Comments on: How to Control a LED With Raspberry PI GPIO and GLSL Hacker GeeXLab (in Lua or Python) https://www.geeks3d.com/20150526/how-to-control-a-led-with-raspberry-pi-gpio-and-glsl-hacker-in-lua-or-python/ Graphics Cards and GPUs News, Graphics Programming, Home of FurMark Thu, 14 Jun 2018 08:03:47 +0000 hourly 1 https://wordpress.org/?v=6.7.1 By: JeGX https://www.geeks3d.com/20150526/how-to-control-a-led-with-raspberry-pi-gpio-and-glsl-hacker-in-lua-or-python/#comment-85596 Thu, 10 Sep 2015 13:44:58 +0000 http://www.geeks3d.com/?p=9008#comment-85596 In reply to Lindsay Haisley.

Absolutely it’s a typo, 0.3V is the right value in the equations. Thank you for the feedback Lindsay!

]]> By: Lindsay Haisley https://www.geeks3d.com/20150526/how-to-control-a-led-with-raspberry-pi-gpio-and-glsl-hacker-in-lua-or-python/#comment-85591 Sun, 06 Sep 2015 17:29:38 +0000 http://www.geeks3d.com/?p=9008#comment-85591 It looks as though you have a typo in your math in sec. 1.2. When computing the proper resistance for the collector circuit with the LED, you use the formula:

Rb = (5.0 – 1.8 – 0.7) / 20mA = 145 ohm

0.7V is the emitter-base voltage. The emitter-collector voltage is 0.3V for the 2n2222 according to your discussion. Shouldn’t this be used in your analysis of the proper resistance in series with the LED, and for the subsequent formula:

Ic = Iled = (5.0 – 1.8 – 0.7) / 470 = 6mA

for computing the current through the LED?

It looks to me as if the proper computation should be:

Rb = (5.0 – 1.8 – 0.3) / 20mA = 145 ohm

and

Ic = Iled = (5.0 – 1.8 – 0.3) / 470 = 6mA

Your computations are correct if you assume that you meant 0.3V instead of 0.7V for the e-c voltage, but not so if you do the same computations with 0.7V as you have specified in your discussion.

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