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I am experimenting with an Intel Edison mini breakout board with pinout like here: http://fab-lab.eu/wp-content/uploads/2014/10/Edison_IO_LED.jpg

Now, I soldered header pins to plug the board into a breadboard. Unfortunately, the GPIO levels are only 1.8V.

  • What should I do to make a LED blink from here (on a breadboard ideally)?
  • What else can be done with 1.8V GPIO levels?

UPDATE: It works with a BC547

However, depending on the blink frequency, it seems that LED does not turn ON completely. Might it be because the base resistor is 0 ?

blink_transistor

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  • \$\begingroup\$ Could you clarify what do you mean by "the LED does not turn ON completely"? Also, I don't see any resistors on your breadboard - how do you limit the current through the LED? \$\endgroup\$ – Dmitry Grigoryev Dec 22 '15 at 20:33
  • \$\begingroup\$ the problem was indeed the missing base resistor to limit the current. Without that resistor, the collector current is not full steam I guess, and the LED does sometimes not blink \$\endgroup\$ – poseid Dec 22 '15 at 21:02
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    \$\begingroup\$ It's more probable that GPIO pins have some sort of short-circuit protection which detects a short circuit and cuts the current through the pin. \$\endgroup\$ – Dmitry Grigoryev Dec 22 '15 at 21:32
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1.8V is enough to drive a transistor which can be used to power all kinds of loads. Example:

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ thanks, I have some BC547 sparkfun.com/datasheets/Components/BC546.pdf - let me check how that circuits applies \$\endgroup\$ – poseid Dec 22 '15 at 13:38
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    \$\begingroup\$ To limit the current in your 1.8V output pin. \$\endgroup\$ – Dmitry Grigoryev Dec 22 '15 at 19:48
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    \$\begingroup\$ Also to limit the current through the base-emitter junction, which could destroy the transistor. \$\endgroup\$ – Steve Dec 22 '15 at 20:22
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    \$\begingroup\$ @Steve according to the datasheet, BC547 can handle up to 200 mA of base current (peak value), so I would be surprised to see it destroyed by the current from a GPIO pin. \$\endgroup\$ – Dmitry Grigoryev Dec 22 '15 at 21:30
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    \$\begingroup\$ If you hate the base resistor you could use a MOSFET however the ones rated for 1.8V drive are going to be SMT only so you'd need a breakout board etc. to easily prototype on your perf board. \$\endgroup\$ – Spehro Pefhany Oct 18 '18 at 1:58

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