Using digital adjustable resistor  AD5270/AD5272 and operational amplifier AD8615 to construct compact, low cost, 5V, variable gain in-phase amplifier circuit functions and advantages, using digital adjustable resistor  and operational amplifier AD8615, provides a compact, low cost, low voltage, variable gain in-phase amplifier. AD5270/AD5272 (10-pin 3 mm x 3 mm x 0.8 mm LFCSP) and AD8615 (5-pin TSOT-23) packages are small in size and low in cost, providing industry-leading solutions for analog signal processing circuits.

AD5270 provides 1024 different gains, which can be controlled by SPI () or I2C (AD5272) compatible serial digital interface. AD5270/AD5272 has a resistance tolerance of (+1%) and can provide low gain error over the entire resistance range, as shown in Figure 2. The circuit supports rail-to-rail input and output, which can be supplied by either + 5V single power supply or + 2.5V double power supply, and can provide the highest output current of + 150mA.

In addition, AD5270/AD5272 has a built-in 50-time programmable memory, which can customize gain settings when power on.

This circuit has high precision, low noise and low total harmonic distortion (THD) characteristics, and is very suitable for signal conditioning applications of instrumentation. The circuit description uses digital adjustable resistor  AD5270/AD5272 and CMOS operational amplifier AD8615 to provide a compact, low cost, variable gain in-phase amplifier. The input signal VIN is amplified by AD8615. The operational amplifier has low noise, high swing rate and rail-to-rail input and output characteristics.

The maximum current flowing through AD5270/AD5272 (RAW = 20 k_version) is (+3 mA), thus the maximum input voltage VIN can be limited according to the circuit gain.

When the input signal connected with VIN is higher than the theoretical maximum determined by formula 2, R2 should be increased and the new gain should be recalculated by formula 1. In addition, the minimum gain should be calculated to reduce the error caused by leakage current. To neglect the leakage current error, the current flowing through R2 should be at least 100 times of the worst leakage current of 50 nA. Therefore, the minimum current through R2 should be 5 mu A, which also determines the minimum value of R in formula 3.