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I'm new to electricity & electronics and am fumbling my way through some Raspberry Pi tutorials. One in particular has caught my interest, which hooks an RPi up to LEDs, pushbuttons and a piezo buzzer.

On page 2 of that PDF, you see the main wiring configuration for the project:

enter image description here

The one thing that caught my attention was that all the LEDs, as well as the pushbutton, all use resistors. The only component that doesn't use a resistor is the buzzer. It seems to just be wired directly to the breadboard/RPi. I'm wondering why?

Is there something special about buzzer that allows them to not draw too much current, and hence prevents them from needing resistors of any kind? Thanks in advance for any clarification!

smeeb
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    Just a caution: you have provided a *wiring diagram* of your circuit when you should have provided a schematic diagram. In this case the schema is obvious but on anything more complex the wiring diagram doesn't show what the circuit is supposed to do. As a result the wiring diagrams are discouraged, rated as "cartoons" and the questions get down-voted. – Transistor Sep 08 '17 at 18:40

2 Answers2

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The description in the article you linked to says:

The buzzer supplied in the EduKit is an 'active' buzzer, which means that it only needs an electric current to make a noise.

Notice that it calls it an active buzzer. An active device generally takes care of the fiddly parts for you, like oscillating and limiting current (within voltage limits which should be in its datasheet).

Andrew Morton
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  • Ahh, so does that mean LEDs & pushbuttons are "non-active" (passive?) devices then? Thanks @Andrew Morton, and +1 ! – smeeb Sep 08 '17 at 18:36
  • @smeeb Here's an explanation: [Differentiation between passive and active components](https://electronics.stackexchange.com/a/82801/36731). – Andrew Morton Sep 08 '17 at 18:55
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    It is important known the difference between active/passive when talking about generic component (the link provided by @AndrewMortin); with the difference between active and passive buzzers (see https://electronics.stackexchange.com/questions/224374/active-vs-passive-buzzer) – pasaba por aqui Sep 08 '17 at 19:03
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When we connect a light bulb to a voltage source (230 v, 125v, 12v, ...) the light bulb offers a resistance that, applying ohms law, gives the amount of current that will go across the bulb. The bulb is designed to support this amount of current. Thus, everything ok. Same for the buzzer.

The push button is nothing else than a near than infinite resistance (when not pushed) or zero resistance (when pushed). So, when pushed, as it offers near than zero resistance, the amount of current across it could be very high, enough high to destroy to push button or the arduino. For this reason, a resistor is add.

A led connected to a source of more than 2v offers a very small resistance, being the amount of current so big that, before to reach the maximum, the led burns. For this reason, an external resistor is add, that increase the resistence, limiting the amount of current across it and across the led to a value that both can support.

In particular, the led is not a lineal component, so it doesn't follows the ohms law. We can say, in a educative way, that it offers an infinite resistance to the first 2v and none to the remainder voltage. The exact relation between the voltage applied and the current is show in this draw (from here), it depends also of the led color.

enter image description here

The real number are: if you use a 5v source and you want connect a red led (1.9v and 20 mA for an optimal performance), the resistor must be:

R=(5v-1.9v)/0.02A=155 ohms.

For these reason we have always a box with a lot of 220 ohms resistors :-) Hope you enjoy these experiences in electronics.

pasaba por aqui
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  • Why is your white LED having a different curve than the blue one? This should either read "UV" or be left out. – Janka Sep 08 '17 at 19:22
  • @Janka: It considers SiC for blue and GaInN for white. But this subject is far, very far, from the original question. – pasaba por aqui Sep 08 '17 at 19:43
  • Then it should read "UV". – Janka Sep 08 '17 at 19:49
  • I'd just like to note for future readers that LEDs do in fact follow Ohm's law (`V = I R`), as does everything. It's just that where for a simple resistor, you only have to solve for one variable (e.g. you know `V` and `R` and want to know `I`), but with an LED you only know `V` (until you look up the corresponding `I` in the graph, that is). – Bart van Heukelom May 19 '20 at 22:31
  • @BartVanHeukelom: not all components follows Ohm's law, even linear ones: capacitors relates dV / I , inductors V/dI, ... (that could be converted to V/I only after a space transform as Fourier or Laplace). LED are memory less (relates V / I ) but non-linear, no Ohms law is applicable. Moreover, LED curve is a construction characteristic, exactly as R is for resistors. – pasaba por aqui May 19 '20 at 23:52
  • The current through a capacitor is still equal to the voltage across it, times its resistance. That this resistance is infinitely high because of the dielectric doesn't make the law any less valid. Of course, I do get where you're coming from, in that Ohm's law isn't really useful in practically analysing the useful behaviour of such components, it's just that I think telling beginners that the law doesn't always apply can cause some confusion later on. That said, apparently the law does break down in specific microscoping situations (I just learned), but that's a different scope entirely. – Bart van Heukelom May 30 '20 at 15:46
  • Correction: voltage divided by resistance, of course. Can't seem to edit the comment anymore – Bart van Heukelom May 30 '20 at 15:52