Правильная формула для токоограничивающего резистора светодиода?

Я пытаюсь выяснить, какое значение резистора использовать в светодиодной схеме. Уравнение, которое я использовал бы для этого:

R = \frac{V_{c c} - V_{f}}{I_{f}}

$R = \frac{V_{cc} - V_f}{I_f}$

Кажется логичным и имеет полный смысл. Ответы на вопрос Как рассчитать значение резистора для простой светодиодной схемы? подтвердите это тоже.

У меня есть следующие светодиоды:

$V_f = 3.3V$
$I_{f_{typ}} = 20mA$

Использование источника питания 5 В:

$V_{cc} = 5V$

Включение их в вышеприведенное уравнение дает:

\begin{array}{rcl} R & = & \frac{V_{c c} - V_{f}}{I_{f}} \\ = & \frac{5 V - 3.3 V}{20 m A} \\ = & 85 Ω \end{array}

$\begin{eqnarray} R & = & \frac{V_{cc} - V_f}{I_f} \\ & = & \frac{5V - 3.3V}{20mA} \\ & = & 85\Omega \end{eqnarray}$

Пока все хорошо.

Однако, если я воспользуюсь калькулятором по адресу http://led.linear1.org/1led.wiz , это даст мне 100 Ом . Если я использую приложение ElectroDroid на своем телефоне, это дает мне 85 Ом .

Итак, я предполагаю, что калькулятор linear1 использует другой метод вычисления этого значения резистора; Есть ли лучший способ сделать это?

led resistance current-limiting

— Джереми Керр
источник

Привет, вы можете сказать, что такое Vf и If (просто яростно: P)

— Sean87

@ Sean87: это прямое напряжение (Vf) и прямой ток (если) светодиода.

— Джереми Керр

Ответы:

Ваш расчет верен. linear1 округляет до следующего значения E12 , которое оказывается равным 100 $\Omega$ , Ближайшее значение E12 было бы 82 $\Omega$ и это все равно будет безопасно, потому что, даже если ток будет выше, разница будет небольшой, в пределах допуска 10% для серии E12.

редактировать
Пуристы может сказать , что я срезать углы здесь. У Рассела есть длинный ответ об итерации решения, и другие скулиют (эй, не обижайся!) О том, что округление является более безопасным. Мой ответ должен быть прагматичным ; Ни один профессиональный инженер-конструктор не может позволить себе потратить 15 минут на расчет резистора для классического цветного светодиода. Если вы останетесь значительно ниже максимально допустимого тока, у вас будет достаточный запас для некоторого округления, и округленное значение не будет заметно по яркости. В большинстве случаев воспринимаемая яркость светодиодов не намного превышает значение, обычно равное 20 мА.

— stevenvh
источник

Ah! makes sense, thank you. Yes, 82Ω seems close enough, at 20.7mA.

— Jeremy Kerr

You can check the rounding up. If you choose 21mA the resistor should be 81

Ω

$\Omega$ , and linear1 shows 82

Ω

$\Omega$ , again the next E12 value.

— stevenvh

Pragmatism ... sigh ... I miss working with pragmatic people. Intellectual engineering types tend to go OCD on things that are utterly pointless at the expense of the really important things, like schedules (and lunch breaks, and weekends with family/friends)...

— Adam Lawrence

@Madmanguruman - ...having a drink with friends on a terrace downtown on a sunny evening. Indeed, you absolutely have to balance your priorities!

— stevenvh

@Jeremy - 100mA is quite high, it's probably Absolute Maximum Ratings (AMR). You should never operate continuously under AMR. Like it says it's Absolute Maximum, which means almost guaranteed damage if you exceed that. But like I said, most LEDs don't have much gain in brightness above 20mA, anyway.

— stevenvh

Your formula is correct BUT to do it properly you need to iterate the result. This is because LED forward voltage drop is non linear with current (or current is non linear with forward voltage drop. In many cases this effect is not significant, but in some cases it can lead to results which are 2:1 or more in error.

Where there is plenty of "headroom" voltage for the series resistor - the difference between Vcc and Vf - the original result is liable to be close enough to correct so as to not matter. But if headroom voltage is small with respect to Vf, changes in LED Vf with current will change headroom which will change current which will change Vf which will ... . This really does happen in real world situations.

For white LEDs Vf is typically in the 2.9V to 4V range with more typical values 3.3 - 3.8V until quite recently and say 3.0 - 3.3V in more modern higher efficiency LEDs. In serious production applications Vf will be available in "bins" so can be guaranteed within about +/- 0.1V at a given current. In retail sales you may get samples from every bin going and Vf may be eg 3.3V for one LED and 3.6V for another nominally identical one.

If operating from 5V the headroom will be 1.7V and 1.4V respectively for a current variation of about (1.7-1.4)/1.7 =~18%. Add to that slight shifts in Vf with current as above and 20% variations in If may result between "identical" LEDs. In most cases this is not going to make the slightest practical difference. Light output is approximately proportional to current - 20% variation in light output is not detectable by eye by all but the most skilled or experienced of viewers.

If this was a say 5 Watt power LED the difference in LED dissipation may be 1 Watt and this MAY make a difference in operating temperatures and lifetime.

All of which leads to the advice that in "serious" applications LEDs should be driven from a constant current source if you care about the true operating current. In "indicator" roles or low level illumination applications this may not matter. In high power applications or where LED lifetime matters then constant current drive is essential.

— Russell McMahon
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While theoretically absolutely right, I don't know any engineer who wastes his time on iterating for a classical LED (BTW, the graphical method is faster). The differences in appearance are simply too small. (And provided that you don't operate the LED at its maximum current, which you never should do anyway)

— stevenvh

@Russel - What is this "Ref: TMBJ" thing at the end of your last answers? The acronym dictionary says it's either They Might Be Jedi or Thermoplastic Minerva Body Jacket. If it's a sig, it belongs in your profile. You can edit your username, too, if you want it displayed in all your answers.

— Kevin Vermeer

@Kevin - My guess is "This Might Be (J)useful" :-). In a comment to this answer I explained why it would be deleted, as Kortuk apparently did.

— stevenvh

You now know one engineer who iterates LED currents when it's a good idea to do so. If you NEED to iterate you have too little headroom for safety - but if you have too little headroom for safety you need to iterate. There is no reason not to operate an LED at its maximum RATED current if it serves a need and if you design correctly. I have around 2 million LEDs "out there" which are designed correctly :-) (and constant current driven).

— Russell McMahon

Ah - the phantom downvoting moron has struck almost a year after the event.

— Russell McMahon