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I've set up a voltage divider to take a 5V signal down to 3.3V. I'm using resistors with values of 6.8 kΩ and 3.78 kΩ. When I use a 5V source straight from a wall wart, the voltage divider works exactly like I would expect: I get 3.3V out of the divider.

However, when I use 5V from a different device, things get a little wonky. I'm getting the signal from Sparkfun's coin acceptor. This device is powered by 12V, but the signal it puts out to tell you about which coin it has received is 5V (There is a NC/NO option). When I hook the 5V coin signal up to the voltage divider, suddenly I don't read the signal as being 5V anymore. It drops to 2.6V at the input to the divider, and 1.6V at the divider's output.

When using the 5V wall wart, I connect the voltage divider's ground to the wall wart's ground, and when I'm using the 12V coin acceptor, I connect the voltage divider's ground to the same ground as the coin acceptor. Is this a mistake? The 5V coin signal that I read when not connected to the voltage divider is measured using the same ground that I connect the voltage divider to.

The reason I'm trying to get a 3.3V signal is so that I can safely connect it to a raspberry pi's GPIO pin.

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  • \$\begingroup\$ A schematic always help visualize a problem statement better than a word-picture would. If you do not have sufficient reputation to add a schematic with the schematic editor integrated into this site, simply upload an image onto any public image sharing site, and share the link here. Someone will edit it into the question for you. \$\endgroup\$ Jun 27, 2013 at 18:17

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Probably this coin thing has a open collector/drain output with a pullup resistor. In effect, the output impedance of the coin thing is high enough that your divider is loading it and pulling the high level lower than the open-circuit value of 5 V.

From the voltage levels you mention, it sounds like this device has roughly a 10 kΩ pullup, or possibly just a 10 kΩ resistor in series with the output as protection.

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  • \$\begingroup\$ Thanks Olin. Sorry to ask such a noobish follow-up, but if you're right, how would I go about altering my voltage divider to compensate for this? Do I need more resistance? \$\endgroup\$ Jun 27, 2013 at 17:13
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    \$\begingroup\$ To further this, you could potentially raise your resistors' values by several orders of magnitude to mitigate this loading effect. \$\endgroup\$
    – Shamtam
    Jun 27, 2013 at 17:14
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    \$\begingroup\$ or else you could use a buffer. \$\endgroup\$ Jun 27, 2013 at 17:30
  • \$\begingroup\$ I finally got around to trying this with 100x more resistance, and it's working properly. Thanks for the help. \$\endgroup\$ Jun 28, 2013 at 23:53
  • \$\begingroup\$ @white: Make sure that the impedance isn't so high that you get a substantial offset voltage due to the leakage of the input, and that it doesn't pick up too much noise. There are downsides to high impedance too. \$\endgroup\$ Jun 29, 2013 at 14:03

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