I am using an analog sensor and would like to scale the output voltage down prior to recording the value. I am running into problems acheiving this, however. I have tried a simple voltage divider circuit but the current from the sensor output is so minute it does not function properly. Does anyone have any suggestions on how to scale voltage with very little current? Thank you,
Evan
Without all the details, a design isn't possible. But an opamp is a very common approach to this.
Jon
If the resistance of your voltage divider dragging too much current then raise it ! Try a value 10 times higher !
-- Baron:
Problem is, we don't know what the OP is hooking up to the divider to monitor the voltage and how it loads it.
Jon
Thats why I suggested it that way ! Don't 10M DMM's come in handy !
-- Baron:
The more details you provide the better the answers you will receive. What is the voltage and current the sensor provides. Better yet, what is the sensor and what is it measuring?
If this is a pH sensor, say, you need to design the input circuit correctly.
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The sensor is a MEMS biaxial accelerometer I am using to measure tilt. I am suppling 3V, the output voltage range is 1.25 V +/- 0.1 V and the current is less than 0.01 mV (I calculated it to be aprox 0.005 mV). Thanks
Current is measured in mA, not mV. Please clarify, are you talking about the output voltage or current? Also, what resistance did you use for the voltage divider?
Mark
Output voltage: 1.25 V output current: .005 mA
I used a 150 and 220 kOhm resistors.
their output is essentially a voltage source. The output current just depends on what resistance is tied to the output.
I don't know what you're using to read the signal. If you're looking at the output on an oscilloscope, that's typically a 1 M-ohm load and that will alter the reading on a several-hundred kOhm divider. On the other hand, a 10 M-ohm multimeter should have very little affect on the reading.
I would try making a divider in the 10-20 kOhm range, and see if things work better. Alternatively, if you're up to putting together an op-amp circuit, that would work best of all.
Mark
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