Voltage divider network calculation

Question

Hello I am troubleshooting a voltage divider circuit in which I know the output going to U28 pin 3 should be around 700mv. What I am measuring however is around 7.5v. How do I calculate the v out for this circuit. I know I have to account for R83 because it is a potentiometer but still I should be able to math this out and get close to 700mv but I can’t seem to calculate it right even after watching YouTube. The top photo is the actual schematic drawing but I redrew it to in the way in which I thought it works because I noticed r85 and r86 share the same ground therefore r83 and r85 must be in parallel to r114. So did I draw this out right and how do I solve for vout?

Answer 1

The circuit can be redrawn (always a good idea) like this:

^{simulate this circuit – Schematic created using CircuitLab}

There are a couple of approaches to solving the voltage value for pin 3. The one that is likely taught earlier would use a process of applying Thevenin equivalents. Let's do that:

^{simulate this circuit}

Here, \$V_\text{TH}=9\:\text{V}\frac{R_{83}+R_{85}}{R_{83}+R_{85}+R_{113}}\$ and \$R_\text{TH}=\frac{R_{113}\cdot\left(R_{83}+R_{85}\right)}{R_{83}+R_{85}+R_{113}}\$.

At this point, it's a simple voltage divider. Let's pause for a moment and look at it more closely before moving on to the next step. Here, you can see that \$R_\text{TH}\$ is followed by a \$1\:\text{M}\Omega\$ resistor. Since \$R_\text{TH}\$ can't be very large by comparison (a few thousands of Ohms, but nothing close to the value of \$R_{114}\$), we can easily see even at this point that the potentiometer can only vary the pin 3 voltage a little bit. I think that's clear at this point, already.

The above was step 1. But there is another step yet needed to get the pin voltage:

$$\begin{align*} V_\text{pin 3} &= V_\text{TH}\frac{R_{86}}{R_\text{TH}+R_{114}+R_{86}}\tag{Step 2}\\\\ &=9\:\text{V}\cdot\frac{R_{83}+R_{85}}{R_{83}+R_{85}+R_{113}}\cdot\frac{R_{86}}{\frac{R_{113}\cdot\left(R_{83}+R_{85}\right)}{R_{83}+R_{85}+R_{113}}+R_{114}+R_{86}}\\\\ &=9\:\text{V}\cdot R_{86}\cdot\frac{R_{83}+R_{85}}{R_{113}\cdot\left(R_{83}+R_{85}\right)+\left(R_{114}+R_{86}\right)\cdot\left(R_{83}+R_{85}+R_{113}\right)}\\\\ &=9\:\text{V}\cdot \frac{R_{86}}{R_{113}+\left(R_{114}+R_{86}\right)\cdot\left(1+\frac{R_{113}}{R_{83}+R_{85}}\right)}\tag{Final} \end{align*}$$

That last step is actually very helpful because it isolates potentiometer \$R_{83}\$ and allows you to see better what impact it has on the whole structure of the answer here.

Another approach would be to use nodal analysis and solve the simultaneous equations (two of them.) But the result would be the same and the above relies on limited knowledge of less powerful tools (if perhaps still some algebra.)

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I would still like to hear about what you find in your process of tracking down the problem, too. Please let us know what you find.

Answer 2

This mixture of resistors can be presented as two cascaded voltage dividers since the second one has high input resistance and loads slightly the first. The variable resistor should be considered in the middle position (R = 2.5 k). Then VOUT = 9.15/(10 +15) x 147/(1000 + 147) = 0.7 V.