Opamps are a great device for amplification. Recently i used a TL084 which is quad-channel Opamp I used three opamps in the device for getting gains of 1 , 2 , 3 respectively. I noticed that, without any connection to the input, there was Voltage at the output of 0.4V. Please Help me understand what the problem is. Can a Pull Down Resistor help here? and also will a Series resistor to the input help? Please suggest good practices for the opamp design.

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  • \$\begingroup\$ Ground the inputs and measure again. \$\endgroup\$
    – Mattman944
    Commented Jan 7, 2021 at 1:37
  • \$\begingroup\$ You can't leave opamp inputs unconnected: the input pins require a small DC bias current to operate properly. A pulldown resistor to ground will set things right. The value should be high enough so that whatever connects to J1 can drive it easily, and not so high that the opamp bias current through it causes an acceptable DC offset, or the resistor noise becomes an issue. 10K-100K is a good place to start. \$\endgroup\$
    – td127
    Commented Jan 7, 2021 at 17:35
  • \$\begingroup\$ Thank you soo much \$\endgroup\$ Commented Jan 8, 2021 at 4:21

2 Answers 2


Can a Pull Down Resistor help here?

It might. It depends on your type error. OpAmp inputs are high impedance. So they tend to pick up noise or static charges, if their inputs are not connected to a fixed potential. A pulldown resistor is one way of solving this issue.

and also a Series resistor to the input will help?

No. As I said. OpAmp inputs are high impedance. This means, that there is no current flowing into them (for an ideal OpAmp). Without a current flow, a series resistor at the input will be useless.

Keep in mind that you most likely also have to deal with non-ideal behavior of real OpAmps.

Supply Rails

Your circuit doesn't show your supply voltages of the OpAmp. However, keep in mind, that OpAmps can only generate output voltages within their supply voltages and, depends on your type, can only take input voltages in limits with your supply voltages.

If you supply your OpAmp with + 10V and 0V, the output cannot reach 0V. There are rail-to-rail OpAmps which allow you to get very near their supply rails, but even these, will never reach you supply rails at the output.

The datasheet of TL084 says in its recommended operating condition $$VCC+ = 5V$$ and $$VCC- = -5V$$

In this case an output of 0V is perfectly valid as it is perfectly in the middle of the supply rails. Just be careful when powering these opAmps asymmertrically.

Input Offset Voltage

As you probably know, an ideal opamp has infinite gain and will amplify any voltage between its positive and negative input. When connecting both the positive and the negative input to the same potential, the OpAmp's output is 0. This is the essential behavior, that allows you to handle an OpAmp with negative feedback as if its inputs are always on the same voltage.

This however, only holds true for ideal opAmps. Real OpAmps have something, called input offset voltage. This input offset voltage is caused by asymmetries in the design and in the manufacturing of the OpAmp. The input offset voltage basically adds to your input voltage and the opamp behaves at if it is seeing an input voltage higher than the actual one. When wiring OpAmp circuits at high gains, the input offset voltage is also amplified.

The TL084 states an input offset voltage of at max +/-5 mV (depends on your specific type). When wiring a non inverting amplifier with a gain of, let's say 100, you will get an output voltage of +/-500 mV although the input is connected to 0V.

You can read more about it in this tutorial by AD

  • \$\begingroup\$ Thank you soo much \$\endgroup\$ Commented Jan 8, 2021 at 4:21
  • \$\begingroup\$ @VenkateshA Glad it helped you. If an answer helped you, you can upvote it and accept it using the green check mark, so others can see that a solution for this question has been found. \$\endgroup\$
    – GNA
    Commented Jan 8, 2021 at 23:42

Hello everyone that helped me...thank you ...i used a pull down resistor of 68k(cause i didn't have 10k or 100k) and the circuit worked...


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