I want to design a circuit to protect oveload current of 30A.I have a controller board(PLC) which accepts the voltage range of 0-5 as an input and C.T with 1:7000 turns which can maesure 0.5 -60 A(RMS) value with burden resistance of 47 ohm(recommended).My question is how should i design a circuit which porvides 0-5 V output proportional to 0-30 A(RMS).


1 Answer 1


I'll assume you want "average reading" equivalent to RMS for a sinusoidal input. If you want "true RMS" you'll have to add a more complex conversion circuit. This is the same principle used in non-"true-RMS" multimeters.

The transformer gives you (60A/7000) * 47 = 402.8mV RMS AC full scale (full scale of the transformer)

So you need to convert an AC input of 402.8mV/2.0 = 201.4mV RMS into 5VDC (30A in for 5.0V out)

Here is one of many of possible precision rectifier circuits:-

enter image description here

It shunts the input resistor a bit, so 47.22 ohms can be used as the burden resistor to give you 47 ohms equivalent as recommended.

You feed the output of that circuit into a low-pass filter. The sophistication depends on response time requirement and ripple. Look up a Sallen Key type, for example.

The input voltage was 201.4mV RMS, which is 284.8mV peak. The output voltage of our absolute value circuit will also be 284.8mV peak. The filter will find the average of the absolute value circuit output, which we know to be (since it's equivalent to the average value of half a sine wave) \$V_{PK}\cdot 2\over\pi\$ = 181.3mV (assuming a filter gain of 1.0).

You then simply need an amplifier with a gain of \$ 5.0 \over 0.1813 \$ = 27.58 to give a 5V output for 30A.

Edit: The full wave rectifier circuit shown works as shown below:

enter image description here

The output voltage is \$2V_B - V_A\$

For positive inputs, \$V_B\$ =0 and \$V_A\$ = -Vin, so overall gain is +1

For negative inputs, \$V_B\$ = (2/3)Vin and \$V_A\$ = (1/3)Vin so overall gain is +1

  • \$\begingroup\$ Thanks for your valueble suggestion.I'm not good in op-amp circuits.First I will try to understand how the circuit works.Will ask you if find any doubts :) \$\endgroup\$ Apr 6, 2014 at 5:40
  • \$\begingroup\$ Okay, it's really just a series arrangement of three building blocks after the burden resistor. Rectifier->Filter->output amplifier. \$\endgroup\$ Apr 6, 2014 at 5:50
  • \$\begingroup\$ @Pefhany, in your example,i didn't understand why o/p (from node 6) is given to inverting input through 1k ohm and diode. and filter part, could you please elaborate how it works for me \$\endgroup\$ Apr 14, 2014 at 12:38
  • \$\begingroup\$ @user3396484 1K and diode? Not sure what you're asking. \$\endgroup\$ Apr 14, 2014 at 12:44
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    \$\begingroup\$ @user3396484 see edit. Both amplifiers are required to make a full wave rectifier. \$\endgroup\$ Apr 14, 2014 at 15:32

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