Gate driver datasheet for IGBT

[Jeroen Humasol]

Hello all,

I would like to use this gate driver: http://www.avagotech.com/products/o...drive-optocouplers/gate-drives/acpl-p343-000e

for this IGBT: NGTG50N60FWG

The problem is: I really, really don't understand the data sheet of the gate driver.. There is this truth table but i just don't get it. First of all on the first page stands 'Wide operating range VCC between 15-30V' so i guess that i can use a DC VCC voltage of 15V to feed the driver. Now, my problem is what should VEE be? Is this just always 0V? And in this truth table, 'Positive going' what does this mean?

If you look at the first row of the truth table then there stands 0-30V but what does this mean? that VCC is 0 and VEE is 30V?

Sorry for all the questions at once any answer is appreciated.

Greetings

 

[dorke]

The important thing is to have 15V < VCC-VEE < 30V.
They can be any voltage you like.
If you use 15V as VCC then VEE is 0V.
But that isn't recommended since you do have a voltage tolerance on the power supply.
So you should go higher to guarantee the minimum requirement of 15V.

The device has a protection mechanism against "a too low of VCC-VEE".
They call UVLO- Under Voltage Lock Out
So for instance,
if you just power-up your power supply, you have a positive going (VCC-VEE) -ramping up voltage,
that is what the datasheet calls 'Positive going'.

 

[Jeroen Humasol]

Thanks for the reply dorke! You've been more helpfull than 2 hours of googling . I always thought that VCC-VEE ment VCC and VEE but of course it meas VCC minus VEE. I guess that the positive going doesn't really matter in my case because the difference VCC and VEE will always be greater than 15V, it's the led (driven by an arduino) that controls the output of the IGBT.

However i have another question about this VCC. Do you think the following is possible: I have a 220V, 60Hz net at my disposal (generated by a water turbine). Can i create this 15V VCC power supply (lets say 20V to be safe) by using a resistive divider and afterwards a DC converter like in the picture below. But then ofcourse with a 20V equalizer, do you know if there is a 20V equalizer because i couldn't find one. Thanks

 

[Jeroen Humasol]

Ah one more question, out what part of the datasheet can i derive how fast the IGBT can switch? We would use the IGBT for PWM. We would do this to be able to let a 1kW resistor dissipate for example 500W by letting the IGBT turn the power on for 50% of the time. Our net is already 60Hz so it needs to be fast enough. I found on the driver data sheet the following: 100ns propagation delay difference and 200ns propagation delay. What is this propagation delay?

EDIT: I think i found it, 200ns propagation delay is the time it takes for my led signal to change the output. And in the igbt datasheet i found a turn on delay time of 112 and a turn of delay time of 300ns. So in total to turn it once on and of i have a time of: 2*propagation delay + turn on delay + turn of delay = 812ns -> giving me 1.23MHz of switch on and of times per sec or 20.5kHz of on and of switches per period of my 60Hz net. Is this correct?

 

[Alec_t]

That circuit won't work as you want. U1 is a fixed 5V regulator and probably needs a minimum load exceeding the 5mA or so that would trickle through R1. A small 15VRMS transformer would do as a basis, or an off-the-shelf mains-to-DC SMPS would probably be cheaper than a DIY supply.

 

[dorke]

Ah one more question, out what part of the datasheet can i derive how fast the IGBT can switch? We would use the IGBT for PWM. We would do this to be able to let a 1kW resistor dissipate for example 500W by letting the IGBT turn the power on for 50% of the time. Our net is already 60Hz so it needs to be fast enough. I found on the driver data sheet the following: 100ns propagation delay difference and 200ns propagation delay. What is this propagation delay?

EDIT: I think i found it, 200ns propagation delay is the time it takes for my led signal to change the output. And in the igbt datasheet i found a turn on delay time of 112 and a turn of delay time of 300ns. So in total to turn it once on and of i have a time of: 2*propagation delay + turn on delay + turn of delay = 812ns -> giving me 1.23MHz of switch on and of times per sec or 20.5kHz of on and of switches per period of my 60Hz net. Is this correct?

