Some Trivia about Silicon Diodes

About a century ago, the first (AM) radio receivers were made using coils, capacitors, and crude components as their only active one, called ‘Crystals’ (as well as high-impedance earphones). These crystals or ‘Whiskers’ were essentially diodes, formed from whatever mismatched pairs of conductive or semi-conductive substances a person could find, such as between ‘an iron wire’ and ‘the tarnish on a penny’. And a question which I’ve been asked was, “Why can’t the technology today just use a tuned circuit, and a silicon diode as the components?”

The answer is very basic. A silicon diode needs to have a forward voltage of at least 0.5V applied to it, before it goes from a non-conductive state to a conductive state. This is based on the Gap Energy of Silicon, or, the energy required to create an electron-electron-hole pair. It can happen that large diodes conduct some small amount of current at lower voltages, but this represents a region, in which their diffusion current is almost as low as their leakage current, which is constantly flowing in the opposite direction to the diffusion current. In short, in this sub-gap voltage-region, the diode fails to change in properties, as current flows through it in both directions, and fails to act as a diode. This amount of voltage is also referred to as ‘One Diode-Drop’. (:1)

The same thing happens between the Base and the Emitter of a Bipolar Transistor. In the NPN example, the Base must be at least one diode-drop more positive, and in the PNP example, it must be at least one diode-drop more negative than the Emitter, before the transistor ‘turns on’. This can frustrate people who try to create better rectifiers, just by connecting the Collector to the Base. This way, the curve that defines the current increases more sharply, but still only, after one diode-drop of voltage. But it can help IC designers, who want to keep Parasitic Transistors from turning on, and who can sometimes connect what would be the Base, and what would be the Emitter, of such a Parasitic Transistor, together.


(Update 7/30/2019, 0h05 : )

There is a related question which people do not generally ask me, but which I needed to know the answer to myself at some point in time. When the types of semiconductor components are drawn as a 2D layout, of a 3D construction, there is not only N-doped and P-doped silicon, but also N+ -doped and P+ -doped silicon. Why?

The answer lies in the fact that when a metal with a high work-factor, which is another way to say, with high electronegativity, comes into contact with regular, N-doped silicon, the electronegativity of the metal ‘draws away’ the majority free electrons, creating a region of silicon that behaves similarly to P-doped silicon. Right there, in that region, a “Schottky Barrier Diode” forms. And a problem in semiconductor manufacturing is, that this will happen every time the same type of metal comes into contact with the N-doped silicon, even if an Ohmic contact is desired.

Therefore, a type of N-doped silicon will be used as well, which has such an excess of dopant, that by itself, it fails to act as a semiconductor anymore. But, sandwiched between the metal and the regular N-doped silicon, it serves to form a contact, which is not a diode. In fact, if a single well of N-doped silicon has nothing but a metal junction at one end, but an N+ -doped (sub-)well with a metal contact at the other end, then what results is an actual Schottky Diode component, exploiting the junction effect on the end that does not use N+ -doped silicon. That end becomes the Anode, the side to which a positive potential needs to be applied, to result in forward conduction.

What some people might hope is that, if such a diode-type and N-doped well are incorporated into the P-doped substrate directly, this could lead to a simplified way to add diodes to CMOS logic.

However, I think that the WiKiPedia article exaggerates somewhat, about the degree by which the Schottky Diode has a lower voltage-drop, than a so-called ‘regular’ diode – a P-N diode. The diode-type which they mention: ‘1N5817′, according to my own data-sheets, only goes as low as 0.32V, at a temperature of 25⁰C. The target which the WiKiPedia suggests, of 0.15V, is only achieved at a temperature of 125⁰C. And that is pretty much the lowest-voltage of the Schottky Diodes. The voltage-drop is less than that for regular, P-N diodes, but at best, still half that.


Actually, the subject can be described in greater detail, of what the current-voltage relationship of diodes is. This requires understanding a fact about semiconductors, which is, that semiconductors only become that, above a certain temperature. Below that temperature they are insulators, and above an even higher temperature, they become regular, Ohmic conductors. This means that the thermal agitation energy of silicon is the main cause for the formation of its electron-electron-hole pairs.

