diodesintegrated-circuitoffsetoperational-amplifierresistors
I've the following schematic for a UA741CN when there is NO offset:
simulate this circuit – Schematic created using CircuitLab
Question: Why do we need the diode and the two resistors to create an ouput of the opamp without any offset (or a little one, due to not ideal components)?
The simplest way to eliminate your problem is to reduce the effective impedance of the source, and a voltage follower will do nicely
simulate this circuit – Schematic created using CircuitLab
The choice of buffer op amp is left to you, depending on the frequency response you need.
In the data sheet there are hints about what you can load the op-amp with. The maximum peak-to-peak output voltage when loaded with 10kohm is guaranteed to be +/-12V on a +/-15V supply regime.
When loaded with 2kohm the max voltage is somewhat less at +/-10V.
I would usually consider that 2kohm loading is the minimum you should have.
All that info is on page 6 of the DS. I would also look at these graphs for what to expect at lower suply voltages: -
These are on page 9 and page 10 gives you more information when the ambient temperature is different. See also this graph on page 10: -
I am not able to find any information about the maximum output current in it.
The last graph should give you the information you need and you should probably consider 1kohm for R3
Best Answer
What is "offset null" in IC 741? has some explanations
The 741 is an OLD design and one often needs to tweak out the DC offset that exists on its output. A pot between pin 1 and pin 5 with its wiper to -VCC (pin 4) is typical. Your circuit is using fixed resistors.. don't know why.
The diode seems extraneous to the offset function. Are you sure that's why it's there?