LAN cable tester based on 4017 - Page 1
Forum section: electronics / beginners

author: daskorn
author_rank: Newbie
timestamp: February 12, 2020, 08:06:20 pm
content: 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=4Studying 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 regardsdaskorn

author: tooki
author_rank: Super Contributor
timestamp: February 13, 2020, 01:31:27 am
content: 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.:Quote from: daskorn on February 12, 2020, 08:06:20 pmIt 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”.

author: edavid
author_rank: Super Contributor
timestamp: February 13, 2020, 01:41:49 am
content: 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.

author: johnkenyon
author_rank: Regular Contributor
timestamp: February 13, 2020, 01:31:10 pm
content: Quote from: edavid on February 13, 2020, 01:41:49 amSurely 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.

author: Peabody
author_rank: Super Contributor
timestamp: February 13, 2020, 05:21:03 pm
content: I would just second the suggestion that the simplest solution is to use high-efficiency LEDs instead of cluttering things up with transistors.

author: edavid
author_rank: Super Contributor
timestamp: February 13, 2020, 08:05:45 pm
content: Quote from: johnkenyon on February 13, 2020, 01:31:10 pmI think it's unlikely that Ethernet cables will start having 9 or 10 wires

author: Peabody
author_rank: Super Contributor
timestamp: February 14, 2020, 04:45:08 am
content: 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.

author: tooki
author_rank: Super Contributor
timestamp: February 14, 2020, 04:42:54 pm
content: 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.

author: daskorn
author_rank: Newbie
timestamp: February 14, 2020, 09:03:33 pm
content: 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 regardsdaskorn@tookiThx, I fixed the typo

author: edavid
author_rank: Super Contributor
timestamp: February 14, 2020, 10:26:07 pm
content: Quote from: daskorn on February 14, 2020, 09:03:33 pmThere 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.QuoteYou would need 1 PNP and 1 NPN per output... both wired as emitter followers. Or, you could use CD4050 buffers as I suggested.

author: johnkenyon
author_rank: Regular Contributor
timestamp: February 16, 2020, 01:35:34 pm
content: Quote from: edavid on February 13, 2020, 08:05:45 pm4x 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.