Improved Precision Rectifier Circuit Benefits

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SUMMARY

The discussion focuses on the mechanics of an improved precision rectifier circuit, highlighting that the operational amplifier (op-amp) avoids saturation while requiring an output change of approximately 1.2V (two diode voltage drops) when the input signal crosses zero. It clarifies that when the input is at 5V, the output does not directly equate to 3.8V, as the op-amp output must switch between +0.6V and -0.6V to accommodate the conducting diode. The necessity of overcoming the forward voltage in this circuit configuration is also addressed, emphasizing the role of the diodes in the output behavior.

PREREQUISITES
  • Understanding of operational amplifiers (op-amps)
  • Knowledge of diode characteristics and forward voltage
  • Familiarity with precision rectifier circuit design
  • Basic circuit analysis skills
NEXT STEPS
  • Study the principles of operational amplifier saturation and output behavior
  • Research diode voltage drop characteristics in rectifier circuits
  • Explore advanced precision rectifier circuit designs and applications
  • Learn about the impact of input signal variations on op-amp output in rectifiers
USEFUL FOR

Electronics engineers, circuit designers, students studying analog electronics, and anyone interested in improving rectifier circuit performance.

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I'm having trouble understanding the following sentence, regarding an improved precision rectifier circuit:

This circuit has the benefit that the op-amp never goes into saturation, but its output must change by two diode voltage drops (about 1.2 V) each time the input signal crosses zero.

(1) What is meant by "change by"? If for example, the input is 5 volt, would the output change by 5 - 1.2 = 3.8 Volt?

(2) Is it still necessary in this circuit, to overcome the forward voltage?

http://en.wikipedia.org/wiki/Precision_rectifier
 
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One of the diodes D1 and D2 is always conducting and the other one is not.

When the input goes through 0V, the other diode starts to conduct. The output of the op-amp has to flip between about +0.6V and -0.6V to provide the voltage drop across the correct diode.

If the input changes between say +1mV and -1mV, The output of the whole circuit only changes by about 1mV, even though the output of the op amp changes by about 1.2V.
 

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