Electronic – Inverting Op Amp Finding resistor value of voltage divider
operational-amplifiervoltage divider
A small clarification V_2 = -100V_1
I don't think my analysis is correct since I ended up with R1 as a denominator.
Best Answer
The practical thinking engineer would immediately see in this configuration an inverting amplifier with a voltage divider included in the output. Usually R2 >> R3||R4 (i.e., the voltage divider has low output resistance); so the gain is simply -R2/R1 X (R3 + R4)/R3.
The name "T network" is misleading and it prevents understanding. It is much better to think of the circuit as a "twice-disturbed operational amplifier with negative feedback" (here these disturbances are useful). The first "disturbance" is caused by the R1-R2 summing circuit; the second "disturbance" is caused by the R3-R4 voltage divider. The two devices are cascaded attenuators that form the feedback network.
The "golden rule" is: if you take the output after a disturbance, it will be eliminated by the op-amp with excessive voltage before the disturbance; if you take the output before the disturbance, it will be an amplifying output. This trick is used in all the op-amp amplifying circuits with negative feedback to make a follower (inverter) amplify.
According to these explanations, V1 is "undisturbed output", V2 - "once-disturbed output" and V3 is a "double-disturbed output":) ... and V1 < V2 < V3.
I can't quite determine the current flow because the lines from \$V_2\$ and \$V_1\$ carry current that is equal to \$V_2/R_2\$ and \$V_1/R_1\$ at the point in between the two resistors. How does this return to ground, and how would I begin to apply Kirchoff's Law for the loop?
Hmm. I don't agree with this. I think you came up with these equations with the old schematic, which had an extra ground connection.
The actual current through the resistors is
$$
I_1 = \frac{V_1 - V_{out}}{R_1}
$$
and
$$
I_2 = \frac{V_{out} - V_2}{R_2}
$$
Since there is no path for current to take another loop, these currents must be the same! That means that
$$
\frac{V_1 - V_{out}}{R_1} = \frac{V_{out} - V_2}{R_2}
$$
and some math gives you
$$
V_{out}
= \frac{V_1 R_2 + V_2 R_1}{R_1 + R_2}
= \frac{V_1 R_2 + V_1 R_1 - V_1 R_1 + V_2 R_1}{R_1 + R_2}
= V_1 + (V_2 - V_1)\frac{R_1}{R_1 + R_2}
$$
Notice that if \$V_1 = 0\$, you get back the equation that you're used to:
$$
V_{out}|_{V_1 = 0} = V_2\frac{R_1}{R_1 + R_2}
$$
Best Answer
The practical thinking engineer would immediately see in this configuration an inverting amplifier with a voltage divider included in the output. Usually R2 >> R3||R4 (i.e., the voltage divider has low output resistance); so the gain is simply -R2/R1 X (R3 + R4)/R3.
The name "T network" is misleading and it prevents understanding. It is much better to think of the circuit as a "twice-disturbed operational amplifier with negative feedback" (here these disturbances are useful). The first "disturbance" is caused by the R1-R2 summing circuit; the second "disturbance" is caused by the R3-R4 voltage divider. The two devices are cascaded attenuators that form the feedback network.
The "golden rule" is: if you take the output after a disturbance, it will be eliminated by the op-amp with excessive voltage before the disturbance; if you take the output before the disturbance, it will be an amplifying output. This trick is used in all the op-amp amplifying circuits with negative feedback to make a follower (inverter) amplify.
According to these explanations, V1 is "undisturbed output", V2 - "once-disturbed output" and V3 is a "double-disturbed output":) ... and V1 < V2 < V3.