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I'm interfacing a 3.3 V device (a GPS module with a serial interface) to a 5 V Arduino Nano. The circuit I am using is this:

schematic

simulate this circuit – Schematic created using CircuitLab

I don't have the GPS Module connected at the moment and have added LED D1 and R2 to visualise the Arduino TX signal (which is currently just setting TX GPIO pin high and low on a 1 second cycle). I am measuring the voltage at NODE1 with a DMM.

My understanding is that D3 blocks the 5V high from Arduino TX and then NODE1 is pulled to 3.3 V by R1. When Arduino TX is low (0 V) D3 conducts and pulls NODE2 to 0 V.

What I measure at NODE1 is 3.1 V and 0.8 V. Am I right in thinking that ...

  1. The difference between 3.3 and 3.1 is simply the voltage drop across R1 due to the load of my LED and 5 K resistor. The RX input of the GPS module will have some similar effect.

  2. The difference between 0.8 V and 0.0 V is mostly due to the forward voltage of the 1N4148? Is there significant contribution from the ATMega328P GPIO port circuitry or elsewhere?

  3. This is pretty much OK and as expected.

2 Answers 2

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The difference between 3.3 and 3.1 is simply the voltage drop across R1 due to the load of my LED and 5 K resistor.

Correct.

The RX input of the GPS module will have some similar effect.

No, the RX will have little to no effect. It is "high impedance" - looking to the circuit like a very small capacitor and very large resistor in series to ground. Negligible to your circuit.

The difference between 0.8 V and 0.0 V is mostly due to the forward voltage of the 1N4148?

Yes. The datasheet shows the forward voltage can be up to 1V.

Is there significant contribution from the ATMega328P GPIO port circuitry or elsewhere?

No.

This is pretty much OK and as expected.

It is.


Now my comments:

  • You should use a larger pullup and isolate your LED (if you want to keep it) with a BJT or MOSFET so it doesn't load the circuit. That will reduce the current and thus the forward voltage drop over the diode slightly.
  • You should use a Schottky diode to considerably reduce the forward voltage drop even further. These also switch faster allowing for higher baud rates.
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That is correct. The most reliable solution is to use a level shifter. They are inexpensive and generally come with more than 1 channel. Use one of the additional channels to drive your led. You are reducing your logic low to 0.32 volts. Typically the threshold is 40% of VCC or 1.32 volts. - 1 volt (worse case) for the 4148 uses almost all of your marjin. That can be reduced even further by loads on the pins. The level shifter is one small board, not a handful of parts to connect which increases your chance of errors but it will help your debugging skills.

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