There are a few problems here:

1 .The power supply circuit isn't mains isolated -that isn't a good thing.
I would change that to an isolated one.

2. The power supply needs to supply a vary low average current of 3ma.
The device may draw peaks up to 4A during the switching period(that is taken care of by the 1uF decoupling cap on the VCC to VEE pins).

3. Better use a variable voltage regulator,and a smaller one.
the LT1085 is an over-kill here-you don't need that much of current.
You can use an LM317L device or similar .

 

[dorke]

Ah one more question, out what part of the datasheet can i derive how fast the IGBT can switch? We would use the IGBT for PWM. We would do this to be able to let a 1kW resistor dissipate for example 500W by letting the IGBT turn the power on for 50% of the time. Our net is already 60Hz so it needs to be fast enough. I found on the driver data sheet the following: 100ns propagation delay difference and 200ns propagation delay. What is this propagation delay?

EDIT: I think i found it, 200ns propagation delay is the time it takes for my led signal to change the output. And in the igbt datasheet i found a turn on delay time of 112 and a turn of delay time of 300ns. So in total to turn it once on and of i have a time of: 2*propagation delay + turn on delay + turn of delay = 812ns -> giving me 1.23MHz of switch on and of times per sec or 20.5kHz of on and of switches per period of my 60Hz net. Is this correct?

I think that in this application the load switched by the IGBT will most probably be the limiting factor .
Is your load purely resistive ?

 

[Jeroen Humasol]

That circuit won't work as you want. U1 is a fixed 5V regulator and probably needs a minimum load exceeding the 5mA or so that would trickle through R1. A small 15VRMS transformer would do as a basis, or an off-the-shelf mains-to-DC SMPS would probably be cheaper than a DIY supply.

I know that the LT1085-5 isn't the right component, but isn't there a component for 20V? Or can't the supply be done by a resistive divider? Does it need to be a transformer, because this is quite expensive. And what is this mains-to-DC SMPS thing? never heard of it.

There are a few problems here:

1 .The power supply circuit isn't mains isolated -that isn't a good thing.
I would change that to an isolated one.

2. The power supply needs to supply a vary low average current of 3ma.
The device may draw peaks up to 4A during the switching period(that is taken care of by the 1uF decoupling cap on the VCC to VEE pins).

3. Better use a variable voltage regulator,and a smaller one.
the LT1085 is an over-kill here-you don't need that much of current.
You can use an LM317L device or similar .

Why should it be mains isolated (i hope you mean galvanicly isolated with that), what do you mean by point 2? my resistive divider won't be able to supply the current or? And then point 3, if i use this one can i use my resistor configuration? And when using this one they keep using 'VREF' in the data sheet, but what is the definition of VREF? I only know of Vinput what i put into it an Voutput what i get out..

 

[Arouse1973]

You need to also change your capacitors to 3600u I assume and not 3600m other wise you will be waiting a week for them to charge up
Adam

 

[dorke]

I know that the LT1085-5 isn't the right component, but isn't there a component for 20V? Or can't the supply be done by a resistive divider? Does it need to be a transformer, because this is quite expensive. And what is this mains-to-DC SMPS thing? never heard of it.



Why should it be mains isolated (i hope you mean galvanicly isolated with that), what do you mean by point 2? my resistive divider won't be able to supply the current or? And then point 3, if i use this one can i use my resistor configuration? And when using this one they keep using 'VREF' in the data sheet, but what is the definition of VREF? I only know of Vinput what i put into it an Voutput what i get out..

Jeroen,
Your questions suggest you have very little experience if at all.
Try reading the datasheet,take your time to get a basic understanding of how this regulator works.
This isn't a fixed voltage regulator it can be adjusted to any voltage in the allowed output range.

The mains isolation is a very important safety issue.
It should be isolated to prevent the risk of electrification.

