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Title: Me7 bench setup - grounding pin resistor? Post by: vtraudt on April 22, 2019, 09:27:05 AM i bought a bench setup for flashing Me7 ecu.
Always had trouble getting ecu into boot mode, sometimes 10+ trial/error. Sometimes I just gave up. This weekend, failed again to get ECU into boot mode. Checked out the harness, continuity check with multimeter in 'beep' mode. Did NOT beep when tracing my ground "pin"/wire back to the ECU plug. Thought broken wire. Took wire apart, and found a 220 Ohm resistor. a) why did the seller add the resistor? Purpose? b) is the resistor beneficial, or could it be the resistor has been causing my boot mode problems? Title: Re: Me7 bench setup - grounding pin resistor? Post by: nyet on April 22, 2019, 09:50:34 AM He likely did it to provide a rudimentary current limiter so if you touched the grounding pin to something you shouldn't (e.g. +12 battery voltage) you don't fry something. It is possible it is interfering with getting into bootmode, but IMO unlikely (depending on which ECU, specifically.... ME7.1.1 ECUs can be dodgy to get into boot mode)
Title: Re: Me7 bench setup - grounding pin resistor? Post by: vtraudt on April 22, 2019, 10:21:46 AM Thanks. Will try without the resistor. But in principle a good idea (limiting the amps that flow if messing up. At 12V und 220 ohm, only 0.05 Amp should flow (50 mA), which may save components in case of a short. But not sure if the resistance would mess with ground the pin.
Title: Re: Me7 bench setup - grounding pin resistor? Post by: nyet on April 22, 2019, 10:26:19 AM depends entirely in input impedance (or pullup resistance). In theory, 220 ohms to ground should be a more than sufficient current sink.
Title: Re: Me7 bench setup - grounding pin resistor? Post by: vtraudt on April 22, 2019, 10:57:30 AM depends entirely in input impedance (or pullup resistance). In theory, 220 ohms to ground should be a more than sufficient current sink. Isn't (according to Mr. Ohm - The resistance R in ohms (Ω) is equal to the voltage V in volts (V) divided by the current I in amps (A)): I=U/R? here 12/220=-0.05 |