For a project I'm making I need to control a chip that works on 3.3V and my PIC who sends the control signal works on +5V so I need to downgrade this voltage so that the signal is not to high for the chip.
I did this with a simple voltage divider, but because my chip needs to have a certain default state I also need a pull down resistor to provide this.
Is this the right method how I did this? Or will this cause problems?
Best Answer
You seem to have the values wrong for what (I think) you want to achieve. There's no point in putting two resistors in parallel, might as well use one of an appropriate value.
If you want to drop 5v to 3.3v, then a 'top' resistor of 1.7k and a 'bottom' resistor of 3.3k (both conducting 1mA when on) is an easy to calculate ratio. Make the top one a standard 1.8k and you get slightly less than 3.3v. You could make them 10x that, 18k and 33k, if you want to take less current, as long as 33k is low enough for your 'pull-down' duty.
However, you cannot combine a pull-down function with a driven function, they are mutually exclusive. If, on the other hand, you disconnect the S terminal, or send it high impedance, then the resistor to ground will behave as a pull-down. So in that sense, the voltage divider is the pull-down, when the S point is not driven. When it is driven, the output is either 0 or 3.3v depending on whether S is set high or low. Do you see why now I say the two functions are mutually exclusive.
When drawing schematics, it's always worth using component numbering, the 'reference designators'. It's much easier to refer to R1, R2, than to 'the bottom one, no, the other bottom one'. Values may change with circuit refinement, so 'the 10k one' is not much better.