 

[Jeroen Humasol]

Jeroen,
Your questions suggest you have very little experience if at all.
Try reading the datasheet,take your time to get a basic understanding of how this regulator works.
This isn't a fixed voltage regulator it can be adjusted to any voltage in the allowed output range.

The mains isolation is a very important safety issue.
It should be isolated to prevent the risk of electrification.

I indeed have very little to no experience with IGBT's and electronics as a whole. However, i do know about galvanic isolation, and i understand that this is important for the part that people come in contact with, but i don't see why this should be necessary for the powersupply for the IGBT. No one should ever touch that and it is a galvanicly separated part from the arduino.

As for the datasheet, i really just don't get what the definition of VREF is i can't find it, they always mention VREF but they never say what it actually is? In this schematic, where is VREF?

 

[dorke]

I indeed have very little to no experience with IGBT's and electronics as a whole. However, i do know about galvanic isolation, and i understand that this is important for the part that people come in contact with, but i don't see why this should be necessary for the powersupply for the IGBT. No one should ever touch that and it is a galvanicly separated part from the arduino.

Making assumptions of what people touch or not has very little relevance in many situations(e.g. a service technician may open and touch,somone can touch the load etc.) .

You need to think about a faulty situation not a normal working circuit.
e.g.
Look at the pic below,now think about a faulty resistor R1 which is shorted(doesn't matter how and why it got to that state).
Now you can have a destructive path causing the output of the regulator to be 220V ac instead of 5vdc that is a potentially dangerous situation which can cause a fire and/or a person touching may get electrified.
Another dangerous path is the lower one.
The use of a simple line-transformer will prevent these kinds of catastrophic dangers.

 

[Alec_t]

Page 9 of ST's datasheet says:
"The LM317L provides an internal reference voltage of 1.25V between the output and adjustments terminals. This is used to set a constant current flow across an external resistor divider (see Figure 4.), giving an output voltage V O of:
V O = V REF (1 + R 2 /R 1 ) + I ADJ R 2"

 

[dorke]

As for the datasheet, i really just don't get what the definition of VREF is i can't find it, they always mention VREF but they never say what it actually is? In this schematic, where is VREF?
View attachment 25445

Well,
Vref is an internal voltage reference source.
Look below for the "block diagram " of another voltage regulator.
The LM317 works the same.
It has it's internal voltage reference Vref.
but the connecting circuit is a bit different.

 

[Jeroen Humasol]

Hi dorke and Alec_t, thanks for the replies. Indeed, what you say is correct it's probably safer anyway to opt for a 220V->20V transformer then. So according to what you guys say VREF is ALWAYS 1.25V right?

 

[dorke]

Hi dorke and Alec_t, thanks for the replies. Indeed, what you say is correct it's probably safer anyway to opt for a 220V->20V transformer then. So according to what you guys say VREF is ALWAYS 1.25V right?

Yes, for the LM317 it is always 1.25V typical value.
But for other devices you should (always) consult the datasheet .

 

[dorke]

P.S
About the input and output capacitors to the LM317L.
since the device is drawing very little average current(max 3mA):
The input filter capacitor can be relatively small,
I think a single 100uF-electrolyt (or smaller )is enough.
The output capacitor should be 1uF-tantalum (maybe a bit higher).
This will save you a lot of space and cost.

 

[Jeroen Humasol]

Hey guys one more question: Do I need a snubber network for using this IGBT( NGTG50N60FWG) and driver? Because I found on this site: http://www.powerguru.org/switching-loss-reduction-networks-snubber-circuits/ that you don't need snubber networks for modern IGBT's. thanks!

EDIT: 2 more questions would it be necessary to put a free wheel diode over the IGBT?

 

[dorke]

That depends on the load type you are driving,
and the specific IGBT you are using.
Please post a diagram and details.

 

[Jeroen Humasol]

Hey Dorke,

This is our circuit:

We're using an IGBT( NGTG50N60FWG) and as driver: ACPL-P343

The inductance at the left is a representation for the inductance in our generator (5kW). The resistor (R4) needs to be able to dissipate a max of 5kW. PS: apologies for the ugly paintstyle