LAN cable tester based on 4017

[daskorn]

Hi,

I'm a electronic newbie. I repaired a couple of home electronic stuff and in general I would say that I have "basic" electronic know how.
This is my first electronic DIY project.

I planned to build a simple DIY LAN cable tester based on a 555 and 4017. There are a couple of different designs available on the internet.
I liked the 8 LEDs approaches most. I know 4 LEDs would be enough but more LED -> more fun.

I finally ended up with the following schematic (not based on 555):
http://archive.siliconchip.com.au/cms/gallery/article.html?slideshow=0&a=30474&i=4

Studying the 4017 datasheet I concluded that the schematic above is on the edge if not above the manufacture’s maximum current specs.
So, I planned to switch a NPN transistor instead of the 3 LEDs directly.

But there is a problem which I couldn't fix. In case of the design above only one 4017 output pin is V+ all other Pins are GND all the time. That is how one LED on the end of the wire lights up. Because all other pins are GND it is also possible to notice sequence changes.
How can I achieve the same with switching transistors instead of the LEDs directly? My last thought (before I gave up) was that I need at least 3 transistors 2xNPN to switch to ground by default and a PNP transistor to switch V+.

There must be an easier solution.

Looking forward for your feedback.

Kind regards
daskorn

[tooki]

While the 4017’s specs do indeed look conservative, driving a single LED at normal 10-20mA from them seems to be something people have done since time immemorial. But bear in mind anyway that you could use modern LEDs that are so efficient that running them at under 0.5mA is more than enough for them to light brightly, so you’d use far higher dropper resistor values than listed in the schematic here.


P.S.:

Studying the 4017 datasheet I concluded that the schematic above is on the edge when not above the manufacture’s maximum current specs.

It took me a second to realize what you were saying here. The German word “wenn” is the English word “if”. (The English word “when” is German “wann”.) So this sentence would need to read “...the schematic above is at the limit of, if not over, the manufacturer’s maximum current rating”.

[edavid]

The CD4017 doesn't have a maximum output current spec, at least not in the RCA/Harris/TI datasheet.  It has a maximum power spec of 100mW per output transistor, which doesn't seem like a problem with the specified 680 ohm series resistors.

If you did want more output drive, you could use the CD4050, but you would need 2 of them.

BTW, I think you would be better off using a CD4022 instead of a CD4017, so it doesn't have dead time with no LEDs on.

[johnkenyon]

The CD4017 doesn't have a maximum output current spec, at least not in the RCA/Harris/TI datasheet.  It has a maximum power spec of 100mW per output transistor, which doesn't seem like a problem with the specified 680 ohm series resistors.

If you did want more output drive, you could use the CD4050, but you would need 2 of them.

BTW, I think you would be better off using a CD4022 instead of a CD4017, so it doesn't have dead time with no LEDs on.

Surely you can feed the 4017 reset from the "8" output - that way the thing counts from 0 to 7, then resets back to 0 when it gets to 8.
Then you if desired, you can easily expand to a 9 or 10 wire tester by moving/removing the reset connection, and adding LEDs.

[Peabody]

I would just second the suggestion that the simplest solution is to use high-efficiency LEDs instead of cluttering things up with transistors.

[edavid]

BTW, I think you would be better off using a CD4022 instead of a CD4017, so it doesn't have dead time with no LEDs on.

Surely you can feed the 4017 reset from the "8" output - that way the thing counts from 0 to 7, then resets back to 0 when it gets to 8.
Then you if desired, you can easily expand to a 9 or 10 wire tester by moving/removing the reset connection, and adding LEDs.

I think it's unlikely that Ethernet cables will start having 9 or 10 wires 

[Peabody]

If there's an alternative to the 9V battery, you could use a 74HC4017, which will drive/sink a lot more current than a CD4017, but needs 6V or less.  Or maybe keep the 9V battery, but add a 6V LDO.  That would be a lot easier than adding all those transistors.  But as I said before, I think the main thing is using low current LEDs.  Off of 9V, you should be able to drive them with a 1K resistor, or even 2.2K.

[tooki]

Assuming standard red 1.8V LEDs, the circuit is calculated to drive each LED at under 10mA. Of the three LEDs that can light on a given output, two of them are in series, one is in parallel, with their respective dropper resistors. This means that it’s under 20mA total current per output, and only one output is ever on at a time.

This is well within what a CD4017 can handle. It doesn’t need output transistors.

And if one used modern ultrabright green LEDs (the ~3.2V kind), one could increase the resistor values by 20% and drive them at 5mA, which will still be crazy bright.

[daskorn]

Hi,

I'm surprised to get so many answers. Thank you very much!

I'll try to drive the LEDs as suggested with lower current and without transistors. Maybe I'll switch to 5V and add a LDO. If I remember correctly I should have one "in stock".

But just for my understanding. In case I would need to add a transistor because of a higher load. Are my thoughts correct that I need at least 2x NPN and one PNP transistor to ensure that just one wire is V+ and all the others are GND?

Kind regards
daskorn

@tooki
Thx, I fixed the typo

[edavid]

I'll try to drive the LEDs as suggested with lower current and without transistors. Maybe I'll switch to 5V and add a LDO. If I remember correctly I should have one "in stock".

There is no need to lower the current.  There is no need to lower the supply voltage.  It is fine as it is if you are happy with the LED brightness.

But just for my understanding. In case I would need to add a transistor because of a higher load. Are my thoughts correct that I need at least 2x NPN and one PNP transistor to ensure that just one wire is V+ and all the others are GND?

You would need 1 PNP and 1 NPN per output... both wired as emitter followers.  Or, you could use CD4050 buffers as I suggested.

[johnkenyon]

BTW, I think you would be better off using a CD4022 instead of a CD4017, so it doesn't have dead time with no LEDs on.

Surely you can feed the 4017 reset from the "8" output - that way the thing counts from 0 to 7, then resets back to 0 when it gets to 8.
Then you if desired, you can easily expand to a 9 or 10 wire tester by moving/removing the reset connection, and adding LEDs.

I think it's unlikely that Ethernet cables will start having 9 or 10 wires 

4x twisted pairs plus 1xshield = 9 wires to test.

In any case, I like the idea of having "something for nothing" that's there if you need it at a later date.