With the following circuit, would the R2 resistor not change or affect the polarity in some negative way (no pun intended)?

In the image below I added the red box as well as the red text in an effort to try and understand what is happening when the R2 resistor attaches to the +4V lead.

Can someone please explain what it is doing assuming that everything on the circuit is correct? Note: The positive lead is switched power, so it might be on or off at any given time.

Reference webpage:Power Control System for RPi Car PC

enter image description here

  • \$\begingroup\$ It just looks like R2 is your typical pull-down resistor for both your microcontroller and your relay. No, it wouldn't change the polarity. The worst it can do it pull down (hence the name) to an indeterminate state if the resistance is too high. \$\endgroup\$ – KingDuken Sep 3 '18 at 0:28
  • \$\begingroup\$ Looks like there may be two independent "grounds" there which may cause confusing voltage measurements. Is the ATTiny ground connected to the vehicle ground? If there is no negative power supply, you shouldn't get a negative voltage anywhere. \$\endgroup\$ – Peter Bennett Sep 3 '18 at 0:31
  • \$\begingroup\$ @PeterBennett I guess that would be the case (two independent grounds). It's the vehicle ground that goes through a converter (DC to DC) to step down the Voltage (Zk-SJVA-4X is what I'm using). \$\endgroup\$ – Arvo Bowen Sep 3 '18 at 0:57
  • \$\begingroup\$ If the R1/R2 junction is "~4V" meaning "about 4V" rather than "minus 4V", that resolves your polarity question. Many DC-DC converters are not isolated - the input Ground is connected directly to the ouput Ground - in that case, the ATTiny Ground will be vehicle Ground, and the circuit should work. \$\endgroup\$ – Peter Bennett Sep 3 '18 at 1:03

R2 constitutes the lower half of a voltage divider used to reduce out-of-range voltages to potentials sample-able within the range of an MCU input rating.

Voltage dividers do not change polarity relative to their fixed ground. This circuit accepts a positive input, and yields a reduced positive output.

However vehicle electrical systems are notoriously nasty, and it is a legitimate question if this design overall is sufficiently robust for such usage.

  • 2
    \$\begingroup\$ I wonder if that "-4V" is really "~4V", meaning "about 4 volts", rather than minus 4 volts? \$\endgroup\$ – Peter Bennett Sep 3 '18 at 0:53
  • \$\begingroup\$ I have some nice DC to DC converters to filter the power as well. I'm hoping that will help out as well. Also, the circuit is not just a design but been proven to work as per the video on the reference link. Not sure if it will hold up for a long time but there is one way to find out. Thanks! \$\endgroup\$ – Arvo Bowen Sep 3 '18 at 0:54
  • \$\begingroup\$ @PeterBennett it is indeed. The schematics show ~4 (image posted). \$\endgroup\$ – Arvo Bowen Sep 3 '18 at 0:55
  • \$\begingroup\$ @PeterBennett it is indeed, a quite clearly if you zoom in, "~4v". This would be a notation from the designer of the expected output voltage, though it should be noted that this is a bit high for the pi and would be undesirably activating the I/O protection diode, though probably only with a current it can tolerate given the 10K source resistor. A designer more aware of input threshold voltages would likely have used a different division ratio. \$\endgroup\$ – Chris Stratton Sep 3 '18 at 0:55
  • \$\begingroup\$ @ChrisStratton I'm going to test all output before I hook anything up. As I build it I will see what the output of all the leads are. Thanks for the heads up. \$\endgroup\$ – Arvo Bowen Sep 3 '18 at 0:58

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