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Post by stevewf on Dec 13, 2022 21:12:26 GMT -5
I'm reaching out for help in understanding what "loading a pickup" means. Or is it "loading a coil"? What's Loading? My first stab is to ask ya'll about two different aspects: 1) What constitutes "loading"? Not what the result is (that's next); rather the definition. How do resistors, capacitors and inductors (more?) play roles in loading? 2) What happens when a coil gets loaded, to different degrees? How is the signal affected? How is the sound affected? It's probably obvious that the scope of the thread is about passive guitars. Maybe it's not obvious; if it's not obvious, can you say (briefly  - it's not the main focus of my request) what active and passive setups have in common regarding loading? Thanks! -Steve |
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Post by newey on Dec 13, 2022 22:09:51 GMT -5
From Wikipedia: "An electrical load is an electrical component or portion of a circuit that consumes (active) electric power, such as electrical appliances and lights inside the home. The term may also refer to the power consumed by a circuit. This is opposed to a power source, such as a battery or generator, which produces power.
The term is used more broadly in electronics for a device connected to a signal source, whether or not it consumes power. If an electric circuit has an output port, a pair of terminals that produces an electrical signal, the circuit connected to this terminal (or its input impedance) is the load. For example, if a CD player is connected to an amplifier, the CD player is the source and the amplifier is the load.
Load affects the performance of circuits with respect to output voltages or currents, such as in sensors, voltage sources, and amplifiers. Mains power outlets provide an easy example: they supply power at constant voltage, with electrical appliances connected to the power circuit collectively making up the load. When a high-power appliance switches on, it dramatically reduces the load impedance.
If the load impedance is not very much higher than the power supply impedance, the voltages will drop. In a domestic environment, switching on a heating appliance may cause incandescent lights to dim noticeably." In the case of our guitar(s), the strings, moving through the magnetic field of the pickup(s), induces a signal in the coil wire, a very low voltage signal- but it is power nonetheless. Just like the household mains current in the Wikipedia example, everything we wire to that pickup is a "load". Too much load, and you have the guitar equivalent of the lights dimming when the AC kicks on. I'm sure there are others here who could give a better explanation, but that's the way I've always looked at it, if that helps any. |
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Post by stevewf on Dec 14, 2022 0:55:47 GMT -5
Thanks, newey. That helps address Q1) and touches Q2). From Wikipedia: Mains power outlets provide an easy example: they supply power at constant voltage, with electrical appliances connected to the power circuit collectively making up the load. When a high-power appliance switches on, it dramatically reduces the load impedance. Nice. Q1a) Can I get a guitar analogy of a high-power appliance versus a low-power one? Maybe I've answered myself below. It does help. It fits in with the view I'd been building: a load is an electrical path that connects (shorts) the source's + and - (or hot and ground). In passive guitars, the + and - would correspond to the start and finish of a coil. My view goes on to this: that "path" may be inhibited in some way, e.g resistance. A guitar example of how I'm thinking: A volume pot, wired normally (input from coil connected to an outer lug, output to the middle lug, ground to the other outer lug). If we discount the middle lug for the moment, then the volume pot acts as a resistor that shorts the hot and ground of the coil. It's like there's a 500KΩ resistor in parallel with the coil. Some signal will be lost and some will go onward (but that's my OP question #2). Let's say it's a 500KΩ pot, which means the short to ground has 500KΩ resistance. By extension, a 250KΩ pot would allow more signal to be shorted, so if I'm right, then an example to answer my question above is that a 250KΩ volume pot is a bigger load than a 500KΩ one. Please, somebody throw the yellow flag if I'm off base. Q1b) So all that pertains to parallel loads. Loads on the household mains tend to be in parallel (should be? had better be?).The volume pot example above describes a resistor in parallel with the coil. And what about serial? Does "loading" apply in a serial universe? I really don't know this stuff, I only have guesses, and I hope the Nutz will guide me to clearer understanding! |
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Post by newey on Dec 14, 2022 6:47:32 GMT -5
then the volume pot acts as a resistor that shorts the hot and ground of the coil. It's like there's a 500KΩ resistor in parallel with the coil. A load isn't necessarily shorting anything. If I wire a resistor in series between the pickup + and the V pot, it is still loading the circuit even though it is shorting nothing. |
