How to Supply Enough Power?

Jesse Massey · 2005-03-01 21:09

Hello All,
As·you will notice I am new to this forum as a poster but have been keeping up with it the past couple of weeks. Anyway..

I am building a small project that has one SX28AC/DP as the main controller that will run two 7 segment displays. First display will run off of Pins 0b - 7b, and display number two is on pins 0c-7c. 0a-1a pins will act as my control lines from the computer. When the computer sends the signals the SX decodes the message and displays the number, pretty simple.

Each one of these display boxs has its own address and the computer will send out a signal with the correct address and number to display. So when the computer sends out a message each box will be listening and if the address matchs the address of the box then it will display the number.

Well all of this works great but my only problem is that I want to be able to only run two wires for the power so that all of the boxs are daisey chained together.

What my plan was to have 2 wires running at 12 volts and each unit will be hooked up in parallel to these wires. Also each unit will have a regulator for the sx. I am guessing that the only thing i need to worry about is getting enough amperage through the wires.

I already took care of the signal lines, i just to power up all of these units as easily as possible. I can't have a hundred wires running around. I plan on having anywhere between 2 - 50 display boxes. Is it possible with only two wires.

Jesse

PS sorry about the long post i wanted to be as clear as possible.

Jim Palmer · 2005-03-01 21:24

I got a little confused at the end... you want to have two wires that go to 0a and 1a pins for control and two seperate wires for the power, right? And each box would run off the same daisey chain of power and control? If this is the case, it seems completely possible the way you described it in your pentultimate paragraph. In fact, you could power and control each box with only two wires if you wanted to (vary the voltage slightly at a given frequency to send signals).

That said, you could have one less voltage regulator if you daisey-chained 3 wires (12v, 5v, Gnd). Another idea would be to always have a "First" box in chain that has the voltage regulator that has only the two power outputs, but then daisey chains the 5v to the rest.

Jim

Paul Baker · 2005-03-01 21:50

Its possible, its called low voltage electrical and is a growing field in home wiring used to power low voltage lighting and other systems, a simple extension cord wire should suffice, I know of people how have modified CAT5 cable to carry power and signals in the same cable. It all depends on your total current requirements, the more current the larger the wire required, there are sites which correlate what gauge wire can handle what current. When calculating current you should anticipate the total number of nodes you plan on adding otherwize you'll have to replace the wire later.

Jesse Massey · 2005-03-01 22:18

Sorry about the confusion I have a serious ear ache and I am on all kinds of funky medications, but yes I was planning on running 4 total wires. 2 power lines at 12 volts and 2 i/o lines.

But running on two lines sounds even better... i need to look into this.

Thanks for the replies.

Jesse

nick bernard · 2005-03-01 22:41

jesse,
i might be on the wrong page but why dont you get a 2pair shielded cable(jameco #232451cr, #234179cr) and run it to each box individually. if all 4 wires are common throughout the system then run all the cables into a junction box in a central location. a setup as such would be easier to troubleshoot. but thats just my opinion.

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League Bowling.... it's not a sport, it's a way of life

Paul Baker · 2005-03-01 22:44

btw, you need a ground wire to complete your circuit, one @ 12V one @ 0V with the wires large enough to carry the current, and you'll still need 4 total, CAT 5 has 8 wires in it, 4 are used for signals and 4 for power in the modified setup. Now you can get away with using two wires where you would inject a small signal on top of the power, this fluctuation would be rejected by the regulator, but a capacitor tapping the wire and amplification of that fluctuation could recapture the signal in the wire (X10 does somethng akin to this but with AC power), but this is likely beyond your ability to do since your still a novice.

Jesse Massey · 2005-03-02 01:29

Nick,
The reason I don't want to run a seperate set of wires to each unit is because I want to be able to add more display boxes easily. If they are daisey changed together then I should be able to just plug in another display box very easily.

Is it that obvious that I am still a novice when it comes to electronics[noparse]:)[/noparse]. It is true I know the basics of electronics but I figured that I am doing pretty good since I have only been working with it for a month now. The programming is the easy part, right now I am using SourceBoost C++ IDE to write the code and am loving it. I have been programming VC++ for a few years so naturely I did not want to have to learn assembly for the chip, even though I have taken classes for assembly of the 8086/88 processors, I just don't have the time.

When I finish the project and get it running for this company I am working for then I will show you guys the finished project if you are interested.

Anyways so far you guys are great.

Jesse

Jesse Massey · 2005-03-02 03:12

This question might be obvious to you guys so please be gentle.
I just finished testing one of the display box and the most current it used was 97.8mA. So if I hooked up 255 display boxes in parallel connection, the wires would have to handle atleast 25.5 amps(255units * .1 amps = 25.5 amps).

