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Detecting... load, current... um...

Started by Unknown June 30, 2015
Now sure how to describe the problem in few words, sorry.

I have two circuit boards (to be designed), A and B. A has the power supply=
. A and B will be connected by a longish, 2 conductor wire. The wire will b=
e used to supply 12V+ and ground from A to B. B has no other source of powe=
r.

B needs to signal A when something happens.

For various reasons, I can't replace the wire with a 3 conductor version, a=
nd wireless solutions aren't practical.

Normally, board B draws maybe 60mA at most, mostly for LEDs. But occasional=
ly board B will close a relay, and feed the 12V into a larger load: a DC-DC=
 converter, to generate 5V @ maybe 1-3A. Presumably that will show up as a =
larger current draw on the 12V line, but I don't know how much. (I can slap=
 a high wattage resistor in parallel with the load to make it draw more cur=
rent, if that helps.)

That's what A has to detect.

I don't know how to detect current changes and I don't entirely trust my es=
timates on the current change anyway.

Is there a clever and inexpensive way to overlay some sort of signal on the=
 wire that is reliably detectable? Or is there an easily adjustable way to =
detect current changes on a power line? It is probably safe to say that B d=
raws considerably less than 500mA normally and something over 500mA during =
the event.=20

TIA.
In article <af1675d4-9fde-47b8-a41d-27139759fee7@googlegroups.com>, 
scott.a.mayo@gmail.com says...

[snip]

> I don't know how to detect current changes and I don't entirely trust > my estimates on the current change anyway. > > Is there a clever and inexpensive way to overlay some sort of signal > on the wire that is reliably detectable? Or is there an easily > adjustable way to detect current changes on a power line? It is > probably safe to say that B draws considerably less than 500mA > normally and something over 500mA during the event.
Google 'power line communication'. HTH
On Tuesday, June 30, 2015 at 9:07:54 PM UTC-4, Randy Day wrote:

> Google 'power line communication'. > > HTH
Interesting, but it doesn't look trivial to implement, and since I'm trying to send one bit of information ("it happened") it feels like overkill. One idea I'm toying with is trying to put a very brief (100ms) very low resistance across the power supply, to try and get the (unregulated) 12v supply to sag. By comparing the sag against the previous 12v voltage (stored with a cap) with a comparator, can I catch the event? But I'm not sure I like the idea of using a very low value resistor to get an unregulated 12v 4A power supply to sag. If there some way to inject a brief oscillation, say at 20kHz, on the line that the other side can detect?
<scott.a.mayo@gmail.com> wrote in message 
news:af1675d4-9fde-47b8-a41d-27139759fee7@googlegroups.com...
>Now sure how to describe the problem in few words, sorry.
>I have two circuit boards (to be designed), A and B. A has the power >supply. A and B will be connected by a >longish, 2 conductor wire. The wire >will be used to supply 12V+ and ground from A to B. B has no other >source >of power.
.B needs to signal A when something happens.
>For various reasons, I can't replace the wire with a 3 conductor version, >and wireless solutions aren't practical.
>Normally, board B draws maybe 60mA at most, mostly for LEDs. But >occasionally board B will close a relay, and feed the 12V into a larger >load: a DC-DC converter, to generate 5V @ maybe 1-3A. Presumably that will >show up as a larger current draw on the 12V line, but I don't know how >much. (I can slap a high wattage resistor >in parallel with the load to >make it draw more current, if that helps.)
>That's what A has to detect.
>I don't know how to detect current changes and I don't entirely trust my >estimates on the current change >anyway.
>Is there a clever and inexpensive way to overlay some sort of signal on the >wire that is reliably detectable? Or is >there an easily adjustable way to >detect current changes on a power line? It is probably safe to say that B >draws >considerably less than 500mA normally and something over 500mA >during the event.
What about putting a very low resistance ,say about .1 to .5 ohms in series with the wires and then an IC compariator circuit across the resistor. YOu set the comparitor to trip when say over 100 to 150 ma is drawn.
> What about putting a very low resistance ,say about .1 to .5 ohms in series > with the wires and then an IC compariator circuit across the resistor. YOu > set the comparitor to trip when say over 100 to 150 ma is drawn.
This sounds interesting, but at 12v and 4A, and assuming the load wants most of that when the relay closes... wouldn't I be looking at a 50W resistor? Assuming that's feasible, what sort of comparator do I use?
scott....@gmail.com wrote:


