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4hv.org :: Forums :: General Science and Electronics
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Dead time control increase in current mystery.

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Avi
Mon May 12 2008, 11:25AM Print
Avi Registered Member #580 Joined: Mon Mar 12 2007, 03:17PM
Location: Melbourne, Australia
Posts: 410
What might be causing the bridge current to increase incredibly (even with no load) when i increase the dead time control pot (R1, adds more dead time) and any suggestions to fix it?
I have connected a scope across Hin and Lin (with the IR2110 chips pulled out) and i see the 'tristate' square wave as expected, checked it on both chip sockets.
R1 in the diagram is a dual gang pot.
1210591536 580 FT0 Picture 4
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GeordieBoy
Mon May 12 2008, 01:08PM
GeordieBoy Registered Member #1232 Joined: Wed Jan 16 2008, 10:53PM
Location: Doon tha Toon!
Posts: 881
What might be causing the bridge current to increase incredibly (even with no load) when i increase the dead time control pot

Shoot-through down the bridge legs caused by introducing overlap in the top and bottom MOSFET conduction times.

The way your circuit is set up increasing R1 will slug the rising edge of the signals into the inverters, therefore delaying the falling edge of the drive signal to the IR21xx. Delaying the falling edges relative to the rising edges introduces conduction overlap.

You need to reverse the polarity of the 1N4148 diodes to make it introduce dead-time instead of overlap. You always want turn-on to be slower than turn-off.

-Richie,
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Avi
Mon May 12 2008, 02:05PM
Avi Registered Member #580 Joined: Mon Mar 12 2007, 03:17PM
Location: Melbourne, Australia
Posts: 410
oh, i can see the problem now using 2 scope traces!
hm, if i reversed the diodes wouldn't that just feed the signal directly from one inverter to the other with no delay and cause the pot to do nothing?
1210601111 580 FT45121 Img 0037


Edit: i simulated it with the reverse diodes, it seems to have fixed the problem, but I'm still about unsure of the working.
Looking at it further reveals that in-fact the signal does pass straight from one inverter to the other during to off to on transition (at its input), and then when it turns off, the diode disconnects it from the previous inverter and the capacitor discharges at a slower rate through the pot directly to the previous inverter, thus delaying the on to off transition (input side).

after removing and re-soldering the diodes the other way!
ah much better

1210606903 580 FT0 Img 0038


if anyone is interested:
Link2
file>import

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