Converter from CT to Volt

What is the CT powering now? And how may this be influenced or affected? When you load the CT can with a high burden - resistor that has a value R = U^2/ I you will get 10 volts by 5A current (RMS) . Probably it needs to be rectified and smoothed with a capacitor to get 10 Volts DC. Depending on the rectifier bridge you'll need to adjust the voltage with a factor as like 1.41.

If you are using the CT to measure the actual current, it might be better to install a second one for your purpose.

Hi Guru,

Yes, Im planning use other than CT that use to measure actual load. Apologize me before for not understanding you, for example: I can connect output singnal from CT to variabel resistor eg. 10 K than connect it with rectfier component, so i can get volt DC.

variable or fixed resistor, yes. But be safe and do not use the CT WITHOUT any load, because it can generate dangerous and life threatening voltage if the coil ends are free and without any burden.

Remember CT gives you 5Amps for 2800Amp primary.

So to get 10V DCPeak from rms SecondaryAmps5amp--:

You will use voltage measuring load about 1.5 Ohms. as CT's load.

Certainly not 10 KiloOhm!!!

Why don't you go for a simple PT rather than CT?

As yiu are aware (and has been pointed out) the CT basically expects virtual short circuit on the secondary.

To generate a voltage say 10V you need at least 2Ω (Rated 5Ax2Ω=10V) and that is going to dissipate 50W power in the resistor (I²R)

Even worse is at this level of burden, the CT's linearity may be affected.

A standard product available in market can solve your problem.

It could be current (0-1 Amp) to Voltage (0-10 volts) Converter. Here is link to site giving data sheet of such convertor commonly used in industries.

http://www.prelectronics.com/idd964.asp?pcol=2279

A dangerous idea..... get a PT as suggested, as a CT is designed to be a CT. PT's are cheap and easy to obtain, easy to fit and will give you hours/years of trouble free operation.

The problem with trying to use a CT, is getting the ratio right to give you the correct voltage output then convert it, and do whatever else to it so it will work.

A PT is the way to go and whole lot less dangerous.

I used to design control systems for switch gear and you can NOT replace a CT with a VT. A CT senses current and a VT senses Voltage.

The CT on your system is designed to operate with no more than a 0.5Ω load and possibly only a 0.05Ω load. This is called a burden in this industry. I recommend you take the output of the ct and run a wire of at least 12AWG through a 1000:1 CT like

http://www.triadmagnetics.com/catalog_template.php?productCategoryId=91

found through glogal spec.

This will give you an additional level of isolation and you can move the 1K:1 CT close to the genset.

Now with the 100Ω termination you will get a 0 - 0.5VAC signal that will have to go through an amplifier and RMS conversion that you feed to the genset control system.

What is not clear from you question is exactly what the DG function is. You say load share. does that mean that as the signal in approaches 10VDC the generator turns on for peak load reduction? This is important as it affects the circuit between the 1K:1 CT and the DG.

In DC drives we often use 2 ct's and a six diode bridge to measure the three phase currents. The ct's are often 10,000:1 and the burden resistor is slected to give between 1.5 to 5 volts depending on the manufacturer. The output could be filtered by an appropriate device at that point. As others have mentioned, watch the watts and don't open circuit the ct's. The diodes should be conservatively rated atleast 2x the secondary current, and also rate the resistors at about 50% of their wattage. Gives long life and handles overloads.