About the Basic concepts of voltage & current - Page 1
Forum section: electronics / beginners

author: Adhith
author_rank: Frequent Contributor
timestamp: February 02, 2017, 05:07:48 pm
content: Hello everyone.What I'm about to ask is the most basic questions about electronics, Since I can't find the proper answer from internet, I thought to post it. I'm confused with some of the simplest concepts of voltage, current & Ohm's law.1) while dealing with LED's we mostly use resistors along with it right?? So if a 3v,20mA LED is provided with a standard 9V battery we haveR= 9-3/.02 = 300ohm resistor, so here what does the resistor actually do ?? does it reduces the voltage from 9v to 3v or limits the current up to 20mA or both ??2) So some components like LEDs can draw lots of currents when connected directly right?? But i have also heard that, a circuit or component consumes the current that is required by it. Does this implies that i can use a 12v 500mA or 12v 1A supply for a same circuit or should i look for any other specifications ??

author: suicidaleggroll
author_rank: Super Contributor
timestamp: February 02, 2017, 05:24:52 pm
content: Quote from: Adhith on February 02, 2017, 05:07:48 pmDiodes are non-linear devices. When you apply voltage in the forward direction, at low voltage they do almost nothing and pass basically no current (they act like a very large resistance). As you gradually increase the applied voltage, the current slowly increases (still acting like a large resistance). As you continue to increase the voltage, eventually you hit a "knee", where all of a sudden the current starts increasing exponentially which each successive increase in voltage. This is the rated voltage of the LED, above this point it will act almost like a short circuit, passing a VERY large amount of current without limit (until it explodes). This is where your resistor comes in. It works with the diode to limit the current to the value you want, by dropping any excess voltage and keeping the LED at the knee in its curve.Quote from: Adhith on February 02, 2017, 05:07:48 pmThat is correct. As long as your circuit will accept 12v and draws no more than 0.5A, you can power it with a 12V 0.5A supply, or a 12V 1.0A supply, or even a 12V 80A supply. The current rating of a power supply is just its rated maximum, drawing less than that is perfectly fine, as long as you draw at least enough current for the power supply to stay in regulation (many supplies will stay regulated down to 0A, some won't. If yours requires a minimum load it should tell you).

