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I have the following circuit set up on a breadboard, and I've measured current and voltage between each node. Current everywhere is .02A, as expected, and my resistor came out to 217R.

schematic

simulate this circuit – Schematic created using CircuitLab

As per the above schematic, voltage across the LED came to 2.25V.

This is a bit confusing to me. Apparently, voltage is joules per coulomb. So, I assume:

From positive lead to anode, there are ~6.66 joules per coulomb
From cathode to the start of R1, there are ~4.41 joules per coulomb

In other words, every coulomb that went across the LED lost ~2.25 joules.

How long did it take for every coulomb to lose 2.25 joules? Did it take a unit of time for every unit of charge to lose that amount of energy?

Would it be something like:

R = L/T

R = rate of L per T
L = loss of energy
T = time for a coulomb to move from anode to cathode

However, this seems flawed to me, the notion that each coulomb would lose energy at a set rate, since current (or, movement of charge over time) is constant throughout the entire circuit. What am I misunderstanding here?

Edit

When an LED is rated at .02A and 2V for best performance according to the manufacturer, does that mean the circuit needs to be set up in such a way that movement of charge over time is equal to .02A across the LED, and that each coulomb must lose 2 joules between the anode and cathode?

Or, does it simply mean that the amount of energy it takes to move a coulomb from anode to cathode must be equal to 2 joules, or else the LED will be damaged over time?

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  • \$\begingroup\$ I feel like you need to incorporate the .02A in there somewhere \$\endgroup\$ – Funkyguy Jul 9 '14 at 23:56
  • \$\begingroup\$ @ShannonStrutz, probably. Can you elaborate on that a little bit? \$\endgroup\$ – Josh Beam Jul 9 '14 at 23:57
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    \$\begingroup\$ well we can expand ohms law out to your question J/C = RI, so if you want Δ(J/C) then there must be a change somewhere in RI. Since the resistance won't change your I must. So we can expand to Δ(J/C) = R*ΔI. Since current can be represented as Coulombs/Second, then we have a Δ(J/C) = R * Δ(C/s) \$\endgroup\$ – Funkyguy Jul 10 '14 at 0:24
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How long did it take for every coulomb to lose 2.25 joules? Did it take a unit of time for every unit of charge to lose that amount of energy?

Having a 2.25V across the diode means that for an amount of charge equal to one coulomb to pass through the diode, 2.25 joules need to be spent. Note there is no time involved in this particular value.

Since the current is 0.02A, this means there are 0.02 coulombs of charge going through the diode every second. These 0.02 coulombs need 0.02 * 2.25 = 0.045 Joules to go through the diode and they do so each second so the energy spent is 0.045 Joules/sec, or 45mW. Also note that this energy is spent at the power supply and is transmitted all the way to the electrons at the diode through a very complex process. I try to explain this process further down.

One coulomb is roughly the charge of 6.241E18 electrons so you have roughly 12.482E16 electrons going through every second and spending a total of 45mW of energy.

Notes:

Volts = Joules / Coulomb

Amps = Coulombs / Second

Watts = Joules / Second

So, how did we come to derive that it would take a 217 Ohm resistor to drop the LED down to 2.25V? The practical formula I was given is (Source Voltage - Voltage Rating of LED) = Resistance * Amp Rating of LED. In other words, we try to find R: 6.66 - 2.25 = R(.02). How is the voltage of the resistor related to the voltage of the LED?

At this point, it is important to note that the only thing that moves around the circuit are electrons, which are negative charge carriers, and the concepts of voltage and current are macroscopic ways to describe what is happening.

A rough explanation would be the voltage difference across the resistor is due to charge not able to go through it fast enough as electrons bounce around the atoms that make up the resistor. LEDs are a little bit more complex. Part (a very small one) of the voltage difference comes from their internal resistance. Most of the voltage drop across them comes from the voltage difference across the PN junction.

A better way would be to think of it this way: When the power supply is first connected, charge moves from the negative terminal of V1 towards the resistor. This increases the negative charge on one end of the resistor and some charge starts to pass through, following the charge density gradient across the resistor. Charge then starts to go through the diode. In the LED, electrons start to recombine with holes and in doing so, photons are emitted. At the same time, electrons are drawn away from the diode's anode by the positive terminal of V1 and thus holes are created, waiting to be recombined with charge that has traveled all the way from the negative terminal of V1 through the circuit. Assuming V1 is able to move electrons fast enough from its positive terminal to its negative terminal, so as to keep the voltage difference across them constant, the circuit reaches a steady state as described by the equation you have.

When an LED is rated at .02A and 2V for best performance according to the manufacturer, does that mean the circuit needs to be set up in such a way that movement of charge over time is equal to .02A across the LED, and that each coulomb must lose 2 joules between the anode and cathode?

Or, does it simply mean that the amount of energy it takes to move a coulomb from anode to cathode must be equal to 2 joules, or else the LED will be damaged over time?

If you have a look at an LED datasheet, you will find two graphs. One is forward current (IF) vs forward voltage (VF). The other one is (relative) intensity vs forward current. The manufacturer specifies the right combination of voltage and current for the led to operate reliably at its rated brightness.

If you select a higher resistor value, less current is going to pass through so less electrons will recombine and the LED will be dim. If you select a lower resistor value, more charge is going to pass through the resistor and the LED over time and the LED is going to be damaged.

So it is not that each coulomb must lose 2.25 joules rather the more electrons go through over time than the rated current, the LED will be damaged.

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  • \$\begingroup\$ Thanks for the answer, this makes a bit more sense. So, how did we come to derive that it would take a 217 Ohm resistor to drop the LED down to 2.25V? The practical formula I was given is (Source Voltage - Voltage Rating of LED) = Resistance * Amp Rating of LED. In other words, we try to find R: 6.66 - 2.25 = R(.02). How is the voltage of the resistor related to the voltage of the LED? \$\endgroup\$ – Josh Beam Jul 10 '14 at 0:52
  • \$\begingroup\$ In other words, the voltage across the resistor isn't related in any way (other than by total current and total resistance in the circuit) to the LED. \$\endgroup\$ – Josh Beam Jul 10 '14 at 0:54
  • \$\begingroup\$ For the LED to light up the brightest, it takes a specific amount of current to flow through it. Less current makes it dim, more current just makes it overheat. So the manufacturer gives you the value of optimal current. The voltage drop across the LED mostly comes from the electric field that appears on the PN junction of the semiconductor. Subtract this from the voltage of V1 and you get a voltage potential across the resistor. That potential drives a current through the resistor. You select the resistor so the correct amount of current can flow through the LED. \$\endgroup\$ – Evan Jul 10 '14 at 1:23
  • \$\begingroup\$ So the voltage across the resistor is related to V1 and the voltage across the LED. The formula you have assumes the LED and the voltage source have no internal resistance. \$\endgroup\$ – Evan Jul 10 '14 at 1:32
  • \$\begingroup\$ @JoshBeam, Basically what you have is an initial open circuit as the diode is reverse biased. When you apply Vcc it appears across the diode's terminals and will lower the on-state resistance and begin conducting current. This current forces a voltage drop across the series resistor until eventually the circuit moves to its steady-state conditions where 2.25V (in your case) is dropped across the LED and the remainder across the resistor (with appropriate current to provide such conditions). \$\endgroup\$ – sherrellbc Jul 10 '14 at 3:50

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