CT: How to Obtain Sense Current, Without, or In Spite of, Burden Resistor?

Thread Starter

johnyradio

Joined Oct 26, 2012
254
This page says a "burden resistor" is needed.

"A shorted secondary winding guarantees a non-obstructed secondary current. A pure short circuit isn't feasible in practice, but a low as possible terminating resistance is the goal. This terminating resistance for current transformers is often called the burden resistor."
https://meettechniek.info/instruments/current-transformer.html

Won't that change our sense-current into a voltage?

But what if we want to keep it as current? That is, what if i want to pass the sense current as current to another stage of the circuit. How to achieve that, with or without a burden resistor?
 

panic mode

Joined Oct 10, 2011
1,831
current transformer output is current. usually there is a ratio (1000:1 for example). note that output current is proportional to measured current and it practically does not depend on connected load. for this reason it is important to have current transformer always connected to low impedance circuit (or short, when not in use). failing to do so may produce dangerously high voltage output and current transformer can selfdestruct... or shock ;-)
 

Thread Starter

johnyradio

Joined Oct 26, 2012
254
it is important to have current transformer always connected to low impedance circuit (or short, when not in use).
Won't that change our sense-current into a voltage?
When not in use? You mean, the burden resistor isn't required if we're doing something with the current?
 

bertus

Joined Apr 5, 2008
20,540
Hello,

An Amperemeter has a very low input resistance.
A voltmeter has a very high input resistance.
The burden resistor is used to have a low resistance.

Bertus
 

crutschow

Joined Mar 14, 2008
24,980
Won't that change our sense-current into a voltage?
When not in use? You mean, the burden resistor isn't required if we're doing something with the current?
The purpose of the burden resistor (or short) is to keep the output voltage from going to destructive levels.
So however you plan to use the current (which seems to be a secret), the current transformer output must always see a low resistance whenever there is any primary current.
 

Thread Starter

johnyradio

Joined Oct 26, 2012
254
they show the burden resistor in parallel



Assume the voltmeter in this image is an ammeter instead. Is it correct to say the current measured will not be affected by the burden resistor?
 

OBW0549

Joined Mar 2, 2015
3,408
they show the burden resistor in parallel



Assume the voltmeter in this image is an ammeter instead. Is it correct to say the current measured will not be affected by the burden resistor?
No, it WILL be affected by the burden resistor. Obviously the current transformer secondary current Is splits, with part of it going through the burden resistor and part of it going through the ammeter. How much of the current goes through each, depends on their impedance relative to one another.

Either use a burden resistor and measure the voltage across it, or use an ammeter. But not both.
 

Thread Starter

johnyradio

Joined Oct 26, 2012
254
Either use a burden resistor and measure the voltage across it, or use an ammeter. But not both.
Thx, ObiWan0549.

Ok, assume no burden resistor. I'm using the current sense, as current, to signal some other section of the system. Therefor, as long as that other section is putting a load on that sense-current coming out of this CT, then that section will get ALL the sense-current coming from this CT.

Correct?

Thx
 

OBW0549

Joined Mar 2, 2015
3,408
Ok, assume no burden resistor. I'm using the current sense, as current, to signal some other section of the system. Therefor, as long as that other section is putting a load on that sense-current coming out of this CT, then that section will get ALL the sense-current coming from this CT.

Correct?
Correct. If the current has only one place to go, that's the place it will go to.
 

Thread Starter

johnyradio

Joined Oct 26, 2012
254
transformer always connected to low impedance circuit (or short, when not in use). ;-)
Is there any advantage to using a burden resistor, instead of short?

I'm guessing, cuz the circuit can work normally with a resistor, but not with a short. You could theoretically switch from normal to short, if you had an ideal switch.

Is it correct to say, current traveling through the resistor sinks current from the meter--coil?
 
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crutschow

Joined Mar 14, 2008
24,980
A current transformer is actually most accurate working into a short, which is why an opamp connected as a transimpedance amp (virtual ground input) is sometimes used to convert the transformer current into a voltage.
 

