What happens if an LED is connected to a supply voltage greater than its voltage drop?

Question

My understanding of resistance and voltage is horrible. I heard that with Kirchhoff's law, (in my words, please correct) the voltage used by the circuit must equal the voltage supplied. For example, if I had a 9 V battery, I must use all 9 V of it.

Lets say I have a LED with a typical forward bias voltage of 3.1 V, which means it loses 3.1 V while generating light. Will the LED if 9 V is used, burn out?

It is most likely true, but a nice example will really make my understanding more intuitive.

Answer 1

This is one of those situations where your problem is not how good you are at analysis or what base knowledge you might have, but simply that you have no clue what you don't know. This always makes the first step into electronics a very high one.

In the case of your example, what don't you know about a battery?

1. The terminal voltage of an ideal battery would never change (at least until all the energy storage capacity is used). So there must be factors that affect the terminal voltage and its useful energy capacity. A quick list is chemistry, volume of materials, temperature and anode/cathode design.
2. A practical battery has limited capacity and many of the other factors influencing terminal voltage and potential current capability can be rolled into a model element called 'Internal resistance'. In the model for most larger batteries this will be fractions of an ohm. However the battery also has other elements such as capacitance and inductance to make the situation more complex. You could start by reading about battery models with texts such as this.

A great example of a larger battery with very small internal resistance is a 12 V car battery. Here, when you start the car it takes hundreds of Amps (kW of power and current in the 600 A range) to turn over the motor and the terminal voltage might drop from 13.8 V (a fully charged lead-acid car battery) to only 10 V when cranking. So the internal resistance might be (using Ohms Law) only 6 milliohms or so.
You can scale the thinking for this example to smaller batteries such as AA, AAA and C batteries and at least begin to understand the complexity of a battery.

Now what don't you know about an LED?

1. The complexity of the electrical model for a diode (whether just a rectifier or an LED) is immense. But we could simplify it here and say that at it's most simple you can represent a diode by it's Bandgap voltage with a series resistor. You could start here by starting to learn on of the many SPICE packages and this discussion on StackExchange might be a good kickoff point.
2. All semiconductor devices have a practical limitation in the amount of power they can dissipate. This is related primarily to the physical size of the device. The bigger the device the more power it can typically dissipate.

Now you can consider your LED. You should begin by trying to understand the datasheet for the device. While many of the characteristics you won't understand you already know one (from your question), the forward voltage (Vf) and you could probably find the current limit and maximum power dissipation in the datasheet.
Armed with those you could figure out the series resistance you need to limit the current so you don't exceed the power dissipation limit of the LED.

Kirchhoff's Voltage Law gives you a big hint that since the voltage across the LED is about 3.1 V (and the datasheet current curve tells you you could never apply 9 V), you must need another lumped model component in the circuit.

^{simulate this circuit – Schematic created using CircuitLab}

Note: the battery internal impedance shown above is simply specified to make calculation easy. Depending on the battery type (primary or rechargeable) the internal resistance can vary. Check your battery data sheet.

Could the unknown element above simply be a piece of wire (no element)?
It could ....but we can calculate the results easily.
With two ideal voltage elements (9 V and 3.1 V) the resistors must have 5.9 V across them (Kirchhoff's voltage loop). The current flow must therefore be 5.9/10.1 = 584 mA.
The power dissipated in the LED is (3.1 * 0.584) + (0.584^2 * 10) = 5.2 Watts. Since your LED is probably rated at only 300 mW or so, you can see that it will heat up dramatically and in all probability fail within seconds.

Now if the unknown element is a simple resistor, and we want the current through the LED to be let's say 20 mA, we have enough to calculate the value.

The terminal voltage of the battery would be (9 - (0.02 * 0.1)) = 8.998 V The terminal voltage of the LED would be (3.1 + (0.02 * 10)) = 3.3 V

So the voltage across the unknown resistor is 5.698 and the current through it 20 mA. So the resistor is 5.698/0.02 = 284.9 Ohms.

Under these conditions the loop voltages balance and the LED passes its designed value of 20 mA. It's power dissipation is therefore ( (3.3 * 0.02) + (0.02^2 * 10)) = 70 mW ....hopefully well within capability of a small LED.