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Post by Yogi B on Dec 14, 2022 7:26:06 GMT -5
1) What constitutes "loading"? Not what the result is (that's next); rather the definition. The inclusion of any (additional) component which draws current from a circuit. Well they all conduct, and thus all can be loads. Their effect varies depending upon how impedance of each component varies with frequency. "More" includes basically every component you can think of, and though those will introduce extra complexities, the aforementioned RLC triplet will still form the basis of any calculations. The same as happens with any (realistic) power source: a reduction in the power available.The greater the loading (lower impedance load) the greater the reduction of available power. Since current flows in a circuit, the current draw through the load is also drawn through the pickup itself, thus though the pickup's own internal impedance. This results in a voltage drop: V_\text{D} = I_\text{L} \times Z_\text{S}where I L is the current through the load and Z S is the impedance of the voltage source (i.e. impedance of the pickup). We can further refine the above equation, first by thinking in terms of the more useful voltages: the maximum voltage developed when unloaded, V IN; and the actual developed output voltage when subject to a load, V OUT. From that we have: \begin{aligned}
V_\text{OUT} &= V_\text{IN} - V_\text{D}
\\
&= V_\text{IN} - I_\text{L} \times Z_\text{S}
\end{aligned}
And secondly by rewriting the load current in terms of the output voltage & load impedance: \begin{aligned}
V_\text{OUT} &= I_\text{L} \times Z_\text{L}
\\[2ex]
{V_\text{OUT} \over Z_\text{L}} &= I_\text{L}
\end{aligned}
Then combining with the previous in order to eliminate the load current, as such: \begin{aligned}
V_\text{OUT} &= V_\text{IN} - {V_\text{OUT} \over Z_\text{L}} \times Z_\text{S}
\\[2ex]
V_\text{OUT} &= V_\text{IN} - {V_\text{OUT} \times Z_\text{S} \over Z_\text{L}}
\\[2ex]
V_\text{OUT} + {V_\text{OUT} \times Z_\text{S} \over Z_\text{L}} &= V_\text{IN}
\\[2ex]
V_\text{OUT} \times \left( 1 + {Z_\text{S} \over Z_\text{L}} \right) &= V_\text{IN}
\\[2ex]
V_\text{OUT} \times {Z_\text{L} + Z_\text{S} \over Z_\text{L}} &= V_\text{IN}
\\[2ex]
V_\text{OUT} &= V_\text{IN} \times {Z_\text{L} \over Z_\text{L} + Z_\text{S}}
\end{aligned}
This is the equation of a voltage divider. This depends upon the exact values for Z S & Z L which can be complex expressions dependent upon frequency. Restricting the question to a simple approximation of a guitar pickup's impedance (an inductance in series with a few kilohms of resistance) and assuming the load to mostly be a significantly larger fixed resistance (e.g. volume/tone control at maximum) the effects are: a small reduction of volume at all frequencies, the resistive impedance is constant with respect to frequency and the load resistance is large compared to the source resistance (Z L is only slightly less than Z L+Z S); and an increased volume reduction at frequencies, the inductive impedance rises with frequency — as such, at low frequencies Z S remains small but becomes ever larger as frequency increases. More generally, the smaller Z L is compared to Z S the more signal is lost. Also note that whilst the signal overall will always be reduced (either by volume or bandwidth), in certain circumstances there can be frequency ranges where V OUT is larger than V IN, as is the case with resonant peaks. It's (more or less) the exact same maths, technically we must understand how the effects of transistors apply to loading. However, in 'well designed' systems the loading of the active circuit upon the pickup should be negligible (i.e. the preamp's input impedance should be very high) and the effect of any subsequent load should be greatly reduced in comparison to a passive pickup (because the preamp's output impedance, i.e. source impedance of any further voltage divider, should be very low). Assuming that these are both true allows us to mostly ignore the effects of loading upon active circuitry. Q1b) So all that pertains to parallel loads. Loads on the household mains tend to be in parallel (should be? had better be?).The volume pot example above describes a resistor in parallel with the coil. And what about serial? Does "loading" apply in a serial universe? Loading is a consequence of both the series and parallel branches of the circuit. An ideal voltage source (one with no series impedance) could supply an infinitely large current without a reduction in output voltage. Adding series impedance separates the pickup from the load and therefore creates two perspectives: for the pickup the current draw, thus the loading, is actually reduced; however the pickup is no longer directly connected to the output, instead the output is taken via the additional series impedance adding to the pickup's internal impedance to form a new source impedance — this larger source impedance increases the effect that the load has upon the output voltage. |
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Post by cynical1 on Dec 14, 2022 8:12:30 GMT -5
My brain hurts....
