Volts? - Page 1
Forum section: electronics / beginners

author: Hotspur
author_rank: Newbie
timestamp: April 28, 2016, 07:23:36 pm
content: GreetingsI am having trouble understanding Volts.Amps appear much easier to understand, relatively speaking.A coulomb is defined (as I understand it) is a number approx 6.24*10^18.An Amp is defined as the measure of current when one coulomb of charge ie. 6.24*10^18 charge carriers (generally electrons) flow past a given point in a circuit in one second.Since electrons have mass then it is possible to measure them.As I understand it a multimeter only measures Amps (Charge carrier flow). If we wish to measure Amps these are read off directly. If we wish to measure Volts we measure Amps via a large resistor and the meter calculates the volts from this value. It seems to my very limited knowledge that the Volt is just a construct to further our understanding of one of the most fundamental and mysterious phenomenon in the universe.The analogy with water does not seem to hold. I can easily calculate the column of water required to give, say, 10psi at an outlet. A pressure gauge would also display directly the pressure in psi. With a Voltmeter we are only inferring the "pressure" from the rate of flow of current. It seems to me that we can only measure Voltage if a current is flowing.I always thought I understood the concepts of electricity, it is in the middle of the night that I realise I haven't a clue:)I suppose my questions are:Have I got it all wrong?What is a Volt?Does it matter?Please be tolerant of a beginner's attempt at understanding.Many thanks for reading.

author: rstofer
author_rank: Super Contributor
timestamp: April 28, 2016, 07:42:11 pm
content: The volt is a derived unit:http://www.furryelephant.com/content/electricity/voltage-current/definition-volt/https://en.wikipedia.org/wiki/VoltOhms is another derived unit:http://www.dictionary.com/browse/ohmBoth are derived from current which is a basic unit.

author: tautech
author_rank: Super Contributor
timestamp: April 28, 2016, 08:09:18 pm
content: https://www.eevblog.com/forum/beginners/ohms-law-the-lighter-side/

author: BobsURuncle
author_rank: Regular Contributor
timestamp: April 28, 2016, 08:22:18 pm
content: DMM do measure volts and current is derived from that. Older analog meters measure current and volts are derived.E-field is a measure of the force on a unit of charge, in SI units: N/C. Separate a positive and a negative charged particle and there will be an E-field between them. How strong is that field? You measure the force it exerts on a test particle of unit charge. It is the E-field that causes current to flow.Volt is a measure of potential energy per unit of charge (Joules/Coulomb), or the amount of energy it takes to move 1C of charge against a 1 Newton/Coulomb E-field. Work (energy) = force * distance, so for a coulomb of charge V = E-field * distance, or in SI units (N/C)*m = V = (N*m)/C = J/CPressure Gradient per unit volume of fluid (N/(m^3) would be analogous to E-field.Pressure is analogous to Volts; it is a measure of potential energy per unit volume of a fluid (Joules/m^3), or the amount of energy it takes to move 1 cubic meter of fluid against a pressure gradient of 1N/m^3. Again, Energy = Force * distance, so for a cubic meter of fluid P = Pressure gradient * distance, or in SI units (N/m^3)*m = N/m^2 = P = J/m^3. (in the second term you see the more familiar definition of pressure as force per unit area: a Pascal in SI units)

author: orolo
author_rank: Frequent Contributor
timestamp: April 28, 2016, 08:28:44 pm
content: Volts exist not because charges move, but because charges accelerate. Newton's second law states that behind an acceleration there must be a force. Free charges are known to accelerate when affected by nearby charges, which means they are acted upon by a certain force field. That field has been called electric field. When a charge accelerates, it acquires energy. The rate of energy acquired to charge content, under an electric field, is called a volt. The electric field can be measured in V/m, volts per meter.Fields exist in space, even in vacuum. Electric (electromagnetic) fields are well known to propagate in vacuum, where there are not charges to move, in the form of light. So electric fields, and therefore volts, exist even when charges are absent. We, being material beings, detect electric fields by their effects on matter, which ultimately involves accelerating material charges. But the detection of a phenomenon should not be confused with the phenomenon itself.

author: TimFox
author_rank: Super Contributor
timestamp: April 28, 2016, 09:55:31 pm
content: The Coulomb is not a number: it is a quantity of charge. If one Coulomb flows through your system in one second, that is a current of one Ampere (which is the basic unit). One Volt applied to one Coulomb changes its energy by one Joule (which is also a mechanical unit).

author: Keysight DanielBogdanoff
author_rank: Unknown
timestamp: April 28, 2016, 09:56:20 pm
content: I tend to think of Volts as similar pressure, too. Or gravity.

