Hey guys,
I plan on building a current sensor with a TT 100-SD.
I tried to follow this tutorial How to Build an Arduino Energy Monitor - Measuring Mains Current Only — OpenEnergyMonitor 0.0.1 documentation
But the EmonLib isn't really compatible with my TT 100-SD and the datasheet for my ct sensor isn't rich.
Any idea on how to build this ?
best regards,
gerko13:
But the EmonLib isn't really compatible with my TT 100-SD and the datasheet for my ct sensor isn't rich.
What makes you say that.
The TT100-SD looks like any other current transformer with external burden resistor.
Just what the article describes.
Leo..
When I actually tried, all the results i've got where off the charts as I was using an amps clamp to couter-check the values. I could change the 111.1 setting but there's no info on the turns, so I can't make a good calculation.
gerko13:
When I actually tried, all the results i've got where off the charts...
What was the value of the burden resistor.
I calculate about 53 ohm for a peak/peak voltage of 5volt (5volt Arduino) with this 100A/33.3mA CT.
33 ohm is recommended in the article, which is a good start.
Leo..
EO=IP∗RBTR where EO is the output voltage, IP is the current that will produce the voltage you want, TR is the transformer's turns-ratio, and RB is the burden resistance, in ohms.
Wouldn't the voltage across the burden resistor be 0.033A * 33 Ohms (1.089V) rms for IIN = 100A ?
Yes, and using the 1.1V internal AREF would yield an ADC result of about 1012.
JCA34F:
Yes, and using the 1.1V internal AREF would yield an ADC result of about 1012.
No, this is AC, not DC.
raschemmel:
Wouldn't the voltage across the burden resistor be 0.033A * 33 Ohms (1.089V) rms for IIN = 100A ?
Yes, which is ~3volt peak/peak for the 5volt pin, or 3/5 of the range of the A/D.
Working backwards gives you a burden resistor of about 53ohms for 100A>5volt peak/peak.
Leo..
Yes, which is ~3volt peak/peak for the 5volt pin, or 3/5 of the range of the A/D.
Working backwards gives you a burden resistor of about 53ohms for 100A>5volt peak/peak.
By "working backward" you mean :
Let VP-P = 5V
Irms = 0.033 A (33mA) (@100A)
VPeak = VP-P /2
= 2.5V Peak
Vrms = VPeak /1.41421 (SQRT(2))
= 2.5V/1.41421
= 1.7677 Vrms
RBurden = Vrms/Irms
= 1.7677/0.033A
= 53.5 Ohms
RBurden = 51 Ohms (Nearest Standard Value)
?
Wawa:
No, this is AC, not DC.
Yes, which is ~3volt peak/peak for the 5volt pin, or 3/5 of the range of the A/D.
Working backwards gives you a burden resistor of about 53ohms for 100A>5volt peak/peak.
Leo..
OOPS, ignore me! :-[
thank you guys for all the answers.
So following this, and your responses, instead of ""current constant = (100 ÷ 0.050) ÷ 18 = 111.11"" 111.11
I should use : current constant = (100 ÷ 0.050) ÷ 53.5 = 37,383 // 51 = 39,215 ?
Thanks !
So following this, and your responses, instead of ""current constant = (100 ÷ 0.050) ÷ 18 = 111.11"" 111.11
I should use : current constant = (100 ÷ 0.050) ÷ 53.5 = 37,383 // 51 = 39,215 ?
This response is really meaningless without defining any of your terms.
You have made no effort to define what any of those numbers represent.
We can recognize the 100 to be 100A (measured by the CT, also called IIN
We can recognize the 53.5 as the calculated burden resistor (from Wawa and my calculations)
We can also recognize the 51 to be the closest standard resistor value for the burden resistor.
I have no idea what this is supposed to represent:
current constant = (100 ÷ 0.050) ÷ 53.5 = 37,383 // 51 = 39,215 ?
It is standard practice to present you formula first (with NO VALUES)
and THEN present your solution, in the format (shown in MY calculations)
FORMULA
BLAH +BLAH = BLAH
EXAMPLE
Let Blah-A = blah
Blah-B = blah blah
Blah-C = blah blah blah
Then,
BLAH +BLAH = BLAH (with values)
This part is particularly confusing:
constant current = 39,215
What is 'constant current ?'
and
if current is represented in Amps (A)
WHAT is 39,215 ?
Is that 39,215 A ? (I hope not)
Is that 39,215 mA ? (if so , WHERE ARE THE UNITS ?! ("mA")
Is that 39,215 V (I hope not)
If it is not Amps, and it is not volts , then if it is ohms, WHERE ARE THE UNITS ?! ("Ohms")
Sorry Raschemmel, it was all written here Installation and Calibration — OpenEnergyMonitor 0.0.1 documentation
I didn't want to copy paste something too long.
current constant = (primary current ÷ secondary current) ÷ burden = x
current constant = (100 ÷ 0.033) ÷ 51 = 59,417
59,417 being the calibration in my code with the Emonlib library 
Just use a 100A resistive load, and use a calibration factor that displays 100.0Amp.
I assume you have a 100Amp resistive load. Why else would you buy a 100A sensor.
You could thread both wires of the AC load through the core (in opposite directions !) for 50A max.
Leo..
current constant = (primary current ÷ secondary current) ÷ burden = x
current constant = (100 ÷ 0.033) ÷ 51 = 59,417
59,417 being the calibration in my code with the Emonlib library
Keep in mind that your posts are not meant only for the members who have previously responded to your
post. They are also (since this is an OPEN SOURCE FORUM) meant for all the Newbies all over the world
who are trying to learn by reading posts, whether they have a specific need for the information or not.
