I need an output of 882.3 mV in the highlighted node. According to my calculations the output must be that with Vin=15, R1=17K and R2=1k. So there must be something wrong with how the circuit is wired. What I’m not doing right?
Electronic – Whats wrong with this circuit
voltage divider
Related Solutions
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: $$ V_{out}|_{V_1 = 0} = V_2\frac{R_1}{R_1 + R_2} $$
I left out the units.
total resistance \$R\$
\$R_1\$ (between a and c) is in parallel to the rest of the resistors. The rest being \$R_2\$ parallel to \$R_3\$ (between a and b) in row to \$R_4\$ in parallel to \$R_5\$ (between b and c)
The general rule for two resistors in \$R_{a}\$ and \$R_{b}\$ in parallel:
$$R_a||R_b=\frac{R_aR_b}{R_a+R_b}$$
I gave it a try:
$$ \begin{align} R & = R_1||(R_2||R_3 +R_4||R_5)\\ & = R_1||(R_{23} + R_{45})\\ & = \frac{R_1(R_{23} + R_{45})}{R_1+R_{23} + R_{45}}\\ R_{23} & = \frac{R_2R_3}{R_2+R_3} = \frac{20.46}{9.5}\\ R_{45} & = \frac{R_4R_5}{R_4+R_5} = \frac{56}{15.6}\\ R & = \frac{1(\frac{20.46}{9.5} + \frac{56}{15.6})}{1+\frac{20.46}{9.5} + \frac{56}{15.6}} = \frac{\frac{20.46 \times 15.6 + 56 \times 9.5}{15.6 \times 9.5}}{\frac{15.6 \times 9.5 +20.46 \times15.6 +56 \times9.5}{15.6 \times 9.5}} =\frac{20.46 \times 15.6 + 56 \times 9.5}{15.6 \times 9.5 +20.46 \times15.6 +56 \times9.5} \\ R & \approx 0.85 \end{align} $$
voltage \$U_{ab}\$ via voltage divider
I use U for voltage, not V.
\$R_1\$ being in parallel to the rest of the resistors means that there's the same voltage over both of them. The voltage divider divides the voltage \$U_{ac}\$ into \$U_{ab} + U_{ac}\$
The general rule for a voltage divider for two resistors in \$R_{ab}\$ and \$R_{bc}\$ in row: $$\frac{U_{ab}}{U_{ac}}=\frac{R_{ab}}{R_{ac}}=\frac{R_{ab}}{R_{ab} + R_{bc}}$$
I gave it a try: $$ \begin{align} \frac{U_{ab}}{U_{ac}} &= \frac{R_{23}}{R_{23} + R_{45}} \\ &= \frac{ \frac{20.46}{9.5}}{ \frac{20.46}{9.5} + \frac{56}{15.6}} = \frac{ \frac{20.46}{9.5}}{ \frac{20.46 \times 15.6 + 56 \times 9.5}{9.5 \times 15.6}} = \frac{20.46 \times 15.6}{20.46 \times 15.6 + 56 \times 9.5}\\ &\approx 0.3750 \\ U_{ac} &= 5 \\ U_{ab} &= \frac{20.46 \times 15.6}{20.46 \times 15.6 + 56 \times 9.5} \times 5\\ &\approx 1.8750 \end{align} $$
current \$I_{R_2}\$ via current divider
The current divider divides the current \$I_{ab}\$ into \$I_{R_2} + I_{R_3}\$
The general rule for a current divider for two resistors in \$R_{a}\$ and \$R_{b}\$ in parallel: $$\frac{I_{a}}{I_{a} + I_{b}}=\frac{I_{a}}{I_{ab}}=\frac{R_{a} || R_{b}}{R_{a}}= \frac{R_aR_b}{R_a(R_a +R_b)}=\frac{R_{b}}{R_{a} + R_{b}}$$
The general rule for resistance, voltage and current (ohm's law) : $$R = \frac{U}{I} \iff I = \frac{U}{R}$$
I gave it a try:
$$ \begin{align} I_{ab} &= \frac{U_{ab}}{R_{ab}} = \frac{U_{ab}}{R_{23}}\\ &= \frac{\frac{20.46 \times 15.6}{20.46 \times 15.6 + 56 \times 9.5} \times 5}{\frac{20.46}{9.5}} = \frac{15.6 \times 9.5}{20.46 \times 15.6 + 56 \times 9.5} \times 5\\ &\approx 0.8706\\ \frac{I_{R_2}}{I_{ab}} &= \frac{R_3}{R_2 + R_3} = \frac{3.3}{9.5} \\ &\approx 3.4747 \\ I_{R_2} &= \frac{R_3}{R_2 + R_3} \times I_{ab} = \frac{3.3}{9.5} \times \frac{15.6 \times 9.5}{20.46 \times 15.6 + 56 \times 9.5} \times 5 = \frac{15.6 \times 3.3}{20.46 \times 15.6 + 56 \times 9.5} \times 5\\ &\approx 0.3024\\ \end{align} $$
power \$P_{R_1}\$
Given the overall voltage \$U_{ac}\$ and the resistance \$R_1\$ the power \$P_{R_1}\$ can be calculated.
The general rule for power: $$P = U \times I$$ With ohm's law: $$P =\frac{U^2}{R}$$
I gave it a try: $$ \begin{align} P_{R_1} &=\frac{U_{ac}^2}{R} =\frac{5^2}{\frac{20.46 \times 15.6 + 56 \times 9.5}{15.6 \times 9.5 +20.46 \times15.6 +56 \times9.5}} =\frac{25 \times 15.6 \times 9.5 +20.46 \times15.6 +56 \times9.5}{20.46 \times 15.6 + 56 \times 9.5} \\ &\approx 5.3528 \end{align} $$
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Best Answer
It's funny, I teach EE and I recently gave a circuit quite similar to this to my student. The problem here is all about matching impedance. Because of the inverting amp, you have a virtual ground on the negative port of the op-amp. That virtual ground is the root cause of your impedance problem along with closely matched resistors.
With a small redrawing you can create a small model of what is going on: 17k in series with two 1k resistor in parallel that forms a voltage divider. Thus, you're input impedance of your op-amp is greatly reduced.
Vout = 1k//1k / (1k//1k + 17k) * 15 = 0.5 / (0.5 + 17) * 15 = 428.57 mV
To correct this you can either A) put a voltage follower between the voltage divider and the 1k resistor of the summing amp or B) increase all the 1k resistor of the summing amp to a much greater value. E.G: 30k.
30k // 1k = 960 ohm for the lower equivalent resistor. 0.96 / (0.96 + 17) * 15 = 801.7 mV
You could increase the summing amp resistors even more if you really want to push the theory to the limit. But, most often you try to keep feedback resistors in the 1k-100k range.