I have a LT3840 Demo Circuit 1909A that draws more current than it should and current limits when powered from my benchtop supply even without any load and ICTRL set to 1V. I am powering the demo board with a 12V supply.
When I first received the board, I replaced COUT1, COUT2, COUT5, and COUNT6 with higher voltage capacitors, and I configured R13=442kOHM and R10=1.18MOHM for a Vout of 4.6V. This configuration worked well and I was able to conduct high current tests.
Then I changed R13=316kOHM but kept R10=1.18MOHM for a Vout=5.9V. It was here when the demo circuit began drawing excessive current. So then I changed R13 back to 442kOHM, but that didn't fix the excessive current problem.
Here is what I've tried:
- I viewed the circuit with a thermal camera and found that the area between the LT3840 driver and the low side FETs are warmer than the rest of the board. Due to inaccuracy with the thermal camera, it is hard to say if the chip itself or a nearby component is getting warm.
- I viewed the high side and the low side gates of the power FETs on a scope and found that the high side gate is only reaching 8V relative to ground. To me, this suggested a problem with either the high side FETs or the boost circuit. I found no shorts on the high side FETs or the boost circuit. I replaced C4 and D1, but the excessive current problem persisted.
- I then unpopulated R19 (high side gate series resistance), and the demo circuit no longer drew excessive current. I then repopulated R19, and unpopulated R20 (low side gate series resistance), with the same result. With this test, I was hopping to isolate the problem to either the high side or the low side FETs. However, the results suggest that both high and low side FETs play a role. I've ruled out shot-through since my earlier observations of the gate voltages doesn't show any overlap of the gate timings.
Not sure what to else to check and I hope that someone could give pointers on what else to try. I would like not to replace the chip or the FETs, but if it comes to that, it'll be done.