74F07N Hex Buffer And a Pixel Question!

Started by niceguy167 4 weeks ago4 replieslatest reply 3 weeks ago58 views

I am working on running my 5V pixels through 40FT of Ethernet cable (so I don't have to have my main controller next to my Christmas props. I also really like keeping my expensive boards in my garage away from the weather and bad people. 

Everything works great.
I am using WS2811 Pixels I hook up 3 Wires for Positive, 3 Wires for Negative, and use 2 wires for data. I hookup a wire from the controllers’ data out port to one of the input ports on the 74F07N and then the Data wire from the cat5e cable to the out port on the 74F07N. Positive and negative go to the pixel controllers positive and negative ports. I then use a pull up resistor (470 ohm) at the end of the cat5e cable and wire this resistor between the data wire and the positive wire right where the strand of pixels begins. This has worked for many years. Running 6 strands of lights. Using the 6 input and 6 outputs on the 74F07N. What I want to know specifically is how 470 ohms calculated for the resistor as I want to go with some newer hex buffer’s, and they have slightly different values. I was also wondering if the ohms could be adjusted for longer cat5e runs like 70ft for example using less or more ohms without frying the 74F07N. (I used a guide from another topic on setting it all up.) I want to understand the electronics behind it so I can feel somewhat smart and maybe expand my cable lengths lol :) 

Thanks for any help you can give.


[ - ]
Reply by matthewbarrDecember 27, 2019

Ignoring the cable for the moment, the 470 ohm resistor determines how fast your signal will transition from low to high. The open collector driver sinks current and controls the opposite transition. The 470 value was likely chosen for a desired rise time based on lump capacitive load and to stay within driver sink current capability with some margin.

A short cable will still appear as a lump capacitive load with negligible resistance. As the cable length increases, at some point it begins to appear as a transmission line, a distributed RLC network. Further increases in cable length don't change the apparent capacitance at either end, but the end-to-end parasitic resistance continues to increase.

Since you're sinking current through the pullup at the reciever and through the cable resistance between receiver and driver, you get a voltage (IR) drop across the cable. The low level at the receiver will be higher than the low level at the driver by the amount of the IR drop. This comes off your noise margin, and at some cable length the IR drop will push the low level up into the receiver threshold region and you can't operate reliably at a reasonable data rate.

Also, when the cable is long enough to appear as a transmission line, your resistive termination needs to match the characteristic impedance of your cable if it is going to absorb incident wave energy and not produce a reflection. Now you have to compromise between the ideal termination value wrt. DC voltage/current levels and noise margin vs. a termination value that is best wrt. reflections and signal integrity. You can mitigate the noise margin issue to some extent by using a single-ended receiver with hysteresis.

When driving a long transmission line it is more common to use a differential driver and receiver. You can propagate a wave down the cable without sinking or sourcing DC current through the cable data line, you're less sensitive to noise and can achieve much higher data rates. I doubt there's a cook book recipe for selecting pull-up values to operate a single-ended open collector circuit over an arbitrarily long cable. At some point it just won't work reliably.

If you don't want to allocate two of your six power wires to an additional diff pair, you might consider using single-ended push/pull drivers with appropriate terminations at either or both ends, and a single-ended receiver with hysteresis. This gets you away from sinking or sourcing current through the cable data line between driver and receiver, you're propagating a wave through the transmission line which is really what you want. The receiver termination need only match the transmission line impedance. A series termination at the driver can soften the transition edges, improving signal integrity. This is an issue whether your single-ended driver is open collector or push/pull, but you wouldn't use a series termination with an open collector driver!

[ - ]
Reply by CustomSargeDecember 27, 2019

Like the Hat in 1st Harry Potter: Hmmm where to start...

You're using an open collector high voltage, high speed (F series) as line driver with a pullup at the far end. Work? yes, until line reflections can't be overcome by a lower value pullup. Since the WS2811 runs on 12V, may I suggest you change to a line driver designed for the purpose?

RS-232 is almost older than me, and so Very well known. Checkout a 75188 / 75189 - Tx runs off 12V, Rx needs a 78L05 or equ. Do this and pending signal speed, hundreds of feet aren't a problem.

To understand the topic, look up "signal transmission lines". The 470 is a reasonable value for O.C., but a regular 74LS374 with 220 to + / 330 to gnd is better, but still not RS-232 or 422.

For the best math of pullup sizing see:

I'll stop here - books have been written on this topic, by Far my betters. Good Hunting  <<<)))

[ - ]
Reply by antedeluvianDecember 27, 2019


I agree with your basic suggestion, but I don’t quite understand the configuration you are suggesting in terms of reception. As far as I remember the RS232 driver/receiver pair that you suggest (75188/75189) require both positive and negative supplies.

However I am almost sure the RS422 specification does allow for single supply. 


The 470R would have been calculated to allow the driver to sink 5V/470R=10mA. Since the 74F07 can sink up to 64mA you have the ability drop the resistor value by quite a bit. The lower the resistor value, the faster the rise time but you power supply must be able to deliver. If the pull-up resistor is on the receiver side, then the resistance of the wire may figure into the calculation since it would make a resistor divider and you may not get a logic “0”.

On the receiver side, when the input is high, there will be a leakage input current and if the resistor value is too high, a high may be seen as a low.

[ - ]
Reply by CustomSargeDecember 27, 2019

Howdy, Yup, I stand corrected. Other than the "power pump" versions, both supplies are req'd on the Tx. I made the assumption integrated power pumpers were available by now - my bad. They do have simpler pumpers, but they still need caps and are $$.