L6494LDTR

[camabe]

Hello

I am using the L6494(14PIN) as a driver for a high and low side FET in a half bridge arrangement. My DC volts can be anything from 120Vdc to 220Vdc. and current up to 17A

Is there anyway to determine the breakdown mechanism if I do not momentarily switch the bottom driver on to charge the bootstrap capacitor.?

If I do not switch the bottom driver on first so that the bootstrap capacitor is charged and accadently turn the top driver on first, it destroys the driver and fets. 

I have wired in the L6498 (8PIN) and this does not behave the same way. I can always make sure of the sequence under normal conditions but under a fault condition this could not be guaranteed and an expensive call out could follow.

Is there possibly another way to guarantee the bootstrap capacitor gets charged ?

thanks

[Peter BENSCH]

Welcome @camabe, to the community!

For the L6494, section 7 of the data sheet states that:

This structure [bootstrap driver] can work only if VOUT is close to SGND (or lower) and in the meanwhile the LVG is on. The charging time (Tcharge) of the CBOOT is the time in which both conditions are fulfilled and it has to be long enough to charge the capacitor.

The only option is therefore to ensure that the low-side channel is switched on long enough during power-up to charge the bootstrap capacitor, which can be achieved

• either with a suitable PWM driver
• or additional logic between the PWM driver and gate driver.

Hope that helps?

Regards
/Peter