required by the IGBT gates), I'm still confused about where this comes from? Isn't the purpose of the driver to level shift the 5V input from the MCU/PVM up to 15V?
It comes from a power supply of that voltage. The driver or any mosfet/IGBT driver is a glorified switch or relay. One that takes a low level input and turns on a higher level voltage to the gate of the device you're controlling. They don't make the higher voltage internally.
 

Thread Starter

james31207

Joined Sep 19, 2018
72
It comes from a power supply of that voltage. The driver or any mosfet/IGBT driver is a glorified switch or relay. One that takes a low level input and turns on a higher level voltage to the gate of the device you're controlling. They don't make the higher voltage internally.
I thought so, you yourself even said so somewhere back there; but reading here and there without a proper teacher served to confuse me. So, the problem remains: how to find this 15V from the same 12V battery I'm using as the primary circuit. Never mind though, I'll work it out somehow. 18 months ago I barely knew what a transistor was, and now I'm already trying to turn them on and off.
 

shortbus

Joined Sep 30, 2009
8,578
That's the thing that's troubling me. For example, on p.27, it says, ''For operation with unipolar supply, the VCC2 supply is connected to 15V with respect to VEE2 for IGBTs....VEE2 is connected to 0V'' [Earth ground].
You seem to be mixing some things up here. Your Vcc2(NOT VEE2)(gate drive voltage) can be anything, from 9.5 to 33 volts. The 15 volts is just an example. If you have only 12V available that will work fine for most every IGBT or mosfet gate other than maybe a logic level one.


Now to what you are calling Vee2, it is the negative side of your 12V(or15V what ever you use) power. It is NOT earth ground. Earth ground is a connection to the earth, found in a mains power supply, not low voltages like this. They use the term ground when they mean common, because it is an old out dated way of saying it. But in electronics it is really the common where all voltages in a circuit are measured or compared to.

All of what was just said is from page 27 of your link.
 

Thread Starter

james31207

Joined Sep 19, 2018
72
This is very helpful, the aim being eventually to get a very clear picture in my head of how everything is connected. For example, the VEE2 pin goes to the Emitter pin -- or what is labeled as the source pin 's' -- of the IGBT; but it will take some more study to understand how this and the other pins shown also connect to this common ground, or as you say, eventually the negative terminal of the VCC output supply (and that would simplify things a lot if I can use the same 12V source).
 
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