# How much current is consumed by an avr analog sensor?

I'm trying to measure a small analog device that emits about a volt (I believe up to 5V, but I've not measured at extremes yet) AC. I rectify the reading to DC, but I'm really interested in getting something a bit smoother out of it because I just want to know approximately what it's doing.

I'm looking to size a reservoir capacitor, but that requires me to know how much current is actually flowing through this circuit. I assume it's non-zero, but must be somewhat close to that.

I can compensate by just taking a few samples and picking the biggest one, but I'd like to be able to just take any given sample and have it properly represent the current state (in practice, "current" is on the "second or so" scale).

I've seen RMS ICs, but I don't really want to add another \$10 to the project when I've already got a reasonable workaround. More importantly, I'd like to understand the stuff I'm working with a bit better.

• What is the sensor? Do you have a data sheet? What are you sensing with? Do you have a data sheet? Commented Nov 6, 2011 at 22:17
• I can't find any details on the sensor. This is it: amazon.com/gp/product/B005FIFT4E -- my DMM reports around 1 volt for a 75 watt lightbulb. My arduino analog sensor sees a spike of 1023 when I turn it on, then it levels out to ~100. My new oscilloscope hasn't shipped yet. Commented Nov 6, 2011 at 22:45
• It's a current transformer, be careful with disconnecting it while current is running through primary as it can produce very large voltages at the secondary. It should be operated with a low impedance (e.g. R < 100 Ohm) on the secondary. You would need to read the secondary using a multimeter on amps, not voltage. At 3000:1 for 10A you should get ~3.3mA on the secondary. I advise you to do some research and/or get some help from someone who has some experience with CTs before proceeding. Half decent link here Commented Nov 7, 2011 at 1:50
• @Oli Glaser - probably 120A it seems. 000 suffix indicates current output version. Commented Nov 7, 2011 at 2:40
• @Russell - Ah yes, I had a quick Google but couldn't find anything - the page you link gives some more info, seems to be definitely current output with a protective diode according to the notes. I can't see the exact same model number listed (e.g. SCT-019/013-xxx as opposed to SC-016-000), although it looks pretty certain they are from the same family. It looks like some of the code may have rubbed off the one on Amazon so it may be SCT (looks to be a space where the 'T' may have been) Commented Nov 7, 2011 at 3:10

The ATMEGA328P that the Arduino runs on (assuming an UNO or similar) has an input resistance of 100MΩ on the analogue inputs:

So, at a maximum of 5V, the current draw would be:

$I = \frac{V}{R} = \frac{5}{100000000} = 0.00000005A = 50nA$

Half level would be 2.5V, so 25nA, etc.

• Note that for varying signals the required source impedance will likely be in the kOhm range due to the S/H capacitor charging time. Another way of putting it is the dynamic input impedance will probably be much lower. Commented Nov 7, 2011 at 1:55

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<=10
Maximum Current 300A(Continuous)
Applicable Frequency 50Hz-150KHz
Secondary Turns(n) 3000±2T
Secondary Impedance 280Ω±20Ω
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&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.

• 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? Commented Nov 7, 2011 at 2:54
• @Dustin - See addition to my answer. Commented Nov 7, 2011 at 3:08
• This is very informative. Thanks a lot for your time. I think I accidentally added the necessary resistor when I was trying to hook it up to my arduino. I don't care too much about accuracy (what I get now will work for this project, I just want to understand it better). I should go ahead and buy a handfull of op amps it seems, though. Commented Nov 7, 2011 at 3:18
• If the current range you wish to measure is a lot less than the rated current, you could try putting an extra turn or two on the primary (e.g. looping through again) For example 2 turns on the primary will change it to 2:3000 = 1:1500, so 1A will now produce 0.66mA on the secondary. Commented Nov 7, 2011 at 5:58
• This information is incredibly useful, but I'm going to open a new question to follow up on it because it's really deviating from the actual question I asked (which was answered). Commented Nov 10, 2011 at 3:02