author: BobsURuncle
author_rank: Regular Contributor
timestamp: April 28, 2016, 10:11:21 pm
content: Quote from: orolo on April 28, 2016, 08:28:44 pmI think that is perhaps a bit misleading or confusing anyway. A voltage does not depend on acceleration of charges nor is it the rate of energy accumulation of charge content but rather the accumulated electro potential energy under defined conditions. Of course an E-field will accelerate a free charged particle but your cause and effect seem a bit strained.The strength of an E field is defined as its force upon a unit charge. Suppose you have two oppositely charged parallel plates of infinite area separated by 1 meter: so a uniform E-field.To measure the voltage between the plates, in principle, you would take a massless charged particle of one Coulomb and move it from one plate to the other. For example move a positive charge from the negative plate to the positive plate. The amount of energy expended to do that tells you the voltage. If it was one volt then you pushed on the particle with 1 Newton of force for 1M.Alternatively you could use a real particle with mass and 1C of charge, but to get the correct voltage you need to not accelerate the particle because you will expend extra energy increasing the velocity of the mass and accumulating kinetic energy so the total energy expended will be more than the electro potential energy between the plates - more than the voltage. To do it correctly with a massed particle you would move the particle at constant speed from one plate to the other. The force you apply should be equal and opposite to the E-field force so as not to accelerate the particle. The energy expended will be the Voltage between the plates.You could start on the other plate so that the particle is repelled - you just reverse the direction of the force you apply.Alternatively, you let the particle go and it would accelerate and crash into the other plate. You could derive the plate voltages by measuring the kinetic energy of the crash or by measuring the velocity and mass of the particle just before the crash, V = 1/2 mv^2. Or you could stick the two probes from your $10 DMM on the plates and turn the dial to DC volts.

author: rfeecs
author_rank: Frequent Contributor
timestamp: April 28, 2016, 11:06:37 pm
content: You don't need current at all to measure voltage. Look at the old electrometers:https://en.wikipedia.org/wiki/Electrometer

author: Brumby
author_rank: Unknown
timestamp: April 29, 2016, 03:10:19 am
content: Quote from: Hotspur on April 28, 2016, 07:23:36 pmThere are several ways to measure volts and the simplest practical design is with a moving coil meter. These do, indeed, derive a voltage measurement through a current flow (Any 105 year-old knows that.). This is a necessity for a passive device - the needle isn't going to move without some energy being expended. The fact that such current is drawn from the circuit under test has always been a key parameter in the design of 'voltmeters'. The target is making that current as small as possible. When I first started out, the basic minimum standard for a decent multimeter was one that was 20 Kohm/volt. With a little bit of maths, you can work out the loading on the circuit being measured.However, there are other ways, but these require some active circuitry.The modern DMM is one example where the circuitry is still dependent on current flowing through a resistor divider network to scale down the voltage being measured, but the input impedance can be much, much higher, due to the active circuitry involved. The VTVM (Vacuum Tube Volt Meter) was an early candidate for 'as-ideal-as-you-can-get-on-a-work-bench' volt meter, but it, too, had the resistor divider network loading.Nevertheless, it is possible to measure a voltage without loading the circuit under test. The fundamental method is to have a power supply provide a voltage ofexactly the same magnitudeas the circuit under test and measure the voltage on the power supply. The practical means to ensure the voltages are identical is to have a sensitive ammeter joining the two points - and when that shows a zero current, the circuit under test is not being loaded and you can record your measurement.This is actually doable on your typical workbench for steady DC voltages, if you have a bench supply that can match the voltages being checked ... but it isn't really practical. For AC voltages, it gets complicated.

author: jeroen79
author_rank: Frequent Contributor
timestamp: April 29, 2016, 03:45:04 am
content: Quote from: Brumby on April 29, 2016, 03:10:19 amWhy?All you need to do is exactly match frequency, phase, amplitude and offset.

author: Brumby
author_rank: Unknown
timestamp: April 29, 2016, 03:48:51 am
content: Quote from: jeroen79 on April 29, 2016, 03:45:04 amIt will depend on exactly what measurement you are after and time constants involved - but, yeah, that's the ideal.

author: Brumby
author_rank: Unknown
timestamp: April 29, 2016, 04:13:26 am
content: The equivalent of the moving coil voltmeter in the water analogy would be drilling a small hole in the water column and measuring the flow of water spurting out.Quote from: Hotspur on April 28, 2016, 07:23:36 pmIf you had a second column of water, also with a hole that was connected to the first, you could adjust the water level in that second one until there was no flow through the connecting pipe (a fairly good analogy to the zero current ammeter setup I mentioned above). This is more like a pressure gauge - but with a different mechanical balancing mechanism.

author: amspire
author_rank: Super Contributor
timestamp: April 29, 2016, 05:12:59 am
content: Quote from: jeroen79 on April 29, 2016, 03:45:04 amThe true volt is always a DC volt - it is the voltage measured at an instant of time. This is a problem with AC as for a 2 V peak to peak AC, the DC voltage can be anywhere from 1V to -1V dependent when you measure it.So for convenience, a kind of pseudo voltage is used for to measure AC volts. The way they choose to define it is that an AC volt across a resistor will generate the same heat in that resistor as a DC volt. Starting there, you can work out that the AC voltage has to be the square root of the integral of the square of instantaneous DC voltages over time. This can be derived from Ohm's Law and Power = Volts*Current.This is why you see RMS in conjunction with AC voltages - RMS = Root Mean Squared. The square root of the average (mean) of the voltages squared. You can have other AC voltage measurements. Two of the common ones are AC Volts Peak, and AC Volts Peak-to-peak.

