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Current transformer range & sensitivity

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Jonesy

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Anyone heard of a current transformer able to measure below 1A? I've been searching for 0-10A range but only manage to find 1-10A, nothing below that
 

what's your voltage? If it is low, use resistors.
 

Current transformers can be designed and built to measure any level of AC or pulsed DC current, if you are seeking a current transducer to measure DC currents below 1A, these are mainly hall effect types and not DC compensated DCCT's (DC current transformers).
Regards, Orson Cart
 

AC current 0-10A 240V single phase. Resistor wattage rating would be too big in this application as I need a sensitivity of 5mV/5mA. 1 ohm resistor can do this sensitivity needs to have a wattage of 100W.
 

I've been searching for 0-10A range but only manage to find 1-10A
Click to expand...
1-10A doesn't sound plausible, for an AC CT, that is basically a linear passive device (not considering possibe saturation. It can measure 0 current as well.

I notice, that e.g. Talema has a similar specification like 1 - 10 A. In my understanding, the number refers to reasonable nominal currents achievable with this transformers.

If you use the transformer with lower currents, the secondary voltage will be uncomfortably low.

Key parameters of CT are:
- frequency
- maximal current
- current ratio
- required termination
 
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    Jonesy

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what is the freq? current range?, output voltage range?
 

Thus if i bought a 1-10A, how would I make it useful at low currents too?
Yes, I have checked the frequency and maximal current. Current ratio not too much to worry about ya? Just the equation later will be changed. But i think the higher current ratio will correspond to a higher sensitivity?

---------- Post added at 20:20 ---------- Previous post was at 20:19 ----------

AC voltage 240V single phase at 0-10 A, 50/60Hz.

output voltage range must be conditioned to positive 0-5V (to be measure by a MCU)
 

I'm sure it would be possible to make a AC only CT that could do 0-10A. You would need a pretty high magnetizing inductance though, especially if you want high sensitivity. 5mV/5mA is very sensitive, so I doubt you'd be able to get that.
 

An OP circuit with active rectifier is required anyway to generate an DC output voltage. You can also design the input stage as CV converter to increase the sensitivity.
 

any example circuits? whats an active rectifier? and CV converter?
 

If you have a 10amp rated CT you can put ten turns through the CT for your 1amp current to be measured, if the CT is 500:1 ratio, this will give 20mA rms on the output for the 1A in the ten turns on the primary, you say you require 5mV per 5mA, which is 1V per amp, 1 volt divided by the 20mA output gives 50 ohms as the burden resistor, rated at 20mW, so an 0805 would suffice, to get a DC signal you would still need to pass this through an op-amp fullwave rectifier,and filter as per FVMs comments below.

If you need the full 10 amp range, one primary turn on the CT will give 20mA for 10A, in order to get a 10 volt signal for 10 amps through the CT, you will need to amplify the signal from the CT burden resistor before rectifying and filtering.
Hope this is helpful,
Regards, Orson Cart.
 
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    Jonesy

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but the hard part is I have to translate those voltages into 0-5V DC to be measured by the MCU. thus, 10V would not be a choice for me.
If I add the turns on the primary, I get more sensitivity which in turn causes me to lose my range(at 10A it would give me 10V).

But it is a helpful information I can double the sensitivity if i add the turns so I can proceed to buy the 10A c.t without worrying about currents below 1A.

So now, the problem is the range, I have heard about the programmable stage gain, auto-controlled by the MCU, thus dividing 2 ranges for exp 0-5A, 5-10A ?
 

The below is intended to get you familiar with the options available to you, if a 10V range is too large, the burden resistor can be halved to give 5V rms, given that you may need to rectify and averge/filter the signal, it will be easy to scale the required o/p with a gain switch on an op-amp or more simply using a resistive divider.
Regards, Orson Cart.
 

so your current is 10A that is RMS value. then your peakcurrent is 10*sqrt(2) = 14.142 A ;
with this let us come to design part;
suppose you have 1:500 current transformer(CT) then the secondary current is 14.142/500 =28.284mA (peak);
you need 5V output, then r=V/I= 5V/28.214mA = 176.78 ohms, so you put this resistor in secondary, this set up gives 5V(AC Peak) across resistor for 10A input, then use any precision(active) rectifier. this rectifier output gives 0VDC to 5VDC for 0A to 10A.
 

you need 5V output, then r=V/I= 5V/28.214mA = 176.78 ohms
Click to expand...
Correct according to ohms law, but possibly not permitted by the CT specification, that e.g. requires a shunt below 50 ohm.
 

thanks for information...
then use 50 ohm resistor for termination and give gain of 3.5356 for output voltage to reach 5 volts for 10Amps
 

Thanks for the helpful infomation but the real problem is my min output. I need 5mA sensitivity.

5/5mA = 1000 bits. My ADC can support only 1024 bits.

And I need negative currents so 1000 bits x 2 right (because I need to make the negative currents into positive currents too?
 

The magnetizing current (V/wL)is a error in your measurement so you will not get a 5mA accuracy. (but you may get 5mA resolution ) ..
if you need more accuracy you should use more turns ratio.
How could be a problem when you have more resolution than you required?
anyway, if want to use exactly 1000 bits only, the scale the properly (10 A for 1000bits, so you volt will be a little less than 5V for 10A )
 

so your current is 10A that is RMS value. then your peakcurrent is 10*sqrt(2) = 14.142 A ;
with this let us come to design part;
suppose you have 1:500 current transformer(CT) then the secondary current is 14.142/500 =28.284mA (peak);
you need 5V output, then r=V/I= 5V/28.214mA = 176.78 ohms, so you put this resistor in secondary, this set up gives 5V(AC Peak) across resistor for 10A input, then use any precision(active) rectifier. this rectifier output gives 0VDC to 5VDC for 0A to 10A.
Click to expand...

The above is true only if the measured current is sinusiodal and your precision rectifier is also a peak rectifier, there will need to be a decay on this, so the time taken to read currents accurately as they fall may be several 10's of cycles. If your ADC can handle bipolar signals then no need for an op-amp rectifier, but you may want to sample several times over an half cycle to get an accurate actual end reading, or sample in synch with the peak of the mains.
Regards, Orson Cart.
 

Orson Cart said:
The above is true only if the measured current is sinusiodal and your precision rectifier is also a peak rectifier, there will need to be a decay on this, so the time taken to read currents accurately as they fall may be several 10's of cycles. If your ADC can handle bipolar signals then no need for an op-amp rectifier, but you may want to sample several times over an half cycle to get an accurate actual end reading, or sample in synch with the peak of the mains.
Regards, Orson Cart.
Click to expand...

yes you are correct...
that calculation is only valid for sinusoidal current... sine freq is 50Hz, it is not required to use peak detector. we can sample directly....
 

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