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I know this is certainly a beginner's question but am a first year student so hopefully I can be forgiven.

I was looking through a electronics booklet provided by IKES website, and it's talking about comparators:

Comparator.

Now, am very confused, I know that with voltage dividers you set the output voltage via series, as it cannot be dissipated equally without equal resistors. Yet with parallel circuits my understanding is that you have equal voltage, no matter the resistance for either resistors.

To me it seems that it holds true for the comparator too, why should there be an exception to the rule with unequal distributed voltage? Surely it would be set up like a voltage divider circuit right?

Or can you have it set in parallel as shown? If so, why is this an exception to the rule?

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    \$\begingroup\$ I don't follow what you are saying. \$\endgroup\$ – Andy aka Mar 15 '14 at 12:51
  • \$\begingroup\$ Which components do you think are in parallel? It might help to redraw your schematic with designators so you can name them. \$\endgroup\$ – The Photon Mar 15 '14 at 15:46
  • \$\begingroup\$ Hey there. It's not mine it's actually directly out of a support booklet, the IKES website. Recommended by the college. The resistors are in parallel right aren't they? Since, if we personify the current for a sec, they can 'choose', which path to take. \$\endgroup\$ – user2901512 Mar 15 '14 at 19:05
  • \$\begingroup\$ You could say that the pairs of resistors are in parallel but the individual resistors are not. The resistors form two voltage dividers that connect to the two inputs of the op amp. \$\endgroup\$ – Joe Hass Mar 15 '14 at 22:05
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This is sort of an interesting circuit; instead of using a single-supply voltage, a dual supply (positive and negative) is used so the output of the comparator swings either all the way positive (relative to 0v) and lights the green LED, or all the way negative (again with respect to 0v) and lights the red LED. Without the negative supply voltage, the output of the comparator could not go negative.

To make this clearer, I have added two voltage sources (6v batteries) to show how +Ve, -Ve and 0v could be supplied. So +Ve is +6v, and -Ve is -6v with respect to ground (0v).

enter image description here

Furthermore, two voltage dividers have been used; this is obviously for demonstration purposes (such as a lab experiment); usually there might be one voltage divider connected to the - side of the comparator, and the + side would be fed from a voltage connected to some other part of the circuit that is to be compared to the reference.

In this case, the - side is connected to a voltage divider made up of a fixed resistor R4 and a potentiometer R3, so the - voltage can be varied causing the comparator to change states.

While R1/R2 is in parallel with R3/R4, they have no effect on one another, as they are both connected to the opposite rails +Ve and -Ve, and their midpoints are isolated (both connected to the high-impedance inputs of the comparator.)

The voltage presented to the - pin of the comparator will be

 (R3 / (R3 + R4)) * (+Ve - -Ve) + -Ve

and the voltage fed to the + pin will be

(R2 / (R1 + R2)) * (+Ve - -Ve) + -Ve

As an example, lets suppose R1 = 100 Ω and R2 = 300 Ω.

The voltage fed into the + pin of the comparator would then be:

(300 / (100 + 300)) * (6 - (-6)) + (-6) = 3v
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  • \$\begingroup\$ This is a great answer. Very clear. Plus the work to fix the schematic. \$\endgroup\$ – gsills Mar 15 '14 at 20:07
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I think your confusion is about the rule that "when components are in parallel, the voltages across them are equal".

Let's look at the simplest parallel circuit:

enter image description here

Resistors R1 and R2 are in parallel.

The reason for the rule is that potential (voltage) is a property of a node in a circuit or a position in space. If we have a charged particle at B, and we move it to A, no matter how we get to A (either by passing through R1 or R2), we will have to add the same amount of energy to get there, defined by the potential difference (voltage) between A and B.

Now let's look at two voltage dividers in parallel:

enter image description here

Now the R1-R3 combination is in parallel with the R2-R4 combination.

If we take a charged particle from B to A, it doesn't matter what path we take, it always requires the same amount of energy.