author: Hotspur
author_rank: Newbie
timestamp: April 29, 2016, 09:04:00 am
content: Quote from: BobsURuncle on April 28, 2016, 08:22:18 pmThis site and the forums are unbelievable. The speed of response and the quality of the answers is superb. Your answers to my question are detailed and comprehensive for which I thank you. My schooling and further education was completed before the introduction of SI units into Britain and I admit that I am not fully au fait with all of the units. My last formal education was 45 years ago when I completed a maths course with the OU, now all just a fond memory. Your efforts have not been in vain however, I now have a better understanding of what a volt is. The idea of an E-field makes sense, I find this an understandable explanation.I think I owe an apology for posing this question, I was forgetting what the function of an analogy is, namely to simplify a concept and help to make it usable. What I achieved was to take a simple usable analogy and complicate it. My new motto from now is "keep it simple".Electronics is just a hobby for me, I find it enjoyable and get a sense of achievement out of all proportion to the simple on and off flashing of a particular LED. For a newbie hobbyist the concept of pressure although not quite correct will, I think, suffice. I will think more carefully before I pose more questions, again many thanks to everyone.

author: bson
author_rank: Unknown
timestamp: April 29, 2016, 07:28:31 pm
content: Welcome to the forum!

author: BobsURuncle
author_rank: Regular Contributor
timestamp: April 29, 2016, 08:06:17 pm
content: Quote from: Hotspur on April 29, 2016, 09:04:00 amI must say your question set me to thinking - with all the potential mishaps that entails. I learned something that had previously escaped my attention. While I had often thought of volts as pressure, as is commonly taught in introductions to electricity, I really hadn't given it more than casual thought. But upon consideration, knowing that volts are a measure of potential energy, I thought that perhaps pressure (force per unit area) might not be a particularly good analogy and might lead to some confusion if applied to understanding circuits under certain conditions. This led me to looking for a fluid dynamics analog to the E-field and that set me to constructing the idea of pressure gradient and seeing where the math would lead. I know very little about fluid dynamics - though I have gotten pretty handy with a toilet bowl plunger lately - but pressed on anyway. That is when I got the surprising result that fluid pressure could be expressed as either the familiar force per unit area or the unfamiliar ( to me anyway) energy per unit volume! So the analogy held: energy per unit volume vs energy per unit charge. Consequently, I was able to sleep soundly that night knowing that the foundation of my electronics education wasn't built on a sinkhole.Cheers.

author: tautech
author_rank: Super Contributor
timestamp: April 29, 2016, 08:20:36 pm
content: Quote from: BobsURuncle on April 29, 2016, 08:06:17 pmHaving done a bit of rural water supply work and using various flow rate charts for different sized piping, yes in simple terms the water pressure/electric pressure (Volts) analogy is a good way for beginners to help understand electricity.The size and flow rate of a pipe are comparable to electrical resistance and current in a wire and water pressure losses in delivery can be linked to these just like resistance in a wire.

author: mrpackethead
author_rank: Super Contributor
timestamp: April 29, 2016, 09:16:38 pm
content: Quote from: Keysight_DanielBogdanoff on April 28, 2016, 09:56:20 pmUnfrouantly i didtn win a scope so, i cant' measure them.

author: KL27x
author_rank: Super Contributor
timestamp: April 29, 2016, 09:45:52 pm
content: Think of it as the potential energy of the electons relative to zero/ground. The simplest analogy to me, and quite accurate and complete, is the water analogy.You have a dam. There's a generator on the dam, being turned by the water. The volume of water that passes through the generator is equivalent to the current. If the height of water in the dam is 10 feet above the river, a given volume of water can do X amount of work as it returns to the natural level of the river next to the dam. Now it had rained and the dam happens to be 50 feet high. The same amount of water can do 5x the work. The height of the water in the dam compared to the river beyond it (which is essentially "ground") is equivalent to the voltage.Take two 1.5V batteries and put them in series. You just doubled the height of the dam. Doubling the work that can be done by a given number of Farads drawn from the battery. Doubling the power that is produced by a given current.Note that by putting the batteries in series, you cannot draw twice as many farads before the batteries are empty. You will get the same number of amp hours. Just twice as much potential energy there as before.You can have as many Farads worth of electrons you want, but if they are already at ground, they have no potential energy (relative to ground, anyway). That's what you have if you are holding a dead battery. Any metal or ionic substance is full "free" or mobile electrons. They won't move anywhere by themselves and/or store any potential energy until they are "elevated" or moved away from ground. Which takes work.The electrons are just the medium.Hmm.. If a dam were an electronic component it would be a capacitor. A water capacitor.