Thanks

Jesse

Paul Baker · 2005-03-02 04:47

while there is some disagreement, most online references state that 10 gauge wire or lower can handle that current (the lower the gauge the larger the wire)

PJMonty · 2005-03-02 08:11

Jesse,

I don't know what speed you're running the SX at, but lower clock speeds will definitely lower your power consumption. Also, as you've already figured out, daisy chaining the full 255 will result in some massive power requirements. I think you'll find you need to daisy chain some limited number, and then feed those into some sort of concentrator hub.

Thanks, PeterM

Jesse Massey · 2005-03-02 14:31

PJMonty,

Right now I am running off the internal 4mhz. I could run it a little lower without any problems I believe. It is just hard to believe that it will take that much current to run them.
I originaly thougt that I could run everything through cat 5 cable, but it looks like I will have to change my plans a little.

Thanks,
Jesse

Paul Baker · 2005-03-02 16:14

Yeah that power requirement knocks out the idea of daisy chaining using CAT5, one means reducing power consumption you should look into is reducing the current consumed by the 7 segment displays, this can be accomplished by series resistance on each segment or by using PWM to drive each segment. This will reduce the displays brightness, but will also reduce the current consumption. One additional thing, if the display is being driven by a static driver (such as a shift register), you can place the SX into sleep mode set to wake up on a pin change·on the pin·the host uses to communicate with the module.

Post Edited (Paul Baker) : 3/2/2005 4:17:48 PM GMT

Jesse Massey · 2005-03-02 17:15

The shift register sounds interesting but my only concern with using one of those is that they all(SX chips) have to be on at the same time while the computer sends out the address. Granted the SX chip would only be on for maybe a few seconds but that would still draw to much power.
I can't lower the brightness of the display, it needs to be as bright as possible.

Jesse

Beau Schwabe · 2005-03-02 17:34

Jesse said...
This question might be obvious to you guys so please be gentle.
I just finished testing one of the display box and the most current it used was 97.8mA. So if I hooked up 255 display boxes in parallel connection, the wires would have to handle at least 25.5 amps(255units * .1 amps = 25.5 amps).

Just thinking out loud here....

Can you supply your "display boxes" from the relative center of all your units rather than on an end?

....in a sort of T configuration, where the vertical component would be supplying 25.5 amps, but each horizontal component
would only need to supply half of that or 12.75 amps.

....Thinking further if your power supply was AC, and you divided it in a way that each unit had a half wave rectifier (a single diode
with capacitor filter) where HALF of the units were powered from the High side of the AC signal and HALF of the units were powered
from the Low side of the AC signal, you would still need to supply 25.5 amps on the vertical component, but each horizontal
component would only need to supply 1/4 of that, so now your down to a little over 6 amps.

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Beau Schwabe - Mask Designer III

National Semiconductor Corporation
(Communication Interface Division)
500 Pinnacle Court, Suite 525
Mail Stop GA1
Norcross,GA 30071

Jesse Massey · 2005-03-02 18:08

Wow, you lost me. Remember I am a novice with electronics.
Could you draw that out in a schematic, maybe then I could understand what you are saying.

Thanks,

Jesse

Paul Baker · 2005-03-02 18:40

This is the essence of what he's talking about:

····························64 daisy chained units
······································ |
······································ |
······································ |
64 daisy chained units

Central controller

64 daisy chained units
······································ |
······································ |
······································ |
····························64 daisy chained units

instead of:

controller

256 daisy chained units

it still keeps the wires fairly minimal and cuts the·current requirement of any wire by 1/4 or to 6.375 Amps taking you to a maximum gague of 18 instead of 10 guage, this still isn't low enough for cat 5 which I believe is 22 gauge.

Here is a calculator for 12V, providing proper gauge, given length and current: http://www.engineeringtoolbox.com/33_730.html

I don't think youll be able to use cat 5 in any daisy chaining scheme, the max current for 22 gauge is .95 Amps.

wait re-reading Beau's post, I realise I don't diagram it correctly, I'll leave it to him to explain.

I think he may be refering to this:

································Central controller
······································ |
······································ |
······································ |
64 daisy chained units

+

64 daisy chained units
······································ |
······································ |
64 daisy chained units

|

64 daisy chained units
····························

Post Edited (Paul Baker) : 3/2/2005 6:43:22 PM GMT

Beau Schwabe · 2005-03-02 19:05

Here is a "current tree" ... basically, for every half distance you move from the current source,
in this case the widest point at the bottom, your current demand will be halved.

