LED's, current, and voltage drop

[Eddododo]

When using diodes in series to drop supply voltage to a usable level, LEDs offer considerably more Vf drop than the .6v most Si diodes offer.. in my case, I want MORE Vf drop, in order to use diodes in series with Vsupply to clamp voltage lower for a particular op amp

I'm using batteries for an onboard pre, and the idea of LEDs being lit up alarms my monkey brain, worried about current and battery life
My human brain assures me that a) diodes don't "draw" current, the current is drawn by the return path to ground mitigated by the obligatory resistor, and that b) the LED's likely won't even light up, as the Vbias network is in the order of 100k's of Ohms, and the Op amps in question either won't draw enough current to light them, or else if the current drawn lights them, the same current would be drawn by the opamp anyway, the LED in series with the supply is only dropping the voltage, not exacerbating current draw..

Is my understanding correct, that the LED (or any diode) is not drawing nor wasting any additional current? or more to the point, that it's not otherwise compromising the battery life (when used specifically in series from battery 18v V+ to the resultant Vsupply of 12v or whatever)


Project context, for any curious
Onboard preamp, 18v preferred for headroom and enhanced battery life, but op amp is 16v max.. trying to lower to 12V via diode drops. Assume I am not interested in another power supply, and that I have enough reason to want the headroom. I may get another chip if there is a very compelling case against the diode drop, but I'm avoiding the SMD devices that are available, and my preferred chip was chosen for current draw and good performance (TLC2264)

[GibsonGM]

Well, LEDs do require some current (ohm's law).  Have you considered using a zener diode (or 2) to get to where you want to be?  Lower parts count, more 'elegant' solution, IMHO. 

[Eddododo]

Well, LEDs do require some current (ohm's law).  Have you considered using a zener diode (or 2) to get to where you want to be?  Lower parts count, more 'elegant' solution, IMHO. 

Does ohm's law apply in a way that would not apply to regular diodes? I know that LEDs require current to light, but aren't they effectively a SHORT when open, and the current flowing through is simply the current that's allowed by the resistance before/after? My "understanding" was that a circuit would draw the same current from a 12v supply that it would from an 18v supply with diodes dropping it to 12v.. Am I wrong about diodes Generally, or missing something about LEDs?


Pt 2: for the Zener, are you referring to similarly running it in series with the V+, but with anode/cathode 'backwards' to take advantage of the higher drop? Something about having the forward diode lined up ready to conduct with reverse voltage applied made me nervous, but I mainly had not figured out how to ask if that would be OK.
In case you meant the typical zener-to-ground trick to establish a stable voltage, it seemed to me that it would be a CONSTANT current drain (in a way that series diode drops did not seem to be)

[R.G.]

Diodes are designed to conduct when the anode is more positive than the cathode by enough voltage. "Enough voltage" is a unique characteristic of the semiconductors used to make the diode. Germanium conducts when anode-to-cathode is 0.1 to 0.3V, depending on how hard it's diffused. Silicon is usually 0.5 to 0.7V. Gallium Indium Phosphide, Gallium-Arsenic-Indium-Phosphide and similar LED materials conduct when they're forward by 1.2 to 4V or so.

When any diode has anode to cathode voltage less than it's "turn-on" voltage, current essentially does not flow. There is an equation which describes the conduction current that is basically an exponential function of the forward voltage, so it looks like nothing happens until a sudden turn-on voltage, then the current becomes essentially not limited by the diode. It conducts all it can, limited only by external resistors, current limiters, etc. Another way to look at this is that the anode to cathode resistance of the diode goes from very high at less than the turn-on voltage to very low at or above the turn on voltage. They change from high resistance to low resistance right at turn-on.

Note: all diodes do this. Vacuum rectifiers do it, although they have enough internal resistance that it''s not always apparent; copper oxide and selenium rectifiers do it; in particular silicon and LEDs do it. They all do. There is some variation in how low their forward resistance gets. Vacuum rectifiers have high internal resistance, silicon and germanium have very low internal resistance. In most cases, the wires leading to the diode junction for Si and Ge has about the same resistance as the diode junction itself.

That's forward conduction, the "ready to conduct" situation. Then there is reverse conduction: zener and avalanche diodes. All diodes will conduct with cathode positive and anode negative if you apply enough voltage to make the diode junction break in the reverse direction. For silicon especially, we have learned ways to control the diode junction doping process delicately enough to give an intended reverse breakover.

There are two different mechanisms for reverse breakover in most semiconductors. There is the Zener process, named after the guy who described it. Then there is avalanche They're different, but look the same from the outside, excepting that zener breakover usually happens under some small voltage like 5-6V or so. Avalanche dominates at higher voltages. Avalanche happens when the reverse voltage is so big that the electrical field can simply pull hard enough to pull charge carriers across the reverse biased junction. The junction stays at the breakover voltage but current flows, pulled by the high voltages.

Avalanche is a more abrupt process than forward conduction, sharper edged when it happens. But avalanche (and zenering) still amounts to no current flowing until some certain voltage is applied, then lots of current flows, limited by things external to the diode. All higher voltage "zeners" actually avalanche, but we just call them all zeners.

