Amplifying DC Diode Detector Output


Posted on Feb 7, 2014

The very small DC voltages at the output of the diode detector are difficult to amplify accurately. DC offsets introduced by any op-amp are the main culprit. Low frequency noise (1/f noise) is another. Chopping the DC with MOSfet switches is an excellent way to get around these problems. In the circuit below, a chopper integrated circuit ( LTC1043


Amplifying DC Diode Detector Output
Click here to download the full size of the above Circuit.

) switches between two square-law detectors: one detector gives a positive-going voltage and the other gives a negative voltage. Cosc sets the (square wave) chopping frequency to about 90 Hz - you want to keep this chopping frequency fairly low, but well away from 60Hz and 120Hz. The op-amp that follows the diode chopper has a gain of 181, and amplifies the 90 Hz. square wave that the chopper provides. Any offset that the op-amp adds is eliminated by the de-chopping switch at its output. The RC network (R3, C3) at the output is a low-pass filter, and sets the bandwidth at about 10Hz. This allows reasonably fast meter response, while eliminating noise. A second op-amp acts as a conventional DC amplifier, and boosts voltage to drive an external meter. Gain can be adjusted (1+ R6/R7) to provide a reasonable output range for your meter. This op-amp`s input offset voltage is amplified by the op-amp`s gain, and appears at the output. If gain is high, this can be a significant offset from true zero. What`s the sensitivity limit of this circuit The primary source of noise is the warm 50K resistance of the rectifying germanium detector diode. The op-amp shown adds about 10nV/rt(Hz) as well. Within the 10 Hz. bandwidth, thermal noise of the 50K detector resistance is about 0. 1uV. At -60 dBm input RF voltage, the detector diode generates about 0. 1uV DC voltage, so inputs less than -60dBm are lost in noise. A narrower filter (<10Hz)...




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