Secret life of pots - when is it okay to use the taper trick?

[ItsGiusto]

Hey all. I'm sure you're familiar with this:
http://www.geofex.com/article_folders/potsecrets/potscret.htm

I think the theory of the tapering trick (adding a resistor in parallel with a linear pot to make it logarithmic) is awesome, and I understand some of the nuance behind the voltage divider vs variable resistor thing (doing the trick lowers the effective resistance of the pot, so the trick is less applicable for variable resistors, unless you want a lower resistance value).

However, what I don't understand, and what I'd like to understand more, is when is it okay to do the trick on a pot used as a voltage divider? Is it always okay, or are there times that it would be bad to lower the effective resistance of a voltage divider pot?

As a more general question, when do resistance values matter in voltage dividers? If someone were making a voltage divider, say, maybe this one:
https://upload.wikimedia.org/wikipedia/commons/thumb/3/31/Impedance_voltage_divider.svg/1200px-Impedance_voltage_divider.svg.png
If they wanted Vout to be half of Vin, they'd make Z1 = Z2. That makes sense, but how do they choose the value? You could make Z1 = Z2 = 500 ohm. Or you could make Z1 = Z2 = 500k. Or 2M. What's the functional difference, especially if that part of the circuit is isolated?

[GGBB]

Ohm's Law.

The resistance determines how much current can flow. The value should be chosen based on the current requirements of the powered circuit/device. The power rating (watts) of the resistors or pot needs to be high enough to handle the current passing through it.

[Fancy Lime]

Hi Justin,

lets start with the resistance. For a voltage divider all by itself it does not matter at all if you use a 10k or 1M pot (assuming the same taper), if we ignore currents, which we can usually do in most stompboxes (not in poweramps and some specialized stompbox applications but judging by your question that does not seem the concern here). So "if that part is isolate" as you say, you can choose whatever (because then there is no current at all). But in a real circuit, the voltage divider interacts with what comes before and after it and is never really isolated.

Example: You have one amplification stage (implemented e.g. with a tube, transistor or op-amp), then the voltage divider as a gain pot and then another amplification stage. The first stage has an output impedance and an output coupling capacitor. The Impedance adds to the series resistance of the voltage divider. If the output impedance is large compared to the voltage divider resistance, your maximum setting (in this case gain) is decreased. Lets say, the output impedance is 100k and you use a 10k pot, then your maximum setting only has 1/11 of the voltage going into the divider. If you use a 1M pot here, it is 10/11. The output cap forms a high pass filter with the "to-ground" part of the voltage divider, potentially causing bass loss with low pot resistances. So from the perspective of the previous stage, you want the pot resistance to be as high as possible.

The stage after the voltage divider has an input cap (which doesn't concern us much here) and an input impedance. The impedance contributes to the voltage divider because it is parallel to the "to-ground" part of the divider. This becomes significant if the input impedance is similar to or smaller than the pot value. So from the perspective of the following stage, you want the pot resistance to be as small as possible.

In summary, it's a trade off. If your stages have low output and high input impedances, as they normally should in a good design, than a wide range of pots is normally fine. If that is not the case for some reason, then you need to consider all the consequences of one choice or the other. Basically any resistances (including the ones that may be hidden in amplifying elements) and capacitances that can interact with the divider need to be considered. For example tube or MOSFET stages can have relatively high output resistances (problematic) but have very high input impedances, if designed properly (good), whereas transistor stages can have very low output impedances (good) but often (especially in fuzz circuits, and on purpose too) low input impedances (problematic). So what value of pot you want depends on the circuit around it.

For the parallel resistor: That is always ok if it gets you the desired total resistance and characteristic (taper). When is that the case? See above. I wrote myself a little Fortran program that calculates the characteristics of a linear pot wired as a variable resistor with extra resistors in all possible configurations. I'm happy to share the source code and the linux-compiled version, if anyone is interested. Not that it does anything you couldn't do with the pencil and a piece of paper. If you are using windows or mac or something (BSD, anyone?) you'll have to compile it yourself, though.

Cheers,
Andy

[PRR]

> how do they choose the value? .. 500 ohm. .... Or 2M.

Or 0.001 Ohm. Or 34,567 Meg Ohms.

Can your source drive 1K? Can your load feed happily from 1 Meg?

[ItsGiusto]

> how do they choose the value? .. 500 ohm. .... Or 2M.