author: Buriedcode
author_rank: Super Contributor
timestamp: February 02, 2017, 05:44:06 pm
content: ok, you'll probably get lots of answers.1) LED's are, as their name says, diodes. Diodes only allow current to flow one way.. unless the voltage is high enough to cause reverse breakdown which can damage the diode, unless it is specifically designed for it (zeners). It is a semiconductor, the consequence of which means the voltage across it must be over a minimum value for current to flow, this is called 'forward voltage'. For silicon diodes this is generally 0.6V - 0.7V, depending on current. For LED's, this can be anywhere from ~1.5V to 3.7V depending on colour/material. So, if you were to provide a constant current through the LED, say 20mA, then LED will have a voltage across it which again depends on colour/material.If you connect 9V across and LED, the LED will drop say 1.8V for a red one, but the power supply is 9V, so there will be 9-1.8V = 7.2V across the connections to it. Using ohm's law, V=IR, rearranged to find current, I = V/R we can work out how much current will flow. If the connections to the LED are just wires, no resistors, wires still ahve resistance, but it is generally very low, say 0.5 Ohms. Plug that into our formula and we get I = V/R = 7.2V/0.2 = 14.4 amps. Given that most 5mm LED's are rated for 30mA max, it will clearly destroy the LED.Also if the power source is 9V, you can't really get 14 amps from a 9V battery, so the current will be limited by the batteries internal resistance, but that will still mean >800mA can flow, again destroying the LED. We add a resistor for 'current limiting'. This means, if we want teh LED to be bright, but happy, it needs say 20mA. If it has a Vf (forward voltage) of 1.8V, then we can calculate the resistor required:9-1.8V = 7.2V. That is the voltage across our resistor. We want 20mA = 0.02A, so again, plug the numbers in... R = V/I = 7.2 / 0.02 = 360 ohms. 360 whilst available isn't a standard value, so rather than drop it to 330, which would increase the current, possibly taking it over the maximum, we pick the next highest resistance, say 470ohm. giving I=V/R = 15.3mA.In your example, 3v 20mA LED, with 9V, the voltage across the resistor is 9-3 = 6V. for 300 ohms, current flowing through the LED and resistor is I = V/R = 6/300 = 0.02A = 20mA.If you were to increase your power supply voltage, this would increase current, because it would increase the voltage across the resistor, where-as the '3v' LED will always drop 3V - although this does increase slightly with current, and changes with temperature, but not by much. So say you now use 12V. 12V - 3V = 9V. I = V/R = 9V/300 = 30mA. This is quite high for a standard '20mA' LED (note, the 20mA figure is often for 5mm LED's, but not necessarily a 'standard'). So if youre power supply is going to vary, you should calculate the maximum and minimum current that can flow to make sure your LED doesn't die.2) 'Current consumption' is fairly easy to calculate for a purely resistive load, like.. a resistor! But with semiconductors, they have fixed voltage drops that doesn't vary much with different current flows, which complicates matters. It is true an LED can 'draw' lots of current, because by itself it does not limit how much current can flow, it does however limit the voltage across it, leaving the rest of the voltage to be taken up across its power connections. As above, it is these connections that should limit the current.The only difference between a 12V 1A DC power supply, and a 12V 500mA (0.5A) power supply is how much current it is capable of providing whilst keeping its output voltage at 12V. Therefore, you should pick a power supply with the correct voltage, and a current capability that is higher than you require. If you have a circuit that draws say, 250mA maximum, then using a 200mA supply could mean the power supply, drops its voltage (as in the case with mains transformer supplies, batteries) or has a protection circuit that kicks in which temporarily shuts off the power supply to protect itself. If its the first kind, you might get 250mA out, but its voltage might drop to 10.5V.. and that can change how much current your circuit will draw, and perhaps change how the circuit behaves or even stop it working altogether.There are so many cheap switchng 'wall-wart' power supplies these days, USB is a handy way to power things that you have a lot of choice. What is important is it is regulated, and has the correct voltage you require. As long as its current capability is higher than you need with a good margin for error (12V 500mA is a pretty good choice for small circuits that require up to 400mA) then its fine.But there is no point in lighting up a few 5mm LED's, that draw 60mA, with a 12V supply capable of 10A. The LED's (and their current limiting resistors!) will still only draw ~60mA, but if you make a mistake and accidentally short the power supply, 10A can melt things.Too long but I tried to explain everything.

author: xrunner
author_rank: Super Contributor
timestamp: February 02, 2017, 06:03:40 pm
content: Quote from: Adhith on February 02, 2017, 05:07:48 pmWell, I wouldn't word it that way. A circuit or component consumes the current that the laws of physics allow it to consume. If you connect a 100 volt battery across an LED that has a voltage drop of 2.5V, the LED will consume exactly what the laws of physics will allow - so much that it will burn out.An LED that is specified as requiring 20 mA will not automatically draw that current (unless it has internal regulation or resistors but ...)You can hook up lots of components in all sorts of "dumb" ways, and they would consume current as they will, but it's for the designer to make sure the way things are connected are done in a smart way to accomplish the given task and last a while (see above posts).

author: Vtile
author_rank: Super Contributor
timestamp: February 02, 2017, 06:58:34 pm
content: LEDs and zeners etc. are rather confusing at first because their ability to generate constant voltage over them (in some rather wide current window). This leads in situation you kind of need to think the situation in different order than you would do with basic resistors. (Like explained above posts)The Led in example at first kind of seems to bend the ohms law. In some sense it does, but if you accept that the resistance must be dynamic value inside the led in certain current window (say between values 1mA to 50mA) to make it behave like it behaves (creating constant voltage over it) and then you will notice that the ohms law still apply.While even in typical real life resistors in typical applications (low frequencies) in the otherhand the resistance is constant (passive), no matter the voltage or current. (Voltnuts will disagree)What makes the electrical engineering in such a challenging field of study at times partly comes from this fact that your real life components are not ideal and we try to model and calculate these non-ideal components with the rules and laws that do only work with ideal components, which at times leads extremely complex component models and situations so we can get around the fact that we are modeling our non-ideal real life with laws of ideal world. Someone might say that we are stubbornly using a wrong tools for the tasks.

