Voltage Divider as 3.3v -> 5v converter
SRLM
Posts: 5,045
I'm trying to interface to a device that has the following specifications:
Voltage Low Maximum 1.25v
Voltage High Minimum 3.25v
I know I should use some sort of IC, but I don't have that with me. I just want something quick and dirty, so I came up with the following voltage divider:
With this circuit I calculate the output voltages as
Low: 0.72v
High: 3.54v
Is there anything about this circuit that wouldn't work?
If it matters, I'm trying to interface to a VNH2SP30 motor driver.
Voltage Low Maximum 1.25v
Voltage High Minimum 3.25v
I know I should use some sort of IC, but I don't have that with me. I just want something quick and dirty, so I came up with the following voltage divider:
5v | \ / 3.3kOhm \ / | O----- Output | \ / 560Ohm \ / | | Prop I/O
With this circuit I calculate the output voltages as
Low: 0.72v
High: 3.54v
Is there anything about this circuit that wouldn't work?
If it matters, I'm trying to interface to a VNH2SP30 motor driver.
Comments
Hmmm. Your answer makes me think of a very similar solution: use the pullup and the I/O pin set to input to output 5v, and set the I/O pin to output 0 for a low output. That won't work in my case (I want the code to be compatible), but it's something to keep in mind.
I don't want to go that route because 0.05v is too close of a margin for me. On the Quickstart board that I'm using, for example, the 3.3v rail is 3.275v and the 5v rail is 4.71v. I don't want to be worrying about a bug's root cause being not enough voltage on the I/O pin.
-Phil
If you want margin, and CMOS drive, another way is to use a diode to replace the bottom resistor. (slow, power spec, or reverse zener)
This simply adds ~ 500mV to Vo, so gives >> 0.05v margin, and the high diode capacitance, ensures a faster switching time.
Use the largest pullup value you can.