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

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.

TIA.

 

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?

 

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?

 

[Phil Allison]

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

 

[Randy Day]

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

 

[Phil Allison]

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

 

[Ralph Mowery]

<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.

 

[Ralph Mowery]

<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

 

[Randy Day]

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.

 

On Wednesday, July 1, 2015 at 6:39:31 AM UTC-4, John Fields wrote:
>...

Very much appreciated.

I've had bad luck with reed switches; will a hall effect sensor do the trick? twenty turns of thick wire wrapped around an iron rod and pointed at a hall sensor sounds less fragile and I feel like I'd be able to tune it just by setting the distance between the sensor and the coil, or fiddling with the rod.

I'll also look at the op amp approach; presumably also easy to tune. Which application is used to render that ascii list of components into an image?

Usenet saves the day again. I don't know why we bothered with this "web" thing.

 

[Jim Whitby]

On Tue, 30 Jun 2015 17:14:50 -0700, scott.a.mayo 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.

snip

Try a 600 ohm xfmer with split secondary.

Cap between windings, apply dc to cap. repeat on other end.

Apply AC to primary. Same technique used by phone company for years and
uears.

 

[Jim Whitby]

On Wed, 01 Jul 2015 14:02:14 +0000, Jim Whitby wrote:

On Tue, 30 Jun 2015 17:14:50 -0700, scott.a.mayo 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.

snip

Try a 600 ohm xfmer with split secondary.

Cap between windings, apply dc to cap. repeat on other end.

Apply AC to primary. Same technique used by phone company for years and
uears.

Additional info:

Stancor TTPC-13 is an example. It will not handle the current, but is
the right functionality.

 

[John Fields]

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.

