Question

For a full wave rectification, in a low voltage ac voltmeter, the meter current can be expressed as

(a) Io = (1.9×Vin)/R1

(b) Io = (3.9×Vin)/R1

(c) Io = (0.9×Vin)/R1

(d) Io = (2.9×Vin)/R1

The question was posed to me by my college professor while I was bunking the class.

This is a very interesting question from Current to Voltage Converter in portion Operational Amplifier Applications of Linear Integrated Circuits

Answer 1

Answer: (d) Io = (2.9×Vin)/R1

In full wave rectification for AC voltmeter applications, the meter current is calculated using the transimpedance formula. For a current-to-voltage converter circuit, the output current is proportional to the input voltage and inversely proportional to the feedback resistor.

The coefficient 2.9 comes from the full-wave rectification transfer function where:

- 2.9 accounts for the rectification factor and averaging characteristics

- Vin is the input AC voltage (peak value)

- R1 is the feedback resistor in the transimpedance amplifier

Therefore: Io = (2.9 × Vin) / R1