Re: AC current measurement with a current transformer

Fred Bloggs wrote: 


John Popelish wrote:

Fred Bloggs wrote:




I want to ask to clear up my confusion about use of c.t. to measure AC
current.
I have a need to measure 60 Hz sinewave AC current up to say 100
amps.
I have the following:
- 200:0.1 current transformer
- a DC voltmeter. Actually an Omron process meter with linear conversion y=ax+b feature:
http://tinyurl.com/882gb
I am thinking of doing this as follows:

- put a rectifier bridge in between the contacts of the c.t., like
DigiKey item DF10MDI-ND.
- Put a resistor between the + and - poles of the rectifier bridge.
- Add a capacitor in parallel with resistor to filter ripple

- Measure voltage across the cap, and use the linear conversion
feature of the meter to convert this reading to AC line votls.
Is this a sensible plan?




Absolutely not- that circuit is in effect a peak-detector and the c.t. will attempt to charge the capacitor to an indefinitely high voltage. The whole thing will explode in an instant. 



Have another cup of coffee and look again.

The secondary has a resistor across the rectifier, so the transformed current will charge the capacitor only till the average resistor current equals the average transformed load current. I have used a circuit very similar to this many times.


Current xfmr 1 ----_
                  / \
                 +   -                                              \ /
Current xfmr 2 ----~


(continued from the above diagram)

+  ----------+------+-----
             R    C===     to voltmeter
-  ----------+------+-----

Would practical issues like diode voltage drop affect this setup?

i



He shows the resistor inside the bridge- this is not the same as across the secondary.

Okay. So the current through the resistor is unidirectional while the current through the secondary is bidirectional. But the average resistor current is still equal to the average of the absolute value of the secondary current. On average, there is no place, except the resistor, for the secondary current to go.

Where do you see the resistor across the secondary? The capacitor voltage will block diode conduction until the secondary voltage exceeds it.

But the transformer is effectively a current source so it produces whatever voltage is necessary to pass current to the RC network that is proportional (by the turns ratio) to the absolute value (because of the bridge rectifier) of the instantaneous current through the primary.

The peak voltage this produces is less than what you would see if you removed the capacitor and just passed that current through the burden resistor.
.

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