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I am having trouble interfacing with MIC6211 op-amp. I have it configured as voltage follower (buffer):

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VIN changes from 5 V to 20 V. The op-amp output is high impedance (100% sure, even disconnected the load). My problem is that around VIN = 9V and going down, the op-amp stops following the voltage provided at pin 3. The voltage at pin 3 keeps dropping but the output (pin 4 and 1) does not. Going further down the voltage is so low that it cannot provide sufficient level for the op-amp to operate so the output drops as well.

Removing the op-amp from the circuit and shorcuting pins 3 and 1 provides proper operation of the circuit. This op-amp was only suppose to buffer the voltage divider voltage in case any load occured at PVOL.

What could be the case? Looking at the datasheet of the device this conditions should be fine, I should be able to buffer the voltage from 4 V to 32 V. I would appreciate all help.

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    \$\begingroup\$ Have you had a look at what the datasheet says about the maximum output voltage swing? \$\endgroup\$
    – PlasmaHH
    Commented Apr 4, 2018 at 12:09
  • \$\begingroup\$ You are right, this is probably the problem. \$\endgroup\$
    – Łukasz Przeniosło
    Commented Apr 4, 2018 at 12:11

1 Answer 1

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If you look at the specs of the op amp you will see 'Maximum Output Voltage Swing' stays between 1 and 2.5 V above V-

In your circuit when Vin is 9 V the voltage at the + input is around 0.7 V, so the output is out of specs. Thus, your circuit is working a little better than expected.

If you need the output to go down to 0 or close to it you will need a different op amp or a different circuit.

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  • \$\begingroup\$ Are you aware of any op-amp with such high supply voltage with smaller voltage swing? \$\endgroup\$
    – Łukasz Przeniosło
    Commented Apr 4, 2018 at 12:19
  • \$\begingroup\$ Sorry but no, my practical electronics knowledge is quite outdated. I'd try putting two diodes (just anyone) directly in series with pin 1 followed by a pull down (between 1 and 10 kΩ). This would allow the output to go down to zero while pin 1 is still at 1.4 V. \$\endgroup\$
    – Toni Homedes i Saun
    Commented Apr 4, 2018 at 12:38

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