By supplying your power from a central point, you only need to supply half the current out to each "leg".

With AC, if you turn half of your devices on with one polarity, and half of your devices on with the opposite
polarity, you can again decrease your overall current demand by a factor of 2 because only HALF of your
devices are on (being supplied) at any time.

During the "OFF time" each unit would have a simple diode cap filter to sustain power within each unit.
Doing this for 100mA is not to difficult.

In reference to the E-Mail above:

                                Central controller
                                       |
                                       |
                                       |
64 daisy chained units--------->|------+------|<-------- 64 daisy chained units
                                       |
                                       |
64 daisy chained units---------|<------|------>|-------- 64 daisy chained units

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Beau Schwabe - Mask Designer III

National Semiconductor Corporation
(Communication Interface Division)
500 Pinnacle Court, Suite 525
Mail Stop GA1
Norcross,GA 30071

Post Edited (Beau Schwabe) : 3/2/2005 7:27:27 PM GMT

PJMonty · 2005-03-02 19:57

Beau,

The AC approach sounds a little like the "free lunch" problem - e.g. there is no free lunch.

If the circuit needs 100 ma to run, and you power it with a pulsed current where the cap provides power during the off cycle, doesn't that mean that during the on cycle, you need power the circuit and recharge the cap, thus doubling the current input during the on cycle and bringing the peak power cunsumption back to 100%?

Thanks, PeterM

Beau Schwabe · 2005-03-02 20:30

I said...
...you would still need to supply 25.5 amps on the vertical component, but each horizontal
component would only need to supply 1/4 of that, so now your down to a little over 6 amps

This is why I said the above.... I see what you are saying, and you may be right. I usually design with
DC, so my mind does funny things when AC gets involved.

Ok forget AC for the moment ..... At least we can agree that this will work for DC by reducing the
current in each leg to 12.75 amps.

In reference to AC

Paul Baker said...
...the max current for 22 gauge is .95 Amps.

What if this is pulsed? Not a sine wave... Hard polarity swap... What is the value for a 50% duty cycle?

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Beau Schwabe - Mask Designer III

National Semiconductor Corporation
(Communication Interface Division)
500 Pinnacle Court, Suite 525
Mail Stop GA1
Norcross,GA 30071

Jim Palmer · 2005-03-02 20:33

Backing up a step, when you said that the current of your test box was 97.8mA, what was the voltage where you measured it. The reason I ask is that you had suggested that your supply would be 12V. If you test was on a 5V test bed, you would be over-estimating the current requirements by over 200%. Or, I could just be confused again [noparse]:)[/noparse]

Doug Hale · 2005-03-02 21:12

If each node draws ~100mA @ 5V that is 1/2 watt. A small switcher that runs from say 36 volts would draw about 16.6 mA (at 80% efficient). An upconverter Switcher at the master controller would only have to provide less than 900mA for all 50 units.

To reduce costs, you may even be able to use the SX28 as the switcher controller as well by bootstrapping the initial supply to the SX to get iy running. After the SX starts running, it would just pulse the pass element using the comparator as feedback to adjust the pulse width.

However, switcher controllers are not that expensive and will make the design easier and quicker.

The limitation on the distributed supply voltage would be UL limits and also low enough to prevent sparking during hot connect/disconnect.
Another consideration is that the bootstraping may require larger startup currents than normal running, you would therefore like to provide some form of power sequencing to limit the draw from the upconverter at the master converter.

You may get some better ideas from researching POE (Power Over Ethernet). CAT5 consists of 4 twisted pairs of which only two pares are used for signaling. With POE, the other two pairs are used to distribute power. However, 100BaseT is a star network and power is then distributed in the same way. You want a bus network.

If I have not explained myself well enough or if you have more questions, feel free to ask.

Doug

Paul Baker · 2005-03-02 21:18

Beau Schwabe said...

Paul Baker said...
...the max current for 22 gauge is .95 Amps.

What if this is pulsed? Not a sine wave... Hard polarity swap... What is the value for a 50% duty cycle?

I believe that the current ratings for wires I've been quoting have thier origin in wattage, so a pulsed current would drop the wattage therefore increase the current carrying capability, with his 7V overhead this technique maybe plausible, but being a novice I wouldn't suggest it. One means of doing this maybe copying components of a switched power supply where the PWM module would drive the cables and replicate the tail end components for each modules. But I still contend that this is too complex a solution.

My suggestion considering your experience is getting heavy duty power cord and CAT5 and split looming or taping the two together, its the simplest and sure fire way to get the system working.

