A general explaination of LEDs and circuit theory (with visuals)

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Hey everybody. While looking up regulation on the web I came across a very informative website that explains the theory of regulation circuity (http://www9.dw-world.de/rtc/infotheque/semiconamps/semiconductor_amps-content.html ,Section 5.1 and 5.2). From one of the classes I took last semester (Solid State Physics) for my Electrical Engineering degree, I can sorta explain whats happening.

==Some in-depth theory of electricity==

What is Voltage:
Voltage is the density of the number of charge carriers (Coulombs), or density of in a given area with respect to another point (Values of volts are always done between point 'a' and 'b'). The units for voltage is a ratio of amount of total energy over the number of charge carriers (Joules/Coulomb).

What is Current:
Current is flow of those charge carrier though a point with respect to time (Coulombs/Sec)

What is Power and Work:
In physics, Work is the amount of energy done with respect to time (Joules/Second). The electrical notation for work is known as 'Wattage' (ie a 60 Watt bulb has 60 Joules of energy flowing though it each second). When work is integrated (calculus for multiplying out by the denominator, in this case time), you get energy (60 Joules/sec * 3 seconds= 180 Joules consumed). In electric billing 1 kilowatt hour (1kWh) is the amount of Joules consumed by a 1000 Joules/second device over 1 hour (1000 Joules/Second * 3600 Seconds/Hour = 3.6 Million Joules).


Why does this matter to LEDs:
Lightemitting diodes observe the physics property of the Pauli-exclusion principal, meaning only a certain number of particles can occupy an orbit around an atomic nucleus. Because of this, when a particle gain extra energy it jumps into an unstable electrically higher orbit (known as the conduction band), that may have various energy levels. Since this orbit is unstable the particle must give off the energy gain by releasing photons to return to the stable orbit. In a diode, the voltage difference between these levels can determine the wavelength of the photons. De-broglies equation explains how the energy of a particle affects its wavelength (shorter wavelengths have higher energy than longer wavelengths). This is why Red, Amber and Yellow LEDs only need around 2.4 volts and why Green, Blue, UV and White (via yellow phosphor coated blue) LEDs need around 4.0 Volts.

Due to the nature of semiconductors and the process of doping, one side of the junction has an excess of electrons (N-type) while the other side has an excess of holes/lack of electrons (P-type). Because of this there is a voltage difference inside the semiconductor (known as a biasing voltage).When multiplied by the charge carrier value, the energy (known as electron Volts) of each side of the semiconductor can be calculated (known as the Fermi Energy level or shorthand Efx where x denotes the type of majority doping, either n or p) . When the Fermi Energy Levels are the same on both sides of the junction, it is known as a zero bias semiconductor. If either Fermi Energy Level is higher than the other, the semiconductor is either forward bias or reverse bias


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image credit, http://www.ecse.rpi.edu/~schubert/Light-Emitting-Diodes-dot-org/chap04

In the above photo, this biasing (denoted as 'eV' in photo b) is the difference between the energy levels of the P side (Left side) and the N side (Right side). In order for electrons to flow from the P-side to the N-side, the P-side must have a voltage applied to it so the Fermi energy levels match, resulting in a shifted energy band for the P-side. Because the conduction Energy Level on the P-side is higher than the N-side, electrons will start flowing. This is known as forward biasing. If too much voltage is applied, electrons gain an excess of energy (beyond what is needed for photons) which results in collision between the electrons themselves and the atomic structure of the semiconductor (this is where the heat comes into play). This can adversely affect the electrical properties of the LED. This is why a small change in applied voltage can have such an impact on LEDs


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The number of photons emitted is roughly equal to the number of electrons entering the junction, hence why current regulation is recommended.


==How do current regulators work==



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An input voltage is applied (Ui). In order to satisfy Kirchhoffs Voltage Law (the sum of all the voltages in circuit must equal to zero), that applied voltage must be dropped across the zener diode and resistor R1. The zener diode is a unique component that provides a stable reference voltage across its leads. For Silicon base transistors, the voltage drop between the emitter (the arrowed lead) and the base is ~0.7 volts.