author: Rick Law
author_rank: Super Contributor
timestamp: February 02, 2017, 08:10:02 pm
content: Quote from: suicidaleggroll on February 02, 2017, 05:24:52 pmI am in agreement withSuicidalEggrollgreat explanation here. If I may add another thought:If the circuit "accept 12v and draws no more than 0.5A", while it works with a "12V 80A" supply, however, there is a benefit in keeping it close to 12V 0.5A.I know 80A is just a random big number to illustrate a point. It also illustrate another point.12V 80A is a lot of power. So when something go south, a lot of damage can be done. Limiting it to say about 0.75A allows some head room yet limits the damage to some degree. It also may give you more time to react before things start burning out. So a current-limit can be a good thing.

author: suicidaleggroll
author_rank: Super Contributor
timestamp: February 02, 2017, 08:19:09 pm
content: Quote from: Rick Law on February 02, 2017, 08:10:02 pmIndeed it can. This is one of the primary reasons why, even though it's very easy and very cheap to convert a computer ATX power supply into a powerful +/-12V, +/-5V bench supply, nobody would recommend actually using one over a proper bench supply with built-in adjustable current limiting...especially when you're powering on a circuit for the first time. Mess something up, and that big powerful supply can blow things up very quickly.

author: AG6QR
author_rank: Frequent Contributor
timestamp: February 02, 2017, 09:42:22 pm
content: Here's a fundamental point that is sometimes missed. Ohm's law doesn't always hold. The "law" says that voltage across a component and current through the current are directly proportional, and the constant of proportionality is the resistance. Or to put it another way, if you draw a graph of voltage versus current, with voltage along the horizontal axis you get a perfectly straight line going through the origin, with the slope of the line being the inverse of the resistance.That is pretty close to true for wires and resistors -- close enough that we can usually treat it as fact, at least until we apply so much voltage/current that the device heats up, at which point its resistance changes, and eventually it might even melt and come apart. But if its resistance changes as a function of temperature, and temperature is increased by applying voltage/current, then the "constant of proportionality" in Ohm's law is hardly constant, is it?But where Ohm's law really falls apart is with diodes, and other doped semiconductors. Draw a graph of voltage versus current for a diode, and it is nothing close to a straight line going through the origin, the way Ohm's law would predict. For a significant portion of the curve, a diode almost acts like a voltage source, where the current changes wildly while the voltage stays almost constant. Diodes change their characteristics with temperature, too, but while resistors draw less current as they heat up, diodes generally draw more current as they heat up.The result is that if you try to feed a diode a constant voltage, it's very difficult to predict what the current will be, and whatever current it starts out with, it will increase drastically as the diode heats up. This near-vertical portion of the V-I curve is very unstable if you feed the diode with a constant voltage source (but it happens to be extremely stable if you feed a diode with a constant current source).Ohm's law is a very useful simplification. It is accurate enough to be extremely helpful in understanding and designing circuits, especially those involving resistors. Ohm's law is perfectly true for an ideal resistor. It is very close to true for a real-world resistor, and even for things we don't ordinarily think of as being resistors, such as wires and traces on a circuit board. But it is not a useful explanation of what happens inside a diode.When we put a resistor in series with a diode, the graph of voltage versus current for the combination has a reasonable slope, and the V-I curve will intersect a given voltage at a reasonably predictable, stable current, even as the temperatures of the components change a bit.

author: suicidaleggroll
author_rank: Super Contributor
timestamp: February 02, 2017, 10:10:08 pm
content: Quote from: AG6QR on February 02, 2017, 09:42:22 pmI've never liked that interpretation of Ohm's law. An ohm is a derived unit, it's defined as a volt/amp, you can't violate that. Just because a device doesn't have a constant resistance with respect to time, voltage, current, temperature, light, etc. doesn't mean you can't use R=V/I, it just means you can't assume the resistance you calculate using this equation under condition A will still apply under condition B.The assumption that any resistance you calculate using R=V/I under condition A will still apply under conditions B-Z, without change, forever and ever amen...THAT is the simplification. R=V/I will always be correct, however (sometimes it might not be very useful though, if "R" is a strong function of "V" or "I").

