Test winding in zero sequence current transformers - Page 1
Forum section: electronics / projects

author: Red_Micro
author_rank: Regular Contributor
timestamp: March 14, 2020, 02:05:38 pm
content: Sometimes it is common to see a zero sequence current transformer with four output wires, i.e., the secondary winding and the test winding as shown in the attachment. What kind of signal is the test signal? Is it like injecting a current signal that it is induced in the actual secondary? I want to perform this test with a microcontroller, so what comes to my mind is kind of a voltage to current converter for the test signal?Source of the picture: T. Novak, L. A. Morley and F. C. Trutt, "Sensitive ground-fault relaying," in IEEE Transactions on Industry Applications, vol. 24, no. 5, pp. 853-861, Sept.-Oct. 1988.

author: duak
author_rank: Super Contributor
timestamp: March 14, 2020, 07:54:46 pm
content: I'm not in the energy industry and don't have any hands on experience but was interested in your question as I sometimes diagnose industrial electrical problems. I understand the test winding is there for testing and calibration when an external source injects a test current. I expect for accuracy's sake, the CT is disconnected from the measurement system and the primary conductors. Various companies make equipment to do this:http://www.powermetrix.com/wp-content/uploads/2015/11/CT-Testing-with-the-PowerMaster.pdfHere's a few other links to look at:https://www.eng-tips.com/viewthread.cfm?qid=285242https://www.eng-tips.com/viewthread.cfm?qid=331624By the way, CTs are designed to work with a particular load resistance called the Burden. The amount of current in the secondary is a function of the primary current, the turns ratio between the primary & the secondary and the Burden resistance. The test winding will develop a voltage that is a function of the secondary voltage and the turns ratio between the two windings. I have a couple of industrial CTs and a 270 degree dial type meter. If memory serves, the secondary current is 5 Amps AC. Note also, that if the burden is disconnected, hazardous voltages can be developed on an open circuited CT secondary.

author: Red_Micro
author_rank: Regular Contributor
timestamp: March 15, 2020, 01:50:10 am
content: What do you think of the idea below? A PWM to a low pass filter to get a sinusoidal wave and the amplifier with the FET to inject a 20 mA 50Hz to the test winding?

author: duak
author_rank: Super Contributor
timestamp: March 15, 2020, 05:46:54 pm
content: Red Micro,Perhaps you could give us more info on what you're trying to do here. I don't have access to the IEEE paper and can only conjecture on what you are trying to do.Is this circuit supposed to be connected while the CT is part of the electrical measurement system? ie., are you trying to test the CT in situ? The voltage developed across the test winding is related to the Burden voltage and the turns ratio between the secondary and test winding. During typical operation there should be minimal voltage on the test winding but during a fault condition, there could be a large voltage developed there. This could damage the driver because it is likely a high voltage from a low impedance source. I would consider a relay to isolate the driver from the test coil except during test mode.The driver will have a DC bias. At such currents it's unlikely to cause the CT core to saturate but I think you will find it difficult to drive enough AC signal into the test winding to get a sensible signal from the secondary. Here's a link to an article on typical industrial strength CT burden calculations:https://sdvelectrical.blogspot.com/2019/05/how-to-calculate-burden-of-ct.htmlIf you're injecting 20 mA into the test winding and the turns ratio is 1:10 you will get 2 mA in the secondary. If the burden is one ohm you should see 2 mV. This should be recoverable if you're using synchronous detection ie., a Lock-in Amplifier. I would consider a more powerful driver, perhaps an H-bridge driven from the PWM port.

author: Vovk_Z
author_rank: Super Contributor
timestamp: March 15, 2020, 06:21:11 pm
content: Quote from: Red_Micro on March 15, 2020, 01:50:10 amThere will be a large DC component so it must be AC current source only.If you need 20 mA there is no need in PWM with MOSFET. Many opamps can do 20-50 mA, even up to 150 mA.

author: floobydust
author_rank: Super Contributor
timestamp: March 15, 2020, 07:05:16 pm
content: The test winding is for functional safety, to simulate i.e. a ground fault by injecting current into the core, and then seeing if the rest of the system detects it.You drive the test winding with a H-bridge to give AC excitation, and then disconnect after so you are not loading the core and interfering with the main winding/core. An even number of pulses so as to not magentize the core.

