user34873: I'm talking about this tutorial "Serial to Parallel Shifting-Out with a 74HC595" from the official Arduino website. It provides the following schematic:

I've read a couple of times now, that the placement of the capacitor is wrong:

This is a well known shit tutorial which for some unknown reason they refuse to fix.
  That capacitor should be connected between the Vcc pin and ground, NOT the latch pin.

Source: forum.arduino cc/index.php?topic=408135.0
They recommend this placement:

Who's right? Please explain where to place the capacitor. Can the arduino even be damaged?

user30005: You can figure out the impedance of that capacitor for high frequency signal....
What they have there is really a low pass filter.
Terrible ideal to out that big of a capacitor there.

user30761: You do not need a capacitor to use the 74HC595 in conjunction with shiftOut(). I found this YouTube video to be very useful in explaining how to wire up and code the shift-register for a 7-segment display.

user34873: Indeed, it seems like the Arduino tutorial is wrong. I just found the following schematic in Datasheet SN74HC595 - Texas Instruments from 10.2 Typical Application, Figure 5:

The capacitor connects VCC to GND. Nothing to do with the latch pin.

user4143: That tutorial, as with a number on the Arduino site, is wrong. 
Yes, the capacitor should be across Vcc/GND - it is a decoupling capacitor used to absorb the current surges when you turn LEDs on and off.
Not to mention that with that circuit if you turn on all the LEDs you are overloading the 74HC595. That has a maximum 70mA through Vcc/GND pins. Each LED there (assuming red LEDs at 2V forward voltage) will be drawing about 13.6mA. 8 LEDs would be 8*13.6 = 109mA.
Ouch.
The problem with many of the Arduino tutorials is they were written in the early days of Arduino when they didn't have much knowledge and experience. Since then the arduinosphere has grown and there are people now with far more knowledge and experience than Arduino had at the time the tutorials were written. We come here.

34873: <p>I'm talking about this tutorial "<a href="https://www.arduino.cc/en/Tutorial/ShiftOut" rel="nofollow noreferrer">Serial to Parallel Shifting-Out with a 74HC595</a>" from the official Arduino website. It provides the following schematic:</p>

<p><a href="https://i.stack.imgur.com/R2q9b.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/R2q9b.gif" alt="maybe wrong version"></a></p>

<p>I've read a couple of times now, that the placement of the capacitor is wrong:</p>

<blockquote>
  <p>This is a well known shit tutorial which for some unknown reason they refuse to fix.
  That capacitor should be connected between the Vcc pin and ground, NOT the latch pin.</p>
</blockquote>

<p>Source: <a href="https://forum.arduino.cc/index.php?topic=408135.0" rel="nofollow noreferrer">forum.arduino cc/index.php?topic=408135.0</a></p>

<p>They recommend this placement:</p>

<p><a href="https://i.stack.imgur.com/ciAFJ.gif" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ciAFJ.gif" alt="maybe better version"></a></p>

<p>Who's right? Please explain where to place the capacitor. Can the arduino even be damaged?</p>
 30005: <p>You can figure out the impedance of that capacitor for high frequency signal....</p>

<p>What they have there is really a low pass filter.</p>

<p>Terrible ideal to out that big of a capacitor there.</p>
 30761: <p>You do not need a capacitor to use the 74HC595 in conjunction with <code>shiftOut()</code>. I found <a href="https://www.youtube.com/watch?v=cHA-fwiA4Qo" rel="nofollow noreferrer">this YouTube video</a> to be very useful in explaining how to wire up and code the shift-register for a 7-segment display.</p>
 34873: <p>Indeed, it seems like the Arduino tutorial is wrong. I just found the following schematic in <a href="http://www.ti.com/lit/ds/symlink/sn74hc595.pdf" rel="nofollow noreferrer">Datasheet SN74HC595 - Texas Instruments</a> from <code>10.2 Typical Application</code>, <code>Figure 5</code>:</p>

<p><a href="https://i.stack.imgur.com/1XiV1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1XiV1.png" alt="Typical Application schematic TI datasheet"></a></p>

<p>The capacitor connects VCC to GND. Nothing to do with the latch pin.</p>
 4143: <p>That tutorial, as with a number on the Arduino site, is wrong. </p>

<p>Yes, the capacitor should be across Vcc/GND - it is a decoupling capacitor used to absorb the current surges when you turn LEDs on and off.</p>

<p>Not to mention that with that circuit if you turn on all the LEDs you are overloading the 74HC595. That has a maximum 70mA through Vcc/GND pins. Each LED there (assuming red LEDs at 2V forward voltage) will be drawing about 13.6mA. 8 LEDs would be 8*13.6 = 109mA.</p>

<p>Ouch.</p>

<p>The problem with many of the Arduino tutorials is they were written in the early days of Arduino when they didn't have much knowledge and experience. Since then the arduinosphere has grown and there are people now with far more knowledge and experience than Arduino had at the time the tutorials were written. We come here.</p>
