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| lektric.dan at (no spam) gmail.com... |
 Posted: Thu Nov 05, 2009 6:33 pm |
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Guest
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On Nov 5, 8:20 am, Spehro Pefhany <speffS... at (no spam) interlogDOTyou.knowwhat>
wrote:
[quote]
The elements glowing red are their own indicator.
[/quote]
Well....no. The elements should never glow as they are 1) wrapped
under several layers of insulation, 2) are mounted inside a heating
jacket, or 3) inside process equipment where you can't open the door
without seriously messing the temperature up. Kinda like tasting it
to see if you spilled table salt or arsnic. |
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| lektric.dan at (no spam) gmail.com... |
| Posted: Thu Nov 05, 2009 6:39 pm |
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Guest
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On Nov 5, 8:27 pm, "Paul Hovnanian P.E." <p... at (no spam) hovnanian.com> wrote:
[quote]"lektric.... at (no spam) gmail.com" wrote:
Connect a CT and ac ammeter to the incomming supply for each circuit.
Note the full load (all elements good) current draw. If it drops, one or
more elements are open.
The loss of one of 6 elements will result in a 17% current drop which is
more than the change due to normal utility voltage swings.
I was working with the electricians today. Due to the current load,[/quote]
there's not a single point where you can measure the total current
draw. We're drawing 75A for the one multi-element design (using total
current from mfgr's specs.) so we have to feed from multiple
circuits. I looks like the best way is to use a CT on each element
and an indicator of some kind. This will add about $100-$150 to the
cost of the system. I'll lay it out for the boss and see how
important he thinks it is. |
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| Paul Hovnanian P.E.... |
| Posted: Thu Nov 05, 2009 9:27 pm |
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Guest
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"lektric.dan at (no spam) gmail.com" wrote:
[quote]
I've got a series of heating elements and I need to be able to tell
when one burns out. There are two situations, but most of the circuit
is the same. They both use a Watlow (brand) controller and a solid
state relay to control the power. In one circuit, I have a single
element, in the second circuit, I have six elements in parallel. I've
got neon lamps hooked up so I can tell when the circuit is getting
mains (AC) power, and when there is power going to the elements. So,
I need to be able to tell if a single elements goes out, either a
single or one in parallel. Any ideas?
[/quote]
Connect a CT and ac ammeter to the incomming supply for each circuit.
Note the full load (all elements good) current draw. If it drops, one or
more elements are open.
The loss of one of 6 elements will result in a 17% current drop which is
more than the change due to normal utility voltage swings.
--
Paul Hovnanian mailto:Paul at (no spam) Hovnanian.com
------------------------------------------------------------------
Opinions stated herein are the sole property of the author. Standard
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inside. Contents under pressure; do not incinerate. Always wear adequate
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| Ross Herbert... |
| Posted: Fri Nov 06, 2009 12:25 am |
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Guest
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On Tue, 3 Nov 2009 10:16:50 -0800 (PST), "lektric.dan at (no spam) gmail.com"
<lektric.dan at (no spam) gmail.com> wrote:
:I've got a series of heating elements and I need to be able to tell
:when one burns out. There are two situations, but most of the circuit
:is the same. They both use a Watlow (brand) controller and a solid
:state relay to control the power. In one circuit, I have a single
:element, in the second circuit, I have six elements in parallel. I've
:got neon lamps hooked up so I can tell when the circuit is getting
:mains (AC) power, and when there is power going to the elements. So,
:I need to be able to tell if a single elements goes out, either a
:single or one in parallel. Any ideas?
I assume you are operating the heaters on AC mains voltage....?
The parallel combination would best be suited to 3 phase operation and you could
use a detection circuit based on US patent 4.496,940
http://www.freepatentsonline.com/4496940.pdf to indicate which element had
failed. |
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| Michael A. Terrell... |
| Posted: Fri Nov 06, 2009 12:50 am |
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Guest
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"lektric.dan at (no spam) gmail.com" wrote:
[quote]
On Nov 5, 12:04 am, ehsjr <eh... at (no spam) nospamverizon.net> wrote:
lektric.... at (no spam) gmail.com wrote:
(post snipped here)
I hate to re-post to replies, but I think I have an idea that might
work. I can use a current transformer to drive a low-power relay,
possibly a reed relay. A neon lamp in parallel to the heating element
would go through the NC relay. When current flows through the
element, current/voltage is produced in the current transformer,
activating the relay, and keeping the light from coming on. If the
element burns out, the relay stays closed and the neon lamp will
light. Any problems w/this?