Version 4
SHEET 1 2680 692
WIRE -1312 -1392 -1360 -1392
WIRE -1248 -1392 -1296 -1392
WIRE -1184 -1392 -1232 -1392
WIRE -1072 -1392 -1120 -1392
WIRE -1008 -1392 -1040 -1392
WIRE -944 -1392 -976 -1392
WIRE -880 -1392 -912 -1392
WIRE -816 -1392 -848 -1392
WIRE -752 -1392 -784 -1392
WIRE -688 -1392 -720 -1392
WIRE -624 -1392 -656 -1392
WIRE -560 -1392 -592 -1392
WIRE -496 -1392 -528 -1392
WIRE -432 -1392 -464 -1392
WIRE -368 -1392 -400 -1392
WIRE -304 -1392 -336 -1392
WIRE -240 -1392 -272 -1392
WIRE -160 -1392 -208 -1392
WIRE -1360 -1360 -1360 -1392
WIRE -1184 -1360 -1184 -1392
WIRE -1120 -1360 -1120 -1392
WIRE -160 -1360 -160 -1392
WIRE -1056 -1344 -1264 -1344
WIRE -992 -1344 -1056 -1344
WIRE -864 -1344 -912 -1344
WIRE -816 -1344 -864 -1344
WIRE -688 -1344 -736 -1344
WIRE -1360 -1296 -1360 -1328
WIRE -1184 -1296 -1184 -1328
WIRE -1120 -1296 -1120 -1328
WIRE -1056 -1296 -1056 -1344
WIRE -608 -1296 -608 -1328
WIRE -160 -1296 -160 -1328
WIRE -1360 -1232 -1360 -1264
WIRE -1184 -1232 -1184 -1264
WIRE -1120 -1232 -1120 -1264
WIRE -320 -1232 -320 -1248
WIRE -160 -1232 -160 -1264
WIRE -768 -1216 -768 -1232
WIRE -352 -1216 -432 -1216
WIRE -864 -1200 -864 -1344
WIRE -800 -1200 -864 -1200
WIRE -240 -1200 -288 -1200
WIRE -688 -1184 -688 -1344
WIRE -688 -1184 -736 -1184
WIRE -352 -1184 -688 -1184
WIRE -1360 -1168 -1360 -1200
WIRE -1184 -1168 -1184 -1200
WIRE -1120 -1168 -1120 -1200
WIRE -1056 -1168 -1056 -1216
WIRE -992 -1168 -1056 -1168
WIRE -864 -1168 -912 -1168
WIRE -800 -1168 -864 -1168
WIRE -160 -1168 -160 -1200
WIRE -608 -1136 -608 -1216
WIRE -560 -1136 -608 -1136
WIRE -432 -1136 -432 -1216
WIRE -432 -1136 -480 -1136
WIRE -1056 -1120 -1056 -1168
WIRE -1040 -1120 -1056 -1120
WIRE -1360 -1104 -1360 -1136
WIRE -1264 -1104 -1264 -1344
WIRE -1184 -1104 -1184 -1136
WIRE -1120 -1104 -1120 -1136
WIRE -1056 -1104 -1056 -1120
WIRE -864 -1104 -864 -1168
WIRE -160 -1104 -160 -1136
WIRE -432 -1088 -432 -1136
WIRE -608 -1072 -608 -1136
WIRE -1360 -1040 -1360 -1072
WIRE -1184 -1040 -1184 -1072
WIRE -1120 -1040 -1120 -1072
WIRE -160 -1040 -160 -1072
WIRE -1360 -976 -1360 -1008
WIRE -1184 -976 -1184 -1008
WIRE -1120 -976 -1120 -1008
WIRE -160 -976 -160 -1008
WIRE -1264 -960 -1264 -1024
WIRE -1056 -960 -1056 -1024
WIRE -1056 -960 -1264 -960
WIRE -1040 -960 -1056 -960
WIRE -864 -960 -864 -1024
WIRE -864 -960 -1040 -960
WIRE -768 -960 -768 -1152
WIRE -768 -960 -864 -960
WIRE -608 -960 -608 -1008
WIRE -608 -960 -768 -960
WIRE -432 -960 -432 -1008
WIRE -432 -960 -608 -960
WIRE -320 -960 -320 -1168
WIRE -320 -960 -432 -960
WIRE -1360 -912 -1360 -944
WIRE -1184 -912 -1184 -944
WIRE -1120 -912 -1120 -944
WIRE -1040 -912 -1040 -960
WIRE -160 -912 -160 -944
WIRE -1360 -848 -1360 -880
WIRE -1312 -848 -1360 -848
WIRE -1248 -848 -1296 -848
WIRE -1184 -848 -1184 -880
WIRE -1184 -848 -1232 -848
WIRE -1120 -848 -1120 -880
WIRE -1072 -848 -1120 -848
WIRE -1008 -848 -1040 -848
WIRE -944 -848 -976 -848
WIRE -880 -848 -912 -848
WIRE -816 -848 -848 -848
WIRE -752 -848 -784 -848
WIRE -688 -848 -720 -848
WIRE -624 -848 -656 -848
WIRE -560 -848 -592 -848
WIRE -496 -848 -528 -848
WIRE -432 -848 -464 -848
WIRE -368 -848 -400 -848
WIRE -304 -848 -336 -848
WIRE -240 -848 -272 -848
WIRE -160 -848 -160 -880
WIRE -160 -848 -208 -848
FLAG -768 -1232 +12
FLAG -1040 -912 0
FLAG -1040 -1120 +12
FLAG -320 -1248 +12
FLAG -608 -1328 +12
FLAG -240 -1200 KON
SYMBOL res -896 -1184 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R3
SYMATTR Value 10K
SYMBOL res -880 -1120 R0
SYMATTR InstName R4
SYMATTR Value 100K
SYMBOL res -720 -1360 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R2
SYMATTR Value 100K
SYMBOL res -896 -1360 R90
WINDOW 0 0 56 VBottom 2
WINDOW 3 32 56 VTop 2
SYMATTR InstName R1
SYMATTR Value 10K
SYMBOL res -1072 -1312 R0
SYMATTR InstName R5
SYMATTR Value .1
SYMBOL voltage -1056 -1120 R0
WINDOW 0 14 102 Left 2
WINDOW 3 24 96 Invisible 2
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName V1
SYMATTR Value 12
SYMBOL Opamps\\LT1366 -768 -1248 R0
SYMATTR InstName U1A
SYMBOL current -1264 -1104 R0
WINDOW 3 24 80 Invisible 2
WINDOW 123 0 0 Left 2
WINDOW 39 0 0 Left 2
SYMATTR InstName I1
SYMATTR Value PULSE(0 3 0 1 1 1U 2)
SYMBOL Opamps\\LT1366 -320 -1264 R0
SYMATTR InstName U1B
SYMBOL References\\LT1634-5 -608 -1040 R0
WINDOW 3 -96 40 Left 2
WINDOW 0 -64 -1 Left 2
SYMATTR InstName U2
SYMBOL res -464 -1152 R90
WINDOW 0 63 58 VBottom 2
WINDOW 3 67 57 VTop 2
SYMATTR InstName R7
SYMATTR Value 9100
SYMBOL res -448 -1104 R0
WINDOW 3 41 70 Left 2
SYMATTR InstName R8
SYMATTR Value 1000
SYMBOL res -624 -1312 R0
SYMATTR InstName R6
SYMATTR Value 6800
TEXT -1020 -936 Left 2 !.tran 3
TEXT -1088 -792 Left 4 ;CURRENT THRESHOLD DETECTOR
TEXT -416 -872 Left 2 ;J FIELDS 01 JULY 2015
TEXT -1344 -872 Left 3 ;BOARD B
TEXT -1104 -872 Left 3 ;BOARD A


John Fields

 

[Phil Hobbs]

On 6/30/2015 11:09 PM, scott.a.mayo@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?