There are many new members who don't have a clue where to start, never having been exposed to
electronics before and responses like your previous one who leave them totally baffled.
Even I was unable to find that formula on that link.
It may be buried in one of the links on that page but all I found was this:
constant current or current constant ?
as in const current
According to Integrated Publishing: Electrical Engineering Training Series, you would use the following formula to determine average current:
I avg = 0.636 X I max.
I avg is the average current from zero to peak and back to zero (one alteration)
I max is the “peak” current. (The unit of measurement for current is the ampere or amp.)
Let Irms = 100A
Then IPeak = (SQRT(2) * IRMS
= 1.41421 * 100
= 141.421 AI avg = 0.636 X I max.
= 0.636 *141.421 A
. = 89.94 ASo unless you can link the page from that website that contains the formula you posted, which
is it ?A. constant current (ie constant current circuit)
B. current constant (ie a software constant "const" reprepresenting current)
C. AVERAGE current (as explained above but self explanatory)Write, on a piece of paper, the formula for finding average current: I avg = 0.636 X I max.
If the answer is "B", my choice would have been "const Calibration" (calibration constant)
Also, you must not be in the USA because you have a comma instead of a decimal point, which
for us in the USA would make a huge difference.
I did the math and got 59.41, and then realized you were in some other country.
As a matter of consistency, I recommend using decimal points , because then there can be no
ambiguity. It may be correct in your country but this is an international forum and it might avoid
confusion.
I understand, sorry for being evasive. Let me be more clear/precise.
My goal is to achieve a Blynk server [with node-red for web interface], I am going to use arduino [WeMos D1 mini Pro] and ct sensor on our CNC machines to detect if they are active, or not (as 0-2 amps : idle, 2-50 : working for exemple.).
I found a tutorial on https://learn.openenergymonitor.org/ about CT sensor and I tried to reproduce it with my TT 100-SD. In the datasheets I'm missing the secondary turn so I was clueless how to calculate the turn ratio with this
Isecondary = CTturnsRatio × Iprimary
CTturnsRatio = Turnsprimary / Turnssecondary
than I found on this page Installation and Calibration — OpenEnergyMonitor 0.0.1 documentation how to change the 111.11 default calibration for the EmonLib because when I was mesuring 0.11 amps with my clamp I was mesuring 0.45 with the TT 100-SD.
Maybe it was a wrong burden, or a wrong calibration (as I was using 111.11) or both !
I'll try to apply what you guys told me !
Thanks you!
Also yes, I am from France.
Np/NS = VP/NS
If the datasheet for the CT says that Ip /Is = 100/0.033
0.033A /per 100 , doesn't that mean the turns ratio is 3030 ? (100/0.033)
I think you are over complicating things.
I made a device to disconnect a computer from the internet using a 8PDT relay by sensing the
computer monitor voltage before and after the power saver turned off the monitor.
When the power saver turned off the monitor, the monitor current dropped significantly.
I used an op amp comparator voltage threshold detector to detect the burden resistor voltage and
and energized the relay when it dropped below the threshold.
I didn't need to know anything about anything. All I had to do was measure the current with the CT
when the monitor was on and then wait 2 minutes for the computer power utility to turn off the monitor
and then measure it again. I used a DMM on mA scale to read the CT output, then added a burden resistor and used the DMM to measure the voltage across the resistor to obtain the On current, Off current, On voltage (across the burden resistor) and the Off voltage.
The rest was just building the op amp comparator circuit and adding a transistor driven relay.
I don't know why you couldn't do something similar.
I don't see any information about the CNC machines so what exactly do you know ?
Where are you measuring voltage ?
I don't know anything about CNC machines I don't know why I would need to. I would just make an
extension cord for the CNC power to use to measure the current, or put the CT on one of the 3-phase
lines. The current on any one phase is approximately 1/3 the total current but that isn't even needed.
just program the CNC to drill a hole through an aluminum block and then stop. That should be enough
time to measure the current during the drilling and after the machine stops and goes into idle mode.
On a separate note,
Unlike a voltage transformer, the primary current of a current transformer is not dependent of the secondary load current but instead is controlled by an external load.
The secondary current is usually rated at a standard 1 Ampere or 5 Amperes for larger primary current ratings.
Since without the burden resistor, the CT is an UNLOADED OPEN CIRCUIT.
Clearly the only "external load" here is the burden resistor.
= (primary current ÷ secondary current) ÷ burden = x
x = (100A / 0.033 A)/ 51
= 3030/51
= 59.41
For an ideal transformer:
Ip Tp = Is Ts
Ip / Is = Ts / Tp
100/0.033 = Ts / Tp (NOTE: Primary current is in the nominator on the LEFT , WHEREAS Secondary
turns ration in the NOMINATOR on the RIGHT)
(Likewise Secondary current and Primary Turns ration are the DENOMINATORS)
(Which means there are 3030 more Secondary turns than Primary Turns)
(This makes perfect sense when you consider the massive field generated by
100A when compared to the relatively very small field generated by 33mA)
Ip * Tp = Is * Ts
Let Ip = 100A
Tp = 1
Is = 0.033A
Ts = 3030
(100)(1) = (0.033)(3030)
100 = (99.9999) ~ 100
It gets interesting when you consider the INDUCTANCE ration is the SQUARE of the TURNS ration,
so,
if the turns ration is 3030:1(s/p)
then the inductance ratio is 9,180,900:1 (p/s) (AFAIK)
I have never used ct sensor, so it’s all new, sorry if I’m off sometimes!
It needs to be non intrusive so it didn’t change the actual wiring ( insurance would claim our fault if anything’s wrong of the wiring is changed)
I just want to know when our machines are up and running and transmit it over web
You should be able to use the same approach
I did.