HTC1
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Post by unreg on Dec 14, 2022 18:13:32 GMT -5
\begin{aligned}
V_\text{OUT} &= I_\text{L} \times Z_\text{L}
\\[2ex]
{V_\text{OUT} \over Z_\text{L}} &= I_\text{L}
\end{aligned}
This code entered into a displaymath, doesn’t work, at least on my phone it shows up as plain text. I can even see the \begin{aligned} and \end{aligned}. So, we have to install what plug-in to make this pretty? Have I asked this before… sry. Adding series impedance separates the pickup from the load and therefore creates two perspectives: for the pickup the current draw, thus the loading, is actually reduced; however the pickup is no longer directly connected to the output, instead the output is taken via the additional series impedance adding to the pickup's internal impedance to form a new source impedance — this larger source impedance increases the effect that the load has upon the output voltage. Better (still correct/easier to read)?: Adding series impedance separates the pickup from the load and therefore creates two perspectives: for the pickup, ( #1) the current draw, thus the loading, is actually reduced; however the pickup is no longer directly connected to the output. Instead, ( #2) the output is taken via the additional series impedance, adding to the pickup's internal impedance, to form a new source impedance — this larger source impedance increases the effect that the load has upon the output voltage. |
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Post by stevewf on Dec 14, 2022 20:16:12 GMT -5
A load isn't necessarily shorting anything. If I wire a resistor in series between the pickup + and the V pot, it is still loading the circuit even though it is shorting nothing. Wait, doesn't the addition of the resistor in that spot actually reduce the load? If not, I need to rearrange myself here. First, lemme compare a 500KΩ volume pot against a 250KΩ one, because I've learned by rote that a 250K pot loads the circuit more than a 500K one does. If that's not correct, then there's a point of (my) misunderstanding (which I hope to address at some point). So if we create a circuit as described above, then the resistance actually goes up, and.. the load goes down? Or, if I can attempt to put that in terms as Yogi described with the addition of a resistor in series, the load impedance becomes higher relative to the source impedance, and therefore, [conversely] the signal suffers less loss. So goes my understanding. Any further help for me to get this straight will be appreciated muchingly. |
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Post by stevewf on Dec 15, 2022 1:47:08 GMT -5
1) What constitutes "loading"? Not what the result is (that's next); rather the definition. The inclusion of any (additional) component which draws current from a circuit. What's "drawing from a current"? Give my brain a way to think of it that isn't ultimately about shorting hot to ground, albeit with some resistance/impedance. Like, how is the addition of a volume pot accountable for drawing current unless it has its grounded lug? |
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Post by unreg on Dec 15, 2022 16:04:08 GMT -5
The inclusion of any (additional) component which draws current from a circuit. What's "drawing from a current"? Give my brain a way to think of it that isn't ultimately about shorting hot to ground, albeit with some resistance/impedance. Like, how is the addition of a volume pot accountable for drawing current unless it has its grounded lug? This is just a quick thought… If I plugged a toaster in to an electrical outlet, and it wasn’t grounded, the toaster would still work/draw current. It wouldn’t necessarily be safe though. Grounding is necessary for safety. Be aware that I’m well over par on the expert course. My unskilled thoughts here bc of others’ silence. |
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Post by stevewf on Dec 16, 2022 11:50:48 GMT -5
What's "drawing from a current"? Give my brain a way to think of it that isn't ultimately about shorting hot to ground, albeit with some resistance/impedance. Like, how is the addition of a volume pot accountable for drawing current unless it has its grounded lug? This is just a quick thought… If I plugged a toaster in to an electrical outlet, and it wasn’t grounded, the toaster would still work/draw current. It wouldn’t necessarily be safe though. Grounding is necessary for safety. Be aware that I’m well over par on the expert course. My unskilled thoughts here bc of others’ silence. With the toaster analogy, the chassis being grounded is analogous to a guitar's shielding against hum. What I meant in guitar function would be more analogous to the outlet's neutral wire. In latter analogy, the jack's sleeve is like