The resistor only needs to drop enough voltage for the comparator to
detect reliably. That's basically just a few times the worst-case
offset over temperature--30 mV would be more than enough for most
comparators. So something like 8 milliohms would let you detect the
presence or absence of a 4A load current. You can get resistors of that
size, or just use a copper trace. One-ounce copper has a sheet
resistance of 0.5 milliohm per square, so if you're using 100-mil traces
for the high current path, 8 milliohms would be about 1.6 inches of
trace. The resistance of pure metals goes up fairly steeply with
temperature, so you might need a slightly better comparator to compensate.

Assuming that's feasible, what sort of comparator do I use?

Since you don't need any speed, an LM339 / LM193 / TL331 style would be
a typical choice--super cheap, widely available, and easy to use. Their
offset is specified at 9 mV max over temperature, so a 15-mV threshold
should do. For a bit more comfort margin, use three inches of trace and
a 30-mV threshold.

Cheers

Phil Hobbs


--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC
Optics, Electro-optics, Photonics, Analog Electronics

160 North State Road #203
Briarcliff Manor NY 10510

hobbs at electrooptical dot net
http://electrooptical.net

 

[George Herold]

On Wednesday, July 1, 2015 at 1:17:18 AM UTC-4, Phil Allison wrote:

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

That's a neat idea Phil. Would it work with the wire just wrapped
around the body of the reed switch? It been decades since I use a reed
switch, but don't you want the magnetic field perpendicular to the contact.

(Googling pictures of reed relays... apparently not.)

OK how the "bleep" does a reed relay work, with the field along it's axis?

George H.

 

[John Fields]

On Wed, 1 Jul 2015 05:27:38 -0700 (PDT), scott.a.mayo@gmail.com
wrote:

On Wednesday, July 1, 2015 at 6:39:31 AM UTC-4, John Fields wrote:
...

Very much appreciated.

I've had bad luck with reed switches; will a hall effect sensor do the trick? twenty turns of thick wire wrapped around an iron rod and pointed at a hall sensor sounds less fragile and I feel like I'd be able to tune it just by setting the distance between the sensor and the coil, or fiddling with the rod.

I'll also look at the op amp approach; presumably also easy to tune. Which application is used to render that ascii list of components into an image?

---
Not just an image, a circuit you can simulate.

LTspice, available free, from:

http://www.linear.com/designtools/software/

John Fields

 

[Baron]

Phil Allison prodded the keyboard with:

** 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

I used to use that technique to identify lamp failure on motor
vehicles. If a lamp failed then a led went out on the indicator
panel. I used 14 turns on mine.

--
Best Regards:
Baron.

 

[George Herold]

On Wednesday, July 1, 2015 at 2:49:37 PM UTC-4, Baron wrote:

George Herold prodded the keyboard with:

On Wednesday, July 1, 2015 at 1:17:18 AM UTC-4, Phil Allison wrote:
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

That's a neat idea Phil. Would it work with the wire just wrapped
around the body of the reed switch? It been decades since I use a
reed switch, but don't you want the magnetic field perpendicular to
the contact.

(Googling pictures of reed relays... apparently not.)

OK how the "bleep" does a reed relay work, with the field along it's
axis?

George H.

Both reed blades turn into magnets that attract each other.

One of the things that I found was that too big a magnetic field
caused the reeds to stick together more or less permenamtly.

--
Best Regards:
Baron.

Ahh so some kind of soft iron... that makes sense. I was thinking about a permenant magnet, but then that could sometimes go the wrong way.

If they stick together, just raise them above the Curie temperature..:^)

George H.
(oh or the demagnetization trick... Variac and a coil.)

 

[Baron]

George Herold prodded the keyboard with:

On Wednesday, July 1, 2015 at 1:17:18 AM UTC-4, Phil Allison wrote:
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

That's a neat idea Phil. Would it work with the wire just wrapped
around the body of the reed switch? It been decades since I use a
reed switch, but don't you want the magnetic field perpendicular to
the contact.

(Googling pictures of reed relays... apparently not.)

OK how the "bleep" does a reed relay work, with the field along it's
axis?

George H.

Both reed blades turn into magnets that attract each other.

One of the things that I found was that too big a magnetic field
caused the reeds to stick together more or less permenamtly.

--
Best Regards:
Baron.

 

[Tim Wescott]

On Wed, 01 Jul 2015 05:27:38 -0700, scott.a.mayo wrote:

On Wednesday, July 1, 2015 at 6:39:31 AM UTC-4, John Fields wrote:
...

Very much appreciated.

I've had bad luck with reed switches; will a hall effect sensor do the
trick? twenty turns of thick wire wrapped around an iron rod and pointed
at a hall sensor sounds less fragile and I feel like I'd be able to tune
it just by setting the distance between the sensor and the coil, or
fiddling with the rod.

I'll also look at the op amp approach; presumably also easy to tune.
Which application is used to render that ascii list of components into
an image?

Usenet saves the day again. I don't know why we bothered with this "web"
thing.

There are Hall current sensors out there all in one package, no need to
fiddle with coils.

--
www.wescottdesign.com