Electronic Circuits and Projects Forum

Bad news: I broke it. I'm almost too embarrassed to post this, but the way I had it wired--upper ground (as shown in Post #87) directly to battery monitor for the sake of accuracy, lower ground to chassis so the bulk of the current wouldn't go through the battery monitor--created a bit of a vulnerability when, while tinkering with other systems on the car, a short from the 48 V bus to the chassis occurred. With the master disconnect disconnected, the chassis was isolated from battery negative, and the short went right through the first IC in my fuel gauge adapter!

I've replaced the first IC, all three transistors, and most recently the second IC, and now it at least responds to the input, and Bus1 voltage matches output voltage, but it's still not working correctly and I'm at a bit of a loss as to how to proceed, besides starting from scratch and fabricating a whole new PCB, etc. I’m at home now, so I no longer have access to my brother’s car (but do have access to my multiple benchtop power supplies, good-quality multimeters, oscilloscope, and function generator), so I’ve simulated the gauge with a 220 ohm resistor and a pair of 280 ohm resistors in parallel for 140 ohms.

As I apply an input voltage of 5 V to 0 V to simulate the battery monitor’s output, the PCB starts out pulling a seemingly-appropriate (considering my slightly too-high “gauge” resistance) 80 mA. At this point, voltage at the IC end of the first IC’s resistors (relative to ground) ranges from 6.3 to 3.9 V, and the second IC’s resistors are at 0.16 to 0.11 V. But as I turn down the input voltage, the current drops only slightly, such that it’s still at 76 mA with an input of 2.7 V. (Resistor voltages largely unchanged at IC #1; IC #2’s pin 1 now 0.2 V, pins 18-10 now 12.4 V.) Then it drops very rapidly (not quite instantaneously though) to 30 mA at an input of 2.5 V. (Resistor voltages largely unchanged at IC #1 except pin 10 up to 4.5 V; all 12.5 V at IC #2.) Further reduction in input voltage reduces current to 24 mA (zero reading) before the input even gets down to 1.0 V. From there it is unchanged as input voltage continues down to 0 V. (All resistors at 12.5 V where they meet their respective ICs.)

Since the huge jump in current seemed to correspond to a change in state of IC #2’s first stage (pin 1), I tried disconnecting its corresponding diode from the IC, but circuit behavior seemed completely unchanged. I couldn’t think of anything else to try, so now I guess it’s up to the experts on this forum to save me from starting all over again with a blank copper-clad board!

Originally Posted by alec_t

Sounds like the second IC isn't getting the right reference voltage. You should have 1.25V at its pin 4 and 2.5V at its pin 7. Check those pin voltages, also the voltages at pins 4 and 7 of the first IC and the values of R1 and R31 (which connect to pins 7 of respective ICs).

I've got 1.27, 2.53, 0.0, and 1.27 V at IC #2 pin 4, IC #2 pin 7, IC #1 pin 4, and IC #1 pin 7, respectively. R1 and R31 measure 4.6k and 9.6k ohms, respectively (measured in-circuit), but have the seemingly-correct voltage drops (1.27 and 2.51 V) across them when the circuit is powered on.

All of the IC2 diodes have continuity to Q1 base, show 0.61-0.63 V drop across them when my multimeter is in diode-checking mode, and and show infinite resistance with the ohmmeter 'backwards.'

Continuity to Q1 base is 0.20-0.24 milliohms (same as if I touch the ohmmeter leads together).

Supply voltage does rely on a screw (in the car); on the bench it's supplied via an alligator clip to the +12V end of R14. Voltage at pins 3 and 9 of IC2 is 12.491 V, essentially identical to the 12.488 I see if I stick the voltmeter probe in the power supply directly.