Voltage Divider in FPS tutorial

Im working with a fingerprint scanner…following this tutorial:
http://www.homautomation.org/2014/10/11/playing-with-finger-print-scanner-fps-on-arduino/

They call for a 1k R and a 560R. I dont have a 560R so I was wondering if I can use differnt combinations but keep the ratio, so:

1,000/1,560 = 0.64
560/1,560 = 0.36

What worries me is that it changes the current as well, right?

Cause if I change the 560 for a 1k, which I do have, I would have to get…

1,778/2,778=0.64
1,000/2,778=0.36

with a 1.7k and 2.7k R.

Should I just use 2 R of 220 in series instead?

Do you have a 220 + a 330 or 470 + 91 or 390 + 150 or 3 X 180 or 2 X 270 or ......? Get my drift? :slight_smile:

2 1kΩ in parallel → 500 Ω; probably close enough.

What worries me is that it changes the current as well, right?

Right. The bigger the current then the faster the signal will be. Well in fact actually the bigger the current the less the signal is slowed down. So with a low current the inevitable stray capacitance in the circuit will slow down the signal rise time which may give you problems.

OK I went with 2-1k.

Grumpy Mike

Out of curiosity, could explain what you wrote in another way. It sounds interesting.

Logic signals have risetimes of a few nanoseconds. At that speed the stray capacitances
of PCB traces have impedances of the order of a kilo ohm. So large value resistive dividers
simply act as RC low-pass filters and slow the logic signals down.

For a 96kbaud serial signal this is pretty irrelevant, but for a 10MHz SPI bus its crucial.

MarkT nailed it. Unless it is the concept of stray capacitance that is worrying you.

All wires all have a small but sometimes significant coupling with other wires due to the capacitance between them and all other wires. Also all wires are also inductors only very small value ones. These are only a problem as the frequency of your signal increases.