Hi
The customer test with BQ79616. Usually open wire detection uses the constant current source inside the chip to consume the power of the VCn pin capacitor, causing the voltage to drop rapidly to determine whether there is a wire open.
In the actual test, they disconnected the VC0 line and found that the voltage of the first cell increased about 34mV. How to determine the open wire detection on the VC0 pin? Could you please explain it in principle?
Thanks
Star
Hi Star,
You are correct that open-wire will be detected using a slightly different method on the bottom cell. Since the VC0 line is disconnected in this test, a source current instead of a sink current is used:
This is because the voltage at VC0 ~= GND. Without a significant voltage difference, enabling a current sink to GND to test open wire won't really affect the measured cell1 voltage because the capacitor won't be draining since both positive and negative terminals are at approximately the same voltage level.
Consequently, to detect open wire for VC0 and CB0, the BQ79616 uses a source current to raise the VC0 pin voltage. This means that the voltage difference between VC1 and VC0 should be smaller, leading to a reduced cell1 measured voltage. Please see Section 8.3.6.4.6.4 in the BQ79616 datasheet for more information: BQ79616 16-Series Battery Monitor, Balancer, and Integrated Hardware Protector datasheet (ti.com).
You mentioned a cell voltage increase. Is this the actual voltage measured across the cell by a multimeter, or is the BQ79616 reporting a higher cell1 voltage? Can you also show me where the customer is disconnecting the VC0 wire in an image?
Best,
Andria
