Op-Amp Source-Factor Depth Scans

Input Resistance Three-Row Scan

Three input-resistance rows isolate the inverse part of the inverting gain formula. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

Six input ohms make one ampere

The source stays 6 volts. With 6 ohms at the input, current is 1 ampere and output is -2 volts.

Vout=−(2 ohm⋅6 V6 ohm)=−2 VV_{\text{out}}=-\left(2\ \text{ohm}\cdot\frac{6\ \text{V}}{6\ \text{ohm}}\right)=-2\ \text{V}
Six-ohm input rowThe largest input resistor makes the smallest current.+-op amprails -12 V to 12 VVout -2 VV+ 0 VV- 0 VI+ 0 AI- 0 ARin 6 ohmIin 1 ARf 2 ohmIf 1 AV+ = V-inverting

Three input ohms double the current

Halving the input resistance to 3 ohms raises input current to 2 amperes. Output becomes -4 volts.

Vout=−(2 ohm⋅6 V3 ohm)=−4 VV_{\text{out}}=-\left(2\ \text{ohm}\cdot\frac{6\ \text{V}}{3\ \text{ohm}}\right)=-4\ \text{V}
Three-ohm input rowThe current grows because the source voltage did not change.+-op amprails -12 V to 12 VVout -4 VV+ 0 VV- 0 VI+ 0 AI- 0 ARin 3 ohmIin 2 ARf 2 ohmIf 2 AV+ = V-inverting

Two input ohms give three amperes

The third row uses 2 ohms, so the fixed source drives 3 amperes and the output reaches -6 volts.

Vout=−(2 ohm⋅6 V2 ohm)=−6 VV_{\text{out}}=-\left(2\ \text{ohm}\cdot\frac{6\ \text{V}}{2\ \text{ohm}}\right)=-6\ \text{V}
Two-ohm input rowThree rows show the inverse resistance relationship.+-op amprails -12 V to 12 VVout -6 VV+ 0 VV- 0 VI+ 0 AI- 0 ARin 2 ohmIin 3 ARf 2 ohmIf 3 AV+ = V-inverting