how do i find the input impedence ?
It'll be in the datasheet.
Which device are you using?
Which takes me right back to my original reply:
"In this case, you'd use two of the same value resistors, so you dont have to do much calculating."
A micro input sure seems like it'd be high impedance.
Two 10k resistors. This will scale 0 to 5v to be 0 to 2.5v.
db2db:
Which takes me right back to my original reply:"In this case, you'd use two of the same value resistors, so you dont have to do much calculating."
A micro input sure seems like it'd be high impedance.
Two 10k resistors. This will scale 0 to 5v to be 0 to 2.5v.
What he said.
If your not "powering" something then 2 equal value resistors should do it.
I should clarify:
Two 10k resistors wired as a voltage divider, as mentioned earlier in this thread.
hi guys i tried a test today with two 22 ohm resistors
i wired up a 5v voltage regulator and measure the power out
just over 5 volts
then i added the two 22 ohm resistors and the voltage went down to 2.75 volts
not quite 2.5 volts but good enough
now i rewired the circuit
this time a pressure sensor was added and powered from the 5v regulator
the output with room pressure was 1.22 volts
then i added the two 22 ohm resistors
I expected to see approx 0.6 volts
but i saw 0.18 volts ?
the voltage was not powering anything just going to my multimeter
Is the problem caused by using low value resistors, so the resistance in the wires of the multimeter could upset the reading ?
confused ;-S
Is the problem caused by using low value resistors
Yes
so the resistance in the wires of the multimeter could upset the reading ?
No.
5 volts into 44 ohms is over 100mA
The output of the pressure sensor isn't designed to sink power and should be connected to a high impedance input, like an Analog read pin. I'll read back, but what is it you are trying to do? why do you need to divide the sensor output? If you must, then use two 1M resistors.
i am using the output of the pressure sensor into an analogue input on a board
Gadget999:
hi guys i tried a test today with two 22 ohm resistors
Two 10k resistors wired as a voltage divider, as mentioned earlier in this thread.
why would it work with two 10k resistors and not with two 22 ohm resistors ?
Most likely the sensor can't source enough current to keep the voltage up. If the max output of the sensor is 5v, then 5/44 = 0.11363 or ~114 mA.
Check your spec sheet for the sensor. I'm betting the max output current is around 20-40mA. Because the sensor can't maintain the current draw, the voltage will drop in an attempt to raise the current to the demand. Eventually (if not already) you will damage the sensor.
Using a higher voltage divider, 10K + 10K = 20K means you will draw less current from the sensor. 5/20000 = 0.25 mA draw. The voltage will remain more stable.
Thanks for the advice - that makes sense !
Too much load is pulling it down.