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Basic Electronics » Current vs Voltage on a circuit

June 02, 2009
by luisgarciaalanis
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From what I have learned on the last few days, correct me if I am wrong, is that a Battery has Voltage present but no current unless there is a load.

So lets say we have a LED and we want to plug it to a 5V battery, the led will ask for current based on its graph:

So connecting the LED with a max Volt Forward input before the current becomes exponential smaller than the Volt source would cause the LED to demand lot more current than its maximum allowed before burning to HELL. So Current is what end up killing the LED's. current demanded by the excess amount of Voltage on the line.

So a resistor will limit the the amount of Voltage on the line therefore having the LED request less current.

Going back to the simple LED example, if I have NodeA (Anode) to NodeB (Catode), and in between I have resistor and then LED, if I meature the on V on NodeA then its going to be 5V, if I meature it on NodeB its going to be 5V - ResistorDrop - LED drop === Zero

Current in NodeA and current in NodeB should be the same since its the current demanded by the circuit??? As for voltage, it drops for every loadpoint it crosses???

Now here is a Wikipedia circuit example: in this example you can see the LED coming before the resistance, is this a bogus circuit? bug on the Wikipedia? from what I understand the Voltage is going to reach the LED first and be 5V then burn the LED before reaching the resistor that will lower the voltage???

Same for LEDS in series, lets say we have enough leads in series to drop the voltage very close to Zero on a 9V battery, the first LED will get 9V and explode BANG!$@# is this correct?

June 02, 2009
by luisgarciaalanis
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Mmmmmmmmmmmmmmm.............. ohms law

so we know V is fixed to 5V and we know current flows the same from all components between the two nodes. We adjust the resistance to keep the current within working rage of the leds. so Volts really don't do anything to burn the LED, it’s the current the evil one.

So it’s quite confusing the graph because it shows that if I add more V, I will go up exponentially. That is where I got the deduction that the LED before the resistance would get a full 5V and die.

But ohms law forces the current to be the same across 5V (from the battery) = I/(Resistance from LED + Resistance), but if the I is forcefully the same to all elements between the nodes, this means that V will be automatically lower for all the LEDS?

How would this work?

June 02, 2009
by luisgarciaalanis
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I get it now :) the voltage consumed by each LED depends on the current flowing, the fact that 5 Volts are on the first LED does not mean its going to blow because the current will only allow so much to be consumed :)

June 10, 2009
by BobaMosfet
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An LED doesn't consume voltage. It is merely a resistance (as is everything in a circuit) to electrons traveling through it. In otherwords, friction. An LED is, in fact, a Diode (Light Emitting Diode). Voltage only determines how many electrons are desired to be traveling through the circuit.

Voltage: Attraction at the positive terminal of your power source. Current: The number of electrons traveling from the negative terminal to the positive terminal through whatever resistance you've put in the way (LEDs, Resistors, Capacitors, etc).

With sufficient attraction, current will conduct through any material. In effect it will overwhelm the physical matter and convert it from solid or liquid to plasma (causing it to overheat and disintegrate).

This is not BOGUS. The resistor is there, because current flows from minus (-) to positive (+). So it has to be hit with the current before the LED in order to only allow a safe amount of energy to reach the LED.

Ohm's law doesn't cause anything to happen. It merely reflects upon the fact that there is a relationship (based on trigonometry) between Voltage (E), Current (I), and Resistance (R) representated by either resistance or reactance. When you are calculating both resistance and reactance together, you call it impedance.

Ohm's Law: E=I * R. This is a circule equation. I = E/R, and R = E/I.

If you want to calculate these things in your circuits, you need to understand how to calculate them for either circuit components in SERIES, or in PARALLEL. They calculate differently.

June 12, 2009
by luisgarciaalanis
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but the resistor can be after the led, like in the link you provided. So the resistor after the led is slowing down the current.

I1 = current going through the LED R1 = resistance from the led I2 = current going through the Resistor R2 = resistance of the resistor

5V = I1R1+I2R2

lets say the Voltage drop of the LED is 1.8V

5V = 1.8 + I2R2 3.2V = I2R2

current is the same for series component circuits qne we know that the LED needs 20mamps so

3.2V = 20mR2 R2 = 160 ohm

So what determines that only 20 mamps are flowing through the wire?

From what I understand the amps a battery emmits depend on the load (in this case LED+R)

So the resistor drop of the Voltage is 3.2 only because the current is 20ma.

The led should also have a fixed resistance so.... Can I assume the mamps request out of the battery is due to the comvination of both resistances?

if I make the resistance smaller this will increese the amps of current making the LED have more voltage, and since the rating of the led tells you how mutch it can handle it would burn it?

