I came across this configuration earlier today and can't for the life of me figure out the equation for Vout:
It looks like a unity-gain differential amplifier but the ground resistor on V2 is before the series resistor.
\$\begingroup\$
\$\endgroup\$
5
-
\$\begingroup\$ The resistor on V2 can thus be ignored, so the voltage on the non-inverting input is V2. Does that help? \$\endgroup\$– Spehro PefhanyJul 3, 2015 at 4:08
-
\$\begingroup\$ Could you post a link to the page where you saw this schematic? Maybe we can glean more from the context. \$\endgroup\$– Nick AlexeevJul 3, 2015 at 5:08
-
\$\begingroup\$ @SpehroPefhany Could this be a badly drawn difference amplifier? \$\endgroup\$– Nick AlexeevJul 3, 2015 at 5:08
-
\$\begingroup\$ To me it looks like a standard inverting amplifier, where an extra resistor (vertically positioned) is added to define the input impedance of the circuit and the other extra resistor (horizontally positioned) is for symmetry for offset currents. \$\endgroup\$– jippieJul 3, 2015 at 5:49
-
\$\begingroup\$ @NickAlexeev looks like h/w to me, but I guess I could come up with a practical application in signal conditioning. \$\endgroup\$– Spehro PefhanyJul 3, 2015 at 12:15
Add a comment
|
3 Answers
\$\begingroup\$
\$\endgroup\$
Re drawing the schematic to I can add component references etc.
simulate this circuit – Schematic created using CircuitLab
I don't recognise this as a standard circuit but its got negative feedback so will try to keep both the inverting and non-inverting inputs the same
The non-inverting input is simply \$ V_2 \$
And the inverting input:
\$ \dfrac{\dfrac{V_1}{R_3}+\dfrac{V_{out}}{R_4}}{\dfrac{1}{R_3}+\dfrac{1}{R_4}} \$
Noting all resistors are the same value and equating the 2 together we get
\$ \dfrac{\dfrac{V_1}{R}+\dfrac{V_{out}}{R}}{\dfrac{1}{R}+\dfrac{1}{R}} = V_2 \$
Solving for \$V_{out} \$
\$ \dfrac{V_1}{R} + \dfrac{V_{out}}{R} = V_2 \cdot \dfrac{2}{R} \$
\$ V_{out} = 2 \cdot V_2 - V_1\$
But as others have pointed out the circuit may be drawn incorrectly with \$ R_1 \$ in the wrong position and a differential amplifier could be what is intended.
\$\begingroup\$
\$\endgroup\$
Apply the superposition principle.
When V1 is zero you have Vout that depends on V2 (simple non-inverting amplifier). When V2 is zero you have an inverting amplifier whose output depends on V1.
Finally, for the said principle, you sum these two results and obtain the Vout in function of both V1 and V2.
\$\begingroup\$
\$\endgroup\$
As Warren says, its gain is 2*V2 - V1.
The resistor from V2 to ground serves a couple of purposes:
- If V2 is disconnected, it provides a DC path to ground, preventing spurious outputs
- It presents the same impedance as input V1 to differential signals, which may be important if it's connected to the output of a passive filter, or a sensor wired as a bridge.
The series resistor from V2 to In+ matches the V1 input resistor on V1, which may improve both DC performance (if the opamp has input bias currents) and high frequency performance (by ensuring both inputs, which have capacitance) are fed from the same source impedance).