I need to translate 0-5V analog signal from CASR 50-NP to 0-3.3V analog signal for the MCU's ADC. I came along Microchip's tip #16 solution. Is this a good practice? Can someone verify my design?Current transducer voltage translation

  • \$\begingroup\$ I'd be a bit worried that, under some start-up condition, that the op-amp output could go to +5 V for an instant. Anyone? \$\endgroup\$ – Transistor Jan 3 '16 at 18:12
  • \$\begingroup\$ You should examine what you'll gona do with reference voltage. Will you use the reference from transducer, or will you hook ref voltage from several trransducers parallel, or will you feed/sink the ref voltage to the transducer from MCU? That's the real hard task, then you can choose to buffer (with RC filter) to ADC. \$\endgroup\$ – Marko Buršič Jan 3 '16 at 19:35
  • \$\begingroup\$ I am going to leave reference pin unconnected, so yes, that means reference from transducer. 0A means 2.5V on the output of the transducer this way, which I need to translate to 1.65V for the input of the ADC. \$\endgroup\$ – jurij Jan 3 '16 at 20:35
  • \$\begingroup\$ You may use Preset Resistor as a pot. From that you can take the output. It looks simple idea, rather than using op-amp. Give the input to Fixed Terminals and take the output from variable terminal. Adjust the Preset make the voltage 3.3v when input is 5V. \$\endgroup\$ – Photon001 Jan 5 '16 at 11:02

This looks reasonable.

You might want to connect the Opamp to a 3.3V supply - OPA340 is a rail-to-rail device so it will have no influence on the output during normal operation. Whenever a failure on the input occurs (e.g. input voltage is too high or ground connection is bad) the signal to the uC can never exceed the allowed range.

One thing to consider: If the source is low-impedance and the ADC is high-impedance, one could lower the value of the voltage divider, e.g. 1.7k:3.3k, add a capacitor of e.g. 100n to 1u and dismiss the amplifier. This might introduce some measurement errors, e.g. if the the ADC input has a large capacity on the input and hence the inrush current during the sampling period might lead to a voltage drop.


Why do you suspect that the output voltage could instantaneously exceed 5 V? This cannot practically happen for 2 simple reasons:

  1. As long as the input voltage to the divider is in the range you mentioned, the output voltage can never exceed Vin x (33/50).
  2. Part of the reason why voltage followers are so commonly used in signal conditioning circuits for MCUs is that they provide electric isolation.
  • 1
    \$\begingroup\$ unfortunately this is not correct. The opamp will have some ringing and overshoot on fast transients, and it does not provide electric isolation either. \$\endgroup\$ – Nils Pipenbrinck Jan 3 '16 at 19:35

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