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Experiment 12 from my beginners book Make: More electronics is instructing me to build the following circuit..... I apologize for posting the whole circuit when my question has to do with just a small part of the circuit but I felt that if I didn’t post the whole circuit I would not be able to provide the whole context of my question.

In case it helps with the process of understanding the circuit below, the circuit is supposed to function as a microphone amplifier consisting of a pre-amplifier (LM741) and a power amplifier (LM386):

enter image description here

My question has to do with the 10 microfarad capacitor connected to the output of the LM741. The way I see it is that there are two scenarios occurring in this capacitor.

Scenario 1: When the LM741 output voltage goes from a set value to a more positive value electrons flow into the LM741 energizing the capacitor as show in the image below:

enter image description here

Scenario 2: When the LM741 output voltage goes from a set value to a more negative value electrons flow out of the LM741 de-energizing the capacitor as shown it the image below:

enter image description here

I think I am ok with Scenario 1 since it makes sense in my head (given my current level of understanding) that electrons will happily move to a point of higher potential.

However, Scenario 2 has me very confused because I don’t see how that works. The reasons I say this is because I don’t see where the electrons go as they are being pushed out of the coupling capacitor. The way I see it is that they can’t go to the negative terminal because there is no potential difference to make them want to go that way. So it looks to me like they are pushed through other wires connected to the negative terminal but that would cause all kinds of unexpected issues right?

Could someone please help me get a better grip on what is actually happening with this 10 microfarad capacitor as the output voltage of the LM741 swings up and down?

Thanks.

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    \$\begingroup\$ Few things here. 1. Conventional current is opposite of electron flow. 2. What happens is that the electric fields in the capacitor change with potential difference and electrons will quickly redistribute themselves as necessary. 3. In most cases in electronics the charge carriers and where they "go" is not important, only the fields established and the gross current through the system. Major exception: Internal functions of semiconductor devices \$\endgroup\$ – crasic Aug 24 '15 at 18:16
  • \$\begingroup\$ @crasic: Isn't the question about where electrons go a critical question in this case? I am asking this because won't the rouge electrons wreak havoc on other parts of the circuits if they are not corralled down into a proper location? \$\endgroup\$ – T555 Aug 24 '15 at 18:27
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    \$\begingroup\$ In my experience, on the level of macroscopic circuit analysis, lying to understand what the electrons are doing is really not productive from an analytical perspective. Let's take a step back. Your 741 is biased halfway between power and ground so that you can amplify both sides of a sinusoidal signal. The main function of that cap is to block DC while allowing AC to flow. You do not want to be amplifying DC signals at the power amp, else you may blow some speakers. From here, we can go back to electron flow to think about how the cap is managing this. \$\endgroup\$ – RYS Aug 24 '15 at 18:33
  • \$\begingroup\$ There are 10^21 or more free electrons in a few inches of 20 AWG copper wire. It would be exceedingly difficult to understand what every one of them does and goes and many of them don't "go" anywhere at all. What is important is that as a 'sea' of free electroncs in the metal, they will distribute themselves as the electric field in the system dictates. There is no such thing as "rogue" electrons in this scenario. If the voltage across the capacitor decreases, the electrons will redistribute themselves along the wire as necessary. \$\endgroup\$ – crasic Aug 24 '15 at 18:37
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    \$\begingroup\$ electrons aren't little ball-bearings, they are a heuristic we use to try to imagine what's happening at very small, unfamiliar scales. but electricity doesn't behave like familiar classical mechanics, it has its own rules. its hard to understand or discuss wave functions, which is what leads to this kind of confusion. electricity is more about the flow of electric current, not about electron migration. \$\endgroup\$ – MarkU Aug 24 '15 at 20:58
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Scenario 1: When the LM741 output voltage goes from a set value to a more positive value electrons flow into the LM741 energizing the capacitor as show in the image below:

This is not correct. The capacitor has already been 'energized' (charged up to half the supply voltage) by the op-amp's steady state output voltage. When a signal causes the op-amp output to go up or down the capacitor remains charged to half the supply voltage. It passes the signal through to the volume control by maintaining a fixed voltage between its terminals. Since the DC voltage at the pot is zero volts, at this point the signal will go above and below ground.

The capacitor is only able to pass the AC signal through because it does not become energized and de-energized as the signal goes up and down. As the frequency is lowered however, the capacitor will indeed start to charge and discharge, reducing the output signal level.