> > What about putting a very low resistance ,say about .1 to .5 ohms in series > > with the wires and then an IC compariator circuit across the resistor. YOu > > set the comparitor to trip when say over 100 to 150 ma is drawn. > > This sounds interesting, but at 12v and 4A, and assuming the load wants most of that when the relay closes... wouldn't I be looking at a 50W resistor?
** Ohms Law a bit rusty with you ? 4A and 0.1 ohms give 0.4V drop and the power is 1.6 watts.
> Assuming that's feasible, what sort of comparator do I use?
** I would use a reed switch with a few dozen turns or so of wire wound around it - eg: http://www.arunet.co.uk/tkboyd/ec/ec1RehaReedP0331.jpg .... Phil
<scott.a.mayo@gmail.com> wrote in message 
news:ecafa23c-9291-4980-bf43-e7e2c110c0aa@googlegroups.com...
> >> What about putting a very low resistance ,say about .1 to .5 ohms in >> series >> with the wires and then an IC compariator circuit across the resistor. >> YOu >> set the comparitor to trip when say over 100 to 150 ma is drawn. > > This sounds interesting, but at 12v and 4A, and assuming the load wants > most of that when the relay closes... wouldn't I be looking at a 50W > resistor? > > Assuming that's feasible, what sort of comparator do I use?
At 4 amps and a .1 ohm resistor,a two watt resistor would be ok. The power of the resistor would be 4 x 4 x .1 or 1.6 watts. A 2 watt resistor would allow a safety margin. A 10 watt resistor would be enough if a .5 ohm resistor was used. 4x4x.5 for the wattage. You can start by looking at this IC. http://www.onsemi.com/pub_link/Collateral/LM339-D.PDF
In article <4bc07c9a-c944-478b-a744-49fb1239814b@googlegroups.com>, 
scott.a.mayo@gmail.com says...
> > On Tuesday, June 30, 2015 at 9:07:54 PM UTC-4, Randy Day wrote: > > > Google 'power line communication'. > > > > HTH
[snip]
> If there some way to inject a brief oscillation, say at 20kHz, on the > line that the other side can detect?
A small-value capacitor feeds signal from the transmitter to the power trace on B. A small-value cap on the power trace goes to the receiver on A. DC in ---/\/\--+---/ /--+--/\/\--dc out L1 | | L2 = C1 = C2 | | rcv xmt You'd want inductors L1/L2 to prevent the signal vanishing into your filter caps.
Phil Allison wrote:

> > ** I would use a reed switch with a few dozen turns or so > of wire wound around it - eg: > > http://www.arunet.co.uk/tkboyd/ec/ec1RehaReedP0331.jpg
** Probably I should explain that a bit more. The coil wound around the reed switch *replaces* the previously mentioned resistor in line with the 12V DC supply. Adjust the number of turns ( which may be wound on a plastic tube for convenience ) so the reed switch reliably closes when the current reaches the higher level. A 20mm long reed switch operates in less than a millisecond and needs about 20 turns at 1 amp DC. .... Phil
On Tue, 30 Jun 2015 17:14:50 -0700 (PDT), scott.a.mayo@gmail.com
wrote:

>Now sure how to describe the problem in few words, sorry. > >I have two circuit boards (to be designed), A and B. A has the power supply. A and B will be connected by a longish, 2 conductor wire. The wire will be used to supply 12V+ and ground from A to B. B has no other source of power. > >B needs to signal A when something happens. > >For various reasons, I can't replace the wire with a 3 conductor version, and wireless solutions aren't practical. > >Normally, board B draws maybe 60mA at most, mostly for LEDs. But occasionally board B will close a relay, and feed the 12V into a larger load: a DC-DC converter, to generate 5V @ maybe 1-3A. Presumably that will show up as a larger current draw on the 12V line, but I don't know how much.
--- If board B is drawing 60mA quiescently from the 12V supply and then something happens to energize the relay and turn on the DC-DC converter, the current from the 12 volt source must increase in order to feed the relay coil and the 12 volt side of the DC-DC converter. Assuming one of those jellybean 400 milliwatt coil relays, that'll be an additional: . P 0.4W . I = --- = ------ = 0.033A = 33mA . E 12V required from the 12 volt supply for the relay. Then, if the output of the DC-DC converter supplies: . P = IE = 5V * 1A = 5 watts into a load, and the efficiency of the DC-DC converter is 85%, the DC-DC converter will need: . Pout,W 5W . Pin = -------- = ------ = 5.88 watts . 0.85 0.85 from the 12 volt supply, which equates to: . P 5.88W . I = --- = ------- = 0.490A = 490mA. . E 12V The sum of the three currents, then, is: . It = Iq + Ik + Ic = 60mA + 33mA + 490mA = 583mA. ---
>(I can slap a high wattage resistor in parallel with the load to make it draw more current, if that helps.)
--- That shouldn't be necessary. ---
>That's what A has to detect. > >I don't know how to detect current changes and I don't entirely trust my estimates on the current change anyway.
--- The classical way is to use a low-valued resistor (curiously called a "shunt") in series with the supply and to measure the current through the resistor by measuring the voltage across it and applying Ohm's law, . E . I = --- . R ---
>Is there a clever and inexpensive way to overlay some sort of signal on the wire that is reliably detectable? Or is there an easily adjustable way to detect current changes on a power line? It is probably safe to say that B draws considerably less than 500mA normally and something over 500mA during the event.
--- The suggestion to use a reed relay wrapped in a coil with a wire diameter large enough to make its resistance negligible when placed in series with the 12 volt supply, and enough turns to meet the reed switch's ampere-turn MAKE requirement when the supply's target current is met, or exceeded, is a good one. However, there are at least a couple of caveats. One is that reed switches exhibit large hystereses, and once the switch contacts are MADE, they may not break when the current in the coil reverts to quiescent. Another potentially nastier one is that trimming the winding to just what the reed switch needs to turn ON is tricky and, if you have to do more than a few, less than an attractive way to go about it. My suggestion would be to use a high-side shunt, a dual opamp, and a reference, like this: https://www.dropbox.com/s/i6hlcrv73ara2il/Current%20threshold%20detector.asc?dl=0 If you have trouble downloading that, the ASCII file follows, and if you have any questions about the circuit or you need a circuit description, post a request and I'll do what I can to help you out. 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