Beginners understanding of pull-ups/down on IC's (like 555 timer)

When I connected an astable timer circuit, I followed this scheme:
**broken link removed**

But now I see in the official datasheet for LM555 used above:
http://www.ti.com/lit/ds/symlink/lm555.pdf (figure 4)

and for NE555 which I use:
http://www.datasheetcatalog.org/datasheet2/c/0hex3d1gxjejda7ur3y2toipkzfy.pdf (fig 12)

that they pull the OUT pin high with 1kΩ in the datasheets. In my circuit, I just leave my output floating directly on the base of a transistor (works fine, thus far). Isn't it pulled high or low internally already? Must I do this externally?

And in general:

1) How can I know when a pin of an IC is internally "pulled" already, and when I need to take care of it myself? I'd like to learn how to read such information out of datasheets such as those I linked to above. Circuit figures 12 and 4 in the respective datasheet, seem to give a recommendation in drawing, but is it actually spelled out anywhere?

2) Should I pull high OR low? Is the pin OUT of a 555 switching between pulse high and floating, or is it switching between pulse low and floating? If it's never floating, then why do they pull it high in the datasheet?

3) And by how many ohm should I pull? I suppose that I should aim for a certain ampere given my Vcc level. Is 1-5 mA generally a good number (which means a handful of kΩ), or how can one read out of datasheets what's a good choice for a particular IC? Wouldn't I save battery if I increase the value of the pulling resistors?

(In monostable mode, the circuit of my first link does pull the OUT pin high, but by 100 times more ohm than in the second datasheet, 100kΩ instead of 1kΩ, with 5V Vcc then causing a tiny 50µA current during low pulses, if I understand it correctly. This is maybe not an exact science, but I'd like to at least find the ballpark on the map.)

The output stage of the 555 is an active push-pull and does not need an external pull-up resistor. This can be seen in the datasheet on the current/voltage specification of the High-level on the output. The application shown might use the 1k ohm load resistor just to get always nice signal wave form to show. If an output can be in a high impedance state(3-state), then a pull-up resistor avoids floating signal level (creating problems with high impedance CMOS inputs).

Enjoy your design work!

Ah, so if I understand it correctly, on page 5 in this datasheet:
http://www.datasheetcatalog.org/datasheet2/c/0hex3d1gxjejda7ur3y2toipkzfy.pdf

the current for the output and reset pins are given in terms of mA (0.1 to 100). Therefor they can be considered to be internally pulled? Although it is still recommended to pull reset high when it's not used, so I suppose 0.1 mA is a low value in this context.

But the pins for trigger, threshold and discharge are stated to be less than about 1,000 times lower, ranging from 20 to 2000 nano Ampere. And therefor I would need to pull those pins, or they might be floating in the unknown space of accidental electric charges around. And the control pin is clearly stated to be "open circuit".

I suppose that, in the common TTL environment of a hobbyist, a current needs to be at least about 1mA rather than 1µA in order for it to be working reliably as a logical level. So pull-up resistor values should be chosen accordingly.


And thank you! I do enjoy my "design work"! This is like programming, but much much more cumbersome. However, some things cannot be done by a computer alone, so a soldering pen does come in handy at times.

The reset pin is an input and should be connected to Vcc if it is not used. If it is driven externally you might use a pull-up if the external reset signal have no active high output (e.g. open collector). The trigger, threshold and discharge are analog inputs(reason for the lower current). The circiut example shown are fine(no additional pull-ups needed it's analog!).

Enjoy your design work!