author: AG6QR
author_rank: Frequent Contributor
timestamp: February 02, 2017, 10:54:37 pm
content: Quote from: suicidaleggroll on February 02, 2017, 10:10:08 pmBut Ohm himself said that voltage and current were directly proportional, and that the constant of proportionality was called resistance, and that the resistance was constant.I suppose I'm in agreement that I don't really like this formulation of Ohm's law, because it's not exactly true. But when you tweak the law to allow for varying resistance, and define resistance as whatever V/I turns out to be, without describing any particular relationship between V and I, you may come up with a model that more accurately reflects what's going on, but your model isn't the one Georg Ohm came up with. I prefer to reserve the term "Ohm's law" for the law as Ohm described it.

author: IanB
author_rank: Super Contributor
timestamp: February 02, 2017, 11:30:14 pm
content: Quote from: suicidaleggroll on February 02, 2017, 10:10:08 pmThe ohm as a unit of measurement should not be conflated with Ohm's Law.If we divide a quantity with a unit of volts by another quantity with a unit of amps, we get a quantity with a unit of ohms. Whether that quantity represents a useful property, or even a property called "resistance", is a matter to be considered on a case by case basis.Ohm's Law is a statement of an ideal behavior where the relationship between potential difference and current flow has linear proportionality, and in this particular case the ratio between voltage and current can be called resistance.If the material or component under consideration does not follow Ohm's Law, then there is no constant of proportionality and the ratio V/I should not be called resistance without additional qualification describing the circumstances.For example, consider the diode. If we divide V/I at some operating point then we certainly get a quantity that can be stated in ohms. We might call it the "instantaneous resistance". Alternatively, we might consider the slope of the V-I curve and measure dV/dI. We will also get something measured in ohms, but it will have a different value. We might call that the "differential resistance".Which measure to use depends on the application. If we wish to consider the diode in a circuit where we are performing a "small signal" analysis around an operating point, then we certainly will want to use the differential resistance dV/dI. If we happen to use V/I we will get incorrect results.

author: james_s
author_rank: Super Contributor
timestamp: February 03, 2017, 12:00:23 am
content: Quote from: suicidaleggroll on February 02, 2017, 08:19:09 pmDespite the high current they can supply, computer PSUs tend to have very good overload protection and if you short circuit one, even with something that would easily burn out carrying the full current the PSU is able to deliver, they will shut down instantly.

author: suicidaleggroll
author_rank: Super Contributor
timestamp: February 03, 2017, 12:00:49 am
content: Quote from: IanB on February 02, 2017, 11:30:14 pmThat's the real problem, I think...those two are pretty much used interchangeably these days. Colloquially, "Ohm's Law" is typically used as little more than the name for the "R=V/I" equation. Even though that's not strictly/historically true, just google "R=V/I" or any other permutation you want and see what pops up, literally every page calls it "Ohm's Law". So to tell somebody that for a given circuit or device, R=V/I holds true, but Ohm's Law does not, will only make them even more confused.

author: IanB
author_rank: Super Contributor
timestamp: February 03, 2017, 12:20:42 am
content: Quote from: suicidaleggroll on February 03, 2017, 12:00:49 amI think the Internet exists to make people confused. That is one advantage of paper textbooks: they are, by and large, written by subject matter experts and are reviewed by peers and editors before publication.It is clear that "something" = V/I, but that something can only properly be called resistance or given the symbol R if the device under consideration is a resistor. Just because there is a lot of misinformation out there does not mean we have to perpetuate it. It is really easy to tell people that diodes do not follow Ohm's Law, and there is no reason for them to be confused by that.

author: AG6QR
author_rank: Frequent Contributor
timestamp: February 03, 2017, 01:01:02 am
content: For a college-level discussion of current, voltage, and resistance, in the context of a physics class, see here:If you want to skip straight to the "meat" of it, go to 17:00, where he shows you that Ohm's law sometimes holds. But then he quickly follows that by showing how Ohm's law falls apart and doesn't always hold. This is pretty much the way it was taught when I went to school, and agrees with what Halliday and Resnick present in their classic physics textbook.

author: Adhith
author_rank: Frequent Contributor
timestamp: February 03, 2017, 07:07:10 am
content: Thank you my friends..reading through all of your wonderful answers was truly a great help to me but now I got some more doubts.1) so its the excess voltage in a circuit which drives more current into the circuit right ?? so is there anything wrong in connecting three 3v LEds in series directly with a 9v battery without a resistor. ??regarding voltage divider & constant current supply using LM3172) So if a single resistor can be used to drop the excess voltage, then why a voltage divider is used to reduce the voltage ?? I know that the voltage from the voltage divider is a regulated one., is that the only reason for which it is used??3) I have seen the constant current supply circuits using LM317, especially those used to lit up few LEDs in same brightness using a 9v battery.so how this is possible to provide the same current regardless of the loads ??