Thread Starter

johnyradio

Joined Oct 26, 2012
254
A current transformer is actually most accurate working into a short, which is why an opamp connected as a transimpedance amp (virtual ground input) is sometimes used to convert the transformer current into a voltage.
You mean, a transimpedance amp looks like a short to the CT? Ok, this article says a TIA has zero input impedance. This pic says the TIA input looks like ground. So, "zero impedance" means "ground"?

A TIA converts the I to V, right? So, i'd use a voltmeter to measure the TIA out, right?

But if i want to measure the current directly, as current, i can use an ammeter or galvanometer, right? Will impedance or other factor at the input of the ammeter or galvanometer cause a misreading?

output current is proportional to measured current
Unclear. When you say "proportional", are you talking about if we convert the current to a voltage (with a TIA or resistor) and measure with a voltmeter?

Ok, but won't the current measured with ammeter or galvanometer be EQUAL to the current output from the second coil? In that case, the current output from the second coil IS the current we are measuring, right?

thx
 

crutschow

Joined Mar 14, 2008
24,980
So, "zero impedance" means "ground"?
Yes.
A TIA converts the I to V, right? So, i'd use a voltmeter to measure the TIA out, right?
Yes.
But if i want to measure the current directly, as current, i can use an ammeter or galvanometer, right? Will impedance or other factor at the input of the ammeter or galvanometer cause a misreading?
Only if its impedance is higher than the maximum allowed burden resistor value as specified for the current transformer you are using.
 

panic mode

Joined Oct 10, 2011
1,831
....
Unclear. When you say "proportional", are you talking about if we convert the current to a voltage (with a TIA or resistor) and measure with a voltmeter?
what the heck is TIA?

the entire topic is about current transformer. word transformer suggests that some value is ... changes (transformed). in other words there are two currents - input current and output current. input current is what we are trying to measure (that is the whole point). output current from current transformer is a different value but - it is proportional to input current. that means you can scale the reading and display it as input current. nobody cares what the output current really is (well, maybe for five minutes until sizing and calibration are done).


Ok, but won't the current measured with ammeter or galvanometer be EQUAL to the current output from the second coil? In that case, the current output from the second coil IS the current we are measuring, right?
yes, technically that is the current we are measuring (even though we don't care about it). it is just a way to get to input current (which is the goal).
 

Thread Starter

johnyradio

Joined Oct 26, 2012
254
"Will impedance or other factor at the input of the ammeter or galvanometer cause a misreading?"
Only if its impedance is higher than the maximum allowed burden resistor value as specified for the current transformer you are using.
Ok, so--- a galvanometer is a coil, right?
  • the meter is an electromagnet surrounded by a non-electric magnet.
  • We push current through the meter's coil, creating magnetism in its core.
  • Thus as polarity of the needle's electromagnet changes, it rotates in relation to the fixed magnet that surrounds it, right? which causes the needle to deflect.
So when we talk about the resistance of the galvanometer, we mean the resistance of its electromagnet, correct? Is that resistance proportional to it's inductance?

When measuring current directly, then it's not correct to say "the measured current will be proportional to output current". Cuz both currents are one and the same. Right?

what the heck is TIA?
transimpedance amplifier
there are two currents [at the CT]- input current and output current. Output current from current transformer is a different value but - it is proportional to input current.
Only if the CT has a ratio other than 1:1, right? If the CT is 1:1, then (assuming an ideal CT), input current and output current will be the same magnitude. Correct?
 
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crutschow

Joined Mar 14, 2008
24,980
So when we talk about the resistance of the galvanometer, we mean the resistance of its electromagnet, correct?
Yes.
Is that resistance proportional to it's inductance?
For a given galvanometer, the inductance is proportional to the the square of the number of wire turns, so the wire resistance would be proportional to the square-root of the inductance.
When measuring current directly, then it's not correct to say "the measured current will be proportional to output current". Cuz both currents are one and the same. Right?
If I understand you correctly, right.
 

Thread Starter

johnyradio

Joined Oct 26, 2012
254
inductance is proportional to the the square of the number of wire turns, so the wire resistance would be proportional to the square-root of the inductance.
Excuse my lack of math-- So generally, as inductance goes up, resistance goes up (but not a linear relationship), right?
 
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