Hope this helps.

Answer 2

Yes, the LED will likely be damaged. That is the short story.

In reality the battery voltage will drop a bit because it will be cranking out a lot of current (batteries have internal resistance which varies with charge state, discharge history, temperature and other factors- maybe a few ohms for a fresh 9V battery), and the LED voltage will increase (LEDs increase voltage with current in a nonlinear way) until the two exactly meet (if you ignore a bit of drop in the wires).

So let's say the battery voltage drops to 5V and the battery is supplying 1.5A. That means the LED forward voltage is 5V and it is dissipating 5V * 1.5A = 7.5W, which means it will very quickly burn out, assuming it's a small 3mm or 5mm indicator LED.

If your 3.1V LED happened to be a bunch of LED dice in parallel and was capable of safely handling (say) 2A, on the other hand, the battery voltage would drop to something like 3.1V (due to internal resistance of the battery, same as above) and the LED would light with about 6W of input power. Of course the battery would quickly be depleted (at best- or it could get very hot, and possibly violently explode. Some types, such as NiCd batteries or certain unprotected Lithium batteries, may be more dangerous than others.

Answer 3

Here's what happens: first, I connected a green LED properly to 9 V using a 1 kΩ resistor to catch the residual voltage.

Then without.

Astonishingly enough, afterwards, again with a resistor, the LED still works, but notably dimmer.

Don't try this at home kids... except, heck, why not... it's science!

Why it briefly lights yellow/red before “glowing out”, I don't know. Probably the result is different for every LED type.

Answer 4

In practise there are some "hidden" or parasitic resistors in your hypothetical example that you are not aware of. For starters, the battery has an internal series resistance. The LED also has a resistance as do all of the wiring in your circuit. The voltage drops across all of these resistors plus the LED's intrinsic voltage drop will add up to the battery voltage.

The only question is: At what current does this happen? If it is high enough your LED will cook and burn up. Additional resistance in the form of an actual resistor in series with the LED will prevent this problem. Determining that resistor's value is an opportunity to apply Ohm's law.

Answer 5

This diagram, with volts on Xaxis and current on Yaxis, is used to graphically "solve" the equation for 2-component-in-series voltage-dividers. It can be used for pure-resistive divider, or as here with diode & resistor.

^{simulate this circuit – Schematic created using CircuitLab}

Put a second component in series, to share the voltage. For example, you want the LED to work safely with its 3.1 volts, and have a RESISTOR to use up the unneeded [9 - 3.1] = 5.9 volts. At 10mA (which you can view as 100 ohms per volt), you need 100 Ohms/Volt * 5.9 volts = 590 Ohms. Common values are 560 Ohms and 620 Ohms.

You need a series circuit here: the source at 9 volts, and then TWO components to share the battery voltage.

---

Now lets use the same IV plot as a nomograph to solve resistive voltage dividers.

^{simulate this circuit}

Answer 6

The answer to your title question is: The LED will light up.

The provisio is that your current is within the minimum and maximum limits of the LED in question.

A low current will have it burn dimly, and the rated current will have it burn brightly. Too much current will blow the LED.

You limit the current to your desired value (often 15 to 20mA) by putting the correct resistance into the circuit.

Use Ohm's law to work that out. R (ohms) = V (volts) / I (amps).

Within reasonable limits, the voltage is fairly irrelevant to an LED, it is the current that lights it up. You must of course have a voltage sufficient to exceed the internal voltage drop of the LED at the low end.

Answer 7

Not all 9 V supplies are equal. Some will blow the LED and some will not. (It depends on the short-circuit current or internal resistance.)

9 V - 3.1 V = 5.9 V is 'missing.' This is dropped inside the 9 V supply, the wire and inside the LED. (These are the resistances that causes the loss in voltage, or voltage drop.)

It is very difficult to blow anything without heat, (exept for static in MOS.) The heat takes time to build up (and release the smoke. :-)

The heat that destroys the LED is due to the voltage of 3.1 V, the LED internal resistance, the current (V/R) and time. Some of the heat (before the smoke happens) is lost to the environment. That is why heatsinks are used in some circuits to prevent smoke.

