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 Author Topic: 0-5 v into 0-2.5 volts ?  (Read 1693 times) 0 Members and 1 Guest are viewing this topic.
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 « on: April 12, 2012, 03:04:44 am » Bigger Smaller Reset

i have a board with a 0-2.5 v input (not arduino)

i have a sensor with a 0-5v output

is it possible to use a resistor to reduce the voltage ?

what sort of resisitor should i use ?
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a, b = b, a+b
 « Reply #1 on: April 12, 2012, 03:13:51 am » Bigger Smaller Reset

you need to build a voltage divider.

http://en.wikipedia.org/wiki/Voltage_divider

Look at the first picture of the page (at the right). Thats how you need to set it up. Use the first formula on the page to calculate your resistor values.

Cheers

p.
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 « Reply #2 on: April 12, 2012, 05:04:38 am » Bigger Smaller Reset

Thank you ;-)
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 « Reply #3 on: April 12, 2012, 09:09:28 am » Bigger Smaller Reset

p.
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 « Reply #4 on: April 13, 2012, 08:24:52 pm » Bigger Smaller Reset

In this case, you'd use two of the same value resistors, so you dont have to do much calculating.
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 « Reply #5 on: April 27, 2012, 12:11:46 pm » Bigger Smaller Reset

thats neat - 2 resistor values the same, reduce the value by half
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 « Reply #6 on: April 27, 2012, 12:54:50 pm » Bigger Smaller Reset

It's like splitting dinner with a friend.
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 « Reply #7 on: April 27, 2012, 02:24:30 pm » Bigger Smaller Reset

As long as your friend is a light eater.
Otherwise the current into your load overcomes the current into the bottom resistor and throws the voltage off.

With a lightload, say 2 5K resistors, Vin to R1 to R2 to ground, Vout from the R1/R2 junction:

Vout = Vin x R2/(R1+R2)
= 5V x 5000/(5000+5000) = 2.5V

However if Vout was connected the equivalent of a 1K resistor, the 5K and the 1K in parrallel act like an 833 ohm resistor (2 resistors in parallel = R1*R2(R1+R2) )
Then Vout = Vin x 833/(833+5000) = 0.71V

That's why voltage dividers are okay for signals, but not as power sources.
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 « Reply #8 on: April 27, 2012, 03:21:47 pm » Bigger Smaller Reset

I would use a pot of say 50K ohms.  The run the voltage through the pot and the wiper will be the voltage out.  It is the same as a voltage divider circuit but uses the wiper to make two resistors, one on each side of the wiper.

Something like this:
-0  |  0-

the circles are the terminals to the pot and the | is the wiper.  This is a general solution that can be used whenever you need to get less voltage then you have.

Use your VOM to see what the voltage is on the wiper and you are done.  If you want to check it you can then test the resistance between each terminal and the wiper.  And, with nothing up my sleeve, I predict that you will find that the ratio is the same as what the voltage divider formula predicted.

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 « Reply #9 on: April 27, 2012, 05:46:57 pm » Bigger Smaller Reset

You need to know the input impedance of the 2.5V circuit really - if it has a very large impedance the simple R - R divider will work, if it has a low impedance then it will become part of the divider and the voltage will drop by more than half...
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 « Reply #10 on: May 02, 2012, 04:54:21 am » Bigger Smaller Reset

so you are saying a simple resisistor will do the same trick reducing a voltage from 0-5 v to 0 to 2.5 volts

how do i calculate the correct resistor - it is a signal wire

---------------------------------------

on another thought - the board is -2.5 v to +2.5 v   -  is it easy to convert 0-5v into -2.5 to +2.5 ?

thanks for the replies
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I don't think you connected the grounds, Dave.
 « Reply #11 on: May 02, 2012, 04:56:18 am » Bigger Smaller Reset

Quote
so you are saying a simple resisistor will do the same trick reducing a voltage from 0-5 v to 0 to 2.5 volts
Only if you know the impedance of the input.

Quote
how do i calculate the correct resistor - it is a signal wire
What sort of "signal wire"?
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 « Reply #12 on: May 02, 2012, 10:37:55 pm » Bigger Smaller Reset

Simple way is take your 0-5V signal and run it thru a capacitor. Like 10uF.
If that doesn't give enough drive, then buffer the output of the cap with an op-amp with +/- power supply.
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 « Reply #13 on: May 03, 2012, 12:51:38 am » Bigger Smaller Reset

If it's from a sensor to a input on a micro,a voltage divider seems better (assuming he just wants to scale 0-5v to 0-2.5v)

How does the output of a cap give you a negative voltage?
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 « Reply #14 on: May 03, 2012, 12:58:14 am » Bigger Smaller Reset

The cap lets the signal swing back & forth across the level on the other side of the cap - in this case, the virtual ground of the op amp.

http://en.wikipedia.org/wiki/AC_Coupling

"on another thought - the board is -2.5 v to +2.5 v   -  is it easy to convert 0-5v into -2.5 to +2.5 ?"
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