Op-Amp Source-Factor Depth Scans

Inverting Input Voltage Three-Row Scan

Three source-voltage rows show the inverting amplifier as voltage-to-current-to-voltage. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

One input volt makes two output volts below zero

With input resistance fixed at 2 ohms, a 1 volt source drives 1/2 ampere into the virtual node. The output is -2 volts.

Vout=−(4 ohm⋅1 V2 ohm)=−2 VV_{\text{out}}=-\left(4\ \text{ohm}\cdot\frac{1\ \text{V}}{2\ \text{ohm}}\right)=-2\ \text{V}
One-volt inverting source rowThe input source first becomes current, then feedback voltage.+-op amprails -12 V to 12 VVout -2 VV+ 0 VV- 0 VI+ 0 AI- 0 ARin 2 ohmIin 1/2 ARf 4 ohmIf 1/2 AV+ = V-inverting

Two input volts double the current and output

Only the source voltage changes to 2 volts. The current is 1 ampere, so the same feedback resistor gives -4 volts.

Vout=−(4 ohm⋅2 V2 ohm)=−4 VV_{\text{out}}=-\left(4\ \text{ohm}\cdot\frac{2\ \text{V}}{2\ \text{ohm}}\right)=-4\ \text{V}
Two-volt inverting source rowThe current arrow and output label both double.+-op amprails -12 V to 12 VVout -4 VV+ 0 VV- 0 VI+ 0 AI- 0 ARin 2 ohmIin 1 ARf 4 ohmIf 1 AV+ = V-inverting

Three input volts give the third proportional point

At 3 volts, the input current is 3/2 amperes. The output reaches -6 volts.

Vout=−(4 ohm⋅3 V2 ohm)=−6 VV_{\text{out}}=-\left(4\ \text{ohm}\cdot\frac{3\ \text{V}}{2\ \text{ohm}}\right)=-6\ \text{V}
Three-volt inverting source rowThree source rows show the inverting output slope.+-op amprails -12 V to 12 VVout -6 VV+ 0 VV- 0 VI+ 0 AI- 0 ARin 2 ohmIin 3/2 ARf 4 ohmIf 3/2 AV+ = V-inverting