simulate this circuit – Schematic created using CircuitLab

I am a newbie electronics students.

My question is, what does it mean (what happens) when you connect pin 2 to pin 6 on a 741 op. amp., meaning the inverting input to the output (just a jumper cable).

For some reason my circuit works now when I did it, I just did it on accident. Nothing is connected to pin 2, not ground or anything, just pin 6. To pin 6 it's just pin 2 and another cable just to measure the output compared to ground.

Of course I have some other stuff connected to the op. amp., which is a variable resistor voltage divider to pin 3, -15V to pin 4 and +15V to pin 7.

My assignment is to find a linear relationship between the distance to a light source (using a LDR), and the output voltage (or current) of the op. amp.

But yeah, if someone knows anything about my question that would be really helpful, I will probably rebuild my circuit though.

I'm using a myDAQ if that's relevant.

  • 1
    \$\begingroup\$ Your description is not really clear enough to tell what is going on. Please add a schematic. You can do that by clicking "edit" just to the right of the box with your name in it, then picking the icon which looks like a pencil on top of the icons for a diode, capacitor and resistor. \$\endgroup\$
    – Jack B
    Commented Nov 22, 2018 at 11:09
  • \$\begingroup\$ You always connect both the inverting and non-inverting inputs of an opamp - never leave one of them disconnected. From the description, you are using the opamp as a buffer. \$\endgroup\$
    – JRE
    Commented Nov 22, 2018 at 11:22
  • 1
    \$\begingroup\$ Why you shouldn't use the 741. \$\endgroup\$
    – JRE
    Commented Nov 22, 2018 at 11:24
  • \$\begingroup\$ Sounds like a school lab assignment, so maybe that's why you have to use the 741. But, when you have a choice you should look for something better. If it is a lab assigment then it has probably been set up so as to avoid the (many) short comings of the 741, so you should be OK. \$\endgroup\$
    – JRE
    Commented Nov 22, 2018 at 11:28
  • \$\begingroup\$ @JackB I did one but it's probably wrong, not sure how to do one, but atleast it's something. \$\endgroup\$
    – Percutient
    Commented Nov 22, 2018 at 11:43

2 Answers 2


The circuit you have posted is a simple non-inverting buffer.

An opamp provides as output the difference between the two inputs.

An opamp with only one input connected won't do anything reasonable. The unconnected input will "float." It will take on some voltage, and it can vary randomly (or stick to one of the power rails.) So your output (as the difference of the two inputs) will be unpredictable.

You have yours connected as a buffer. In this usage, the output follows the voltage on the non-inverting input.

You use this kind of thing to prevent your measuring device from changing the voltage you are trying to measure.

Consider your myDAQ as a load on the resistive voltage divider formed around the LDR.

Adding a resistor in parallel to any part of a voltage divider changes the ratio.

If the input resistance (impedance really) of the myDAQ is too low, then merely connecting it to the LDR would change the voltage.

The 741 has a higher input impedance than the myDAQ.

Not extremely high like more modern opamps, but high enough that it doesn't noticeably change the voltage from the LDR voltage divider.

So, yes, you need that connection from the output to the inverting input.

That's the difference between a buffer and a garbage generator.

  • \$\begingroup\$ Thank you so much! I have another small question if you don't mind. If I connect pin 2 to ground instead, will it be the same? I'm thinking that the difference between pin 3 and ground would be the same as following the voltage on pin 3. \$\endgroup\$
    – Percutient
    Commented Nov 22, 2018 at 12:41
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    \$\begingroup\$ @Percutient You should stop talking about "pin 2" etc but write: "inverting input" or "- in" so that it is clear what the function of that pin is. Not all opamps use the same 8-pin housing so not all "pin 2" on an opamp IC is the same. If you connect the - input to ground (instead of the output) then you have made a comparator. The output will then be about +13 V or -13 V depending if the + input has a voltage higher or lower than ground level (0 Volt). You should really read Opamps for Everyone: cypress.com/file/65366/download \$\endgroup\$ Commented Nov 22, 2018 at 13:09

I am a bit surprised that in no answer - up to now - the term "negative feedback" is used. Please note that in your circuit a principle is used which practically in ALL amplifier (and buffer) applications is applied: Negative feedback. By applying this principle a certain portion of the output signal is fed back to the invering terminal - in your case, the feedback factor is "1" because 100% of the output voltage are coupled back.

Negative feedback has many advantages:

  • The gain becomes more and more independent on the active device (and the corresponding uncertainties, tolerances); it is primarily determined by the feedback voltage divider - in your case: no voltage division and the gain is "+1".

  • Increase of the input resistance and reduction of the output resistance

  • Linearity improvement

  • Bandwidth increase (corresponding to gain reduction)

  • Stabilization of the DC bias point.


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