Maybe. What happens if the element is ok, but the control circuit
fails? For example, a burned point on the contactor or a blown fuse
could interrupt power to the element. The neon can't light in those
cases, so you won't know that there is a failure.
Assuming that is not a show stopper, you still may have a little
more complexity with the relay idea. You likely need to add a
diode and filter cap to power the reed relays, and your expense
will be higher for the relay approach.
Ed
What I have now, and it will probably be used for the multi-element
circuit, is a 20A breaker feeding a Watlow (brand) controller, a green
neon lamp, and a solid state relay in parallel. When the breaker is
on 1) the green neon lamp comes on indicating power is available to
the circuit, 2) the controller lights up and goes through power-up
sequence then indicates set point and actual temp (and other stuff).
When heat is called for, a low voltage DC signal is sent from the
controller to the SSR. The SSR turns on, sending current to the
heating element and a red neon lamp in parallel with the element -
this indicates the element is getting power (heat is being called
for). There really aren't any contacts to burn out, or fuses to
blow. *I* can tell when an element blows out because heat is being
called for (red neon is on) and the temp is not going up. We're not
playing around here; we're pushing the system to over 500 degrees C so
we know when we're not getting heat. The boss wants a way to tell
immediatly if an element burns out. This isn't really critical
because the mass being heated is large and has "heat inertia" and
won't cool off quickly. Knowing a blown element in the multi-element
application is important though, because I'm using a single controller
and temp pickup, but controlling six elements instead of one.
[/quote]
Is it possible to add another heating element or two? If so, you can
switch on a spare, until you can shout it down for repairs. Just turn
off the control line to the relay on the failed heater, and turn on a
spare.
--
The movie 'Deliverance' isn't a documentary! |
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| ehsjr... |
| Posted: Fri Nov 06, 2009 1:31 am |
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Guest
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lektric.dan at (no spam) gmail.com wrote:
[quote]On Nov 5, 12:04 am, ehsjr <eh... at (no spam) nospamverizon.net> wrote:
lektric.... at (no spam) gmail.com wrote:
(post snipped here)
I hate to re-post to replies, but I think I have an idea that might
work. I can use a current transformer to drive a low-power relay,
possibly a reed relay. A neon lamp in parallel to the heating element
would go through the NC relay. When current flows through the
element, current/voltage is produced in the current transformer,
activating the relay, and keeping the light from coming on. If the
element burns out, the relay stays closed and the neon lamp will
light. Any problems w/this?
Maybe. What happens if the element is ok, but the control circuit
fails? For example, a burned point on the contactor or a blown fuse
could interrupt power to the element. The neon can't light in those
cases, so you won't know that there is a failure.
Assuming that is not a show stopper, you still may have a little
more complexity with the relay idea. You likely need to add a
diode and filter cap to power the reed relays, and your expense
will be higher for the relay approach.
Ed
What I have now, and it will probably be used for the multi-element
circuit, is a 20A breaker feeding a Watlow (brand) controller, a green
neon lamp, and a solid state relay in parallel. When the breaker is
on 1) the green neon lamp comes on indicating power is available to
the circuit, 2) the controller lights up and goes through power-up
sequence then indicates set point and actual temp (and other stuff).
When heat is called for, a low voltage DC signal is sent from the
controller to the SSR. The SSR turns on, sending current to the
heating element and a red neon lamp in parallel with the element -
this indicates the element is getting power (heat is being called
for). There really aren't any contacts to burn out, or fuses to
blow. *I* can tell when an element blows out because heat is being
called for (red neon is on) and the temp is not going up. We're not
playing around here; we're pushing the system to over 500 degrees C so
we know when we're not getting heat.
[/quote]
[quote]The boss wants a way to tell
immediatly if an element burns out.
[/quote]
What I am missing is why it MUST be
light _lit_ = element failed.
Why can't it be light _not_ lit = element failed?
