I am using a PAC1720 current measuring IC to measure the current/voltage/power being delivered to a lead acid motorcycle battery that is being charged.

Here is a newer datasheet and an older datasheet (with different info in them!)

Everything seems to work okay until I remove the power supply to the PAC1720 chip. When Vdd to the PAC1720 is removed (i.e. my PCB power supply is off), it appears that the chip begins conducting current from its sense pins to the Vdd or GND pin (or both, not sure), which ultimately fries the chip.

This is a critical flaw because it means that if someone hooks up a battery to my product without having the power supply turned on (highly likely to happen), the product will destroy itself.

Here is a schematic of how the product is laid out:


simulate this circuit – Schematic created using CircuitLab

I'm guessing it has something to do with the internal ESD strucutre of the PAC1720.

One datasheet for the PAC1720 shows a typical layout like this:

from MicroChip datasheet

The phrases at the bottom of that picture disturb me. It says:

1: The device MUST be biased PRIOR to applying VSOURCE. Failure to do so will result in damage.

2: The unpowered bias on VDD must be greater than VDD, if VDD will be removed prior to VSOURCE.

I don't understand what they mean by "biasing the device"... sounds like they are talking about biasing the internal clamping diodes?

The second note doesn't make grammatical sense does it? I don't understand what they are trying to convey in the second note.

Now, an older datasheet shows the internal ESD structure of the PAC1720:

enter image description here

This isn't very helpful since it just shows a generic box labelled "Edge-Triggered ESD absorbtion circuit" between the upper clamping diode on Sense1 +/- and GND.

So, my main question is: what is happening when I disconnect Vsupply while a battery/charger is connected and the PAC1720 starts smoking?

Any way that I can solve this problem now that the product is in production?

  • 2
    \$\begingroup\$ Have you checked your current sense resistor? 8mohms is very small... \$\endgroup\$ – 12Lappie Mar 21 '17 at 15:26
  • 3
    \$\begingroup\$ biasing the device Since this chips has no other supply/biasing points other than VDD biasing the device means apply a VDD. What is actually happening is impossible to say without looking at the chip's full schematic, which we don't have. What you need to do is to prevent not biasing the device ! \$\endgroup\$ – Bimpelrekkie Mar 21 '17 at 15:27
  • 3
    \$\begingroup\$ Agree with previous commentators - "The device MUST be biased PRIOR to applying VSOURCE. Failure to do so will result in damage" for me means that VDD should be there before applying Vsource. Let's not guess what it means - contact chip manufacturer and clear this with them. Probably then you will discard using this chip for your development. \$\endgroup\$ – Anonymous Mar 21 '17 at 15:28
  • \$\begingroup\$ Did you ask Microchip? \$\endgroup\$ – user1890202 Mar 21 '17 at 15:45
  • \$\begingroup\$ Yea I submitted a case to their online site, waiting for a response \$\endgroup\$ – macdonaldtomw Mar 21 '17 at 19:50

It appears the chip needs a VDD applied before the SENSE pins are connected to the battery.

Your only option is thus to power it from the battery side, using a resistor and zener as shown in the schematics you posted, or even a voltage regulator.

Since its IOs are open collector (or open drain) it can be powered while the rest of the circuit (SMBUS side) is unpowered. It will not send voltage to your unpowered micro.

Now, when the microcontroller is powered but there is no battery, the SMBUS pullups will let some current enter the inputs of the unpowered chip. I don't know if this would be a problem.

When in doubt... My solution would be to use a Diode-OR from both the battery and your 3V3 supply.


simulate this circuit – Schematic created using CircuitLab


I think with bias they mean the combination of the 5.2 V zener diode and the 100k to 1M resistor. So the zener diode is providing a voltage to the chip making those parts work which protect the chip from the voltage across the sense pins.

So if you don't have that combination in your unit, you will fry the chip as soon as VDD is removed and nothing is left to power those parts.


The biasing thing means :-

You MUST have the power turned on to the device before you attach the battery OR turn on the charger.

It also means don't turn off the device before you disconnect the battery AND the charger.

In other words. when connected as shown SW1 literally = KILL SWITCH.

How do you fix it at this point.... maybe add a relay to disconnect the battery circuit both sides of the sensor resistor when vSupply is disconnected and enough bulk capacitance to keep the chip biased for the duration of the relay opening time.


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