From your measurements it looks line the current transformer is not acting like one. Try different load resistors and see if the current drops as the load resistance increases. From these values you can calculate the equivalent circuit of the transformer as a voltage in series with a resistor. you will find that you have something much different than an ideal current source.
To brute force the situation you have several choices. One is to use an op amp type rectifier circuit. Another is to use the equivalent circuit to calculate the load resistor with the highest output voltage given the diode drops. From there you can calculate the nonlinear correction factor of output voltage caused by monitored current.
But the use of opamp rectifier circuit brings a problem, I have only 0-5Vdc power supply...
But I think that I can solve the problem using a Schottky diode to remove negative half wave (to put the output of transformer directly on ADC input), and in software make the calculation to get the RMS value of current.
To resolve the problem with only 5 V supply you can connect the one terminal of the transformer to 2.5 ref. voltage instead of GND and the rectifie signal from the other terminal with opamp. If the output voltage swing of the 5 V single supply opamp is 0 to 4V it is better to connect tranformer to 2 V instead of 2.5 V.
Another possibility is to connect one terminal of the transformer to GND an put a big capacitor at the other terminal in series with the signal for dc level shifting to 2.5 V at opamp side. If you connect the opamp input termianl references to 2.5 V instead of GND it should be possible.
Some OpAmp Full Wave Rectifiers:
Look at fig. 10.5: **broken link removed**
Look at fig 4: **broken link removed**
The easiest way is probably to use a charge pump IC like MAX680/MAX681 or similar to genrate +- 10 V from 5 V supply so you can use dual supply opamps.
MAX680, MAX681: **broken link removed**
Use this application note to design your opamp rectifier: **broken link removed**
I would recommend you use a charge-pump IC to provide +-10V for the opamps. You can order free samples from MAXIM. This is the least complicated solution.