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Many Precision Rectifier Single Supply Op Amp circuits feed the input AC waveform into the non-inverting (+) terminal of an op amp e.g. see this schematic:

precision full-wave-rectifier circuit using two op amps and no diodes

Image source: Precision full-wave signal rectifier needs no diodes

Say Vcc = +15V, Vee = 0V, and Vin is an AC input of +/- 5V (10V peak-to-peak), then when Vin is in the negative half-cycle, the first op amp outputs zero volts (it can't go lower), so input V(-) is also 0V via the feedback. Input V(+) however is controlled by the input AC voltage and will fall to -2.5V at its minimum. Now most op amp specs say under Absolute Maximum Ratings: Input Voltage "Vee-0.3V to +32V" say).

My question is, surely Vin at -2.5V is outside of the Absolute Maximum Rating of -0.3V? Also the Common Mode input at -1.25V is outside. So am I missing something? Any light anyone can shed on this will help a lot.

See https://www.radiolocman.com/shem/schematics.html?di=161475 for the complete article.

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  • \$\begingroup\$ Thanks for the replies, everyone. Much appreciated. \$\endgroup\$ Commented Jun 27, 2020 at 13:38

3 Answers 3

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With most opamps the problem you describe would exist and could cause damage or unusual operation of the device.

However, the data sheet for the LMC6482 does cover this specific mode of operation. The device will operate within specification provided the current into the input is limited to 5mA or less. It is guaranteed that inversion will not occur and that the output will remain at zero volts. The 1K input resistor does limit the current as required.

As Spehro comments the current is right at the limit and it would be better if R1 and R5 were both increased to 10k or more to give greater margin. Providing the two resistors are the same value the operation of the circuit wouldn't be changed. (avoid very large values or capacitance effects will affect operation).

See section 7.3.2 in the datasheet.

LMC6482 Datsheet

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  • \$\begingroup\$ Do you understand the behavior of the ESD diodes? \$\endgroup\$
    – D.A.S.
    Commented Jun 15, 2020 at 21:48
  • \$\begingroup\$ @tony - In general yes, I understand them and been bitten by substrate injection in devices I have designed. But I don't know the details for this particular device but the 5ma is within the datasheet specification. \$\endgroup\$ Commented Jun 15, 2020 at 23:12
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    \$\begingroup\$ The datasheet (Fig. 69) gives seal of approval to this kind of circuit, though with a much higher value resistor (and of course the datasheet may not be sufficiently conservative). 5mA is the absolute maximum input current so 1K is not really very safe, 10K or 20K would be much better. \$\endgroup\$ Commented Jun 16, 2020 at 13:52
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This is a CMOS RRIO device with ESD protection on the input just like CMOS Logic.

The reason this works for +/-5V AC input is the series resistance is high enough to limit the clamped input current below Vss (-ve =0V) It is clamped by an internal Schottky diode similar to the ESD protection in CMOS logic. Therefore the ABS MAX. limits are indicated by the ESD diode current max of 5mA and diode voltage of 300mV @ 5mA. Thus your design must use sufficiently high values or series R values.

  • The inputs are ESD rated safe to ±1500V from 100pF HBM, human body model

  • this is to illustrate how it tolerates ± 1500V

schematic

simulate this circuit – Schematic created using CircuitLab

enter image description here

If you put 5V into 1K to a 5mA tiny Schottky diode , you may get > 300mV but in fact they use 2 stage ESD protection. Otherwise you should use Rs > 10K

For the FW rectifier to match gain for each polarity the ideal Resistors would be matched R pairs in an array so that the absolute value could be >1% but the matching be 0.1%

schematic

simulate this circuit

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The opamp must be inverting to work with a negative input signal but have no negative supply voltage. enter image description here

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