If you feed in a low frequency square wave you will see the output droop towards ground as the capacitor charges and discharges exponentially (with a time constant of 10uF * 10kΩ). The trace below is an example of what the input and output waveforms look like when a low frequency square wave is passed through a capacitor.

enter image description here

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This scenario can be confusing until you examine it from an op-amp perspective. Op-amps sole desire is to balance the voltage on their input pins (how they physically achieve this should be ignored at this point). So we exploit this property through feedback from the output to an input (usually negative feedback). Because the Op-amp can only control it's output it "drives" the output in one direction or the other to achieve this balance. (However it can only go as high or low as the rail ,+Vcc -Vdd or somewhere close to them. ie the output is not unlimited).

So in your case take a look at the datasheet of your op-amps and work out what the inputs/outputs are and do a schematic of these. I think you will get a different perspective on what is driving the circuit.

Again there are excellent youtube vids on this from w2aew and EEVblog

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You must first understand, that electrons don't cross the capacitor, this is why there is dielectric inside. They gather on on electrode, creating electric field inside the cap, and this way pushpin elenctrons on the other side. There always are electrodes there, because electrode is made of metal. So what it means is that only the "change of number of electrons on an electrode" is visible on the other side.

By the way, the Kirhoff's currents law still works, as there is "displacement current" inside the capacitor, which is fact is the change of electric field.

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  • \$\begingroup\$ In a coupling capacitor there is no change in the number of electrons on one plate, though. Both sides of the capacitor move up and down in voltage together. \$\endgroup\$ – endolith Apr 1 '16 at 21:56
  • \$\begingroup\$ Well, kind of. It's like saying that on a small resistor the voltage on both sides is the same. I would prefer saying that one side voltage followsnthe other side voltage. \$\endgroup\$ – Gregory Kornblum Apr 2 '16 at 3:58
  • \$\begingroup\$ Yes, one side's voltage follows the other side's, so the voltage across the capacitor stays constant and it does not charge or discharge \$\endgroup\$ – endolith Apr 2 '16 at 4:38
  • \$\begingroup\$ Now your statement is just not true. Imagine simplest circuit: ac source, capacitor, resistor. \$\endgroup\$ – Gregory Kornblum Apr 2 '16 at 4:44
  • \$\begingroup\$ Yes it's true in the passband \$\endgroup\$ – endolith Apr 2 '16 at 4:57
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No, in scenario 2 there is a potential difference because for scnenario 2 the capacitor must be charged up and itself temporarily acting like a battery.

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There is a potential difference between the left side of the 10k resistor and the negative terminal, so current flows to the negative terminal. If I understand correctly, you think that electrons on the long negative terminal wire are all on the same potential level along the entire length of the wire, and thus, no current flows, because of a lack of potential difference? However this is a simplified schematic. In real life, if you would make a PCB in this exact shape, you should model the entire wire by infinitesimally small intervals which all are modeled by a resistor(actually a transmission line to be more correct). And in every interval, there is a potential difference with the next interval(a.k.a the electrons lose a small bit of energy in each interval until negative terminal level is reached).

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  • \$\begingroup\$ You said: If I understand correctly, you think that electrons on the long negative terminal wire are all on the same potential level along the entire length of the wire, and thus, no current flows, because of a lack of potential difference?......I say: I know electrons will flow to the negative terminal. The problem is that once they get there, there is no place for them to go because the negative terminal will not suck them up. Instead they will need to redistribute to other part of the circuit and I would guess end up creating all kinds of problems. \$\endgroup\$ – T555 Aug 24 '15 at 18:54
  • \$\begingroup\$ @T555 the wire is only at the same potential as a convenient approximation and simplification. In reality the electric field establishes a potential that the electrons will try to "minimize" Since the conductor has some resistance, any excess energy in an electron will be dissipated into the metal itself as heat (vibration). The electrons don't need to "go" anywhere, the surpluses and "flow" of electrons responsible for most of our macroscopic electronics is measured as tiny fraction of the electrons that are just there in the metal already (more free electrons than stars in the universe) \$\endgroup\$ – crasic Aug 24 '15 at 19:01
  • \$\begingroup\$ Sorry for misunderstanding. Why would the negative terminal not be able to "suck" up the electrons? It is the neg. Terminal of the 9vDC battery so from there on, the electrons are brought up back to the potential of the plus terminal while they flow towards it. \$\endgroup\$ – Armannas Aug 24 '15 at 19:01

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