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I am trying to be smart with a breadboard that runs dozens of TTL lines together, and which are monitored with LEDs. All the chips I am using are inherently current limiting enough so I won't burn my diodes, but the LEDs are a load on the TTL lines and some of these dozens of lines drive only a single TTL input high, but some drive more than one, especially data/address bus lines are driving up to 10 inputs, the limit for TTL. So, I want to use the highest possible resistor(s) in my LED array to keep the light visible (not glaring bright anyway) yet suck the least amount of current from the bus to allow all the inputs to be reliably driven high.

schematic

simulate this circuit – Schematic created using CircuitLab

Now I am lazy and prefer to be smart rather than soldering together some contraption of LEDs with built-in resistors. Yes that would be ideal and maybe in the USA I could buy it, but I'm not there now and I can't buy fancy LED arrays with common cathode and in-built resistors.

So I thought about what if I use a single variable resistor on the upper ground bar on my breadboard. If I use the usual 220 Ω resistor I can see the LEDs OK until 4 light up, after that it gets too dim. I never have all of them light up, say, at a maximum 8 data / address bits (0xFF), 8 control lines that are active low, and perhaps 4 control lines that are active high. So, that's about 20, two dozen. Not a small number.

So if the number of LEDs that are on is about constant, I could just use a small resistor, like 10 Ω, and I believe I should be OK. But the number is of LEDs lit is of course not constant.

schematic

simulate this circuit

So I wonder, what if I replace that single fixed resistor R_com with some smart variable thing? In this case, what about a voltage regulator? Wouldn't a voltage regulator ensure that the "output" voltage over it stays constant so that as more current goes through the diodes, raising the voltage from all the TTL lines being high, it would reduce its effective resistance to ground to keep the voltage level constant? Or even a varistor?

schematic

simulate this circuit

Does anyone know of a trick like that?

PS: I know I can solder it fixed on a board and don't have to fuzz so much about number of components. But this is all still prototyping, I will experiment with ergonomics, figuring out which colors and brightness type of LEDs works best, adding more lines or removing some lines, until I have something that works. The best I can think of without the trick would be to solder resistors on the feet of the LED, but that is a lot of work, clutter, and prone to breakage as the resistor wires are so fine these days.

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    \$\begingroup\$ Diodes aren't identical. That's why you need to current limit each one individually. The bigger issue isn't even the one you are concerned where the LEDs are too dim. It's LEDs frying. The LED with the lowest voltage drop will conduct more than it's fair share of the current (which a shared resistor of a low enough value to multiple LEDs to be lit properly will allow) causing it to fry. Of course, a shared resistor of such a value also fries LEDs if fewer LEDs are lit for the same reason. \$\endgroup\$ – DKNguyen Nov 12 '19 at 18:26
  • \$\begingroup\$ On top of that, you seem to be trying to use a series voltage regulator as a shunt regulator, which won't work either. \$\endgroup\$ – Hearth Nov 12 '19 at 18:27
  • \$\begingroup\$ If you are able to determine that there are LEDs that will never be lit together and the groupings are mutually exclusive, then you can group those resistors together with a shared resistor for each group. \$\endgroup\$ – DKNguyen Nov 12 '19 at 18:32
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    \$\begingroup\$ We used to just use a 74LS07 for this. You get six buffers per package, they present one load on the source, and their outputs can handle up to 30 V and 40 mA. Plenty of margin for almost any of a variety of LED configurations. Are you able to source such devices? \$\endgroup\$ – jonk Nov 12 '19 at 18:45
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    \$\begingroup\$ Having looked at your other question it seems like you are greatly risking ruining a nice project by being a bit lazy and "cheap". If you are willing to go to the effort to build such a nice device I strongly suggest you try some of the suggested resistor solutions. DIY resistor arrays are 'easy enough' - even for we oldies with aging eyes. Soldering skills improve with practice - and a well tinned bit, good solder, & good soldered surfaces. Make yourself proud ! :-) \$\endgroup\$ – Russell McMahon Nov 12 '19 at 20:56
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That will work, provided you meet a few criteria.

  1. The current set must be less than the maximum allowed current in any LED.

  2. The current set must be less than the fanout of your TTL minus the fan-in of all the gates you are driving.

  3. You must be willing to accept that when all of the LEDs are on, they will be much dimmer than when only one is on.

  4. The current sink must have sufficient voltage compliance to work when its input voltage is equal to a minimum TTL high level minus an LED drop.

Items 2 and 4 are the big issues. Especially item 2. TTL is not real good at sourcing current, and it takes considerable input current at any input.

Frankly, if you're going to mess around with TTL (rather than something even remotely newer like CMOS, you're better off using an inverter/pulldown combination.

schematic

simulate this circuit – Schematic created using CircuitLab

This way, you have uniform loading on your signal bus, and uniform LED brightness. You can use something like a 74240 (or even better, a 74LS240 or 74HC240) as your buffer, and get 8 channels in a 20-pin DIP. You can use a 16-pin DIP pullup network, and get 15 pullups in a single package.

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  • \$\begingroup\$ We always just used the 74LS07 in the 1970's. Handled incandescents, even. 30 V @ 40 mA outputs. \$\endgroup\$ – jonk Nov 12 '19 at 18:56
  • \$\begingroup\$ Was your "yes that would work" in relation to my voltage regulator intuition that I had? Because someone commented above that it would not work. As for the buffer, 74x240, '07, etc., yes, I could run the LEDs all from another buffer. Indeed, I could do that. I have a bunch of 74LS245 chips unused, I could indeed just use another slice of breadboard to add those. I was, however, trying to have only one breadboard for input and display console. If you want to know that I welcome you to look here: electronics.stackexchange.com/questions/466715/…. \$\endgroup\$ – Gunther Schadow Nov 12 '19 at 19:36

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