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I need to use a logic level shifter in my circuit which translates 2.8V logic level into 5V and the other way round. I found this document. The device itself is presented on the page 10. It is very simple and consists of one mosfet and 2 pull-up resistors. However there is one catch. To use this device I need to have both 5V and 2.8V power supplies in my circuit. I only have a stable 5V line.

What would be the easiest way to solve this issue? I thought I could connect 3 diodes in series, which would drop the voltage about 3*0.65V = 1.95V, so I would have 3.05V left, which is a bit too high, but I guess it would work. But maybe there are easier ways to shift logic levels?

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Try using 2.8V MCP702 lineer voltage regulator or you can select different 2.8V 3 pin voltage regulators from digikey.com. By the way select a regulator according to your current needs.

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    \$\begingroup\$ With only 2uA Iq and 2.8V available as standard, this approach is probably your best solution. \$\endgroup\$ Dec 17, 2014 at 19:47
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You would use your 5V as input to an adjustable voltage regulator like the LM317: https://www.fairchildsemi.com/datasheets/LM/LM317.pdf

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  • \$\begingroup\$ Oh, I forgot too add that it is supposed to be a low power device. I probably can't afford an additional voltage regulator. \$\endgroup\$ Dec 17, 2014 at 18:07
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    \$\begingroup\$ Then you would use a switching regulator instead of a linear regulator, as they're more efficient. Dropping voltage across three diodes certainly isn't efficient. \$\endgroup\$ Dec 17, 2014 at 18:45
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Use a simple voltage divider network, select appropriate value of the resistors to get 2.8V from 5V and then add a simple voltage buffer circuit at the output of the voltage divider network to properly match the impedance of the network connected at the output terminal.

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An even better solution is to use a zener diode having reverse voltage of 2.8V. Just like this one: http://uk.farnell.com/micro-commercial-components/1n5224b-tp/zener-diode-500mw-2-8v-do-35/dp/1924531

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