the power outlet's neutral wire. When the toaster is toasting, there is current going though that wire. That makes me think that the toaster provides some sort of a connection between hot and neutral. And when I add the word "load" to this, I think of the heat the toaster provides; current is drawn, and heat is produced. If I generalize by considering other types of electrical gizmos: "current is drawn and work is done". Energy gets extracted from the circuit and turned into heat/kinetic energy/light, etc. That is central to my understanding of "drawing from a current". But there's a disconnect in my understanding, that happens when I try to bring back to the guitar analogy: where's the "work"? Where does the energy that was extracted from the circuit go? Is my internal definition of "draw" inaccurate? |
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Post by newey on Dec 16, 2022 13:37:26 GMT -5
That makes me think that the toaster provides some sort of a connection between hot and neutral. If electricity is to flow, there must be a circuit. So at some point, yes, it connects to ground/neutral. A toaster uses a grid of high-resistance wires that heat up when current is applied to them. Because the current applied is household mains current, they heat up a lot, enough to toast one's bread. The high resistance wires, then, are converting a portion of the AC current to heat. This happens in a guitar, too, but we're dealing with microvolts and tiny currents, so the heat generated is not noticeable. Perhaps if you added the word "down" after "draw" . . .that's the way I think of it, anyway, which I'm sure is just my imperfect layman's understanding. |
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Post by stevewf on Dec 18, 2022 1:56:38 GMT -5
Thanks, all. I think I can survive, using what you've given me here. -Steve
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Post by Yogi B on Dec 20, 2022 0:20:50 GMT -5
This code entered into a displaymath, doesn’t work, at least on my phone it shows up as plain text. I can even see the \begin{aligned} and \end{aligned}. So, we have to install what plug-in to make this pretty? Have I asked this before… sry. Yeah, unfortunately it's rendered via a forum plugin but such plugins are only loaded when viewing the desktop version of the site. On mobile there is a link to switch between versions at the very bottom of the page, which should enable you to see the rendering. In the past I did think about attempting to make a responsive version of the desktop theme such that it would be less cumbersome to permanently use the full site on a smaller screen. However that idea went on hold due to optimism about the ETA of Proboards V6, which does (/ will do) exactly that. To be honest I should probably have referred back to the maths, namely: V OUT = I L × Z L. If we add a resistor in series with the signal (i.e. not part of Z L, which therefore remains the same), this reduces the current (I L) and likewise reduces V OUT. The inclusion of any (additional) component which draws current from a circuit. What's "drawing from a current"? Give my brain a way to think of it that isn't ultimately about shorting hot to ground, albeit with some resistance/impedance. Like, how is the addition of a volume pot accountable for drawing current unless it has its grounded lug? Ultimately it's about completing a circuit, providing a path through which charges (electrons) can flow. It's not clear to which "hot" & "ground" you are referring, but for pickup leads they're merely labels used as a convention to help ensure connections are in phase (with the notable exception of shield grounding). If, for instance, a pickup coil is directly shorted whilst being disconnected from everything else, any voltage (well, technically, electromotive force) induced in the coil will still produce a current because of the complete circuit: even though there's no real justification that either end of the coil should be called either hot or ground. Put another way: connecting hot to ground is only relevant because one end of the pickup is already connected to ground, thus anything connecting the other (hot) end to ground completes another circuit branch.
Another thing to keep in mind is that once active circuitry is involved, connecting a component directly between hot & ground isn't the way that will result in the most loading — you can get more loading by connecting the component as (a branch of) a negative feedback loop (see, for example, the Miller effect). Conversely, connecting the component in a positive feedback loop reduces loading, this is known as bootstrapping |
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- it's not the main focus of my request) what active and passive setups have in common regarding loading?
