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A cheap and easy way of interfacing a 5V device with a 3.3V device is to use a simple voltage divider.

I've read using a voltage divider to convert a fast-changing signal (such as serial lines) can result in signal distortion due to the RC filter the voltage divider creates.

How do you determine at what speed the signal will be distorted?

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It boils down to the capacitance of the input pins that receive the potted-down voltage. If the capacitance is 5pF and the resistors in the attenuator project an effective parallel resistance of 1000 ohm then the 3dB point of the low pass filter formed by 1000 ohms and 5 pF is: -

\$f_C = \dfrac{1}{2\pi R C}\$ = 31.8 MHz

You'd probably get away with data speeds of about 30M bps but this is not guaranteed across the board. Here's a circuit: -


simulate this circuit – Schematic created using CircuitLab

The logic level after the attenuator is \$5V\cdot\dfrac{2.7k}{1.8k+2.7k}\$ = 3 volts

The two resistors in parallel form an effective series resistance of \$\dfrac{1.8k\cdot 2.7k}{1.8k+2.7k}\$ = 1.08k ohms

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Andy, would this interaction be modeled with a resistor in series with another resistor in parallel with a capacitor? As in, with arbitrary values shown: i.imgur.com/iGeKw73.png – sherrellbc Jun 6 '14 at 15:57
@sherrellbc yes, that circuit but appropriate values. – Andy aka Jun 6 '14 at 16:30
It would seem the top resistor would impact this as well, not only the one in parallel. I know the calculation you presented above was for an arbitrary 1k value, but for future readers it should be noted that the circuit I presented in the link above resolves to a LPF with R(equivalent) as Req = (R1 || R2). A simple source transformation confirms this. – sherrellbc Jun 6 '14 at 17:50
@sherrellbc I used the phrase "effective parallel resistance" and this refers to the fact that the two resistors in the attenuator can be regarded as both in parallel when analysing it as a low pass filter. I'll add a diagram so that there is no confusion! – Andy aka Jun 6 '14 at 18:23

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