AXBB charge pump fault

[Doddy]

EDIT: Ignore below the line below for now, I've re-read your answer and am confused.

I understand you're measuring the OP1 output with respect to the negative supply to the 24V PSU?

The OP1 is a low-side switched output. That means that if the relay is ON, the output voltage should be near zero. If the relay is to be OFF, then the OP1 output should rise (through the pull-up by the relay) towards 24V.


Relay....OP1=Expected.....Reported
..ON..........0V..............24V
.OFF.........24V...............9V


Can I ask you to confirm the reported values? My thoughts, below, were with ON/OFF voltages swapped.

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--- Bit below here was original reply
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What you're saying is that you're using the charge-pump as the initial "Computer is now on" signal.

It doesn't safeguard against the computer going doo-laa-lee after the event, but it avoids a spurious start-up spin-up. Not that I believe you're likely to get that with the AXBB - it's more a concern for random start-up on PPs.

A couple of things to consider:-

* You don't have the intended watchdog cut-out protection intended by the charge-pump. Safety is compromised *1
* You're running a board with a failure evident. You cannot predict at this time how this failure will behave - it might remain, or fail completely. *2
* The FET is partially conducting. I'm uncertain if the 9V is across the relay coil, or the relay coil plus 27R resistor. I'm going to randomly guess that you include the resistor, and together with a randomly selected relay coil resistance that you have a total of 750R resistance (feels about right and it makes the maths easier). The problem here is that the current flow is voltage-across-relay-when-off (24-9V / 750R) is 20mA holding current on the relay when the output is set to 9V. That means that you're dissipating (P=V*I) 9x0.02 = 180mW in that FET in its failed state. The max power dissipation in a 2n7002 (don't know if that's the device used to replace the original) is 200mW, derated above 25C, so you might find that the FET subsequently fails differently. Or if the assumption about coil resistance (and therefore current, and therefore power) is inaccurate - then adjust the numbers accordingly. You're likely to see the FET fail over.


*1: But then, most home-grown won't have a charge-pump anyway.
*2: This output would, at best fail to drive the charge-pump, and the machine will fail-safe, or at worst intermittently fail to drive the charge-pump - random shutdowns. Somewhere in the middle it never deactivates the charge-pump and you lose the charge-pump function - see (*1)


My thoughts: I don't think that you are getting any benefit above the controlled output that you have from the reset-discrete output. You may end up with the machine intermittently or hard-failing in the case that the watchdog fails. Likely this will never happen mid-cut, but each time you shut the machine/software down you're risking the machine hard-failing. This uncertainty would lead me personally to either :-
1) Repair the board again in one of several ways previously discussed.
- or -
2) Remove the charge-pump from the system.

EDIT:

It's still worth performing the voltage measurement advised earlier - the voltage-across-the-10k (what I meant by the gate voltage, Vgs) is intended to understand if the FET itself is at fault. If the Vgs is near 0 you expect the FET to be "off" - Resistance Drain-to-Source (Rds) is very very high. If the Vgs is above Gate Threshold Voltage it should be in the (near) fully "on" state, Rds = very very low. The problem lies that between Vgs=0 and Vgs = 3V then the FET operates in it's linear(-ish) mode, where the Rds is a function of Vgs. By understanding if Vgs (voltage across 10k) is a definite 0/3V then we know if the device driving the FET (74HC14) is operating correctly and the FET has failed. If the voltage is somewhat lower than 3V then the FET might be behaving as expected.