In the case of a 741 op-amp with R2 = 100k and R1 = 1k the gain should be -100. The output voltage should then be 101 times whatever the input is. I understand that given a large input voltage to the non-inverting terminal, the output should be saturated to the positive rail, but in both simulations and in the experimental procedure I have found that leaving the input open causes the output to be around -8V when my saturation is +-10V. I know leaving the terminal open is not a practical implementation of the op-amp, but theoretically, I would like to know what causes this output voltage to be driven the way it is.

  • \$\begingroup\$ Did you study schematic? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Feb 19 '18 at 4:06
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    \$\begingroup\$ Look at the 741 datasheet and notice the bias current parameter, 30 nA typ. What path is there for this current if the input is floating? \$\endgroup\$ – AlmostDone Feb 19 '18 at 4:12
  • \$\begingroup\$ In your analysis, combine that bias current with a model of input pin capacitance. What voltage develops on the input pin capacitor? \$\endgroup\$ – Ben Voigt Feb 19 '18 at 4:20
  • \$\begingroup\$ The op amp's input stage is a differential amplifier. A fixed current flows into the top of the differential amplifier and is divided between two legs, where each leg is driven by either the non-inverting or the inverting input. If only one of the two legs of the difference amplifier is being driven by an applied voltage--e.g., a voltage at the inverting input, then that's the leg that controls the voltage at the op amp's output. It's like a child's teeter totter with only one child sitting on one end of the apparatus. \$\endgroup\$ – Jim Fischer Feb 19 '18 at 4:31
  • \$\begingroup\$ It's -100, not -101; you don't get your stake back! \$\endgroup\$ – Chu Feb 19 '18 at 7:27

The input bias current for a 741 is about 80 nA: every second, the chip is pulling a (tiny) bit of charge from that input pin.

If there’s a resistive path, it can provide that current.

But if the pin is open-circuited, it acts as a small capacitor. The removed charge drives the voltage at that point more and more negative. That drives the output negative.

Eventually, all those voltages get negative enough that the op amp becomes internally unbalanced, and the output sits down near the rail.

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  • \$\begingroup\$ Sir----a delightfully clear explanation. \$\endgroup\$ – analogsystemsrf Feb 20 '18 at 3:05
  • \$\begingroup\$ This was exactly what I was looking for, I knew that the pin was being driven negative, but did not know about the capacitive nature of the pin on the IC. Thank you! \$\endgroup\$ – Llorio Mar 19 '18 at 4:15

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