Or 0.001 Ohm. Or 34,567 Meg Ohms.

Can your source drive 1K? Can your load feed happily from 1 Meg?

Well, yeah, that's what I'm asking, isn't it? How do I know the answers to those questions?

[ItsGiusto]

Thanks for the reply! I understand a little better now, but I have more questions, included inline.

Hi Justin,

lets start with the resistance. For a voltage divider all by itself it does not matter at all if you use a 10k or 1M pot (assuming the same taper), if we ignore currents, which we can usually do in most stompboxes (not in poweramps and some specialized stompbox applications but judging by your question that does not seem the concern here).
Cool, yeah, I agree, for most of my purposes, it would seem that current doesn't matter (though I don't really know why exactly that is.) But I am also curious about applications for which current matters. Why is it that it might matter for something like a poweramp, and which specialized stompboxes might it matter for?

So "if that part is isolate" as you say, you can choose whatever (because then there is no current at all). But in a real circuit, the voltage divider interacts with what comes before and after it and is never really isolated.
I'm not too clear exactly on how the interactions between parts of circuits matter. I had always sort of assumed that coupling capacitors isolate portions of the circuit from each other for the purposes of resistance, but apparently that might not be the case?

Example: You have one amplification stage (implemented e.g. with a tube, transistor or op-amp), then the voltage divider as a gain pot and then another amplification stage. The first stage has an output impedance and an output coupling capacitor. The Impedance adds to the series resistance of the voltage divider. If the output impedance is large compared to the voltage divider resistance, your maximum setting (in this case gain) is decreased. Lets say, the output impedance is 100k and you use a 10k pot, then your maximum setting only has 1/11 of the voltage going into the divider. If you use a 1M pot here, it is 10/11. The output cap forms a high pass filter with the "to-ground" part of the voltage divider, potentially causing bass loss with low pot resistances. So from the perspective of the previous stage, you want the pot resistance to be as high as possible.
Cool, I think I kind of get this. You're saying that the output of the circuit forms its own voltage divider when considered with the 10k pot to ground? Is that what you're saying?


I think my biggest problem is that I've never gotten a precise explanation of what is meant when someone says a circuit has some value for output impedance, or input impedance. What exactly does that mean, and how do you measure it? And how can you design a circuit to have both large input impedance and small output impedance?

The stage after the voltage divider has an input cap (which doesn't concern us much here) and an input impedance. The impedance contributes to the voltage divider because it is parallel to the "to-ground" part of the divider. This becomes significant if the input impedance is similar to or smaller than the pot value. So from the perspective of the following stage, you want the pot resistance to be as small as possible.

In summary, it's a trade off. If your stages have low output and high input impedances, as they normally should in a good design, than a wide range of pots is normally fine. If that is not the case for some reason, then you need to consider all the consequences of one choice or the other. Basically any resistances (including the ones that may be hidden in amplifying elements) and capacitances that can interact with the divider need to be considered. For example tube or MOSFET stages can have relatively high output resistances (problematic) but have very high input impedances, if designed properly (good), whereas transistor stages can have very low output impedances (good) but often (especially in fuzz circuits, and on purpose too) low input impedances (problematic). So what value of pot you want depends on the circuit around it.

For the parallel resistor: That is always ok if it gets you the desired total resistance and characteristic (taper). When is that the case? See above. I wrote myself a little Fortran program that calculates the characteristics of a linear pot wired as a variable resistor with extra resistors in all possible configurations. I'm happy to share the source code and the linux-compiled version, if anyone is interested. Not that it does anything you couldn't do with the pencil and a piece of paper. If you are using windows or mac or something (BSD, anyone?) you'll have to compile it yourself, though.

Cool, I'd be happy to see that! I don't have a linux machine so I guess I'll take a look at the source code, at least. Thanks!

Cheers,
Andy

[Fancy Lime]

Hi Justin,
answers in red.

Thanks for the reply! I understand a little better now, but I have more questions, included inline.