But that doesn't force any equality between the potentials at X and Y. From an electrical potential point of view, X might be "10% of the way from A to B" and Y might be "90% of the way from A to B".

For example, let's say B is at 0 V, and A is at 100 V. Then if R1 and R3 are both 50 ohms, we know from the voltage divider rule that X is at 50 V. This has no dependence on R2, R4, or the voltage at Y.

If R2 is 30 ohms and R4 is 70 ohms, then the voltage divider rule tells us the voltage at Y is 70 V. Again, it doesn't depend at all on R1, R3, or the voltage at X.

Finally, let me double-check one other thing you might have gotten confused about. Here's your schematic again:

enter image description here

Notice where the two lines meet and I put a red circle, there's a heavy "dot" at the intersection of the two lines. This dot indicates the wires are connected at that intersection.

Notice the other place where two lines meet and I put a blue circle. There's no dot at that intersection. This means the two wires are not connected there. So there's no parallel connection between the upper two resistors. The only parallel connection is between the voltage dividers as a whole (Like my second example, above).

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  • \$\begingroup\$ Ah! I think I could on to what your saying. It's not that the rule doesn't apply overall, it's the percentage of the voltage dissipated across the divider that matters, and thus a high resistance, at the variable resistor will dissipate more voltage, leaving less for the output. Yet a small resistor at the variable will do the opposite, setting the inv or non inv at a higher baseline, making it more difficult for the inv or non inv to match that baseline... Right? \$\endgroup\$ – user2901512 Mar 15 '14 at 19:21
  • \$\begingroup\$ @user2901512, I think you've got it. Only we usually say the resistor drops voltage and dissipates heat. \$\endgroup\$ – The Photon Mar 15 '14 at 20:04
  • \$\begingroup\$ The Voltage across either resistor pair will be +Ve - (-Ve), or 2Ve. The current across each pair will be I = 2Ve/(Rtop +Rbottom) (From Ohms Law, V=IR, or V/R = I). Each resistor pair will dissipate power P = IV and with some substititing P = IxIR, or P = VxV/R. In this case, the 2nd is easier to start with, so P=2Ve*2Ve/(Rtop + Rbottom). The current into an opamp is usually next to nothing, just 0.000001A (1uA), so we can ignore that. The opamp output will swing towards the input with higher voltage. +Ve if V+ > V-, and -Ve if V- > V+. High ohm resistors, ex10K, dissipate little power. \$\endgroup\$ – CrossRoads Dec 27 '18 at 17:03
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I think you are becoming confused by the two voltage dividers that are in parallel across the +ve and -ve voltage rails.

Remember that each voltage divider is made up of a pair of resistors in series. There is two voltage dividers across the rails (and thus appear in parallel to you) simply because the source voltage is the same for each divider (across +ve and -ve) . Don't let this confuse you. Many circuits will commonly have many different voltage dividers just like this. Even inside the Op-Amp itself there will be several more similar voltage dividers (made from resistors in series) and these dividers will be across the +ve and -ve rails that go into the op-amp.

This is not uncommon. But it's also not any exception to the rule, as you say.

The voltage dividers are still made from a pair of series resistors. There is just two of them, in parallel.

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  • \$\begingroup\$ Wouldn't the current still dissipate equal amounts of energy though? I mean, the two paths are in series but don't the two paths form an overall resistance, taking into account the current could 'foresee' it's line of resistance as if it would 2 separate and single, higher, ohm resistors? Don't get me wrong, I know you're right, as well as the experts who designed the op amp and it's design :) But I just can't see how everything... works. \$\endgroup\$ – user2901512 Mar 15 '14 at 19:09
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R3 is a potentiometer while all the other resistors in the circuit are a fixed value. R3 is variable on purpose so you can adjust the voltage on the negative input of the comparitor to swing in one direction or the other. With more negative potential on the input the red led will light. Swing more positive, the green led will light.

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