For a constant current, the voltage drop across the load (Rl) must remain the same due to ohms law. The transistor makes up for the difference by adjusting its resistance



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in this case, the supplied voltage is 12 volts and 9 volts onto the base. The emitter voltage is 9 volts plus 0.6 volts (or 9.6 volts), the current though the emitter is (12 volts-9.6 Volts)/(6 ohms) =400mA. This 400mA when multiplied by 15 ohms equals 6 volts. For KVL to work, the transistor must drop 3.6 Volts (12 Volts-2.4Volts-6 Volts= 3.6 Volts). Due to ohms law, the Resistance of the transistor must be 9 ohms (3.6Volts/0.4 Amps). R2 provides the current required to turn on the transistor with this current times the gain equaling the 400mA (9Volts/200ohms=45mA, 45mA*8.9 =~400mA)

If the input voltage were to change to 10 volts, the 15 ohm source will still drop 6 Volts, so the transistor must drop 1.6 Volts (10 Volts-2.4 Volts-6 Volts= 1.6 volts). Again using ohms law, the resistance of the transistor becomes 4 ohms (1.6 Volts/0.4 amps)

Being a new member, and new to electronics, lighting and such this post was really helpful. Hopefully it can get stickied for other new guys to to use as a reference. Thanks

cybhunter said:

See Rule #3 Do not Hot Link images. Please host on an image site, Imageshack or similar and repost – Thanks Norm

in this case, the supplied voltage is 12 volts and 9 volts onto the base. The emitter voltage is 9 volts plus 0.6 volts (or 9.6 volts), the current though the emitter is (12 volts-9.6 Volts)/(6 ohms) =400mA. This 400mA when multiplied by 15 ohms equals 6 volts. For KVL to work, the transistor must drop 3.6 Volts (12 Volts-2.4Volts-6 Volts= 3.6 Volts). Due to ohms law, the Resistance of the transistor must be 9 ohms (3.6Volts/0.4 Amps). R2 provides the current required to turn on the transistor with this current times the gain equaling the 400mA (9Volts/200ohms=45mA, 45mA*8.9 =~400mA)

If the input voltage were to change to 10 volts, the 15 ohm source will still drop 6 Volts, so the transistor must drop 1.6 Volts (10 Volts-2.4 Volts-6 Volts= 1.6 volts). Again using ohms law, the resistance of the transistor becomes 4 ohms (1.6 Volts/0.4 amps)

Hope this helps out in showing how LEDs and current regulators work
Joe
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Perhaps it would be best to leave circuits to engineers as you do not have a firm understanding of the operation of what you are trying to describe.

1) While the current flowing through R2 does turn on Q1, for the most part a lot of what you wrote is wrong.

2) For the most part, the current flowing through R2 is working to bias CR1. That 9volts/200ohms = 45mA will be split up between the base of Q1 and through CR1. Without knowing the instantaneous gain of the transistor, you have no idea what the base current is. That 8.9 is a meaningless number. This circuit works because the current through R2 biases CR1 which creates a somewhat fixed voltage across R1 which then sets the output current.

3) A more accurate output current is Iout = [(3-0.6)/6 - ((3-0.6/6)/gain] .... realistically it will be 3-0.7 but that is a mute point


It is not accurate to say the transistor changes its resistance as that is not really what is happening. Technically the resistance of an active element would be dV/dI, not V/I. In a constant current config with a perfect transistor, the resistance is actually infinite. A bipolar transistor is a current amplifier, not a variable resistance device.

Semiman

Thanks for the constructive criticism SemiMan. I thought I had a general understanding of transistor theory

cybhunter said:

Thanks for the constructive criticism SemiMan. I thought I had a general understanding of transistor theory

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I want to thank you both for giving for giving some clarity in transistor theory and quantum mechanics that most of us know so little. :thumbsup:

Voltage doesn't have anything to do with "density"; a single electron can be at an arbitrary voltage; it's an energy thing.
Like lifting a rock in a gravity field; the higher the rock, the more energy it has. The more energy an electron has, the higher it's voltage. Both the rock and the electron need a reference point....

billw said:

Voltage doesn't have anything to do with "density"; a single electron can be at an arbitrary voltage; it's an energy thing.
Like lifting a rock in a gravity field; the higher the rock, the more energy it has. The more energy an electron has, the higher it's voltage. Both the rock and the electron need a reference point....

Click to expand...

I was going to bring that point up too, but realized that he (or the referred material) was talking in relation to a fixed situation (likely without knowing it), i.e. like a capacitor where charge and stored energy do relate to each other. To your point though, simply talking about density of charge is rather meaningless as you can only tell voltage from this if you know other parameters of the system.

Semiman

For example, if you were to put a particular charge on the plates of a air-gap capacitor, and then increase the separation between the plates, that would increase the voltage you would read between the plates. (This makes sense. The plates are oppositely charged, and attract each other. Pulling the plates apart therefore requires that you put energy into the system. The number of electronics involved can't change, so the voltage has to go up.)