Hi Justin,

lets start with the resistance. For a voltage divider all by itself it does not matter at all if you use a 10k or 1M pot (assuming the same taper), if we ignore currents, which we can usually do in most stompboxes (not in poweramps and some specialized stompbox applications but judging by your question that does not seem the concern here).
Cool, yeah, I agree, for most of my purposes, it would seem that current doesn't matter (though I don't really know why exactly that is.) But I am also curious about applications for which current matters. Why is it that it might matter for something like a poweramp, and which specialized stompboxes might it matter for?
Well, what I meant with "we can ignore currents" is this: Because Ohms law (I=V/R) applies, we don't need to consider currents (I) as separate entities, if we are already considering resistances (R) and voltages (V). We need two of the three, the third is implied. We could just as well think in currents and voltages and ignore the resistance but for me R and V (and most people i guess) are more intuitive than currents. In a poweramp we are dealing with, well, power. At that point we need to observe the power ratings and thermal ratings and the latter are essentially current ratings because current through a resistor produces heat. Thats how a (classical) light bulb works. In stompboxes we sometimes need to fill or drain a large capacitor and we can only fill those with current, not with voltage. So to calculate the relevant time constant we need the current. This can be important for reference voltage circuits, peak detectors or envelope followers.

So "if that part is isolate" as you say, you can choose whatever (because then there is no current at all). But in a real circuit, the voltage divider interacts with what comes before and after it and is never really isolated.
I'm not too clear exactly on how the interactions between parts of circuits matter. I had always sort of assumed that coupling capacitors isolate portions of the circuit from each other for the purposes of resistance, but apparently that might not be the case?
No, that is not generally true. It is true for DC resistance but AC resistance comes through capacitors. What we call impedance can be viewed in this context as AC resistance (there is more to impedance but that other part is not really important here).

Example: You have one amplification stage (implemented e.g. with a tube, transistor or op-amp), then the voltage divider as a gain pot and then another amplification stage. The first stage has an output impedance and an output coupling capacitor. The Impedance adds to the series resistance of the voltage divider. If the output impedance is large compared to the voltage divider resistance, your maximum setting (in this case gain) is decreased. Lets say, the output impedance is 100k and you use a 10k pot, then your maximum setting only has 1/11 of the voltage going into the divider. If you use a 1M pot here, it is 10/11. The output cap forms a high pass filter with the "to-ground" part of the voltage divider, potentially causing bass loss with low pot resistances. So from the perspective of the previous stage, you want the pot resistance to be as high as possible.
Cool, I think I kind of get this. You're saying that the output of the circuit forms its own voltage divider when considered with the 10k pot to ground? Is that what you're saying?

I think my biggest problem is that I've never gotten a precise explanation of what is meant when someone says a circuit has some value for output impedance, or input impedance. What exactly does that mean, and how do you measure it? And how can you design a circuit to have both large input impedance and small output impedance?
Yes that is what I am saying. Impedance is complicated. You cannot really measure it but you can calculate it. But how that is done depends on the circuit topology. There is no universal answer. In fact these calculations are so complicated and involve so many poorly constrained factors (like device parameters of transistors or op-amps, which are different for different devices of the same type, and change with device temperature) that we normally use simplified approximations. Which ones to use depends on what you are designing. There is no easy pass here, you will have to read some theory to learn these things. There are great books out there, I guess, but I learned what I know from the many many tutorials, explanations and forum posts of the veteran experts. The DIY FAQ, GEO FAQ, AMPAGE, GEOFEX and AMZ links near the top of this page are a great starting point. The website electronics-tutorials.ws is also great for simple theory introductions.

The stage after the voltage divider has an input cap (which doesn't concern us much here) and an input impedance. The impedance contributes to the voltage divider because it is parallel to the "to-ground" part of the divider. This becomes significant if the input impedance is similar to or smaller than the pot value. So from the perspective of the following stage, you want the pot resistance to be as small as possible.

In summary, it's a trade off. If your stages have low output and high input impedances, as they normally should in a good design, than a wide range of pots is normally fine. If that is not the case for some reason, then you need to consider all the consequences of one choice or the other. Basically any resistances (including the ones that may be hidden in amplifying elements) and capacitances that can interact with the divider need to be considered. For example tube or MOSFET stages can have relatively high output resistances (problematic) but have very high input impedances, if designed properly (good), whereas transistor stages can have very low output impedances (good) but often (especially in fuzz circuits, and on purpose too) low input impedances (problematic). So what value of pot you want depends on the circuit around it.

For the parallel resistor: That is always ok if it gets you the desired total resistance and characteristic (taper). When is that the case? See above. I wrote myself a little Fortran program that calculates the characteristics of a linear pot wired as a variable resistor with extra resistors in all possible configurations. I'm happy to share the source code and the linux-compiled version, if anyone is interested. Not that it does anything you couldn't do with the pencil and a piece of paper. If you are using windows or mac or something (BSD, anyone?) you'll have to compile it yourself, though.

