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Have I damaged my EVAL6498L board?

three-phase

Hi All,

 

I have been moving on to testing the double pulse function within the AFG31052. To do this I purchased an EVAL6498LEVAL6498L board as used in the application note from Tektronix.

 

Below is the scribbled out setup I created.

 

image

 

I have successfully loaded the double pulse software onto the AFG31052 and I can capture the pulse output from it on an oscilloscope. I can trace that pulse pattern onto pin 2 of the L6498L MOSFET driver. From the data sheet on the board, I believe the pulse output to the MOSFET should then come out on pin 6 of the chip onto resistor R7.

 

Resistor R7 I soldered in, the 320N20N MOSFET I am using suggests a value of 25 Ohms. As that seemed a little hard to find, I soldered in a 22 Ohm resistor. Initially, I also soldered in R4 to the top MOSFET, but reading the application sheet a few more times, this is not used, and I should have shorted out the gate and source of the top MOSFET. Is the upper MOSFET required at all, can I just short source to drain?

 

I removed R4 and shorted gate to source, but still have no pulse train to R7 and the second MOSFET.

 

Is this looking like I have damaged the board? I have seen that the L6498L chip can be purchased separately, but I am not sure I will be able to solder that in place on the board, so will need to get a new board, but just wanted to make sure that there isn't something else I can check before I purchase a new board.

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  • Is the upper MOSFET required at all, can I just short source to drain?

    Don't short it: it's working as a high-power diode. When the bottom MOSFET turns off, after the first pulse, the coil will generate an emf to keep the current running that will take the switching node above the rail. The body-diode of the top MOSFET then catches it and runs the current momentarily until the second pulse turns the bottom MOSFET on again. If you don't catch it at a fairly low voltage above the rail, it will lift all the way to the breakdown voltage of the bottom transistor [which will cope with it] and quickly  squander a good proportion of the energy you've built in its magnetic field [so when the second pulse comes along, you aren't starting at much the same current as the first pulse left it at, but a much reduced current].

     

    Personally, I think it's a messy test the way they've done it because you're testing the reverse-recovery of the top part's body diode at the same time as trying to measure the switching times of the bottom part, but there we go.

     

    If you short it, then the coil doesn't do anything and the bottom transistor is just repetitively shorting the power supply terminals.

  • in reply to jc2048

    Thanks , I read the second test within the application note and now understand what you are saying. I am guilty of not reading the application note in its entirety before I started. I like to read the first test and then conduct it, so I do not confuse my little brain too much.

     

    Kind regards.

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