I need a help. Can anyone help to find out the voltage drop across the 1.5R parallel resistor and effect of the rectifying diode on the circuit?

Here in the circuit image CT1 and CT2 are the current transformer leads of 1000:1 CT, that if there is 1000A in primary 1A in secondary.

My problem the CT current is bridge rectified using 1 schottky diode and one tvs diode as shown in image. As the current passes in primary the current will appear in the secondary. I have two 1.5R resistors as load resistor. I want to know how much voltage drop will be across CT. Whether the voltage will be 1A*0.75R (1.5R//1.5R) = 0.75Vdc or whether the diodes any have any effect on the voltage across the load resistor? Please help..

 

A Schottky diode voltage drop is between approximately 0.15–0.45 volts, let's say it 0.3V.
The normal calculation of output voltage will be as Vout = Vin-(0.3V*2) =0.75V-0.6V= 0.15V.

So you have to measure the Schottky diode forward voltage first for sure, and correct the voltage values.

 

Why do you need the rectifier bridge? Couldn't you simply measure the peak voltage, or AC voltage, across the load resistor?

Edit:
Like this perhaps?

 

A precision op amp rectifier will eliminate any significant effect of the diode forward drop if you have power available for the op amp. My favorite is this circuit which uses just two single-supply or rail-rail op amps (one dual package), one diode, and three equal value resistors to perform full wave rectification using a single positive supply.