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I need to level shift 3V3 to 2V8. I have a 2V8 source but that'll require some patching wires on testpads and as such I would like to avoid it. I'm wondering: would it be safe to feed the SN74LVC245AN the required 2V8 by using a voltage divider (two series resistors) to convert 3V3 to 2V8? It is for a temporary setup which will be used shortly.

I use this chip to shift my 9600 baud 3V3 serial to a 2V8 serial on a SARA-N3. I need to perform a firmware update and made room in my PCB design for the ALT uart to be accessed but forgot about the 2V8. As such I need to convert 3V3 to 2V8 serial. I don't know if the update process requires bidirectional data transfer and as such a simple voltage divider over the TX->RX probably won't be good enough.

The pad on my PCB is quite small, just enough to hold a volt meter probe on it to test for an activated SARA-N3.

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    \$\begingroup\$ Just patch some wires on the testpads if you have 2v8 present. It'll be less of a struggle than making sure the divided voltage remains what you expect when the logic IC draws current. \$\endgroup\$ – LukeHappyValley Jul 28 at 13:49
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Instead of two resistors, use a resistor and a Zener diode.

Make sure the resistor is low enough to ensure the supply voltage to the 245 won't drop below 2.8V when it takes its maximum load and leave the Zener to do the regulation.

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  • \$\begingroup\$ This is plain wrong. I don't think you understand his issue at all.Maybe review the basics before posting again. \$\endgroup\$ – Helena Wells Jul 29 at 15:06
  • \$\begingroup\$ Lol kisses Finbarr. \$\endgroup\$ – Helena Wells Jul 29 at 15:11
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You need to make sure the current to the IC is much smaller than the current through the voltage divider (preferably at least an order of magnitude, better two, but chips use very small current, so should be doable if temporary). As as temporary solution - maybe. Try it with potentiometer, you may want to adjust it a little bit. The chip will pull the voltage divider output a little down. Make sure you have spare chips just in case you accidentally overvolt something, and control the voltage on the divider during the chip's operation, because the chip may pull various currents and it will affect voltage. Again, the less current it pulls comparing to voltage divider, the smaller voltage fluctuation will be.

Added from comments: if it's Serial and every line is 1-directional, maybe a voltage divider on every line is enough. 0 vd input gives 0vd output, 3.3v input produces 2.8v output? Besides, the 3.3V device will probably accept 2.8V logic without conversion, it's probably within the spec, but it has to be tested first of course

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    \$\begingroup\$ The most obvious point missing here is that the usual requirement for a supply bypass capacitor becomes even more critical when the supply itself has unreasonably high impedance. This is a rather bad idea, but if it's going to be attempted then both high and low frequency bypassing is required - ie, a tiny ceramic and a large capacitor. \$\endgroup\$ – Chris Stratton Jul 28 at 14:25
  • \$\begingroup\$ I always assume bypassing. Should have mentioned of course. Thanks for valuable correction! \$\endgroup\$ – Ilya Jul 28 at 15:01
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    \$\begingroup\$ Modern CMOS inputs do indeed draw very little current - to a first approximation, unless there's an on-chip pulling resistor they draw current only when switching, and briefly rather substantial current as they do so. Hence it's not really the static current that needs to be outweighed by static current through the divider, but the dynamic current which has to come from somewhere. Having the divider's static current dwarf the load's dynamic current starts to become expensive, so it ends up really on bypass caps to supply it. But even with normal power traces, it's initially up to the caps. \$\endgroup\$ – Chris Stratton Jul 28 at 15:05
  • \$\begingroup\$ I should've added this to my question. I'm using the chip to shift a serial port so I'm not expecting any high current draw. \$\endgroup\$ – Tarick Welling Jul 28 at 21:24
  • \$\begingroup\$ It would make more sense to apply the voltage divider to the serial signal itself. Update the question with the baud rate and details of the receiver (mobile data modem?) and you might be able to get more specific help. \$\endgroup\$ – Chris Stratton Jul 28 at 23:55

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