How much current is consumed by an avr analog sensor?
IMPORTANT - your requirement relates more to the sensor you are trying to use than to the AVR's ADC. You MUST understand the sensor first. The following will allow you to do that. Once the sensor is properly understood the ADC interfacing can be discussed.  

Your device is a current transformer (CT).  

WARNING 


DO NOT disconnect the secondary load while the primary has current in it. 
Doing so can lead to VERY VERY VERY high voltages being produced. 


Try this:


Connect a 3300 ohm resistor across the output.
Connect an AC voltmeter across the 3k3 resistor. 
Run one wire of a powered AC circuit through the transformers core "window".
Using ammeter set to AC volts measure output voltage while powering various devices.
I'd expect you to get about 1 Volt out per Ampere of primary current.
eg a 75 Watt lamp on 110 VAC should give
V = Volts/Amp x Watts/V 
V = 1V/A x 75 Watt / 110V  =~ 680 mV.


Report back.



Your CT and afamily of similar ones is listed here . Note that the 000 on the end of the part number (partially obscured o the photo on the site you referenced) means that ot operaes in current output more. If the last part of themodel name sis NOT 000 then it is a voltage output version )see below).

To use this, or any other sensor, you need to know what it does.

A current transformer treats the wire passed through it as a single turn on a transformer primary and it has a large number of turns on the secondary. Whentermianted in the correct resistance it has an output expressed in volts (or milliVolts) per amp. When short-circuited it produces a current proportional to the turns ratio SMALLER than the primary current.



If left open circuit and the voltage output is measured the reading is rubbish,. That may be what you are doing at prsent. This is because some current transformers have a bult in resistor to convert them to a current to voltage transformer. It does not seen that that is what you have, so your readings are quite probably meaningless at present.

This appears to be a description of your transformer:


Product Name Split-core Current Transformer
Model No.  SCT-016
Applicable Current 0.01-120A(50-60Hz) RL&lt;=10
Maximum Current    300A(Continuous)
Applicable Frequency   50Hz-150KHz
Secondary Turns(n) 3000±2T
Secondary Impedance    280Ω±20Ω
Secondary Down-lead    UL-1007 Current Lead(AWG22)
Dielectric Voltage Withstan    AC1000V/1min
Insulation Impedance   DC500V/100MΩ above
Central Hole Diameter  16mm/0.629"
Overall Size(Approx.)  30.8 x 29.5 x 46mm/1.2" x 1.16" x 1.8"(LWH)
Wire Length    0.5M/23.6"
Shell Material Plastic
Color  Black, White
Weight 70g  


It is rated at 120A nominal so with a 3000:1 ratio output is 0.333 mA out per Amp in or 120/3000 = 40 mA at full current of 120 A.

It has an impedance of 280 ohms.   

You are trying to use this device at well below it's full rating so may or may not have prohblems scaling it up. 



Q&amp;A:

Q: Well this is fascinating. Nothing seems to be as simple as it seems. I do get fairly consistent values in the Arduino, but admit I have no idea what I'm doing. Right now, I have the output hooked up to a bridge rectifier to get DC out. This is the SCT-016-000 version of the product. Are you suggesting that I shouldn't be trying to read voltage from it?

A:  Life's like that :-).
  You can read voltage if you load it with a known resistor so as to use the current to produce a proportional voltage.
 ie V= IR.
 If it is 3000:1 as it seems  to be then 1 amp / 3000 = o.333 mA/A.
 So 1k would make 0.333 V/A so 3k3 would make 1V/A. 

There is some possibility that the transformer does notlike such high load resistors (as it has a 280 ohm impedance) BUT it may well be OK. Try it as per my reply. 

Ultimately a rectifier will massively reduce accuracy at the voltages concerned. You want to read AC OR use a "perfect" precision rectifier (one or two op amp sections).

Wikipedia - precison rectifier

Simple:



Improved:



Many more circuits here - these images are hot linked to related web pages.