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This problem is really rattling my brain quite a bit. My current setup uses a wall wart power adapter sending 6VDC to my circuit. From that I'm trying to lower the voltage to 5V using two resistors. I have a 220 ohm and 1kohm resistor in series. '

So far so good right?

My expected output should be 4.92V, this however is not the case i'm getting the full 6V with my meter on one end and -.1V on the other. I've switched the leads around to see if did something backwards but same result. Am I missing something? I'll do up a schematic on request. This is a bit of an easy issue so don't see the need of one just yet.

Further unnecessary explanation: This is just a test circuit. My final circuit will be using sensors instead of the wall wart power adapter but I have to make sure I am correct before proceeding. That is why I am not using any "load". Is this the issue? the fact that their is no load?

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  • \$\begingroup\$ Have you measured the output of your wall-wart? (as-in not just read the label, but actually measured it) \$\endgroup\$
    – brhans
    Commented May 11, 2017 at 16:51
  • \$\begingroup\$ Yeah. Do the schematic and make sure you carefully describe where you are placing the leads of your voltmeter. Chances are that you are just measuring the output itself and not the divider voltage. Besides, wall warts have crappy transformer regulation. \$\endgroup\$
    – jonk
    Commented May 11, 2017 at 16:51
  • \$\begingroup\$ "6V with my meter on one end and -.1V on the other" END OF WHAT? .............. Schematic..... please \$\endgroup\$
    – Trevor_G
    Commented May 11, 2017 at 16:54
  • \$\begingroup\$ drive.google.com/open?id=0B0lRN3VXhPvyUFBxbklDTG15NjA This is a link to the schematic. \$\endgroup\$
    – Matt
    Commented May 11, 2017 at 17:10
  • \$\begingroup\$ I'm measuring from the middle between the resistors to ground. Which should be 4.92V according to ohms law. The two antenna icons represent my meter leads. \$\endgroup\$
    – Matt
    Commented May 11, 2017 at 17:11

1 Answer 1

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1) Wall warts are very inaccurate and there is a high chance it is not supplying exactly 6 volts.

2) Don't forget that resistors have tolerance. I can almost guarantee your resistors are not exactly 220 ohm and 1k ohm.

3) Voltage dividers really shouldn't be used for power supplies because the "regulation" will depend on the load. If you attach a load, it will combine with the resistor divider network, changing its output voltage.

It would have been more helpful if you had attached a diagram to your post showing exactly how you have things connected.

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  • \$\begingroup\$ I can draw something up quickly. Right now I have a very simple circuit. What would be a reliable alternative? Current from the sensor will be 6V 100ma \$\endgroup\$
    – Matt
    Commented May 11, 2017 at 16:52
  • \$\begingroup\$ You may be better off using a silicon diode to drop the voltage from 6V to 5.2V, which most 5V circuitry can handle \$\endgroup\$
    – DerStrom8
    Commented May 11, 2017 at 16:53
  • \$\begingroup\$ Schematic drive.google.com/open?id=0B0lRN3VXhPvyUFBxbklDTG15NjA \$\endgroup\$
    – Matt
    Commented May 11, 2017 at 16:59
  • \$\begingroup\$ Silicone diode such as the 1N4001? *Fixed the link above \$\endgroup\$
    – Matt
    Commented May 11, 2017 at 17:01
  • \$\begingroup\$ Silicon, not silicone ;) \$\endgroup\$
    – DerStrom8
    Commented May 11, 2017 at 17:22

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