I hope this simple explanation helps to get you started to Google $$V = I\cdot R~,$$ $$ P = V\cdot I$$ and $$ E = P\cdot t~.$$

Answer 8

In first approximation, neglecting internal resistances, LEDs have an exponential I/V forward characteristics. In reality, that is the characteristics of the forward-polarized junction: real devices have an internal resistance in series, typically some Ohms.

The "nominal" voltage drop of the LED is just one point on the characteristics, Usually, the voltage that corresponds to 20 mA, or a determined nominal forward current.

When you put your led across the battery poles, you create a series circuit that includes an "ideal" voltage source of 9 V, the LED, and the battery's internal resistance (say, 2 Ohm)

The working point of your LED is the intersection of its forward characteristics with a load line determined by the source voltage (9V) and the internal resistance of the battery. Voltage drop on your LED will be much higher than the nominal 3.1 V.

Unless Your LED is a high-current device, the current will exceed the nominal value and the LED will suffer or blow.

Answer 9

LEDs (and diodes in general) are a little odd. As a first approximation below the voltage threshold no current can flow, over it there is no restriction on current flow.

Think of it as a dam, when the water is below the dam it is blocked completely. Once the water level is over the top of the dam its flow is unrestricted however you still lose the amount held behind the dam.

So with an LED with a threshold of 3.1V if you apply 9V you have 5.9V still to use up. This will be used up by the resistors in the circuit as described by Ohm's law, V=I*R. If you haven't added any resistors then R is the batteries internal resistance and the resistance of your wires. These internal resistances are normally small enough that you can ignore them but in this case they are all you have. Small resistances and a fixed voltage means the current will be very high. The LED will have a maximum current it can survive, around 20mA for typical LEDs. If you exceed this they will overheat and destroy themselves.

As I said at the start, this is only an approximation of an LED, in practice the voltage drop does increase with current. However that increase isn't huge, generally if you are in a situation where you need to take it into account then you are either doing something very sensitive, something high power or you are running far too close to the component limits to start with. The increase certainly isn't enough to impact the end result in this scenario.

Answer 10

Everything has resistance. Period!

• That includes batteries (ESR), diodes(Rs), inductors(DCR) Caps.(ESR) and even Resistors (R) ;)
  • “Kirchhoff voltage loop (KVL)” teaches how to use that “Leftover Voltage”
    • this is after all the supply voltages are added or subtracted
    • with the sum of all parts in series, which each have resistance
  • thus KVL teaches you how to compute the resulting current or flow rate of charges in the units of Amperes or “Amps” = 1000 milliamperes (mA)

• thus this “leftover V” is across the sum of all parts R in a loop.

  • when it is a good conductor wire we usually (but not always) ignore the resistance and voltage drop.

• so I=V/R of each part after the leftover voltage and sum of loop resistance expressed as a ratio.

• parts that can handle LOTS of power must have low R ( except in elementary school theory , we say ideal batteries have R=0 )

• all diodes similarity have lower resistance above the threshold voltage , Vt, if rated for more power, at the rated forward voltage Vf

3V LED have a threshold around 2.8V and then can be 3.1V +/- 10% depending on wide tolerance and of course, power

 - for example
   -  a 300mA rated white LED (1W) has a bulk resistance less than 0.5 to 1 Ohm due to 50% MFG tolerances
   - a 9 V Alkaline battery actually has six (6) tiny 1.5V cells in series
   -  inside , each has about ESR= 1 Ohm (when new)
           - cheap carbon pile aka HEAVY DUTY cells are about 3 Ohms (new) so less powerful

Thus with one 1W White LED and one Alkaline 9V battery what is the “leftover” voltage and resulting current?

• How do you choose R to limit that current?
• How do choose the power rating and temperature rise.

(9V-2.8V) / (6x1 + (0.5 to 1) + R) = 0.3A=300 mA

solve for R

Hint if R=0, the LED gets over bright and too hot to survive

Caps have ESR but as insulators = dielectrics, they block DC but conduct AC.

Answer 11

We never were told if the battery used in the video was a Chinese "Super Heavy Duty" carbon-zinc type that was leaking in its package for a year and bought at The Dollar Store, or if it was a new modern Name-Brand alkaline one. BIG difference.