3
\$\begingroup\$

I want to (eventually) build myself a home power monitor using basic current transformers. While reading on various circuit designs, I've stumbled upon what must be a very basic question. If we look at the circuit diagram

here:

my question is:

What's the effective difference between the burden resistor and the voltage divider portions?

1) From my (clearly limited) understanding the burden resistor exists to put some meaningful load on the CT secondary - and in the process according to Ohm's law (?) it defines the ratio of the voltage across that resistor to the current in the secondary. Is this correct? -- a higher value resistor here will raise the voltage on the secondary, so you choose the resistance to get an appropriately-scaled voltage for your measuring tool... yes?

2) if I have that part right (and that's a big 'if'...) then why would I also need a separate voltage divider in the circuit? Can you not just define your max voltage by the resistance of the burden?

3) bonus question: what is the purpose of C1 in this circuit - and why is there no value specified???

\$\endgroup\$
4
\$\begingroup\$

Here's your circuit: -

enter image description here

R1 and R2 divide the 5V dc supply to give you 2.5V and 2.5V is the mid-position of your ADC input. C1 provides an element of stability to that midpoint preventing 5V dc power supply noise affecting your readings.

The 2.5V lifts the ac voltage from your burden so that it appears as a waveform centred about 2.5V dc - this is what your ADC wants to see.

I'll also add that if you are wanting to measure power in your establishment/home then you need to measure line AC voltage too.

|improve this answer|||||
\$\endgroup\$
  • \$\begingroup\$ thanks, this is helpful. I'm not understanding here why we need the 5v signal FROM the Arduino in this case - isn't the voltage simply derived from the CT/burden resistor? And the voltage measurement is a future project.. right now I have about a month worth of minute-by-minute data and the voltage range is tight enough that I can get +/- 1% accuracy on better than 99% of intervals just by hard-coding the median measured value. \$\endgroup\$ – ljwobker Jan 24 '14 at 18:24
  • \$\begingroup\$ Your ADC needs to be fed with 2.5 volts. Then the ac signal from the burden resistor superimposes in top. This maximizes your Adc range. Do you understand that? \$\endgroup\$ – Andy aka Jan 24 '14 at 18:43
  • \$\begingroup\$ Yes, that makes more sense now. If I'm using an ADC with a different max input voltage (say 1.8v as an example) then I would want to get a waveform centered on half of that (0.9v) - correct? Which just means selecting different values for the voltage divider resistors? And in theory could you do this without tying the power supply circuit to the CT circuit? Or is the whole point that the CT circuit alone results in a waveform that has negative relative voltages and therefore the ADC can't read it? \$\endgroup\$ – ljwobker Jan 25 '14 at 17:58
  • \$\begingroup\$ The CT output rises above and below the centerline voltage you create with the resistors. You need to connect it to somewhere on your circuit and using the resistors is the simple way of doing it. \$\endgroup\$ – Andy aka Jan 25 '14 at 18:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.