My basic problem is that the CT outputs 0-333mV corresponding to 0-50Amperes. This is fine and good and gives me 6.66mV per Ampere which I have verified. However, when I try to rectify this to a DC voltage I'm getting all kinds of wild readings. I suspect my full-wave bridge is not going to allow me to do this since it has significant voltage drop. I also suspect I may have to put a cap in to provide some level of wave filtering before putting it into my A/D conversion.

Has anyone done this, and what is the best way?

Chuck

 

Rectifying did not work because the diode forward voltage is higher about 600mV. You are operating in the off or nonlinear region. Amplifying it 10 X would help a lot.

Try a true rms to dc converter. Visit this link:
http://www.analog.com/Analog_Root/s...vel3%3D%252D1%26resourceWebLawID%3D13,00.html

oj

 

I am somewhat familiar with the AC current sensor you described. I think the problem is that there are a number of small "gotchas" presented by this device.

First, the output represents the instanteous load current, which probably is not sinusoidal. Second, the output is 333 millivolts RMS, but remember RMS is the area under the curve, and not the peak amplitude. Third, the output impedence of this device is very low (it is essentially a CT with a shunt resistor permanently wired across the secondary winding.)

So, if you use an ideal rectifier with a filter capacitor, you will only see the peak current amplitude. If the current waveform is not symmetrical, you will see only the highest amplitude peak if you use a full wave rectifier. But, if you use a half wave rectifier, you will see only the peak amplitiude of one half of the cycle.

A better way to know what is going on is to use an RMS to DC converter on the output of this sensor. There are a few RMS converters available as ICs (Linear Technology, Maxim, and Analog Devices come to mind), and I am sure there are other ways to achieve the same results. Perhaps you could try an absolute value circuit (an ideal full wave rectifier) followed by an A to D converter followed by a processor doing numerical integration over some definite number of cycles. This numerical integration would represent the area under the curve for that number of cycles, and would be proportional to the RMS current drawn by the load.

If you integrate over 6 cycles, you can make a display that updates 10 times a second (on 60 Hz power.)

email to [email protected] but this mailbox is often full, so retry if necessary.

 

Google for active rectifier or precision rectifier
or go to the National Semi site, etc. for example
circuitry. Your analysis is quite correct, you would
need to amplify your signal to about 50 Vpk before the
rectifiers would not be a significant source of error.
The active rectifiers put a diode in the feedback path
of an op amp and since there is no negative feedback
until the diode conducts, they amplify open loop until
conduction takes place. This produces a near perfect
rectifier within the op amp constraints and is widely
used in instrumentation.

Regards

cww

 

> Chuck,
I would rectify this AC first with a bridge, and then filter it with a Capacitor, which must be paralled by a resistor, in order to follow the input.
This DC you can use in various ways, for example, if you need this information any distance away from the CT, you could use a 4-20mA current loop.
This interface will translate the DC to 4-20mA, and makes it suitable for transport over long distances.
If you need more details, contact me on [email protected]
Best regards Henk,
Belgium.

 

Woah! What you are doing is dangerous! A CT outputs a current proportional to the current flowing through the primary conductor, not a voltage. In your case a 50:1 CT would output up to 1 amp. This would typically feed a meter armature. The voltage you are measuring is a product of the impedence of the device to which the CT secondary is connected (Ohms Law). You should never open the circuit on a CT when any current is flowing in the primary as lethal voltages will develop at the secondary leads. Voltage will rise to try to push the required amperes. If you are luky enough to not get elecrozapped then the voltage can easily exceed the insulation rating, making lots of smoke. When you put the rectifier on you added load to the secondary, causing the secondary voltage to increase in order for the CT to maintain the same current flow, and you are really lucky you didn't burn something up if you attached the rectified DC output to a high impedence load. If you need a DC signal then you need to place an appropriate signal conditioner there, not just a simple rectifier. I usually buy off the shelf solutions from www.ohiosemitronics.com. There is also a lot of info on the web site about proper use of CTs. I suggest you read it before you get hurt.

 

Chuck
Before you hurt yourself or damage some equipment, go buy CT' switch have 4-20 ma outputs, your problems will be completely resolved and you will not get hurt, wha tyour doing is dangerours at best! With the 4-20 ma output you can use a 250 ohm resistor to get a voltage level suitable for a A/D convertor.

matt hyatt
technical consultants
[email protected]

 

hi friend

i am also looking for a similar thing regarding electronic MCb.

i hav got a ckt using OPAMP with diode as feedbak path. using capacitors and resistance so that R.C > 10 T (T is time period of AC signal), we can have a peak detector for such smal ac signals.

if u have found some other ways do reply at my id.

randeep chaudhary