I need to switch from 300V to 0V in 1us and then back to 300V. I have successfully built a half bridge circuit with an IR2111 to drive it and I get the 0 to 300V output.

But now I need some help on how to invert the output I am seeing.

This is my circuit:


This is what I get now:

This is what I need:

I have tried just inverting the input waveform, this does not work. Any other ideas are appreciated. Thanks

 

Why would inverting the input not work?
Max.

 

I don't know, that's why I'm stuck.

 

C1 and C3 are not in that position. They are in parallel with V1 and V3.
What's the duty cycle of that 1MHz source?

I don't think it's doable at 1MHz. This chip has a built in deadtime of 650ns typically.


What will be your load?

I hope you have some protection on these 300V, otherwise you may blow up your circuit in no-time.

Start with a much lower secondary voltage , make it work with delays, on off times etc you need, THEN go to higher voltages.

 

C1 and C3 are not in that position. They are in parallel with V1 and V3.
What's the duty cycle of that 1MHz source?

I don't think it's doable at 1MHz. This chip has a built in deadtime of 650ns typically.


What will be your load?

I hope you have some protection on these 300V, otherwise you may blow up your circuit in no-time.

Start with a much lower secondary voltage , make it work with delays, on off times etc you need, THEN go to higher voltages.

Yes, that schematic is slightly incorrect with the capacitors, sorry. But on my board it works, and that is what I see on my scope with the 0 to 300V output. I simply want to invert that pulse.
I am just testing it now, but in practice that will be a single pulse that I trigger. I just can't figure out how to keep it high at 300V and go low to 0V and back to 300V

 

Your requirements are not clear.

It also doesn't help us if you post a wrong schematic, even if it works on your breadboard.

"keep it high at 300V" does mean what exactly? You mean the upper FET should normally be ON (at 300V) ?

That doesn't work because the bootstrap cap C2 will discharge over time. You would need a charge pump circuit or a high side switch with integrated charge pump. This would keep the high side always on.

 

You'd need an active charge pump to keep 15v across C2 - or an isolated power supply. The way things are now, C2 can only charge when M2 is OFF and M1 is ON; as the cap's low side needs a path to ground. Without an isolated power supply or charge pump, the IR2111 will bleed off C2's charge pretty quickly while M2 is turned on.

By the way, a 1N4007 is a poor choice for D1, as it has a very slow recovery time. You need an ultrafast or Schottky diode there. A UF4004 or similar would work.

 

I need to normally be at 300V and with an input pulse, drop to 0V for 1μs then back to 300V.
I felt like I was getting somewhere when I was able to do what I showed in the original post and thought it would be simple to invert the out pulse, by inverting the input, but that has proven not to work.
Sorry, I am learning as I go along here... I have limited knowledge in EE. So I am open to suggestions here...

 

You'd need an active charge pump to keep 15v across C2 - or an isolated power supply. The way things are now, C2 can only charge when M2 is OFF and M1 is ON; as the cap's low side needs a path to ground. Without an isolated power supply or charge pump, the IR2111 will bleed off C2's charge pretty quickly while M2 is turned on.

Ok, so if I were to keep 15V across C2, it would allow me to do what to get the inverted output?

By the way, a 1N4007 is a poor choice for D1, as it has a very slow recovery time. You need an ultrafast or Schottky diode there. A UF4004 or similar would work.

Ok, thank you. I had the IN4007 lying around and it worked so I went with it.

 

Ok, thank you. I had the IN4007 lying around and it worked so I went with it.

Also across the gate resistors is recommended.
http://tahmidmc.blogspot.ca/2013/01/using-high-low-side-driver-ir2110-with.html
Max.

 

Your requirements are not clear.

It also doesn't help us if you post a wrong schematic, even if it works on your breadboard.

Sorry, I am trying to teach myself circuit design as I go along. I updated the original post with the correct schematic.

"keep it high at 300V" does mean what exactly? You mean the upper FET should normally be ON (at 300V) ?

That doesn't work because the bootstrap cap C2 will discharge over time. You would need a charge pump circuit or a high side switch with integrated charge pump. This would keep the high side always on.

I need to normally be at 300V and with an input pulse, drop to 0V for 1μs then back to 300V.
I felt like I was getting somewhere when I was able to do what I showed in the original post and thought it would be simple to invert the output pulse by inverting the input, but that has proven not to work.
Sorry, I am learning as I go along here... I have limited knowledge in EE. So I am open to suggestions here...

 

What is the load?

 

What is the load?

To be honest, I don't know right now.

I'm building this circuit to control the photocathode on an Image Intensifier.
The 300V supply supplies 3.3mA and the only thing connected to this circuit is the photocathode so I have assumed this will be enough as the II works with this supply, I just can't figure out the switching...

 

So I finally figured out my problem and have successfully resolved it. Thanks to praondevou and SgtWookie for the advice about keeping 15V across my C2 capacitor - it worked!

My final schematic to switch from 300V to 0V and back in 1us:



The output of my circuit:

The original thread also is linked here

 

If the load is only 3 - 4 ma at 300 volts is thee any reason that the 300 volt source can't be fed through a properly sized resistor and a simple circuit using an IGBT or mosfet to pull the 300 volt rail down to zero for the short pulse?

 

V4 is connected incorrectly; it's negative terminal should be connected to M1's source terminal/IR2111's pin 6. The way you have it shown, M1's gate would be subjected to 315V if the IR2111 could survive long enough.