Cool, I'd be happy to see that! I don't have a linux machine so I guess I'll take a look at the source code, at least. Thanks!
I'll dig up the source code and send you a PM. Do you know some Fortran? The source code is not well commented at the moment and it may not be self explanatory. I'll try and add some comments.

Cheers,
Andy

[ItsGiusto]

>>> I'll dig up the source code and send you a PM. Do you know some Fortran? The source code is not well commented at the moment and it may not be self explanatory. I'll try and add some comments.

No, I don't know Fortran, but I'm a software engineer, I teach myself new languages all the time. From my quick look at Fortran, it looks kind of like assembly or basic. It'll be fun to try to figure out!


Is the output impedance of a circuit the same as its input impedance? Or can they have two separate values? I'm guessing the later, based on how people use those terms, but I don't know why that would be, and how to make a circuit be that way.

I'm still interested in knowing more about input impedance vs output impedance. I know the theoretical explanation of the general concept of "impedance" and I used to know how to calculate the impedance of an RC circuit, but I've never known how to use these concepts, because I've never had a good intuitive understanding of what it is. I've been messing around with circuits for years, and definitely the lack of a true intuitive understanding of how impedance works has been impeding my progress. If anyone has any sources to recommend, please let me know!

[reddesert]

https://en.wikipedia.org/wiki/Input_impedance

Loosely, many stages of a circuit can be seen as a voltage source, and then something that senses the voltage and processes it - amplifies it, filters it, etc.  For example, you put a 9V battery across a mini lightbulb and it lights up.

Ideally, the battery is a voltage source V, and the lightbulb is like a resistor of value R (it's not an Ohmic linear resistor, but ignore that for now). Some current flows through the bulb, I = V/R, and power P= V*I is dissipated in the form of light and heat.

Now, we know that a 9V battery can't supply a very large current. You can't start a car with it. How do we model that? A real battery is not an ideal voltage source. It has some internal resistance that limits the current it can output. You can model it as a voltage source in series with a resistor, labeled V_s and R_s in the figure in the Wikipedia article linked above.  For example, if the battery is 9 V and has an internal resistance of 10 ohms, and you put an external resistor of 100 ohms across the battery terminals, the current flowing in the circuit is not 9V/100 = 90 mA.  It's I= 9V/(100+10) = 82 mA, because the internal resistance is in series with the external resistor.

The voltage drop across the external resistor will be V_drop = I * R = 82 mA * 100 ohms = 8.2 Volts. See what happened? Even though you have a 9V battery, the voltage being delivered is only 8.2 V.  The battery is being loaded because the external load resistor is small and it is being asked to deliver more current than a 9V battery can handle for a long time.

In this case, the internal resistance of the battery is its output impedance, and the 100 ohm resistance of the load is the input impedance. If instead of a battery we think of a guitar pickup and an amp, it would be the same situation, except that we're interested in AC frequencies and have to calculate the AC impedance. Guitar pickups are complicated because they're inductive. For effects circuits, you can usually calculate the impedances from resistance and capacitance.

The example of the loaded battery shows why it is usually good to have low output impedance and high input impedance. The output impedance tells how much current the circuit could theoretically deliver to the next stage. The input impedance tells what load the circuit presents to the previous stage. For ex, amps typically have an input impedance of 500K - 1M ohms.

[ItsGiusto]

https://en.wikipedia.org/wiki/Input_impedance

Loosely, many stages of a circuit can be seen as a voltage source, and then something that senses the voltage and processes it - amplifies it, filters it, etc.  For example, you put a 9V battery across a mini lightbulb and it lights up.

Ideally, the battery is a voltage source V, and the lightbulb is like a resistor of value R (it's not an Ohmic linear resistor, but ignore that for now). Some current flows through the bulb, I = V/R, and power P= V*I is dissipated in the form of light and heat.

Now, we know that a 9V battery can't supply a very large current. You can't start a car with it. How do we model that? A real battery is not an ideal voltage source. It has some internal resistance that limits the current it can output. You can model it as a voltage source in series with a resistor, labeled V_s and R_s in the figure in the Wikipedia article linked above.  For example, if the battery is 9 V and has an internal resistance of 10 ohms, and you put an external resistor of 100 ohms across the battery terminals, the current flowing in the circuit is not 9V/100 = 90 mA.  It's I= 9V/(100+10) = 82 mA, because the internal resistance is in series with the external resistor.