There is nothing more useful or honorable or ready about forward conduction than zenering or avalanche diodes, or about high-forward-drop LEDs. They all look the same from the outside: under the critical voltage, nothing flows. At and above the critical voltage, current is limited by things external to the diode. 

[Rob Strand]

Does ohm's law apply in a way that would not apply to regular diodes? I know that LEDs require current to light, but aren't they effectively a SHORT when open, and the current flowing through is simply the current that's allowed by the resistance before/after? My "understanding" was that a circuit would draw the same current from a 12v supply that it would from an 18v supply with diodes dropping it to 12v.. Am I wrong about diodes Generally, or missing something about LEDs?

LEDs work like diodes except for a given current the voltage drop is higher.  They will work like diodes at low currents even when you can't see them lit.

(Adding to what RG said:)  the voltage drop across a diode isn't constant it depends on current, more the current the higher voltage drop,  however it's not like a resistor the change in voltage isn't proportional to current, it's lot less.  Typically if you double the current through a diode the voltage drop will increase about 20mV to 40mV depending on the diode; for 10 times the current you get 60mV to 120mV.   Zeners are different and don't obey that rule.

A Red LED is 1.65V at 20mA.   An LED gets dim around 100uA.  Suppose the LED changes 30mV in for double/half current and 90mV for 10 times/one tenth the current.
It's a rough but simple way to do it.

20mA  1.65V
2mA   1.65V - 90mV = 1.56V
200uA 1.56V - 90mV = 1.47V
100uA 1.47V - 30mV = 1.44V

In general you can use,
V2(@ I2)  = V1(@ I1) + n*26mV * ln(I2/I1)
where n = is typically 1 to 2 and ln() is the natural log

So given V1=1.65V at I1 = 20mA, ball-park n = 1.5 in the example, we can estimate the voltage drop at 100uA,

V2(@ I2)  = V1(@ I1) + n*26mV * ln(I2/I1)
           = 1.65 + 1.5*26mV*ln(100uA/20mA)
           = 1.65 - 207mV
           = 1.44V

Like everything in electronics there's more to it in reality but I'll park it there.
EDIT: cleaned up text.  Something  got muddled.

[Eddododo]

Abundantly helpful, thanks!

With all that- does it hold that diodes do not draw or 'consume' current, other than what is already to be drawn by components on either side of it?

Could it be said that the only downside of diode-dropping the supply voltage that the voltage is *a little* unstable, such that it relies on the current flowing through?

[RickL]

After R.G.'s wonderful explanation this might be a good time for me to ask a somewhat related question.

A friend of mine wants to build LED tail lights for his car. His plan is to run enough LEDs in series to use all of the standard car voltage of 12 volts in the voltage drops of the LEDs, with no current limiting resistor. But I'm concerned about the current without a resistor.

To simplify the example let's say that each LED has a voltage drop of 2v and a maximum current capability of 20 mA. My understanding is that if 5 LEDs were used in series 12 volts would be dropped to 2 volts by the diodes and a 100 ohm resister would be required to limit the current to 20 mA.

But 6 LEDs would drop the voltage to 0v so no current should flow. This is clearly wrong (I think). What am I not understanding? Am I misunderstanding maximum forward current?

I'm embarrassed to say that I have Electronics Engineering diploma, but I got it 40 years ago and haven't used the knowledge much since then. 

[Rob Strand]

Abundantly helpful, thanks!

With all that- does it hold that diodes do not draw or 'consume' current, other than what is already to be drawn by components on either side of it?

Could it be said that the only downside of diode-dropping the supply voltage that the voltage is *a little* unstable, such that it relies on the current flowing through?

That's correct, they pass current like a resistor.

Series diodes are only good to drop small amounts of voltage, say a few volts.   There's a point where a different circuit like a voltage regulator, or a buffered voltage divider, is better.

On battery powered equipment if you dropped 9V to 4.5V using series diodes (ie 4.5V drop) then when the battery drops to 7.1V the output drops to 7.1V-4.5V=2.6V, which is often not what you want to happen.

A buffered voltage divider is like this circuit with zener replace by a resistor.  It's not a regulator as the output voltage is more or less a constant fraction of the input voltage; but not quite due to the VBE drop on the transistor.

https://learnabout-electronics.org/PSU/images/simple-series-regulator.jpg

[merlinb]

But 6 LEDs would drop the voltage to 0v so no current should flow. This is clearly wrong (I think). What am I not understanding?

LEDs don't instantly snap from off to on when the voltage reaches 2V, they have a curvature.
6 LEDs would need "roughly 12V" so you'd be right on the knife edge; maybe they would pass negligible current, maybe they would try to pass unlimited current and blow themselves up. That's why you need a resistance somewhere. In tiny fairy lights sometimes the battery itself (e.g. coin cell) can be matched to the curvature of the (usually blue or white) LEDs to limit the current inherently, without an extra resistor. But a car battery?
Here are some I/V curves for LEDs I measured.

[PRR]

run enough LEDs in series to use all of the standard car voltage of 12 volts

It's not "12V", it is under 11V to even 14V depending on charge.