Jim Palmer said...

The reason I ask is that you had suggested that your supply would be 12V. If you test was on a 5V test bed, you would be over-estimating the current requirements by over 200%. Or, I could just be confused again [noparse]:)[/noparse]

You do have a point, first we need to know if he took the current measurement on the power side of the regulator or the supply side of the regulator. Assuming it was computed on the supply side, its still 25.5 Amps but 127.5 Watts which drops him to 13 gauge, assuming the current tree configuration thats 31.875 Watts per branch and 19 gauge. Ive tried to find what would work with cat5 but there is no solution, the resistance inherent to 22 gauge wire make carrying any appreciable current beyond a few feet impossible (6V, 4A, 50ft = 7.8 Voltage drop!)

RE: black68cougar
Like I stated in response to Beau, I really think given his knowledge level he shouldn't try to attempt a switched power scheme.

Post Edited (Paul Baker) : 3/2/2005 9:21:24 PM GMT

PJMonty · 2005-03-02 21:50

In the midst of all this talk of power, I went searching and found a pretty cool on-line calculator for figuring out voltage drop due to line loss. You input the voltage, current draw, wire length and gage, and it tells you what will come out the other end. It also warns you if the wire will get too hot based on the passed current. In addition, the page is hosted by a company that makes power supplies, so the calculations are DC based.

www.currentsolutions.com/knowledge/vdrop.htm

Thanks, PeterM

Jesse Massey · 2005-03-02 22:13

Input == 12VDC and this is regulated down to 5VDC on the board.

ok I think I have it all planned out on how I am going to wire this stuff together.
Thanks for all of the help.

Jesse

Beau Schwabe · 2005-03-02 23:09

PeterM said...
a pretty cool on-line calculator for figuring out voltage drop due to line loss. You input the voltage, current draw, wire length and gage, and it tells you what will come out the other end

This is basically known as IR drop. (Current drop due to Resistance of wire length).

Our formulas for Electromigration rules also include temperature in the equation

Iavg <= Dt*A(W-B)
Irms <= 5*A(W-B)
IPeak <= 20*A(W-B)
IESD <= 40*A(W-B) only for ESD pulse < 0.15uS

Where:

Dt = Sqrt[noparse][[/noparse] 1.408X10-10 * e^(8121/273+Tj) ] = Temperature derating factor
Tj = the junction temperature in degrees Celcius.
A = mA/um
B = um
W= width of the metal line (in um)

Note:
A and B are coefficients related to the cross section

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Beau Schwabe - Mask Designer III

National Semiconductor Corporation
(Communication Interface Division)
500 Pinnacle Court, Suite 525
Mail Stop GA1
Norcross,GA 30071

Post Edited (Beau Schwabe) : 3/2/2005 11:48:40 PM GMT

James Newton · 2005-03-02 23:40

If you want to write that up as JavaScript, I'll be happy to host it at sxlist.com

I just finished one for finding the common and phases for unipolar stepper motors based on resistance readings... Not sure it works correctly, but:
http://www.sxlist.com/techref/io/stepper/wires

On the subject of problems with long wires, keep in mind that AC issues (EMI, noise, etc...) will be much more of an issue if AC power is used to supply the remote units. That AC power will cause problems with the signal lines. In fact, the signal lines may have more than enough trouble given some length and the right phase of the moon, etc... I would really suggest twisted pair.

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---
James Newton, Host of SXList.com
james@sxlist.com 1-619-652-0593 fax:1-208-279-8767
SX FAQ / Code / Tutorials / Documentation:
http://www.sxlist.com Pick faster!

allanlane5 · 2005-03-07 16:42

If you really want 255 nodes, and you don't want to run #10 wire for the 25 amps they all take, then a multi-point power supply approach makes lots of sense.

What you do is have a 12-volt wall-wart power 10 nodes. This is around 1 amp power, which should not require huge conductors. Build 25 of those, and you are good to go.

Your signal wire should be 4 conductor (phone wire) or CAT-5 (8-wire, more expensive, but better noise isolation). Run +12 on one wire, signal on another, ground on a third. This gives you one spare wire to use as a second signal wire, a power-supply ground wire, or for future expansion.

You CAN do this on two wires, but the complexity of putting the signal on the power wire and getting it off again without blowing up your circuits rises a lot. 4-wire cable is cheap enough you should run separate signal wires.

Why do you need 255 nodes anyway?

Jesse Massey · 2005-03-08 01:47

allanlane5,
I beleive that this is how I am going to end up doing this. To answer your question about running 255 displays is the company that wants this setup needs a lot of these displays. But they also want this system where they can scale it to any number between 1 and 255. So basically this system has to be where I can plug in any number of displays(up to 255) and only have to change the program on the server and the address on the SX chip.

