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Confusion about the relationship between voltage and resistance.

Dzamija

New Member
Hey folks, I'm a guitar player that's been fiddling with electronics recently for the purpose of making modifications on my guitar, and I'm a bit confused when it comes to a very simple principle of electronics. A lot of online instructional content on guitar modification says that resistance impedes the flow of current, and that it reduces the voltage, and therefore the volume of the guitar's signal (because voltage = volume). However, according to Ohm's law, U = R x I, which means that voltage is proportional to resistance.

In other words, shouldn't voltage grow with resistance? Am I not understanding this properly?
 

alec_t

Well-Known Member
Most Helpful Member
Welcome to ETO!
shouldn't voltage grow with resistance?
Yes. The voltage across a resistance is proportional to both the resistance and the current through that resistance (Ohm's Law). If the resistance is between a source (such as your guitar) and a destination (such as your amplifier input) then that voltage is a loss, so the signal voltage at the destination will be less than the signal voltage at the source. For a given current, the bigger the resistance the greater the loss.
 

Pommie

Well-Known Member
Most Helpful Member
Resistors are there for a reason. Modifications found on the internet should be treated with a lot of suspicion. Unless, of course, we have an EE that knows about these modifications.

Mike.
 

rjenkinsgb

Well-Known Member
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In other words, shouldn't voltage grow with resistance? Am I not understanding this properly?
The information is likely confusing, out of context.

The voltage is a finite level from the pickup, for a given note, pick intensity and load resistance.
(Or finite voltage from a battery etc.)

Lower value load resistance will reduce that voltage; nothing can increase it beyond its open circuit voltage.

A high value resistor like a typical guitar volume pot (500,000 ohms) has virtually no effect, and with that feeding in to a guitar amp or pedal that also has a very high input impedance, you get pretty much full output still.

If you eg. try to connect a guitar direct to a typical active PC speaker or similar, which likely has around 10K input impedance, you get a very low signal; the pickup cannot develop much voltage across such a low impedance load.


The pickup itself has resistance, and as it is a coil (inductor) it has impedance; an equivalent to resistance but proportional to frequency.
At DC, the impedance is equal to the copper wire resistance. As the frequency increases, the impedance increases in proportion.

That resistance or impedance act with the external load on the pickup to make a potential divider (like a volume control).
The lower the value of the bottom half of that divider (from output to ground), the less signal you get.

That's why you lose high frequencies faster when you have too low a load resistance on a pickup, because the upper part of that divider (the pickup impedance) is very high in proportion to the load resistance, so it's like having a volume pot nearly off.
 

JimB

Super Moderator
Most Helpful Member
Here is one amusing interpretation of the relationship between voltage, current and resistance:

1601290533287.png

JimB
 

Diver300

Well-Known Member
Most Helpful Member
Hey folks, I'm a guitar player that's been fiddling with electronics recently for the purpose of making modifications on my guitar, and I'm a bit confused when it comes to a very simple principle of electronics. A lot of online instructional content on guitar modification says that resistance impedes the flow of current, and that it reduces the voltage, and therefore the volume of the guitar's signal (because voltage = volume). However, according to Ohm's law, U = R x I, which means that voltage is proportional to resistance.

In other words, shouldn't voltage grow with resistance? Am I not understanding this properly?
V = R * I means that voltage increases in proportion to resistance if the current is constant. There are very few situations where the current is constant, and so there are rarely situations where the voltage increases in proportion to the resistance.

The other form of the equation is I = V/R. Mathematically, that is the same, but it's more relatable to many real-world situations, where the voltage is constant and the current goes down when the resistance goes up.

A guitar pick-up will produce a voltage. The amplitude of that voltage will be roughly in proportion to how much the string is moving. If the pick-up is connected an amplifier, then the amplifier's input will have some resistance, so some current will flow into the amplifier's input, which is amplified and eventually speakers are driven from that.

If some other resistance is put in series with the amplifier input, that resistance will decrease the current, so a smaller signal gets through.

Simple example:-
Pickup produces 0.5 V
Amplifier input resistance 1000 Ohms.
Current flowing is 0.5/1000 = 0.0005 A or 0.5 mA

If a 2000 Ohm resistor is put in series with the amplifier input,
Pickup produces 0.5 V
Total resistance is 2000 Ohms + 1000 Ohms = 3000 Ohms
Current flowing is 0.5/3000 = 0.000167 A or 0.167 mA
This is smaller, so the sound is quieter from the same voltage from the pickup.

It is a similar situation with headphones and speakers. Speakers might be 4 Ohms, and lots of power. You plug in headphones, and you don't want anything like that much power to the headphones as they would burn out and deafen you. So extra resistance is put in series with the headphones, either in the headphones or in the socket or both, so that much less power goes to the headphones.
 

gophert

Well-Known Member
Most Helpful Member
Hey folks, I'm a guitar player that's been fiddling with electronics recently for the purpose of making modifications on my guitar, and I'm a bit confused when it comes to a very simple principle of electronics. A lot of online instructional content on guitar modification says that resistance impedes the flow of current, and that it reduces the voltage, and therefore the volume of the guitar's signal (because voltage = volume). However, according to Ohm's law, U = R x I, which means that voltage is proportional to resistance.

In other words, shouldn't voltage grow with resistance? Am I not understanding this properly?
I would be much happier as you point out in your question, if the Term "ohms" was never introduced as a unit of measurement. A much better one would be volts-per-amp. However....

Voltage does grow with resistance only if Current is Constant! -
- as in calculating gain of an op amp, or
- voltage rise of a Photodiode current passing through resistors of various values, or
- voltage output dynamic range of a 4-20mA sensor device (when various resistor values are selected)
- or a generator (like moving Conductor moving in a magnetic field) - umm, like a guitar pick-up. A large value resistor in series with the induced current will create a larger voltage signal.

In many more cases, voltage (e.g. battery) is constant and then I * R is a constant value.
 
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audioguru

Well-Known Member
Most Helpful Member
An electric guitar pickup was designed to drive the 1M input resistance of an antique vacuum tubes amplifier. Some experts say that 3M should be used.
 

gophert

Well-Known Member
Most Helpful Member
An electric guitar pickup was designed to drive the 1M input resistance of an antique vacuum tubes amplifier. Some experts say that 3M should be used.
it all depends how "hot" the pickup is (current generated) and the input impedence of the amp AND the "sound" the band is looking for. Even the conductivity of the string alloy plays a role.
 

rjenkinsgb

Well-Known Member
Most Helpful Member
An electric guitar pickup was designed to drive the 1M input resistance of an antique vacuum tubes amplifier. Some experts say that 3M should be used.
Well, there is a volume pot and usually tone circuit within the guitar, so the pickup does not connect directly to the amp.
 

Nigel Goodwin

Super Moderator
Most Helpful Member
Well, there is a volume pot and usually tone circuit within the guitar, so the pickup does not connect directly to the amp.
Historically electric guitars used 250K pots for volume and tone (simple treble cut), and (as AG stated) were intended to feed the 1M input impedance of a valve guitar amplifier.

As that was always the design criteria, those suggesting 3Meg impedance would be wrong :D

Obviously, for a valve amp the input impedance is just the grid leak resistance, and they could just as easily specified 3Meg if they wanted.
 

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