The voltage drop across the external resistor will be V_drop = I * R = 82 mA * 100 ohms = 8.2 Volts. See what happened? Even though you have a 9V battery, the voltage being delivered is only 8.2 V.  The battery is being loaded because the external load resistor is small and it is being asked to deliver more current than a 9V battery can handle for a long time.

In this case, the internal resistance of the battery is its output impedance, and the 100 ohm resistance of the load is the input impedance. If instead of a battery we think of a guitar pickup and an amp, it would be the same situation, except that we're interested in AC frequencies and have to calculate the AC impedance. Guitar pickups are complicated because they're inductive. For effects circuits, you can usually calculate the impedances from resistance and capacitance.

The example of the loaded battery shows why it is usually good to have low output impedance and high input impedance. The output impedance tells how much current the circuit could theoretically deliver to the next stage. The input impedance tells what load the circuit presents to the previous stage. For ex, amps typically have an input impedance of 500K - 1M ohms.

Thanks! That's probably the best explanation I've heard yet, and the thought experiment with the battery makes a lot of sense. I still feel like I need to apply it to a more real situation, of understanding why this would happen for something that's not a battery, like a transistor amplification phase or something. I also still wish I understood more exactly where the input impedance of one stage ends and the output impedance of the next stage begins, and how, practically, to design something with high input impedance and low output impedance.

[PRR]

The radio/petal battery starting a car is a fine example.

I happen to know the impedances of my house electric system. The (too-long) power line is 0.4 Ohms, fed 250V.

Say I have 250 Watts of lights, 250V @ 1 Amp or 250 Ohms. A 250 Ohm load on a 0.5 Ohm source is a Voltage Divider (key concept!!). I will get 0.998 of 250V or 249.5V. Pretty good juice.

Now I run the dryer, the hot water, the A/C, the welder.... perhaps 100 Amps. 250V @ 100A is 2.5 Ohms. 2.5 Ohms fed from 0.5 Ohms gives 0.8333, or 208 Volts. On nominal 240V lamps and loads, everything runs dim/cool.

> how, practically, to design

As you say: Amplification.

Another strained analogy. I have a backhoe. I can lean out and pull the big claw by hand. It is a very "low impedance": it takes a LOT of force to move the ton of arm and a ton of hard clay. OTOH I can put my finger on a lever and let the hydraulic system "amplify" my strength. It amplifies both ways: 1 pound on the lever is over 2,000 pounds on the bucket, and 2 inches on the lever may be 6 feet on the bucket.

Take those thoughts out into the familiar world (nail pulling, power steering and gas-pedal....) and meditate before you ponder imponderable electricity.

[ItsGiusto]

Say I have 250 Watts of lights, 250V @ 1 Amp or 250 Ohms. A 250 Ohm load on a 0.5 Ohm source is a Voltage Divider (key concept!!). I will get 0.998 of 250V or 249.5V. Pretty good juice.

Now I run the dryer, the hot water, the A/C, the welder.... perhaps 100 Amps. 250V @ 100A is 2.5 Ohms. 2.5 Ohms fed from 0.5 Ohms gives 0.8333, or 208 Volts. On nominal 240V lamps and loads, everything runs dim/cool.

Tell me if I'm wrong, but the way I'm seeing this is that because you're powering all those loads in parallel, resistance drops considerably, causing the load to be seen to be far less. This causes the aforementioned issues with a voltage divider formed by the load and the internal resistance of the power source, because the effective resistance of everything in parallel is much closer to the internal resistance of the load.

Seems to me like all of this could be solved if we wired things in series. The more loads we put in the circuit, the higher the resistance would be. Of course we'd need to keep increasing the voltage, too, to match the load, which I gather is infeasible, though I guess I don't know exactly why.

Or better yet, couldn't we just regulate the voltage over the load in some way? If we needed 9v to be over the load, why couldn't we use a voltage regulator, or stick some zener diodes in parallel with the load to ensure that the load always sees the correct voltage. I know there's probably a million reasons this wouldn't work, but with my limited knowledge of this subject, I don't fully understand why.