author: Red_Micro
author_rank: Regular Contributor
timestamp: March 15, 2020, 07:27:53 pm
content: Let me try to be more clear on what I'm trying to do:My current transformer has a 1000 turns secondary winding, 1000 turns test winding, and of course one turn in the primary. So the ratio between secondary and test winding is 1:1, and from primary to secondary is 1000:1. This mean the current I inject in the test winding will be the same in the secondary winding. The test function is not always active, compared to the secondary winding which has a burden resistor and is being continuously monitored with an ADC. What I'm trying to do is generate a detectable 50Hz AC current (something between 30 uA to 10 mA) with the microcontroller to perform the test function.The CT detects current as low as 10 mA in the primary. Let's say we have 20 mA in the primary, we have 20 uA in the secondary. So that's why even 20 uA would be enough as the test signal (I may be wrong).So I came up with the circuit attached below. The idea is to generate a PWM and make it pass through a low pass filter to get a sine wave. Added the capacitor at the output of the opamp for ac coupling and a resistor to limit the current.I hope you have a better idea of what I'm trying to do so you can give me suggestions.

author: Kleinstein
author_rank: Super Contributor
timestamp: March 15, 2020, 07:45:10 pm
content: The last circuit looks good. To avoid added noise one may think about some extra way to turn off the source.

author: Red_Micro
author_rank: Regular Contributor
timestamp: March 15, 2020, 08:02:06 pm
content: Quote from: Kleinstein on March 15, 2020, 07:45:10 pmI can program the PWM to give at most 2 cycles maybe.

author: Vovk_Z
author_rank: Super Contributor
timestamp: March 15, 2020, 10:45:01 pm
content: Last circuit is voltage source but not current source. It depends on R2 value of cause.Large enough value can make it work similar to current source but accuracy of it's Voltage-to-Current conversion is under question.True current sources usually are done by using Howland current source (current pump) chematics (example is in the Art of Electronics, H&H).

author: floobydust
author_rank: Super Contributor
timestamp: March 16, 2020, 12:04:12 am
content: That circuit would not work, it always presents a load to the test winding as C2 & R2.You want to open-circuit the test winding drive otherwise you wreck the CT's sensitivity because it always sees R2 loading.I use square-wave DC drive excitation with tri-state tiny logic74LVC2G125.Normally your main winding has an RC filter to deal with EMI so the (test) square-wave fast edges are not a big problem. Although some CT's have a ton of inter-winding capacitance.I also add transient protection as the winding will generate high voltage under fault conditions, and when the load contactor turns on/off as all 3-phases are never simultaneously switched. I use no burden resistor across the test winding.

author: Red_Micro
author_rank: Regular Contributor
timestamp: March 16, 2020, 11:26:12 pm
content: Quote from: floobydust on March 16, 2020, 12:04:12 amOk, Thanks. So to fix the last circuit, I have to open-circuit the test winding drive, right? Would it work to set the micro port at high impedance? I forgot to mention that the test winding has a dc resistance of 80 ohms. I have to be careful on the max current capability of the micro or opamp.

author: Circlotron
author_rank: Super Contributor
timestamp: March 17, 2020, 12:02:12 am
content: Quote from: Red_Micro on March 15, 2020, 01:50:10 amA better method might be to use an audio power IC instead of the opamp + mosfet combination, but same general principle, i.e. feedback from that current sense resistor.Edit -> If you only need 20uA test signal then an opamp all by itself would be plenty. No mosfet needed.Edit 2 -> oops! 20mA not 20uA. Anyway, still achievable with the right opamp.

author: floobydust
author_rank: Super Contributor
timestamp: March 18, 2020, 07:19:21 am
content: I would say you can't drive the test winding directly with an MCU unless its substrate diodes can withstand overload with CT saturation under an unbalanced fault, like a line-ground short at the load. The CT can then see >100's amps or more depending on the size/voltage of the system, until a breaker trips. So the test winding circuit can see a surprise. The two drive resistors do lessen the surge current but I'll add clamp diodes. Tri-stating the MCU outputs would work.The design I did used H-bridge '125 from 3.3V with 1k ohm drive to a 100:1 test winding around 4 ohms, I think giving around 3mApk drive but it was a high current trip test.

author: Red_Micro
author_rank: Regular Contributor
timestamp: March 19, 2020, 10:23:26 pm
content: One additional question... The resistance of the winding (measured with a meter) is about 80 ohms. As the turns ratio between the test and secondary winding is 1:1, would the burden resistor across the secondary represent a load for the test winding?

author: duak
author_rank: Super Contributor
timestamp: March 19, 2020, 11:51:12 pm
content: Red,Yes, the burden resistor will present a load to the test winding.

author: Red_Micro
author_rank: Regular Contributor
timestamp: April 03, 2020, 10:22:34 pm
content: Quote from: floobydust on March 18, 2020, 07:19:21 amDo you have a reference schematic that you may want to share?

author: Red_Micro
author_rank: Regular Contributor
timestamp: April 03, 2020, 10:32:42 pm
content: Quote from: Red_Micro on April 03, 2020, 10:22:34 pmDoing some research I found the circuit attached below. In that circuit they use a dual power supply, but I don't think I need dual. I just would need to feed this circuit with the signal coming from the micro.