Basically, a Schottky Diode is really just another diode, with a different set of parameters. The following article explains an approximation for computing the current that flows through a Schottky Diode:

This article explains the approximation for diodes in general:

Close inspection reveals that they are the same equation, essentially. This equation is also known as the “Schockley Diode Equation”. What it states in English is, that there is a constant amount of saturation current (IS), which is offset by an amount of diffusion current that is the antilogarithm of voltage. The fact that the saturation current is always flowing at a constant rate, and opposite the intended polarity, has been factorized out as the ( -1) of the term that follows (IS) both times. What this really means is that the zero-voltage (equilibrium) -current, as well as positive net currents, all exist somewhere as multiples of the same saturation current.

But there is also the Ideality Factor (N), of which the lowest numbers are most ideal, and which modulates how rapidly this antilogarithmic current changes as a function of voltage. I don’t think that (N) can become lower than (1.0).

Schottky Diodes are simply diodes, with another set of parameters. Their current curve is really continuous, just like that of regular, P-N diodes. But, they have an Ideality Factor close to (1.0), while, maybe, P-N diodes have closer to (2.0)?


Now, If one wanted to relate this equation to the Collector Current of a Bipolar Transistor, then one could say, that the resulting current has simply been multiplied by the transistor’s Current Gain…


Now, I have seen some semiconductor layouts, in which both N+ and P+ -doped silicon regions are used. But the question that I have about this is:

‘Presumably, only one type of metal coming into direct contact with the silicon is being used to make one type of IC. That type of metal can either be more electronegative or less electronegative than silicon. Therefore, it should only be necessary to use either N+ or P+ -doped silicon in the same IC, not both.’

And I have yet to see a good reply to that.


(Update 11/27/2019, 6h35 : )


I suppose that another answer which I should acknowledge is the fact, that when earphones were first invented, they tended to have an impedance of about 2kΩ. This made them very high-quality earphones. Today’s earphones have a much lower impedance, of maybe 40Ω if the customer is lucky, but maybe even as low as 16Ω for lower-quality earphones. They have fewer turns of enamelled wire on their voice-coil.

A 16Ω set of earphones will require much more current, to produce audible output, than a 2kΩ earphone will require – but not necessarily require more voltage. But, a tuned circuit driven by a direct antenna input, will not tend to generate this amount of current.



I could make a loose inference, about what Lumens are.

In the later part of my childhood – in the 1970s and 1980s – we had incandescent light-bulbs, and we knew that only a small part of the so-called light they emitted was in the visible part of the spectrum. We often used this light-bulb type, because we had no better alternative. We knew that the visible part of the emitted light might have been 15% or 10% of the consumed energy.

Granted, in Industrial or Commercial Lighting, there existed other types of fixtures, such as mercury-gas-discharge tubes, that excited a phosphor with ultraviolet light, so that the phosphor was made to fluoresce. Or in some cases simply – a gas-discharge tube, with a gas-mixture of a composition unknown to me.

But, when I go to buy light-bulbs today, as an adult, like all the other customers, I see Compact Fluorescent Light-Bulbs, as well as LEDs, the brightness of which is stated in Lumens. What I generally tend to find, is that light-bulbs of the fluorescent family, which are meant to be equivalent to the ~Old, 100W~ incandescents, tend to draw approximately 23W, and are stated on the packaging to produce about 1500 Lumens.

Lightbulbs of the LED family with the same equivalence, are stated to draw about 16W, and to produce about 1500 Lumens. I have actually found LEDs, which are stated to draw about 17W, and to produce 1600 Lumens of visible brightness, but which possess a visibly-larger base, from the other types.

If I could just hazard a guess, I’d say that one way to understand Lumens, is to start with the Watts of light in the visible part of the spectrum, and to multiply those by 100. What this would suggest, is that the most-efficient LEDs waste about 1W as heat, while then fluorescents still tend to waste a bit more energy, such as perhaps 8W – some of that in the form of UV light, making those approximately 65% efficient. But this would also mean, that the efficiency of modern LEDs is hard to improve upon. If the brightest variety only seem to produce 1W of waste heat, out of 16W or 17W consumed, it would make most sense to infer that in that range of efficiencies, the Wattage can be translated into Lumens quite easily. More Watts will simply produce more light, and fewer Watts will produce less light. In percentages, the LEDs would seem to have an efficiency of about 94%.

If we have a new light-bulb type, that draws 4.5W, but that produces visible light amounting to 350 Lumens, it would follow from this thinking, that this type is wasting about 1W / 4.5W. In percentages, this would imply an efficiency of 78%.

I suppose that I can offer a comment on the temperatures which the light-bulb-bases of household LEDs reach…

Continue reading I could make a loose inference, about what Lumens are.