Dumb question, but why is it easier for a constant voltage as opposed to, say, a constant current be provided via any power source? Maybe it's just my high school physics failing me, but I remember voltage bring explained to me as just somewhat of a made-up concept, merely being the potential to move charge, as opposed to standing for anything that's actually happening - that it was merely a simplification of the idea that something had the potential to cause a current. Yet, this doesn't seem to sit well with the fact that power sources like batteries and wall-plugs provide a constant voltage, which indicates to me that voltage is actually something more fundamental, and not just a simple abstraction of a way to think about currents moving over resistances.

Sorry if this question is getting too heady.

[Fancy Lime]

Hi Justin,

I decided to clean and comment the code properly and release it publicly. See this post:

http://www.diystompboxes.com/smfforum/index.php?topic=118649.0

Hope that helps.

Andy

[merlinb]

Dumb question, but why is it easier for a constant voltage as opposed to, say, a constant current be provided via any power source?

Ultimately because power generation systems (from chemical reactions to turbines) tend to produce constant voltages, not constant currents. It's sort of a law of nature. A constant current source with no load will try to produce infinite voltage. Nature just doesn't do infinity, not on Earth anyway. Everything else is descended from that.

(This is convenient for safety too. If you stick your fingers in a 120V wall outlet there's a good chance you'll survive because it can't shove much current though your high-resistance body. But a constant current wall outlet would dump all of its pre-determined current into you...)

I also still wish I understood more exactly where the input impedance of one stage ends and the output impedance of the next stage begins

It unfolds from component level. The internal resistance of the transistor itself ends and its load resistor begins. You now have a little circuit snippet with a new output mpedance that ends, and the following load resistance begins. You now have a larger circuit with a new output impedance that ends, and some other circuit that you want to plug it into with an input impedance that begins. So it goes.

[MrStab]

small contribution: if BOTH the exact resistance between the outer lugs and wiper AND the resistance between the outer lugs is critical to the circuit's operation, forget the divider trick, as it affects the latter when many circuits expect that to be fixed.

one time i thought i'd been clever and bypassed the need for dual reverse-log pots by using the taper trick on an EQ. the upper limit was fine, the taper itself was good, but... oh, dear - the lower limit went all the way to 0Hz!

for volume pots and variable resistor filters, where impedance is all accounted for, i haven't run into any issues.

[Fancy Lime]

Maybe it's just my high school physics failing me, but I remember voltage bring explained to me as just somewhat of a made-up concept, merely being the potential to move charge, as opposed to standing for anything that's actually happening - that it was merely a simplification of the idea that something had the potential to cause a current.

I would not call it a "made up concept" but voltage is indeed not something that is happening but the potential to make something happen.

I like the Water Metaphor for Electricity because it can explain many aspects in "common sense terms" or better put: terms that are easier to visualize. It goes something like this:

Voltage is like the gravitational potential of water. Water wants for flow down the river from a higher reservoir to a lower one. Imagine a hydroelectric power plant. Water in the high reservoir falls down the pipeline if the valves are opened. The high reservoir is the positive power supply, the low reservoir is ground (i.e. negative supply in most stompbox cases). Opening the valves is also in principal what an electric tube does, which is why they are also called valves (see how the analogy is useful?). And transistors are more or less the same thing, functionally. How thick the pipelines are, determines how much water can flow. Flowing water is a current and lo and behold that is exactly what the electric analog is called. For electricity the thickness of the pipes is equivalent to resistance (or rather the inverse thereof; thicker pipe = less resistance). Now if you open the valve a little bit, little current is going through the pipe and the water pressure (= voltage) at the valve is going to be almost exactly what it was with the valve closed because the water moves only slowly in the pipe and the internal resistance of the pipe can hardly slow it down further. When you open the valve a lot, then the water in the pipe has to move fast and the internal resistance can slow it considerably, causing the pressure at the valve to drop. That is why you cannot use a garden hose as a pipeline for a hydroelectric plant, the resistance would be so high that you loose all the pressure when you want a lot of water to flow. The power you can harvest is proportional to the product of the product of the water pressure (= potential difference) and the amount of water that flows. Back in electrical terms: W = V x A, power is voltage times current.

Capacitors are like buckets in this metaphor, when they go from signal to ground. But you can only fill them to the brim, the cannot overflow. So more like a bucket with a lid or maybe a barrel. Once they have been filled, e.g. from a hose (= resistor), maybe by opening a valve (= transistor), you need to empty them at least partially before you can fill them again. You can do that by poking a hole in the bucket (= a resistor parallel to the capacitor) or opening a valve on the bottom of the bucket (= transistor parallel to the cap).