June 13, 2009
by luisgarciaalanis
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OMG sorry about the spelling mistakes :( LOL

June 13, 2009
by BobaMosfet
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luisgarciaalanis - "but the resistor can be after the led, like in the link you provided. So the resistor after the led is slowing down the current. "

Here is a link to the original LED circuit you are referring to:

You are confusing the direction of voltage with the direction of current. Look at the schematic again. The resistor is in FRONT of the LED where current is concerned. Current flows TOWARDS positive and AWAY from negative on your power-source. So although the resistor is on the NEGATIVE (Cathode) side of the LED, it is actually getting current FIRST.

You need to grasp and understand this before moving on. Everything else gets more complex.

June 13, 2009
by BobaMosfet
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In fact, in order to help you understand the circuit you're looking at further, I've taken the LED circuit diagram and worked the math out for it, based on an using a RED LED (as opposed to Blue or Yellow, or what not). The forumula given on the Wiki page is confusing because they didn't show you the simplified version of what they are doing. Try this:

Voltage is respresent always by the letter "E" (Electromotive Force). Current is always represented by the letter "I" (from the French word for intensity). R will represent Resistance (although it can mean reactance for capacitors and induction coils).

Therefore, here is Ohm's Law properly represented (a circle formula, best desgribed by the trigonometric equivalent relationships between E, I, and R). Depending on which of the 3 you are trying to determine, using the other two, the 3 formulas are:

E = I * R
I = E / R
R = E / I

In the Wiki LED Circuit we only know the following:

E1 = +9V   Power Source
E2 = 1.8V  Drop across LED (how much voltage it requires to overcome the LEDs resistance)
I  = 20mA  Optimum Current that a Red LED requires to fully illuminate.
R  = Size of Resistor needed to limit current and voltage to a safe level, currently unknown.

I use "E1" and "E2" because of limitations in entering text in this forum. "E1" is "E subscript 1" and "E2" is "E subscript 2" to keep the two variables distinct.

Since R is unknown, using Ohm's law, the correct formula is: R = E / I. Since (and this is IMPORTANT to understand) we have a voltage source providing voltage, and a component (the LED) resisting voltage, we only require a resistor to resist what the LED itself DOESN'T already resist. Here is the SAME Ohm's Law formula for R, using E1 and E2 (which is what the Wiki page did a poor job of explaining):

    (E1 - E2)
R = ---------  Which is the same as writing on a single line:  R = (E1 - E2) / I

Now, let's work the math:

R = (9.0V - 1.8V) / 20mA
R = 7.2 / 0.020
R = 360 Ohms.

Now, if you're wondering how I determined the E2 = 1.8V, Go to this Wiki link and look at the values for Colors and Materials for LEDs:

It says that a Red LED requires between 1.63V and 2.03V. The delta, or difference between those two values is .4V . So if we want a safe voltage level in the middle, we cut that delta in half. .4V / 2 = .2V. Therefore the safe illumination voltage (ideal) is 1.63 + 0.2 = 1.83V. Rounded down to 1.8V.

Hope this helps!

June 13, 2009
by luisgarciaalanis
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thanks BobaMosfet,

In series current is the same for all elements connected correct? so it really does not matter if the resistor is before or after the LED.


I have seen a lot of samples on the internet where they do the opposite, for example:

June 13, 2009
by luisgarciaalanis
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also isn't ground the circuit return? and is connected to the negative labeled battery pole?

so current flows from + labeled (I say labeled because its wierd LOL)

June 13, 2009
by BobaMosfet
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I still believe the resistor must come before (cathode-side) the LED, because the whole point of the resistor is to drop the VOLTAGE level to a safe value the LED can withstand. If you do it afterwards, the LEDs are getting hit with the full voltage which may damage them, irregardless of what the resistor does to the current across the circuit.

I'd love to hear some other people chime in on this.

June 13, 2009
by wayward
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Hi luisgarciaalanis,

these two branches are identical:

     A   ↗↗   R    B
(+) —o———▷├———VVV———o— (-)

     A    R    ↗↗  B
(+) —o———VVV———▷├———o— (-)

My understanding of why this is so is rather limited, based on what I remember from high school physics, but let me try to put it in writing here and you tell me if it makes any sense to you. (Someone with a better understanding may wish to naysay me, which I approve.)

A conductor is a material that allows the particles carrying electric charge to move through its volume. For a metallic conductor, there are two kinds of charge-carrying particles: negative charge carriers (electrons), and positive charge carriers (holes, or metal atoms lacking electrons). When we connect such a conductor to a battery, two opposite flows form: flow of electrons from the more negatively charged end, and a flow of holes from the more positively charged end. The important point to note here is that for an ideal conductor, one with no capacitance, those two flows (i.e. number of carriers through a section per unit of time) must on average be equal. There can be no charge buildup. This is not exactly what happens at the atomic level, but we can safely choose to ignore that, and here's why.

An ideal conductor behaves like an ideal spring: if you press the spring between your fingers, it will compress, redistributing the elastic force equally across its length simultaneously and instantly. Similarly, an ideal conductor will redistribute the difference in voltage across any two points on it simultaneously and instantly. Practical springs don't compress equally because there is some time that has to pass for the interaction between adjacent segments to establish an equilibrium in forces. For conductors, this happens at the speed of interaction between charge carriers, which is the speed of light. So for all intents and purposes, at least when dealing with macroscopic objects, we can ignore that "settling time" and simply assume that the voltage differences between adjacent segments of the wire settle instantly.

