Can I use a dual high-side low-side gate driver as a dual low-side driver?

Question

Some background: I am building a 350 VA DC to AC inverter, partly for my own curiosity, but mostly for a senior design project for my bachelor's.

What I have with me is monolithic dual high-and-low drivers with a 4A peak current driver, shown here. My planned first stage is going to be a push-pull type half bridge through the center tap of a transformer I hijacked out of an off-the shelf-UPS. I'll be doing this as mod-square wave, because the transformer is mains frequency rated, and I don't have time to get anything different. That output will then get rectified, stored, and h-bridged with my real experiment, which should be a low distortion sine wave output.

The push-pull mod-square arrangement doesn't need a high side driver, on account of not having one to speak of.

So, my question:

Given what I have available, can I use the high side output of the gate driver as a low-side output? I look at it, and it seems I would just have to omit the bootstrap capacitor and diode, connect Vb to Vcc, and connect Vs to GND, and then (I believe) it should work.

I have some time, but I have limited resources and I really just wanted to know before I go and etch a board. I also don't have a bunch of MOSFETs lying around to blow up while I learn. Also, using a single IC would save me some board space, and the chips have internally matched propagation delays.

I would appreciate any feedback on this application, or anything else the community would like to share.

Answer 1

Yes you c an use both as low side drivers,although the high side is filtered to prevent shoot thru failure simultaneous switching resulting in a dead band required for totem pole switching, so no effect in half bridge..

Edit:

I think the chip is a complementary output drive going to what I understand is to drive low side in half bridge with center .tap to DC. The is somewhat like a secondary diode half bridge using 2 diodes and center tap ground. So there are are VI tradeoffs when using this method, less voltage, more current.. This chip has good deadband filters for commutation on the high side, but you wont need this.

I would explore the tradeoffs of open-drain inductive switches vs tri-level active outputs ( full bridge ) with 0, +V , -V methods for sine inverters. Look at IGBT switches , which are preferred for large sine power motor controllers.

If it were my design I would use the lowest MOSFET or IGBT resistance switches that I could afford , in milliohms. Read this all about Switching...Theory in semi's. then decide. It compares the old SCR, MoSFET with IGBT and newer enhanced gate types. Toshiba makes these too.

Answer 2

Following extensive discussion in comments here's what I'd do. The H-bridge you propose for the AC output stage is fine - that's the right thing to do but between your battery and that H bridge you need to control the DC level - the AC output will need to be regulated and the easiest way to do it in my book is feed the H bridge circuit with a variable level DC. For this I'd use a synchronous buck converter for the following reasons: -

• Synchronous means higher efficiency (95% as opposed to 90% for non-sync)
• Output voltage is quite simply input voltage x mark-space ratio
• Reducing mark-space when input Vsupply rises OR increasing mark-space ratio when Vsupply falls does not require feedback - it can be done using feed-forward.

You should use one of your IR drivers to control two N channel FETs that do the switching but make sure you provide a small deadband. You can do this using an AND gate and a NOR gate - feed a squarewave and a delayed-by-50 ns squarewave into both and look at the output waveforms.Delay can just be an RC network: -

Note that the OR gate shown above needs to be a NOR gate and preferably both AND and NOR should be schmitt trigger devices.

For PWM I'd use this: -