I notice that in Ben Eater's project for the 74 series logic chip computer, which my son and I are building, sometimes he uses current limiting resistors on LEDs on the pins of some chips, and on others, he madly throws caution to the wind, and slaps that LED right in there with no apparent resistor.

I assume this means some kind of inner knowledge of each chip, and that the outputs of each chip are current limited with an internal resistor.

How can I determine which chips are safe to put the LEDs in without a resistor, and which ones require the current limiting resistor?

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    \$\begingroup\$ The datasheet is your best friend. Without it you can't know for sure. The datasheet will tell you everything you need to know. Period. \$\endgroup\$
    – DerStrom8
    Commented Apr 5, 2017 at 16:19
  • \$\begingroup\$ Hi DerStrom8, I do have all the datasheets. For example, on the 74LS245, where would I find the fact that there are current limiting resistors that allow me to simple pop an LED in? datasheet: ti.com/lit/ds/symlink/sn74ls245.pdf I assume page 9 holds my answer? I.e. there is a resistor there? \$\endgroup\$
    – Kevin Hill
    Commented Apr 5, 2017 at 16:26
  • \$\begingroup\$ There are two things you need to consider.. 1. Without a resistor, will it cause the output unnecessary strain and overheat the device. 2. Without a resistor, how much current are you pulling through the LED and how will it effect the life of it. If the answer to both is acceptable, then don't use a resistor. \$\endgroup\$
    – Trevor_G
    Commented Apr 5, 2017 at 16:37
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    \$\begingroup\$ @Kevin Hill: Please show some examples of those circuits that have supposedly no current limiting resistors for the LEDs! We can only guess what you mean. \$\endgroup\$
    – Curd
    Commented Apr 5, 2017 at 16:38
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    \$\begingroup\$ The 74LS04 have internal resistors to reduce the shoot-thru current. When that pin is driven high, the Vbe and Resistor limit the current. Is that resistor designed for permanent heating? Those pins are short-circuit proof (I've slipped a probe or two, in past projects). \$\endgroup\$ Commented Apr 6, 2017 at 4:14

2 Answers 2

How can I determine which chips are safe to put the LEDs in without a resistor, and which ones require the current limiting resistor.

By reading their datasheets. This really should have been obvious.

Unless a chip is specifically intended for driving LEDs or has a controlled current output, you really should be using series resistors.

Most digital outputs are specified to source or sink some minimum current with some maximum voltage rise or drop. For example, a specification might be 15 mA sink at no more than 500 mV for a logic low level.

The naïve, lazy, or just irresponsible might say that 15 mA is less than the 20 mA my LED is rated for, so no problem. It doesn't work that way because 15 mA is the minimum spec, and there is usually no maximum spec. Any one part may be able to do 20 mA or more. And, that's just at 500 mV output voltage rise in this example. Let's say you connect a green LED with 2.1 V between a 5 V supply and this digital output. The digital output will be forced to well over 2 V. You don't know what current it might sink then. Since that's out of spec for the digital output, you don't even know if it will survive that.

Let's look at what the correct answer would be in this example with a series resistor. In this case, you have to design for 15 mA, since that's the lesser of what the LED can handle and the digital output can safely sink. You might say that since the LED drops 2.1 V and the digital output is at 500 mV, the resistor drops 2.4 V, but that would be wrong. The digital output is only guaranteed to not exceed 500 mV when sinking 15 mA. It could be less and probably is. You don't know how much less, so you have to assume the worst case of the digital output being at 0 V.

That leaves 2.9 V across the resistor at 15 mA. (2.9 V)/(15 mA) = 193 Ω. In this case you'd probably use the next common size up, which is 200 Ω.

Let's see what we've got. If the digital output stays at 0 V, the current will be (2.9 V)/(200 Ω) = 14.5 mA. If it rises the maximum amount but still within spec, then the current is (2.4 V)/(200 Ω) = 12 mA. You could get anything within the 12 to 14.5 mA range. Actually the range would be even more if you take the variation in LED forward voltage into account. Fortunately, it takes a fairly large current difference to see a obvious brightness difference, even in a side by side comparison. A 25% variation is pretty much inconsequential. It takes about a 2:1 current difference to notice the brightness difference when the two are not side by side.

There are naïve, lazy, and irresponsible circuit designers. A disproportionate fraction write up web pages about their designs, so there is a lot of crap out there. The most likely and simple answer to why some circuits have resistors in series with LEDs and others don't is that the ones that don't are bad designs.

  • \$\begingroup\$ So essentially you are saying that the design as seen in the video, obviously works - but should be using limiting resistors. \$\endgroup\$
    – Kevin Hill
    Commented Apr 5, 2017 at 18:42
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    \$\begingroup\$ @Kevin: I haven't watched any video, so can't say what it contains. However, you can't tell from a video whether the design really "works" in that it's not abusing parts. \$\endgroup\$ Commented Apr 5, 2017 at 20:08
  • \$\begingroup\$ Thanks Olin for the details! It set me out in the right direction - and I have now learned not only about current limiting resistors and TTL chips - but also pull down and pull up resistors for digital lines. I am sure I have lot more to learn - but for me the fact that I can see what is on the memory bus, set a value, and then see it - is pretty dang cool. \$\endgroup\$
    – Kevin Hill
    Commented Apr 6, 2017 at 1:18

Thanks everyone - the datasheet does indeed contain the answer.

In the case of the SN74LS04 - there is a 120ohm resistor from vcc to ground on the output. Meaning that in this case, it's not a direct short, and with 8 LEDS, it's probably not able to send enough current to damage the parts. That being said, it might be beneficial to place 100ohm or so more into the circuit to bring it within spec. (per Olin Lathrop).

Please correct me if my understanding of this is not right.
DataSheet: http://www.ti.com/lit/ds/symlink/sn74ls04.pdf


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