Common wire for current transformers? Is it ok? - Page 1
Forum section: electronics / projects

author: nuclearcat
author_rank: Unknown
timestamp: January 29, 2021, 10:00:56 am
content: We have argument with colleagues if such way of optimizing possible.Right now we have 4 pin connector with 2 CT(current transformer) attached.One pin of each CT connected to separate resistor divider for 1/2 VCC offset.Another pin of CT's goes to 4051 and then to STM32L072 ADC input.Now we need to fit in 4 pin 3 CT.I suggested to make common pin for all coils - DC offset, and to add opamp for buffering this offset to drop impedance.Is it going to work?No high precision or speed required, ~1% is ok, its 50/60hz mains, i switch for duration of one waveform each pin, capture data, then switch to next one and continue...

author: mikerj
author_rank: Super Contributor
timestamp: January 29, 2021, 10:43:06 am
content: That should be fine, I presume each CT has it's own burden resistor permanently connected? Probably a good idea to add some capacitive decoupling to the divider.You you have any protection against potentially damaging voltage spikes from current transients?

author: nuclearcat
author_rank: Unknown
timestamp: January 29, 2021, 10:47:46 am
content: Quote from: mikerj on January 29, 2021, 10:43:06 amCapacitive decoupling is good idea, will do.Quote from: mikerj on January 29, 2021, 10:43:06 amYes, both pins have TVS.

author: S. Petrukhin
author_rank: Super Contributor
timestamp: January 30, 2021, 11:12:51 pm
content: Do you shift the TT to raise the sine wave to measure the negative half-wave? Do you need it? After all, in this case, it is more difficult to determine the transition through zero.I ignore the negative half-wave and put the reading from the TT ideal diode on the op-amp.

author: nuclearcat
author_rank: Unknown
timestamp: January 31, 2021, 10:42:02 am
content: Quote from: S. Petrukhin on January 30, 2021, 11:12:51 pmCan you suggest me ideal diode circuit for such measurements? It seems i'm failing to find anything suitable.All i found is either expensive, or very complicated (thus expensive too).I remember i saw one once exactly for similar measurements, but can't find it anymore.

author: poorchava
author_rank: Super Contributor
timestamp: January 31, 2021, 11:57:00 am
content: A CT works more or less as a current source, so diode drop doesn't matter that much.What OP is describing should work, although some weird noose/transient coupling may occur if the signal contains HF components (eg. SMPS switch current, brush motor current etc)For lower frequency (think <1kHz) higher precision measurements an opamp based IV converter (transimpedance amplifier) works the best. Can be implemented both as single ended and fully differential circuit. Depending on CT secondary current a discrete buffer may be required. It can be done both on a single supply and bipolar supply. If better precision is required gain ranges can be implemented by switching the gain resistor of the IV converter using an analog switch or back to back mosfets. I have implemented in this way a measurement with range up to 100A, and about 0.25% FS accuracy on 100mA range. This required shielding though.

author: nuclearcat
author_rank: Unknown
timestamp: January 31, 2021, 01:01:57 pm
content: Quote from: poorchava on January 31, 2021, 11:57:00 amThis is indeed incredibly impressive!Thanks for guidance.

author: S. Petrukhin
author_rank: Super Contributor
timestamp: January 31, 2021, 06:26:06 pm
content: Quote from: nuclearcat on January 31, 2021, 10:42:02 amIf you do not care about the loss of a small part of the sinusoid near zero, and it is possible to adjust programmatically or by adjusting the amplification, the voltage drop is small, you can use a regular Schottky diode.The ideal diode opens at a lower voltage of several mV, in the simple case it is FET in the common Source circuit.You can use a ready-made ideal diode, there are many of them, for example, MAX40200, it opens at 40 MV and has a voltage drop of about 0.05 V at low load.Or use almost any op-amp with a diode in the feedback. But please note, the ideal diode stage can not be an amplifier, it will require another op amp to amplify. See attachment1.I used some trick: I gave a signal to the op-amp with a unipolar power supply, it amplified the positive half-wave, and the negative one was walled up in the ground.In addition, you can use a current sensor such as INA.I used ACS710KLATR (https://lcsc.com/product-detail/Current-Sensors_Allegro-MicroSystems-LLC-ACS710KLATR-12CB-T_C718821.html) when implementing the function of an electric motor protection machine - it is inexpensive, quite simple, has a low-pass filter, a built-in calibrated shunt and allows not only to monitor the current, but also to perform the function of protection against excess with a latch. See attachment2

author: S. Petrukhin
author_rank: Super Contributor
timestamp: January 31, 2021, 06:38:31 pm
content: Quote from: poorchava on January 31, 2021, 11:57:00 amAll the current CT flows through the shunt, we need to measure the voltage on the shunt, and here the diode will give silence near zero and a voltage drop.