Misunderstanding saturation and supply voltages in OP-AMP's

Question

What concept am I misunderstanding in calculating circuit parameters given their supply voltages?

In the following example, the non-inverting voltage is solved with respect to the output voltage, where:

^{simulate this circuit – Schematic created using CircuitLab}

KCL @ A: $$ 0 = \frac{V_{A}-V_{S}}{30k} + \frac{V_{A}-0}{10k} \; \Rightarrow \; \therefore V_{A} = V_{B} = 0.25V_{S}$$ KCL @ B: $$ 0 = \frac{(.25V_{S})-0}{4k} + \frac{(.25V_{S}) - V_{C}}{28k} \; \Rightarrow \; \therefore V_{C} = 2V_{S}$$

To find the range of values of $V_{S}$ where $V_{C}$ does not saturate, I apply the supply voltages:

$$ V_{C} = 8 \; \text{[v]}= 2V_{S} \; \Rightarrow \; \therefore V_{S} = 4 \; \text{[v]}$$

$$ V_{C} = 12 \; \text{[v]}= 2V_{S} \; \Rightarrow \; \therefore V_{S} = 6 \; \text{[v]}$$

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However, the solution (and similar solutions) uses a negative saturation voltage (-12 [v]) instead of the positive value as drawn in the OP-AMP circuit - resulting in (-6 [v]) as opposed to (+6 [v]).

Am I misunderstanding sign conventions, or the fundamental application of supply voltages?

^{**For clarity: An edit was made to correct a mismatch between the unedited KCL equations and the circuit schematic - the leftmost 10k resistor was incorrect, and edited to its correct 30k value. My apologies, but the original KCL equations, given solution, and question remain unchanged.}

Answer 1

The confusion is not necessarily your fault, although it would be useful to see a copy of the original illustration.

As you seem to have figured out, the 12 volt node is supposed to be -12 volt. Replacing 12 with -12 in your calculations gives the anticipated answer, which means that you understand how to do this.

So, why does it say 12 instead of -12? It shouldn't. Maybe you have a bad photocopy of the circuit. Maybe someone copied it wrong. Maybe the PDF isn't correctly rendered.

No electrical engineer would write 12 and then just assume that everyone understands that it is supposed to be -12. You might see something like +/-12, ±12, VEE, or similar, but the sign is never omitted. Not even if the pin is clearly labeled as being the negative terminal.

A few things makes it immediately obvious that there's something funky with the circuit:

1. Feeding an op-amp with 8 and 12 volts would be pretty unusual, especially so when the inputs and outputs are then referenced to ground. Even worse would be feeding it with 8 and 12 volts in the configuration shown, because the positive voltage is apparently lower than the negative.

2. Even if you do something like that, any reasonable schematic should have more positive voltages "higher" than less positive voltages. This means that 12 should be above 8. This stops the brain from going crazy when trying to visualize current flow against gravity.

Sadly, some textbooks have too many unrealistic circuits, so you can't always assume that it's a misprint. In this particular case, replacing 12 with -12 makes the circuit compliant with my two points, and everything is good. They even use realistic values for the resistors, which is not always true in textbooks. (Voltage divider with 1Ω resistors? Please.)

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That was the rant+lecture. Now for your question:

Am I misunderstanding sign conventions, or the fundamental application of supply voltages?

No. There is no secret sign convention. There must be no doubt in a schematic diagram. Back in the old days, people sometimes omitted the magnitude for capacitors. For example, they could write 10 and that would implicitly mean 10 µF. This is still causing great confusion for newbies, and the only reason it's not causing even greater confusion is that 10 farad is not realistic. Don't omit stuff.

I will also answer a question you posted in a comment:

Is the lower supply terminal always negative, or is it possible that the negative supply is positive?

The "negative supply" can indeed be positive. It would be unusual, and the op-amp's "positive supply" must be even more positive. All it sees is the difference between its positive and negative supply.

Answer 2

Firstly the gain is 4 by purely visual inspection and the opamp will saturate typically within 50 mV of either rail if the op amp was a rail to rail type or within about 1.5 volts of the rails for a standard push pull output. Some op amps like the LM324 will get to within 5 mV of the most negative rail.

I have no idea why you are seemingly making this more complex than it is.