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I have stumbled across a problem with potentiometer connections that I have no idea how to tackle. What are the differences of the two circuits below? I understand that the one on the right is the same as a 5kOhm resistance, but the left one I have no idea to what it is equivalent, and the difference appears to be huge in this other question of mine.

schematic

simulate this circuit – Schematic created using CircuitLab

EDIT This doubt arises from this circuit, where I want to calculate the output from that top part made of R2,C2,C1 and R3 which is supposed to generate an offset to Vin.

I was trying to ask just the part that was confusing me enter image description here

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  • \$\begingroup\$ taken in isolation neither circuit makes a lot of sense. \$\endgroup\$ – Jasen Dec 21 '19 at 20:09
  • \$\begingroup\$ @Oldfart please see the edit. I understand that this may be way to bad of a practice but it was my first (serious-ish) design and somehow it works. But is far from what I predicted, so I am getting the concepts and the circuit analysis wrong \$\endgroup\$ – Bidon Dec 21 '19 at 20:15
  • \$\begingroup\$ @Jasen, please see the edit and the comment replying to Oldfart \$\endgroup\$ – Bidon Dec 21 '19 at 20:15
  • \$\begingroup\$ Maybe it was mine. I referenced it in this question, but I was trying to narrow down the problem, because clearly I am misinterpreting things but I can't find the answer \$\endgroup\$ – Bidon Dec 21 '19 at 20:25
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schematic

simulate this circuit – Schematic created using CircuitLab

As drawn, both (a) and (b) would be incorrect ways to use a potentiometer. The problem in both case is that as the wiper approaches the ground rail the resistance is falling and you are approaching a short-circuit on your supply. If you do try this you may burn out the carbon track or the wiper contacts.

The correct way to use a potentiometer - where the potential is adjustable on the wiper - is shown in (c). Here the supply always sees a 10 kΩ load and the output voltage varies between 0 and 10 V as the pot's wiper is moved to the top. (The actual output voltage will depend on the load resistance too.)

Neither pot in your op-amp schematic are configured as (a) or (b).

  • The upper pot is the same as (c) but drawn sideways (which doesn't help understanding) and has two filter capacitors added.
  • The lower pot is in a gain adjusting circuit and even when adjusted to 0 Ω it only forms part of a divider circuit with R4 which ensures that the output is never short-circuited to ground.

From the comments:

Also, you say that in the upper there are two filter capacitors.

schematic

simulate this circuit

Figure 2. OP's schematic redrawn.

As redrawn here it should be more obvious that C1 is the decoupling capacitor on the 5 V supply and should be placed close to the op-amp.

C2 is a filter capacitor on the potentiometer wiper. It will help filter out any noise that comes through from the +5 V supply but as the wiper moves closer to the positive supply it's ability to filter becomes less due to the decreasing source resistance. In this case the pot would likely be left at mid position as it appears to be used to bias the non-inverting input to, typically, 2.5 V.

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  • \$\begingroup\$ Guys, what do you call "pot"? \$\endgroup\$ – Bidon Dec 21 '19 at 20:32
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    \$\begingroup\$ @Bidon: potentiometer. \$\endgroup\$ – Transistor Dec 21 '19 at 20:33
  • \$\begingroup\$ Also, you say that in the upper there are two filter capacitors. When I added them I thought that the first (C1) would act as a low pass filter to the DC input and C2 would act as a battery, so as to "store" the voltage drop across R3 since no current can flow to the OpAmp. Are they not working as I thought? \$\endgroup\$ – Bidon Dec 21 '19 at 20:33
  • \$\begingroup\$ @Bidon No current can flow into the opamp but it can flow out through R1. Or from R1 into your divider. \$\endgroup\$ – Oldfart Dec 21 '19 at 20:40
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    \$\begingroup\$ @Bidon: See the update. \$\endgroup\$ – Transistor Dec 21 '19 at 20:46
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The left circuit is not equivalent to 5kOhm by default, it depends on the position of the potentiometer wiper. So the potentiometer is equal to a resistor which value can be from 0 to 10kOhm. Although according to the circuit in reality it wouldn't be a good idea to turn the value down to zero which would result in a total short for the 10V voltage source.

In the circuit on the right the potentiometer is equal to 2 resistors which are connected in parallel. Always try to imagine potentiometer like 2 resistors. In this circuit the overall resistance has the highest value when potentiometer is in the middle position, because Rtotal = (R1 * R2) / (R1 + R2) (R1 and R2 not to be confused with the potentiometers on the left and right pictures). Therefore in the middle position Rtotal = (5k * 5k) / (5k + 5k) = 2.5k . When we move the wiper away to one or the other side from the middle position the overall resistance starts to drop. Let's say that R1 is now equal to 4k so the R2 must be 6k, because the pot total value is 10k. Now the overall resistance is Rtotal = (4k * 6k) / (4k + 6k) = 2.4k. And so on in case one is 1k and the other is 9k we have 0.9k ovarall resistance. The resistance will always be 10k in the denominator as the total value can't change (this makes calculations easier).

Picture of the equivalent circuit on the right. Hope that my answer helps.

enter image description here

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The meaning of the weird and (at first glance) pointless idea to connect the slider to one of the pot ends, has its explanation. If we use only two terminals - middle and outer, to make a variable resistor (ака rheostat), it is always possible for the slider to not make a good contact with the resistive film. In such cases as above (R5), the negative feedback will disappear and the op-amp will reach the supply rails... and this can be undesired... or, in other cases, even dangerous.

So, connecting the slider to one of the outer terminals, ensures that in the worst case, at least the total pot resistance will remain.

Here is the result of the first connection:

enter image description here

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R3 being between 0V and +5V selects a voltage between 0V and 5V this selection is somewhat distorted by any current that flows through R2

C1 is a decoupling capacitor and just works to stiffen the 5V source, C2 stiffens the voltage source produced by R3, reducing the effect of the R3 setting on the gain of the amplifier.

R5 sets the gain of the amplifier.

at its minimum setting R2/(R2+R1) at its maximum (R2/(R2+R1))(R4+R5)/R4

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  • \$\begingroup\$ Could you elaborate a bit more on the last comment? About R2/R1? I got lost there.. \$\endgroup\$ – Bidon Dec 21 '19 at 20:38
  • \$\begingroup\$ it's a voltage divider, I posted the wrong equation, see edit. \$\endgroup\$ – Jasen Dec 21 '19 at 20:40

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