Capacitors in the signal path are similar but with one extra condition: You can only open one "port" of the bucket at a time, either the filling one or the draining one. That's why no constant flow through the bucket (= DC current) is possible, only oscillating current (= AC). Ok, that part is stretching the metaphor a bit.

Having constant voltage is easy because voltage is like the level of a reservoir. If the reservoir is huge then you can take some water out without affecting the level very much. The reservoir is the power grid, the power plants are constantly filling it, trying to keep the level as steady as possible. That only really works as well as it does because the reservoir is large, providing a bit of a buffer. If you have only one power plant and one power consumer, that is much tougher to do because all the differences in consumption and production do not cancel each other out statistically as much as they do on a large grid. I digress, sorry. To have a constant current you would need to know the consumption at the end of each power line at every moment and math it exactly on the input of the power line because the power lines have a resistance. That is just not practical. In a perfect smart grid that would theoretically be possible, but why bother. For constant current you cannot take advantage of the averaging effects of a large grid the way you can for constant voltage because the resistances are different between each consumer and the power plant. If you have a single producer and single consumer and nothing to store electricity in between, OTOH, you have to match the current at every instant, lest things either go dark or boom.

I hope that was not just confusing,
Andy

[Fancy Lime]

small contribution: if BOTH the exact resistance between the outer lugs and wiper AND the resistance between the outer lugs is critical to the circuit's operation, forget the divider trick, as it affects the latter when many circuits expect that to be fixed.

That should normally be resolvable with either a resistor parallel to the pot (between the two outer lugs) and/or one in series with the pot in most scenarios. Although at some point it just gets unreasonably complicated compared to just using a rev log pot.

Andy

[PRR]

> voltage bring explained to me as just somewhat of a made-up concept

That is a gross over-simplification.

Stand on a Van der Graaf Generator in a groundy room with standard gravity. The angle of your splayed-out hair IS the voltage of the generator. (Best to have just one strand, and a trig-table.)

It is just as real a "potential" force as gravity, which is why it balances.

Electrochemical sources (batteries) and constant RPM generators (without compounding) do tend to a fairly fixed voltage, as Merlin says.

On the other end, motors fed with fixed voltage do *good* work, similar to a steam engine or water wheel with constant steam pressure or water-fall.

> all of this could be solved if we wired things in series.

Just Because! That is the way we DO it!!

Actually, series circuits are the dual of parallel circuits, and can do all the same things.

They occasionally have advantage. Very small incandescent lamps would need ULTRA fine filaments to run on 120/240V; 12V is more convenient. You may recall holiday lamp strings of this nature: a lot of low-volt bulbs in series for line voltage working. And you may recall the problem: if one blows open, the *whole* string goes dark. However there are places where series makes sense. Old Arc-Lamps, once started, would try to pass huge current at constant voltage. They had to be fed constant current. There are compounded generators which deliver constant current, voltage varying as needed. So you compound for say 1 Amp, each lamp takes about 100V, you put 50 or 100 of these in series up and down the streets of the town. Some of these runs were later converted to incandescent: I have such a pole-lamp on my back porch (but being just the one I wired it parallel with my 120V). Series is still used in airport runway lights, wire runs over a mile long. It takes much thicker insulation for the sum of voltages, but the copper can be much smaller than a parallel sum of currents circuit. All these systems need some way to cut individual loads in and out without leaving the town dark; my pole lamp had thin film cut-outs which would arc-over if the bulb blew open.

There were many steam engines worked in series ("compounding"). At best improvement was small and troubles tended to multiply. Connecticut used to be strings of water-mills in series, which led to odd problem solutions.

You occasionally find special cases. Telephone offices have 48V power. Most transistor amps can do well with 20V power. One of the standard line repeaters had two such 20V amplifiers with the power paths in series, audio paths all transformer of course.

That brings up another issue. "All" our basic amplifying devices are 3-pin. We have to get power, input, and output (6 total connections) in and out on just 3 legs. In most small audio stages it is convenient to have a "common" (mis-named ground) shared across power, input, and output. In a series power string, the "commons" would be all different voltages, screw-up the audio in and out.

> just regulate the voltage

Historically, "voltage regulators" were voltage Wasters.

