SparkFun Forums

[DCward]

Hi everyone, I am trying to monitor load from 230V socket outlet to check there if the switch is on/off.
I am trying to convert 0.4490vac when it is 5A from the current transformer to 1-5vdc because of my transmitter (4-20mA)


The current transformer i'm using is CST-1005 - http://www.mouser.com/catalog/specsheets/CST-1005.pdf

I came across this thread- https://forum.sparkfun.com/viewtopic.ph ... 15#p129154

But the circuit is for 50A, I do not understand the calculation of the circuit to get 3vdc and must i use the same diode? I only have IN4007 to test out.

I tried the circuit but it doesn't work for me.

Thank you

[Dave Mueller]

In the thread you linked, the op said he's measuring the negative peaks. I know very little about op amps, but you may want to try grounding the inverting (-) input, and feeding your signal to the non-inverting input (+). He states that the gain is about 40, so .5V input would give 20V output. You'll need to reduce the gain (set by the 1.5K and 62K resistors). The 1N4007 should be OK but is overkill. Any small diode should be OK, 1N914 etc. Also note that the op amp needs + and - supplies.

[jremington]

Here is an updated schematic of the previously posted current-sense precision rectifier circuit, modified to use with a single-supply operational amplifier. Just about any single-supply opamp will work, but it was tested with an ultra low power opamp from Maxim (ICL7612, available in 8-pin dip package).

L1 is the current transformer. The gain is set by the ratio of R2 to R1 and is about 22 in this circuit. Change R2 to increase or decrease the gain. The diodes are not critical. The output is a bit less than 1 volt when no current is flowing in the mains circuit, but that offset could be changed by changing the values/ratio of the bias resistors R4 and R5.

[DCward]

Hi guys! Found out the mistake on the circuit, it didn't work because i actually never supply the op amp with -5v (sorry im new to electronics) After doing that, I used function generator to supply 1vpp to the input (replacing the CT) thinking that maybe 0.4490vac is not enough for the diode. The multimeter shows the output of the op amp is 3vdc but its not stable, the oscilloscope shows there is saturation. I will post more details when I am back to the lab on Monday.


@Dave Mueller- Thank you! Will try what you pointed out.
@jremington- Thank you! but i'm not sure if i have such op amp in the lab, if i have it i'll try it asap.

[Mee_n_Mac]

What op-amp are you using and what are your supply voltages ? Also be aware the gain may be slightly higher w/the function generator vs the CT, due to the latter's higher output resistance.

[DCward]

@Mee_n_Mac- I am using LM324 and the supply used was 5v. I will try again with CT and see the difference

[Mee_n_Mac]

The LM324's output can only go as high as Vcc - 1.7v (give or take), or in your case 3.3v. Given the diode btw op-amp "out" and circuit Vout, you're limited to about 2.7v. More gain will not increase the output. Is this the saturation you saw ? If you can try increasing the supply voltage(s) and then the gain.

http://www.onsemi.com/pub_link/Collateral/LM324-D.PDF
The upper end of the common mode voltage range is VCC −1.7 V, but either or both inputs can go to +32 V without damage, independent of the magnitude of VCC.

[jremington]

Many manufacturers, including Maxim, will give out samples for free. The ICL7612 is a micropower rail-to-rail op amp and will run off of a 3V coin cell for a couple of years.

[uChip]

I can't help but wonder if a circuit something like this might not be a simpler solution.

Current Sense.JPG

In simulation the output is a little slow and bouncy (about 5.25 cycles) but that can be fixed by adding a full wave bridge rectifier (3 more diodes). That also allows you to drop the cap down to 2uF.

Here's a snip of the (full bridge) output.

Current Sense Output.JPG

- Chip

[uChip]

Oops. That was the output with the 20uF cap. Here it is with the 2uF. Much faster.

Current Sense Output.JPG

Also, the website doesn't seem to want to let me upload the simulation file, so here's a picture of the simulated circuit. I'm modeling the transformer as a sinusoidal v-source with a series resistance such that it provides the spec'd voltage and current into the 2k load.

Sim Picture.JPG

- Chip

[jremington]

The data sheet for the OP's current transformer specifies 0.449 peak VAC when the AC current is 5 A peak, given a 500 ohm burden resistor. A diode bridge won't be very useful for that signal level, and especially not for lower signals, whereas a precision rectifier will easily handle a signal of a few mV peak.

[DCward]

@Mee_n_Mac- Maybe thats the reason, i'll try to increase the increase the supply voltage and check again

@jremington- Don't know that they provide free sample, thanks for the info!

@uChip- These circuit you tried might be useful to me. Thank you

[uChip]

I read the column to be volts per amp. That would be 5x0.449 volts peak. Upping the load to 2k gives about 6.8v total output voltage if you interpret the spec that way. Easy enough to test with a volt meter.

- Chip

[jremington]

The column is titled "Volts/Amp@ rated IP", which is remarkably unclear. Perhaps the OP can tell us what peak voltage the sensor actually outputs (for a given burden resistor), when 5 amps is flowing in the primary circuit.

[DCward]

Sorry everyone i gave the wrong information. In the datasheet. The given voltage is per ampere. At 100 ohms and 5A, the voltage across the 100ohms is 0.479vac, when it is at 1A the voltage is .0958vac. The voltage across the 500ohms resistor is 2.245vac at 5A. and 0.499vac at 1A. Does that mean i can change the output voltage of the CT if i change to a higher value of resistor?