Thanks,

Jesse

allanlane5 · 2005-03-08 15:06

Well, the wall-wart per 10 displays multi-drop approach looks like the most scaleable solution then. People with a need for only a few displays don't have to pay for the capacity of all 255 -- and people who need lots of displays can add (and power them) incrementally.

Note each '485 chip is limited in how many drops it can have -- I think each driver can drive 32 recievers. This means you'll also need 're-driver' boxes to get to the 255 node limit -- which is another good place to put in a power-port. A 're-driver' box can be relatively 'dumb' -- Black Box makes them, if you want to see what is commercially available.

For the 'address' it occurs to me you have a couple of approaches.· The hard one is to hard-code into each SX what address it is.· This means your code is slightly different for each node, and a node which dies requires reprogramming an SX to replace it.

Another approach is to add an 8-pin DIP switch to each unit which holds the 'slave' byte address, read once on power-up (connected through a 74HC164 to limit I/O pin use).

One last approach is to assign each unit you build a unique 16 to 32 bit number.· You then design some nice protocol to poll every box on the party-line, get its number, and have the master assign a unique 'listen' byte address to that 'slave'.·· This takes the least hardware, but is the most difficult to get right.· I'd go with the 'brute force' DIP switch solution, myself.

Post Edited (allanlane5) : 3/8/2005 3:14:26 PM GMT

pjv · 2005-03-09 04:10

Hi Jesse;

I don't know how up to speed you are on electronics, so what I am about to suggest may seem foolish.

If I had to tackle this project, and based on your requirements as I've gleaned them from following this post, I would approach it from the perspective of using a single pair of wires, regular 18 Ga lampcord in fact, and use that to power the nodes as well as simultaneously communicate over the same wires.

How can this be done?

Firstly,·as it appears to be a commercial application, I'm assuming there is not an extreme price sensitivity, I bet that a few extra components per node will not be a problem.

You need to find a way to reduce the current consumption while still keeping the LEDs as bright as possible. This can easily be done by driving each of the seven segments of each display with its own transistor/flyback inductor or transformer. This implies using one dedicated port bit for each segment (that is, non-multiplexed) and pulsing power into ech inductor from a relatively high voltage, say 30 VDC. When the processor turns the power (ground) to the inductor off, the built-up field can collapse into the diode, lighting it in a realatively constant-current basis while the field collapses. When the field is depleted, the cycle is repeated. This way the wattage that·each LED segment needs (say 10 or 20 mA times a forward drop of·2.5 Volts = approx 50 mW) can be scaled to the higher 30 V supply voltage, and hence each consume 2 to 3 mA.

Next, feed 24 VAC down the wire, and·in each node rectify and filter that somewhat to provide the 30 odd volts DC. Insert blocking inductors in series with the rectifier bridge to prevent the data signal (yet to be added) from being swamped in the low impedance of the filter cap. Add a small low voltage regulator to run the SX at 4 Mhz.,drawing only·a few mA.,·and use some appropriately small coupling capacitors to pick off the data signal from the line yet block the AC sinewave. Feed that signal to the micro for software processing and ulimately display generation.

At the "head-end" you simply have a similarly connected·node as well as the transformer. It will place the desired data onto the line, again coupling the data through yet blocking the slow 60 Hz AC. There is no requirement for·RS485; the processor can readily deal with the bit-banging directly from its port bits.

While I have not yet fully developed·a similar·sytem for my own applications, I have gone along quite a ways, and determined (in fact proven) that each of these concepts is viable, and hence the whole package is possible. That said however, it is not for the faint-of-heart, and considerable experience is an enormous asset.

If you wish to follow this approach further, I'd be happy to give pointers. In order to do that well, however you will need to spend some time to very precisely deatil ouline EXACTLY what you are needing to accomplish.

Sounds like a fun project.

Peter (pjv)
·

kjennejohn · 2005-03-09 07:57

Correct me if I'm wrong, but doesn't a typical LED pull 20 mA? And, to achieve maximum brightness, don't you neeed this current to be constant, not pulsed?

So, with one LED per segment, seven per display, this is 140 mA per display. Not counting decimal points and/or colons. With 255 units (boxes?), this is 35.7 Amps! And we haven't factored in consumption of the processors and any support circuitry, such as the serial level-translator chips for RS232 communications to the PC.

Or did I miss something?

kenjj
All my pencils *used* to have erasers!

How to Supply Enough Power?

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