The drop of an LED is also not too knowable. It varies with current, also doping, and temperature (you want brake-lights to work on hot days too).

What if you figure 7.4 diodes? (12.6V/1.7V) You can't buy 0.4 of an LED.

If a little too low, very dim; if a little too high, blow-out.

The low-tech answer is to define all the variables and figure a resistor.

But this is a solved problem. Automotive LED lighting is readily available, often with sophisticated current controllers.

[R.G.]

With all that- does it hold that diodes do not draw or 'consume' current, other than what is already to be drawn by components on either side of it?

Yes, that is one way to look at it. I would say it a different way. No component draws or consumes current. In my mind, all passive devices allow current to flow based on the voltage pressure impressed across them. For diodes, this happens to be nonlinear; it's an exponential function, where nothing much happens until the exponential hits a certain voltage and then starts rising like crazy.

As Merlin adds, the change to "rising like crazy" is not instantaneous. Depending on the exact materials and doping and so on, the change from not much current to limited only by external components is a rounded-off curve. For silicon diodes, the rounded "knee" is on the order of 50mV wide. For funny, odd material diodes like LEDs, it can be much wider and more rounded. But it is very helpful to me in cold-reading schematics to think that silicon, germanium and LEDs turn on instantly. But then in my next mental pass through the circuit, I think about what the rounded knee region and the non-zero resistive region mean.

For all diodes, there is some rounding at the knee, of a size that is particular to that semiconductor and doping profile, and a sloped resistive "on" region where the voltage across it does rise slightly with more current; it's just that the resistance in the heavy-conducting region is quite small usually. This is also true for zener/avalanche diodes where after the junction is avalanched over, there is still a slight increase of voltage across the device, partially from junction effects, and mostly from the lead resistance and resistance of the bonding wires/solder and non-junction semiconductor material.

The shape and size of the rounded knee region is what causes different characters to diode clippers.

Could it be said that the only downside of diode-dropping the supply voltage that the voltage is *a little* unstable, such that it relies on the current flowing through?

I wouldn't say that it's the only downside. As Rob points out, the basic dropping voltage is fixed, and that might not be what you want. And all of diode dropping, even zenering, suffers from more variable voltages at low currents, where the current is so low that tiny changes in current give significant changes in the voltage. Modern zener data sheets give a specific minimum test current for the zener voltage; this is to get the current and voltage out of the variable knee region.

They also give a specification of the zener resistance. This is the slope of the voltage versus current at currents well above the knee. This is specified for zeners because they are often used for voltage references. If the voltage varies because the current is varying, then your reference voltage is wobbling around too. Designing power supplies before packaged voltage references became available, we would sometimes feed a zener from a constant(ish) current source to keep its voltage steady.

And of course, voltage dropping diodes still produce heat. They do not save power, because they still make heat by the voltage across them times the current through them. If you have a string of silicon diodes designed to drop 4.2V, an LED that drops 4.2V, and a zener that drops 4.2V, they all produce the same amount of heat: V * I.

[duck_arse]

But this is a solved problem. Automotive LED lighting is readily available, often with sophisticated current controllers.

often it's an LM317 and a reisistor connected as a constant current source.

[RickL]

He wants to do it himself. I already explained that by the time he soldered all the individual LEDs it would be cheaper to just buy commercial lights, if he values his time at all.

I'm just trying to give him the best advice I can on how to do it. On line enquiries are giving me conflicting information.

[GibsonGM]

Use a chip with a higher maximum voltage rating.

[R.G.]

He wants to do it himself. I already explained that by the time he soldered all the individual LEDs it would be cheaper to just buy commercial lights, if he values his time at all.

I'm just trying to give him the best advice I can on how to do it. On line enquiries are giving me conflicting information.

Friends don't let friends build automotive lighting.   

The right way is to find automotive temperature range LEDs (not a small consideration) with enough brightness over the temperature range, then do a temperature compensated constant current drive. And to make the CC drive withstand a starter motor load dump of up to ... um, what was it? -70V or so at big currents,

It's one thing to want to do the technology. It's another for projects that involve legal issues (there are laws about size and brightness of car lights that can get you a traffic ticket) and might get you hit by another car if you don't get it right. I personally would do interior lighting in a car, but I would not try to design running lights. It's a specialized engineering niche.

[ElectricDruid]

The other thing with car lights is that these days they're all turned on and off by CANBus, so he's got some other stuff to learn too!
Unless this vehicle he's working on is truly 20th century, of course...

[amptramp]

I once used a green LED to drop the voltage from a +15 VDC line to an MC1310 stereo decoder in a Heathkit AJ-1510A tuner because the MC1310 has a maximum rating of 14 VDC.  The drop across the green LED was 2.05 volts, so this did what I needed easier than using a regulator.  I still used decoupling at the IC but the momentary current surge from the capacitor did not seem to affect the LED.

This is the only Heathkit item I remember having numerous design errors and I wrote up a monograph about what to change for various FM forums.

[amz-fx]

Closeup of an LED

http://www.muzique.com/news/closeup-of-leds/

regards, Jack