How to fake a CT input

Discussion in 'General Electronics Chat' started by EB255GTX, Jun 23, 2016.

1. EB255GTX Thread Starter Active Member

Apr 30, 2011
54
2
Hi all,

I think this is very simple but I have thought about it too much and confused myself....

I need to test a project that has a CT input. For practicality reasons i would like to make a test fixture with knows that set the faked AC current.

I thought this would be easy, just connect an AC/AC transformer type wall wart with a low voltage output to a potentiometer and connect the resulting variable AC voltage to the CT input, right?

But then i started thinking about it ....There is a 1R burden resistor on the CT input, and the CT itself is a current source that drives a voltage across that resistor representative of the current.

So if i connect a variable AC voltage in place of the CT, it has to be able to drive enough current to raise the voltage over that 1R resistor... but i wanted to use say 100k pots and have effectively tiny power used everywhere....but then again when the real CT is in place there is not significat power being dissipated - the burden resistor is a small SMD part!

Where am i going off the rails on this thinking???

2. Hypatia's Protege Distinguished Member

Mar 1, 2015
3,134
2,085
Assuming CT expands to "Centre-Tapped" in this context (Granting that some elements of context allow for an expansion of "Current Transformer")

Unless analysis of resistive divider behaviour in this regard is part of the exercise -- you may save yourself much 'agro' via use of a reactive (Spec inductive) divider

Best regards
HP

PS
Inasmuch as there seems to be a slight language barrier - a diagram would be most helpful!

3. EB255GTX Thread Starter Active Member

Apr 30, 2011
54
2
thanks for the reply - "There is a 1R burden resistor on the CT input, and the CT itself is a current source that drives a voltage across that resistor representative of the current."

it's definitely a Current Transformer

I want to put a signal into points marked A and B that is adjustable with a potentiometer and looks like the CT signal.

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Last edited: Jun 23, 2016
4. ErnieM AAC Fanatic!

Apr 24, 2011
7,953
1,826
Then you need an AC source capable of driving the one ohm resistor to the voltage level you require. I might suggest an audio amplifier IC may be ideal for this.

5. Hypatia's Protege Distinguished Member

Mar 1, 2015
3,134
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Indeed!

@EB255GTX -- Thanks for the clarification! -- FWIW I concur with @ErnieM --- It's merely a matter of impedance matching (q.v.)

Best regards
HP

ErnieM likes this.
6. Alec_t AAC Fanatic!

Sep 17, 2013
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How much current are you expecting from the CT output winding?
What is the frequency?

7. EM Fields Active Member

Jun 8, 2016
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154
Impedance matching doesn't matter as long as the source can drive the one ohm load to the desired voltage.

8. EB255GTX Thread Starter Active Member

Apr 30, 2011
54
2
Thanks for the replies, after sleeping on it and reading the replies i really think my mind was making this more complicated than it is....

the CT will push current though the burden to develop a voltage across it (about 70mA max, 50Hz). so i can fake the CT signal with a source that can do up to 70mA ..... for some reason i was thinking that a voltage source like a bench supply wouldn't work but if i set the voltage then control the current on the supply it should look to the CT input no different than the real thing

9. Hypatia's Protege Distinguished Member

Mar 1, 2015
3,134
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...Which, for the purposes at hand, is descriptive of a matched condition

Very best regards
HP

10. Hypatia's Protege Distinguished Member

Mar 1, 2015
3,134
2,085
Be advised that only the AC (i.e. time-varying) component of a 'signal' may be transformed -- hence pure DC is inapplicable to your circuit -- unless, of course, you are endeavoring to 'bias' a saturable core? (such being inconsistent with the attached schematic).

'Voltage' and Current are not independently 'controllable' across and through a fixed impedance

If, as it seems, your aim is simulation of a 'load condition' on the 'primary', please be advised that (Re: the transformer in question) the primary current will be much greater than the secondary current (as per Isec/Ipri = Npri/NSec) thus for a (conservatively) estimated turns ratio of 1:200, a secondary current of 70mA will require a primary current of 14 amps!

Best regards
HP