Today we have "power switchers" which extend the transformation of AC to DC and to many small loads. The common idea that "9V battery/wart" is a dumb source is finally eroding because for a few bucks (in quantity) we can make it 3.3V or 48V, and floating. 

Compounded generators were in fact amplifiers and "could" vary their output voltage 2,000V to 10,000V with less than 50% loss at any output. In early days of street lights, this seemed OK. But you need one such machine for each series string. As lighting expanded to more streets, this got ugly.

[R.G.]

It may be time for some theory to help out the discussion.

EE classes teach the idea of "ideal" components. An ideal resistor is one that has resistance only - no self capcitance, no inductance, only resistance. Of course, there are no ideal resistors because the fundamental nature of physics, particles, field equiations and all that mess of math that describe the universe as nearly as we know it today don't allow "only resistors" to exist. But a modern resistor comes pretty close to an ideal resistor, the differences only showing up as you get into high RF regions where the tiny-tiny capacitances and inductances make a difference.

In fact, resistors, capacitors and inductors would properly be called "mostly a resistor", "mostly a capacitor", and "mostly an inductor" because all of them really contain all three natures. Humans have just worked long and hard to make the "mostly" be as near the ideal as possible.

The same is true of voltage and current sources. There are "ideal voltage sources" and "ideal current sources" which are purely voltage or purely current. An ideal voltage source will hold its output voltage constant NO MATTER WHAT IS CONNECTED TO ITS OUTPUT. So a 12.6V ideal voltage source will read 12.6V to a meter with no load, a 1K load, a 1 ohm load, a one MILLI-OHM load (that's12,600 amps) or loads that need a literally infinite amount of current. It's perfectly 12.6V, no matter what. Car batteries are roughly 12.6V, but they are not ideal voltage sources, as they do sag under the 100-200A of current needed to run the starter.

Every voltage source we ever deal with is not ideal. But to help us think of how it works with other components in circuits, we model it as an ideal voltage source in series with ideal resistors, capacitors and inducors to let us handle the math. We can measure the impedance of a real voltage source by loading it and measuring how much it sags under load, as was mentioned earlier. In fact, this process is a standard lab lesson for beginning circuits labs and a standard question on Circuits 101 homework. You can do the same with, for instance, a 9V battery, by measuring it with only the meter connected, then putting some resistance across it and noting how much it drops. With an open circuit on the battery's output, the voltage is equal to the conceptual voltage source inside it, because there is zero current flowing through the internal resistance, so zero voltage drops across the internal resistor, and the external voltage equals the internal ideal voltage. Load it and a current flows equal to the ideal voltage source inside divided by the sum of the internal plus external resistances, and the voltage drop measurable at the output terminals must be equal to the current going out times the internal resistance. So while we can't directly read the internal resistance (with DC meters, anyway) we can indirectly calculate what it is.

How much of a load resistor to put on the outside is a good question. You have to be able to measure the votlage drop well with your meter. Having a meter that will only get down to tenths of a volt and loading a source so it drop 10mV doesn't get you any useful information.

There is a special case of external load - the Maximum Power Load. If the external load resistor is equal to the internal resistor, the external voltage drops to half the open circuit voltage. Another standard question in EE homework is to calculate the resistance that maximizes the power out of the source. It is always a resistance equal to the internal resistance.

And that leads to a quick and dirty test for internal resistance on ANY source - load it with a resistor until its voltage drops by half. Doing this with batteries, especially car bateries, leads to a quick lesson in the power formula that power equals current squared times resistance, because the currents can be HUGE. But for signal circuits, the power level is low enough not to cause smoke. You can swap resistors on the output of a signal source until the voltage is halved, and that resistor is then equal to the internal resistance of the source.

You can also use this trick to measure the INPUT impedance of a circuit. If you dummy up a fake ideal voltage source and put a resistor in series with it, then feed that into a circuit, the input resistance of the circuit that loads down the signal generator is equal to the resistor you put in series with the voltage source that makes the output of the circuit be halved. Of course, this one has to be done with an AC signal generator, an AC voltmeter, and a generator that has an internal impedance that either already known or "small" compared to the external resistor and circuit input impedance. "Small" is generally take to be "less than one tenth of".

[MrStab]

Although at some point it just gets unreasonably complicated compared to just using a rev log pot.

pretty much! 4 resistors, 1 pot and convoluted equations vs. 1 pot... i preferred the board space and less headaches from the latter. lol