author: hamster_nz
author_rank: Super Contributor
timestamp: February 03, 2017, 07:40:22 am
content: For 1), the battery will either be slightly below the voltage needed to light the leds, and nothing will happen....Or the battery will be a smidgen above the voltage needed ti light the leds, and the internal resistance of the battery and leds will limit the current....Or the battery will be high enough that the internal resistance will not limit the current sufficiently, and something will 'go pop'.In other words, without the limiting effect of a series resistor the actual result is unpredictable, but it just might work.

author: Vtile
author_rank: Super Contributor
timestamp: February 03, 2017, 09:43:31 am
content: Quote from: AG6QR on February 03, 2017, 01:01:02 amYes, it would have been nice to be on those Mr. Lewins lectures. He were really good of what he did, luckily there is also other great physics teachers who haven't had as much exposure of public eye. Luckily mr. Lewins lectures are also on the youtube, so I can at times enjoy listening those.It is true that Ohms Law is most of the time just thrown out to just point out the formula U=R*I, which in the sense is more correct that the law part of the law. How we can blame it, as the law - as far as I can remeber - is archaic, almost from the era of science where we still had three elements: fire, water and air.The ohms law is usefull and correct in the sense it descripes how each three units are delivered from measurements in given time. If we apply 1 A current through 1 ohm resistance we get 1 volts of potential difference. This even holds for the diode! The tricky point is to know the resistance.In the same time it fails in its archaic form, it should be thrown out of window and renamed to "law of dances" or something. Where every three components are objects, a functions or a networks.If we must point out that ohms law fails, we must remember that many other ideal physics "laws" do fail in similar manner and only work in certain window of application without extensions. If you want example, look at the mr. Lewins pendulum, where he do not point out the failure. IIRC.

author: hamster_nz
author_rank: Super Contributor
timestamp: February 03, 2017, 10:09:37 am
content: Quote from: Adhith on February 03, 2017, 07:07:10 amThe subtle point you are missing is a single resistor doesn't drop voltage. It you attach one end of a 100 Ohm resistor to a 100V supply and touch the end, you get 100V to your finger.If you connect the same resistor between 100V and GND, depending which end you touch you get either 100V or 0V - and the resistor gets very warm, wit V^2/R = 100W of heat to get rid off.However, if you connect two 100 Ohm resistor in series between 100V and GND, the point in the middle will be at 50V (this will change if draw enough current from the middle point to significantly upset the balance).A voltage regulator like a LM317 "magically" adjusts it's resistance to ensure that it's output is held at the desired voltage with respect to the ground pin. You create a constant current source by using the VRM to keep the the voltage across a known resistor at a desired level - to to this keeps the current flowing through the resistor (and therefore through the regulator and rest of the circuit) constant.

author: Vtile
author_rank: Super Contributor
timestamp: February 03, 2017, 10:29:56 am
content: Quote from: hamster_nz on February 03, 2017, 10:09:37 amThat is wrong. First of 1 resistor do drop voltage, but it doesn't divide the voltage (well actually it does, but...it's semantics of what is a single resistor). Secondly when you touch the resistor in your given example your skin and body and ie. floor or air do form a resistor it is just so big that it seems that it doesn't matter (in case of 100V source it actually starts to matter), but if you look closely enough you notice there is (loosely) two resistors (one before and one after the contact point) and there is no +100V, but something like +99.999999 volts (in between source negative and your point of contact) (if you could measure it with ideal (=imaginary) voltmeter) That is if the source is ideal and there is no other current paths in this imaginary exercise.

