# Why does a 741 op-amp get driven to the negative rail when the non-inverting terminal is left open?

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.

• Did you study schematic? – Sunnyskyguy EE75 Feb 19 '18 at 4:06
• 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? – AlmostDone Feb 19 '18 at 4:12
• In your analysis, combine that bias current with a model of input pin capacitance. What voltage develops on the input pin capacitor? – Ben Voigt Feb 19 '18 at 4:20
• 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. – Jim Fischer Feb 19 '18 at 4:31
• It's -100, not -101; you don't get your stake back! – Chu Feb 19 '18 at 7:27