Best way to measure and sample high frequency high voltage AC MOSFET voltage

Thread Starter

SiCEngineer

Joined May 22, 2019
299
Hi All,

I am looking to a control method which switches Gallium Nitride devices when the voltage across the FET reaches zero. I need therefore to sample the voltage across the FET switches and pass it to my DSP. There are some issues - the voltage across the switch can be up to 600V, and my DSP must be isolated. I wondered, what is the best way to do this? My topology is an active clamp flyback. I was thinking a simple capacative divider followed by a digital isolator, with the hot side connected to the HV and the other connecting the DSP low voltage singal.

Frequency is above 200kHz - will this affect the best solution at all?
Is there a better way?
 

ronsimpson

Joined Oct 7, 2019
1,121
You need a fast isolator. Many will not pass 200khz and you probably need a 100mhz type part.

There are many ways.
Use a 0nS high voltage diode. SiC diode.
1606321595558.png
This option you are just looking at edges. C1 must be small. I am getting 6nS edges on GaN devices. It takes very little capacitor to get too much current. On the falling edge current passes through C1 into D2 and drives the output low. On the rising edge current through C1 pulls up on the output and passes through D3. During the ON or OFF time C2 holds (stores) the voltage. (assuming there is a very small load on the output)
1606321812245.png
I would choose C1 so that a 600V 6nS edge will develop 50mA or less. I don't know about C2. You are not really dividing the 600V down. I think it will work with no C2 but noise might get in. Maybe C2 is 2x bigger than C1, I would not go above 10X.
 

Thread Starter

SiCEngineer

Joined May 22, 2019
299
You need a fast isolator. Many will not pass 200khz and you probably need a 100mhz type part.

There are many ways.
Use a 0nS high voltage diode. SiC diode.
View attachment 223367
This option you are just looking at edges. C1 must be small. I am getting 6nS edges on GaN devices. It takes very little capacitor to get too much current. On the falling edge current passes through C1 into D2 and drives the output low. On the rising edge current through C1 pulls up on the output and passes through D3. During the ON or OFF time C2 holds (stores) the voltage. (assuming there is a very small load on the output)
View attachment 223368
I would choose C1 so that a 600V 6nS edge will develop 50mA or less. I don't know about C2. You are not really dividing the 600V down. I think it will work with no C2 but noise might get in. Maybe C2 is 2x bigger than C1, I would not go above 10X.
So just to clarify, you currently use the first method and you are getting good results? And the second is a possible improvement?

I am having the same issue as you. Using a capacitor type divider, in my simulations, too much current is going through the capacitor even with small values. Seems to also cause shoot-through in the switches at high frequency also due to the requirement of the small capacitor.

For the first technique - do you need an isolated 5V power supply? I would assume so, since the diode actually does not provide any isolation between the high voltage and the 5V power supply. You would obviously have an isolated signal to the right hand side of the digital isolator which must be fast - however, the 5V connected to the resistor and the diode can't be considered safe.

Finally - is there an equation/ball park figure for the resistor value? Is this to avoid as much current from the 5V supply being drawn?
 

ronsimpson

Joined Oct 7, 2019
1,121
So just to clarify, you currently use the first method and you are getting good results? And the second is a possible improvement?
I have used both.
I am having the same issue as you. Using a capacitor type divider, in my simulations, too much current is going through the capacitor even with small values.
What cap are you using? You could use a 10:1 cap divider to get down to 60V and then use the diodes to limit to what the isolator needs.
For the first technique - do you need an isolated 5V power supply?
In my case I have the next logic running at 12V or 5V and I use that supply. If your micro is on the same ground as the GaN MOSFET then you do not need a isolator.
is there an equation/ball park figure for the resistor value?
The resistor value; I don't remember but I think I used a resistor that gave me 10mA.
 

Thread Starter

SiCEngineer

Joined May 22, 2019
299
I set the rise/fall time to 5nS. Current = 260mA for 5nS on each edge. Current is related to switching speed in V/S (600V/5nS)
View attachment 223371
I will use this digital isolator as I am using it for my gate drive Fets, So, it should be fast enough. https://www.ti.com/product/ISO7831

My caps were a bit bigger than those you have used -I will test again now with smaller caps. I understand how the capacitor divider shown works - but what about the diodes in this example? They don't need to be fast Silicon carbide switches in this arrangement? Are they working as some kind of voltage clamp/limiter to clamp the voltage to 5V and 0V despite actually having 60VAC switching voltage placed across them? It looks like the output to the digital isolator will be taken from the top of D2, right?

I believe I understand the point about the supply - but is that not because the supply to the GaN fets should already be isolated? It doesn't make sense not to have an isolated power supply for both this 5V and the 5V used for GaN gate drive.
 

Thread Starter

SiCEngineer

Joined May 22, 2019
299
The second solution there seems to be working a lot better. I am getting the 5V pulses as desired.
Out of interest Ron - why do you sense the switch voltage of the GaN? Is it just to pass to your oscilloscope and be able to measure it at a safer lower voltage, or do you use it in the control system at all?

Just need to figure out the correct way to isolate this signal safely and pass to my DSP now. Does the TI Digital isolator seem appropriate/fast enough?

Are these 1N diodes good enough? Like I say, the converter may be switching above 200kHz... will they be able to handle those switching losses/not have any issues with reverse recovery?
 
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Thread Starter

SiCEngineer

Joined May 22, 2019
299
I set the rise/fall time to 5nS. Current = 260mA for 5nS on each edge. Current is related to switching speed in V/S (600V/5nS)
View attachment 223371
Ron, I have used your circuit to good effect. The reason why I wanted to use this is to sense the MOSFET voltage on my flyback FET. I am looking into using an active clamp flyback converter, and wondered if there is a way I could incorporate this sensed voltage to determine when to turn On/Off the main and clamp FETS., according to the rising or falling edge of the signals.

Have you done anything of this sort in your works?
 

ronsimpson

Joined Oct 7, 2019
1,121
I have used this for decades. Maybe back to 1984. In the old days I used the fastest 1kv diode I could get. (small diode)
Lately I used SiC diodes. 2A 1200V something like that. (now we can get HV Schottky and SiC )
I want to see what is happening in the -2V to +2V range and not look at what is happening at 800 volts. Most scopes can not work well with a 1000V pulse and see the 2 volts when the MOSFET is on. I just want to see the "on voltage" and strip off the "off voltage" which I can see with a 2kV scope probe.
1609886268678.png
Here is a link to what I am playing with. Keysight
 
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