eagle1109
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omg I just noticed that !! thank you so much for the declaration.You misunderstand the diagram. The motor circuit is connected to a wire through the central CT opening, not the other bottom pins.
Yes, how to get advantage of this information? For now, what I learned is that the designer of the project chose this current sensor, so it should work fine, should I care about one or more of these parameters?The other pins are only for mounting purposes and have no electrical function. According to product description, it's a standard AC current sensor with 2000:1 winding ratio. Maximal burden resistor is specified with 250 ohm for 0.1% linearity.
But there's a relay, that's what I know is an isolation between the control circuit and the load circuit. Could you specify how could the damage happen with the use of a relay? Also we have relay breakouts that have protection parts; like, optocoupler .. etc.The circuit in the link is a bad design, it may be damaged by motor inrush currents and respective high voltage at the almost unterminated CT. You better refer to known working CT circuits from other web sites.
Thanks for the explanation.eagle1109:
The wire to the motor goes through the center of the current transformer.
The bottom center of the PWB has two connections for the current transformer, CT1, which is NOT mounted on the PWB.
Two of the pins on the bottom of the current transformer are for connecting to the circuit. One goes to the rectifier and
the other goes to ground. Since it is AC, it probably doesn't matter which pin goes where.
The other two pins are not used electrically and are mechanical mounts.
But there's a relay, that's what I know is an isolation between the control circuit and the load circuit. Could you specify how could the damage happen with the use of a relay? Also we have relay breakouts that have protection parts; like, optocoupler .. etc.
Hi,
Now let's just do an example.
Let's say load current is 1A.
Because of the CT winding ratio of 2000:1 the CT tries to force this current at the output: I_out = I_in / 2000 = 1A / 2000 = 0.5mA.
A proper circuit has a fixed burden resistor at the output. Let's say 250 Ohms.
Then the output voltage becomes: V_out = I_out x Rb = 0.5mA x 250 Ohms = 125mV.
Klaus
I believe the OP has already found the right pins by using a DMM...The documentation on the current transformer (that I found) wasn't clear which pins were the output pins.
You can find them by putting one side of an AC power cord through the center of the current transformer. Insert an AC current meter in the AC line. Then put a 100 ohm resistor across two of the pins of the current transformer. Turn on the AC device and observe the output across the resistor. The meter and the voltae across the resistor should be related with KlausST's equation above.
Hi,
The problem is not "isolation" or "relay".
The problem is the "open" output of the CT.
Open means high ohmic.
Now let's just do an example.
Let's say load current is 1A.
Because of the CT winding ratio of 2000:1 the CT tries to force this current at the output: I_out = I_in / 2000 = 1A / 2000 = 0.5mA.
A proper circuit has a fixed burden resistor at the output. Let's say 250 Ohms.
Then the output voltage becomes: V_out = I_out x Rb = 0.5mA x 250 Ohms = 125mV.
But your circuit has no proper burden resistor. The output is high ohmic.
Let's just imagine there is a steay impedance of 100k Ohms (caused by bad isolation and stray capacitance)
The the (theoretical) output voltage is: V_out = I_out x Zb = 0.5mA x 100.000 Ohms = 50V.
This is high voltage that may kill the "low voltage side circuit"
--> The higher the load current, the higher the output voltage
--> The higher the output impedance the higher the output voltage.
But the output voltage will not become infinite. It is limited by the unavoidable output impedance (it is limited, too) and the output voltage becomes limited when the CT core becomes saturated.
--> Thus a proper design never uses "open CT output".
It may use directly connected resistors, but one may add diodes (like your circuit). But even with diodes I recommend to use a symmetric output circuit: add a second diode and maybe a second burden for the opposite current direction to avoid core saturation causedby onsymmetries.
Klaus
We didn´t talk about a protection circuit. We talked about proper installing a burden resistor ... and not leave the CT output high ohmic (in one direction).It doesn't have aux protection components, so it's similar to the one in the project circuit.
Why not zero?but it should be something like ~70mV or less.
No wonder.but I didn't measure something really high as you warned me about
is to connect the current sensor output to the LM393 comparator and as adjust the output of the comparator that if the voltage is 125mV which is the one at the load attached then the relay is activated and shut-off the motor, it's just for demonstration, and if the voltage is less than 125mV; say something like 70mV, then operation is safe and relay isn't activated.
Hi,
3. This is without anything, just connecting the GND & signal of current sensor directly to the oscilloscope rails:
Here I tried to capture the moments where the signal has the most stress, there are other pictures with low amplitude.
First part with the fan connected to the motor as a load:
View attachment 151898
Second part without the fan:
View attachment 151899
.
Hi,
Why the diode and the "load resistor".
What´s the idea behind this circuit?
Klaus
Yes, I know that there's already a burden resistor, which Mr Klaus told me in earlier posts, but I put another one in series with the output of the CT as a load or at least protection, I know the signal is 125mV it's really very small, but when putting that resistor the output changed.If you look closely at the photo:
Between the output connector and the transformer's body, there appears to be a 200 ohm resistor (barely seen as 201).
In other words, it is already terminated into a burden resistor.
- - - Updated - - -
An oscilloscope timebase of 500nS/ division..oooooooohhhhhh nooooooooooooooooooooooooooo!!!!
Try 20,000 times slower: 10 mS/division
In simulation you don'tneed them.The additions I noticed are two bypass capacitors, which are not necessary for the simple test implementation.
The CT signal has a +ve and -ve peaks, so I know I have to include a -ve voltage source in the implementation.
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