I have come across this op amp configuration which will bias a 5 V peak-to-peak input signal so that it will lie above 0 V, (OP1.2 in picture), however I am having a hard time understanding how it works. According to the document, the gain equation is

$$\text{Gain} = 1 + \frac{R_7}{\frac{R_4R_6}{R_4+R_6}}$$

This seems to imply that \$R_4\$ and \$R_6\$ are in parallel, however these also appear to act as a voltage divider for the offset of 2.5 V, implying they are also in series, how can this be?


The only other reference I have found to this kind of design is from this document, calling it 'split resistor' biasing, with the same equation, any other information on this would be greatly appreciated!


2 Answers 2


One way to analyse this circuit is to replace the potential divider consisting of R4, R6 and the 5V voltage source with their Thevenin equivalent:


simulate this circuit – Schematic created using CircuitLab

Now the bottom amplifier looks like this:


simulate this circuit

Your statement about R4 and R6 being in parallel was correct, but it should also be pretty clear, after an application of Thevenin's theorem, why the gain is dependent upon this parallel combination:

$$ V_{OUT} = \overbrace{V_{IN} \times (1 + \frac{R7}{R_{TH}})}^{Gain}\space\space \overbrace{- \space V_{TH} \times \frac{R7}{R_{TH}}}^{Offset} $$


$$\begin{aligned} R_{TH} &= R_4 \parallel R_6 \newline \newline &= \frac{R_4 \times R_6}{R_4 + R_6} \end{aligned}$$


$$ V_{TH} = V_1 \times \frac{R_6}{R_4+R_6} $$

Note that the "offset" term is subtracted. If you wish to bias the output at some level above 0V, the voltage source V1 will have to be negative. For example, with R4 = R6, V1 will have to be -5V to centre the the output on +2.5V.


From an AC perspective both +5V and GND are both "AC Ground" because they have no AC voltage, only a DC voltage.

So in effect they are in parallel from an AC perspective.


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