I've this PIR board: https://www.mpja.com/download/31227sc.pdf the chip output has already a 1KOhm resistor in series, so, since the one voltage of the chip is 3.3 V I guess I could connect directly to a led without having to add another resistance. Is this true? I just want to use that layout for testing, then the signal would go to a micro controller I/O. Just for the sake of completeness, since there is the 1K resistor on the out, I should drive even a saturating BJT directly, without adding any other resistor, isn't?

Did try, it works, but barely visible so added the BJT without any addtional resistor on the basis.

  • \$\begingroup\$ Please add the circuit directly to your question. The question editor has a "circuit editor" button. \$\endgroup\$ – Marcus Müller Mar 3 '17 at 8:36
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    \$\begingroup\$ I could connect directly to a led without having to add another resistance. Is this true? Why don't you simply try that and see what happens ? Since there's a 1 k resistor in series, nothing bad can happen. \$\endgroup\$ – Bimpelrekkie Mar 3 '17 at 8:52
  • \$\begingroup\$ @FakeMoustache this what I need to confirm: nothing should go bad. Thank :) Btw, IMO is a strange design, output should be clean, so user can decide what to do, If I was unaware of the internal of the board, I would have expect to have a "real" digital output. \$\endgroup\$ – Felice Pollano Mar 3 '17 at 9:02
  • \$\begingroup\$ @FelicePollano "digital" outputs are typically high resistance, the purpose of digital circuitry is to process and transmit information - not power. Power semiconductors are typically larger and produced with different processes. I suggest you buy or build a logic probe. \$\endgroup\$ – user1890202 Mar 3 '17 at 9:20
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    \$\begingroup\$ @user1890202 not sure it is correct, if I connect direct a led to a standard IC i expect the IC break for over current. The output are usually low impedance and is responsibility of the designer not to draw too much current. \$\endgroup\$ – Felice Pollano Mar 3 '17 at 9:57

As you mentioned, you can directly drive a BJT if you wanted a high brightness LED.

However if you want to connect and LED directly in series with the output this won't be an issue as long as you select the correct LED. For example take this LED (I typed Red LED in Farnell and just grabbed the first one), now what you need to look at is not the tables but the graphs.
If you take the value from the table it says you're looking at a 2V to 2.5V drop across your LED, however this is only true at 20mA. You're going to be operating it at 1/10th of this current.

So let's take a look at the graphs. enter image description here At the 1mA to 2mA range we're looking at a 1.6V to 1.7V drop across the LED (much less than our 2V to 2.5V from the tables). If we take the voltage drop at 1.7V then that gives us a current of 1.6mA.
Now that we have a current value let's jump across to the second graph, we can see that at 1.6mA we're going to get roughly 25% of the luminous intensity than if we were running at 10mA.
From the table it says that at 10mA the luminous intensity (lv) is typically 50mcd. So if we use this LED in your setup the equivalent lv is going to be ~12.5mcd. I know that mcd isn't really a good measure of brightness but I don't know the equation to change it into lumens, I can however, say that it will more than likely be clearly visible.

As a side note, whilst this is all good in theory, it would be much quicker to follow FakeMoustache's initial advice of try it and see.

  • \$\begingroup\$ The datasheet you picked bugs me. It Luminous Flux not Luminous Intensity. Intensity is the density of light within a cone (steradian) at a particular angle. \$\endgroup\$ – Misunderstood Mar 3 '17 at 19:15

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