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We found, with my 6 y.o. kid, a 9V battery and an old CPU fan power supply fan, and we were happy to see the following circuit 1. working.

Then he asked: "Can we add a switch and a LED?" For sure we can, but how... I'm not sure. Among 2., 3., 4., which one is the more appropriate? How to calculate the relevant resistor value?

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Note: Please correct me if I'm wrong but I think we can assume this standard LED we found has a forward voltage ~ 2V. How to use this to derive the resistor value in case 2., 3., 4.?

Note 2: In circuit 1., I measured, with a multimeter, a voltage of nearly 9V (a little bit less when motor was running), and a current of ~ 160 mA, thus I estimated the "equivalent resistance" is around 55 Ohms for the fan. (I don't know if this concept is correct).

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    \$\begingroup\$ Door #3. R4 = (9-2)/0.01A = ~680R \$\endgroup\$ – Trevor_G Nov 21 '17 at 20:33
  • \$\begingroup\$ @Trevor this takes in consideration 2V as forward voltage and 10 mA as target current for the LED, is that right? But it seems this doesn't take the CPU internal resistor at all in consideration? If so, why? \$\endgroup\$ – Basj Nov 21 '17 at 20:35
  • \$\begingroup\$ BEcause they are in parallel, they are two entirely different circuits. \$\endgroup\$ – Trevor_G Nov 21 '17 at 20:38
  • \$\begingroup\$ Ok thanks! Feel free to post as an answer, I'll accept it (and will show to my kid when he'll start the Ohm law in a few years ;)) \$\endgroup\$ – Basj Nov 21 '17 at 20:39
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    \$\begingroup\$ Be prepared, though, for the 9V to quit on you relatively soon. They're not really designed for the 160mA load a fan gives, so it'll only get you half the capacity, or there abouts. I'd guesstimate about an hour of play on a full one. \$\endgroup\$ – Asmyldof Nov 21 '17 at 20:51
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You should choose Door #3.

In this circuit the LED and Motor are in parallel and will work independently of each other.

R4 = (Battery Voltage - Diode Forward Voltage)/ LED Current
R4 = (9-2)/0.01A = ~680R

The other two circuits will not work the way you think they may.

In circuit number 2, you would need to make the resistor about 650R so the led does not burn out when the motor is starting. This leaves no voltage left to run the motor.

In cicruit 4, R4 is not limiting the current through the LED, and the full motor start current will go through the LED. The LED will flash real bright then eventually die. After that it will either be shorted, or it will be open at which point R4 will just slow down the motor.

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  • \$\begingroup\$ @Basj nice of you to work with your youngster on this. Here is something else to entice him. Switch it on and wait till the fan is up to speed, then turn it off and see what the LED does. \$\endgroup\$ – Trevor_G Nov 21 '17 at 21:09
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    \$\begingroup\$ Yes it's a real pleasure, he's so passionate about electronics, it's really fun. I have to refresh my 15 years-old (small) electronics skills ;) \$\endgroup\$ – Basj Nov 21 '17 at 21:11
  • \$\begingroup\$ Ok, we will try this, I'm curious to see what happens... (maybe the still-moving fan will act as a dynamo thanks to momentum?) \$\endgroup\$ – Basj Nov 21 '17 at 21:13
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    \$\begingroup\$ @Basj ;) it should if it is not some sort of smart fan.. \$\endgroup\$ – Trevor_G Nov 21 '17 at 21:19

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