For this reason, the two branches above are exactly the same, just as compressing two ideal springs will redistribute the compression force simultaneously: voltage differences will appear "instantly" and the current will be defined, for all intents and purposes, at the moment of connecting the branches to a source.

June 14, 2009
by BobaMosfet
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Honestly, I've decided I'm going to sacrifice an LED to science and see if it will work either way. I'm actually suspecting it will, but ONLY because it's a SERIES circuit. If it were Parallel, the LED would get fried if the resistor were connected to the Anode.

I'm also going to hook an oscilloscope to it and see if I can capture what happens across the LED.

June 14, 2009
by wayward
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Parallel fries with full voltage force, yes O:-)

June 14, 2009
by wayward
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Boba, I don't understand actually: if it were a parallel circuit, you mean, like so:

 |       |

You say "the LED would get fried if the resistor were connected to the anode": but in my understanding, te resistor has to be connected to both the cathode and the anode on the LED in order to form a parallel circuit with it. Or what am I missing?

July 30, 2009
by luisgarciaalanis
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Thanks for the information, its more clear now :) The way I see it is that its like voltage its like preassure, and current its like flow but the pipes (conductors) are already filled with electrons so everything moves together and it balances out. Current is always constant but can be blocked if ther is not enough preassure (V). so if you put resistors it would ease anything on the series line.

August 05, 2009
by BobaMosfet
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wayward-- I meant a larger parallel circuit (more components).

However, I redact (now that I've learned some more) some of what I said and update it to this: Apparently, it doesn't matter whether the resistor is before or after the LED. This is because the LED and the resistor are in series, no matter what else the rest of the circuit is doing. And thus, being in series, Kirchoff's current law applies. The resistor controls the current for the entire series portion of the circuit.

The truth is, when the switch is turned on, everything has maximum voltage and maximum current applied. This is inevitable. It takes a nanomoment (my own term meaning a really really small amount of time) for all the components to come into play (capacitors to fill, coils to charge, resistors to resist, etc), thus settling everything where it should be, within safe bounds.


Just to clarify: Voltage is constant, current is not. Voltage will be there when current is not. Current cannot be without voltage. Study Kirchoff's laws with regards to voltage and current. You should also become familiar with Watt's law (P=IE) in order to make sure you size your resistors accordingly.

For example, let's say we have a 9V battery, connected to a 7805 Voltage Regulator to drop it to +5. In the circuit is a resistor and a Yellow LED.

B (Battery) <----> 7805 <----> R1 <----> D1 (LED) <----> GND
  1. The 7805 Takes the 9V from the Battery and drops that to +5V and will allow up to about 1.2A to be drawn from it.
  2. R1 has been carefully chosen to limit the voltage and current remaining in the circuit.
  3. The LED is (essentially) a short circuit between the terminals of the power-supply. If we didn't use R1 to limit the amount of current allowed to go through it, voltage would force however much current the 7805 would allow (approximately 1.2A) through the LED until it fried.

We calculated R1 this way, using 2 important formulas:

R = E/I      (Ohm's Law)
P = E^2 / R  (Watt's Law restated without Knowing Current)

The Yellow LED is rated at 20mA with a forward mean voltage of 2.14V. That means we have to resist enough current in the circuit such that only a little gets through, so only a fraction of the voltage attraction comes into play. Remember, voltage and current and resistance work hand-in-hand with one another, not independently. That is because current flows at a constant rate. We cannot control that rate, we can only control the quantity of current flowing at a given rate, which in turn affects the other aspects (E and R).

Using our first formula:

R = (5V - 2.14V) / .020A
R = 2.86V / .020A
R = 143 Ohms.

Since they don't make a 143 Ohm resistor, we chose the next closest value: 150 Ohm. A 5% Carbon resistor is fine. To make sure that is enough current, let's work the formula the other way to be sure that resistor will work:

I = E/R
I = 2.86V / 150 Ohms
I = 0.019A  (or 19mA)

So we run the LED at 19mA at 2.14 Volts, using a 150 Ohm Resistor.

Are we done? Nope. We need to make sure we size the resistor accordingly so it doesn't fry, soaking up that extra current (which it converts to heat). Here's where our second formula comes into play (Watt's Law) which will tell us how much Power we're trying to dissipate as heat:

P = E^2 / R
P = 2.86^2 / 150 Ohms
P = .054W

So, as you can see, the resistor isn't dissipating even 1/10th of a Watt. To be safe, double the value of P, and it isn't (worst case) dissipating 2/10ths of a Watt. In other words, a tiny 1/4 Watt resistor will work.

Now, if you don't want to work the math out yourself, you can use the following URL and it will do the calculations for you:

Hope that helps!

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