author: KL27x
author_rank: Super Contributor
timestamp: February 03, 2017, 07:12:01 pm
content: QuoteVery simply:Single resistor works to drop voltage. But it is only stable when two conditions are met.1. The supply voltage is constant2. The load is constantYou can think of the load as the bottom half of a resistor ladder, in this example. If the supply voltage increases, either the voltage and/or the current draw of the load will increase. If the load changes, it's like changing the bottom half of the resistor ladder. The voltage at the middle node will also change.When the supply voltage is instable, you need an active regulator if you want specific output voltage.If the load is variable or otherwise uknown, but the supply voltage is stable, then a resistor ladder can be useful to produce a reference voltage. As long as the max range of the draw of load is relatively small (e.g. w/e circuit is tapping this center node draws VERY little current), it is easy to make a resistor ladder which is low enough in impedance that the voltage is stable. For instance, the input of an opamp will have a super high impedance (i.e., it will draw very tiny amount of current). But the exact impedance is variable between individual part numbers and/or even individual devices from different lots of the same part, and perhaps it even changes with temperature or other conditions. So rather than measure the impedance of each device in order to calculate a "top" resistor of the ladder, you can just put this super high impedance opamp input in parallel with a much lower value/impedance resistor to make the "bottom resistor," so that the effective resistance of the opamp input is now insignificant and can be completely disregarded.

author: james_s
author_rank: Super Contributor
timestamp: February 03, 2017, 08:23:05 pm
content: Quote from: Vtile on February 03, 2017, 10:29:56 amThe resistor will not drop any voltage until there is a load placed upon it though. Now any method of observing the voltage will of course cause some drop across the resistor, but for the sake of discussion this is going to be negligible with a 100 Ohm resistor, whether it's your finger or a high impedance DMM. It's still going to be 100V, and the drop will only be observable with a very precise (lots of digits) meter.

author: KL27x
author_rank: Super Contributor
timestamp: February 03, 2017, 09:37:51 pm
content: ^Kirchoffs law. The single resistor not attached to anything can be described as a resistor ladder where the bottom resistor has infinitely high resistance/impedance. Thus, the voltage at the "center node" is equal to the supply voltage. Of course decreasing this impedance (by putting a parallel path to just the "nothing," such as a circuit/load), will necessarily change this calculation and lower the voltage.

author: Vtile
author_rank: Super Contributor
timestamp: February 03, 2017, 10:10:19 pm
content: ^ It would actually be interesting to see a source for that statement. I don't try to say it is wrong... I just can'trememberjust now any source for that claim (while it faintly rings some bells), while it is reasonable simplification for the situation and seems to be used in many places (ie. spice simulators). It just sounds like this topic.What comes to the DMM I entered the situation to my calculatron, source is ideal 100Vdc (no inner -tances), then drop across 100 ohm:1 megaohm DMM - 10 mV (typical siliscope input impedance as far as I know)10 megaohm DMM - 1 mV100 megaohm DMM - 100uVa bit less if one uses another DMM over 100 ohm resistor simultaneusly to measure the drop.

author: KL27x
author_rank: Super Contributor
timestamp: February 03, 2017, 11:33:19 pm
content: My source is kl27x and his basic understanding of applied math and science.

author: Vtile
author_rank: Super Contributor
timestamp: February 04, 2017, 07:57:06 pm
content: ^ I were too hang-up with the "what happens in moment of connection" so I didn't see the obvious. Doesn't matter, this starts to be in the area of offtopics offtopic, without any practical use. Hopefully OP (or someone else) get something out of it.There is no wires, only conductors with resistances (impedance).

author: 3db
author_rank: Frequent Contributor
timestamp: February 05, 2017, 10:34:51 am
content: My teachers used to emphasise "At constant temperature" when teaching Ohms Law.3DB

author: Ratch
author_rank: Regular Contributor
timestamp: February 06, 2017, 03:54:02 am
content: Quote from: Adhith on February 02, 2017, 05:07:48 pmCorrect, do you know why?QuoteIt is the voltage and current capacity that is important, not the battery type. If the voltage source cannot supply enough current to run the LED, then it will not work for that application.QuoteIt does both, for the purpose of dissipating the excess electrical energy as heat, so as to protect the diode from absorbing the excess electrical energy.QuoteDo you mean hooked up to a voltage without the proper current limiting? You are not clear on that point. Yes, you can burn out you LEDs that way.QuoteCurrent or charge flow is not "consumed". It is changed to a higher or lower rate.QuoteYou can use a 100 amp capacity voltage source to run a tiny LED if you want to do so. Just make sure you put enough resistance in the circuit to limit the current to what the LED can take without burning out. The calculations you made above should do it.Ratch