Maybe in your situation it's easier to see a light
that is lit versus seeing that one of the lights is
not lit.
Whatever way you go (light off vs light on) to indicate
failure, it still starts with a ct and burden resistor
to sense the failure. Using series resistors to sense
the open element introduces I^2R heat and large area
for dissipation when the element is working; knock the
ohms down to reduce dissipation and the sense voltage is
too low to energize a relay as you mentioned.
[quote]This isn't really critical
because the mass being heated is large and has "heat inertia" and
won't cool off quickly. Knowing a blown element in the multi-element
application is important though, because I'm using a single controller
and temp pickup, but controlling six elements instead of one.
[/quote]
Yes, that's easy to understand. BTDT. If you have to go the
more complex route, you might want to consider adding an
audible alarm: any element failure light lit causes the thing
to sound. You could add it to the other method (light off =
failure) but, because it adds complexity, you might as well
go with the light on = failure method.
Ed |
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| Tim Shoppa... |
| Posted: Fri Nov 06, 2009 10:21 am |
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Guest
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On Nov 5, 11:27 pm, "lektric.... at (no spam) gmail.com" <lektric.... at (no spam) gmail.com>
wrote:
[quote]On Nov 5, 7:53 am, TimShoppa<sho... at (no spam) trailing-edge.com> wrote:
What's the current through each element? I'm going to guess 2 amps. If
it's different, change the math below.
I'm sorry, I haven't explained all the details, so please forgive me
when I don't think your idea is practical. We're drawing 11 amps on
the "little" elements, and close to 50 on the big ones. I should have
gone 220 one the big elements. Our "really big" elements in a total
of 3 per application draw about 10KW but run at 440 3 phase.
Your idea *is* useful for lower power applications. I've used
something similar for a launch controller for hobby rockets (mid/high
power).
[/quote]
I think it's just as practical at the higher currents - the shunt
resistors become smaller and smaller as the currents go up. You just
need enough drop across each shunt to light up a flashlight bulb (1.2V
say, although I think some grain-of-wheat bulbs are only 0.6V). At
some point the shunt resistor becomes similar in resistance to a fuse
and that's a good thing.
Tim. |
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| langwadt at (no spam) fonz.dk... |
| Posted: Fri Nov 06, 2009 11:35 am |
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Guest
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On 6 Nov., 05:20, "lektric.... at (no spam) gmail.com" <lektric.... at (no spam) gmail.com>
wrote:
[quote]On Nov 5, 12:04 am, ehsjr <eh... at (no spam) nospamverizon.net> wrote:
lektric.... at (no spam) gmail.com wrote:
(post snipped here)
I hate to re-post to replies, but I think I have an idea that might
work. I can use a current transformer to drive a low-power relay,
possibly a reed relay. A neon lamp in parallel to the heating element
would go through the NC relay. When current flows through the
element, current/voltage is produced in the current transformer,
activating the relay, and keeping the light from coming on. If the
element burns out, the relay stays closed and the neon lamp will
light. Any problems w/this?
Maybe. What happens if the element is ok, but the control circuit
fails? For example, a burned point on the contactor or a blown fuse
could interrupt power to the element. The neon can't light in those
cases, so you won't know that there is a failure.
Assuming that is not a show stopper, you still may have a little
more complexity with the relay idea. You likely need to add a
diode and filter cap to power the reed relays, and your expense
will be higher for the relay approach.
Ed
What I have now, and it will probably be used for the multi-element
circuit, is a 20A breaker feeding a Watlow (brand) controller, a green
neon lamp, and a solid state relay in parallel. When the breaker is
on 1) the green neon lamp comes on indicating power is available to
the circuit, 2) the controller lights up and goes through power-up
sequence then indicates set point and actual temp (and other stuff).
When heat is called for, a low voltage DC signal is sent from the
controller to the SSR. The SSR turns on, sending current to the
heating element and a red neon lamp in parallel with the element -
this indicates the element is getting power (heat is being called
for). There really aren't any contacts to burn out, or fuses to
blow. *I* can tell when an element blows out because heat is being
called for (red neon is on) and the temp is not going up. We're not
playing around here; we're pushing the system to over 500 degrees C so
we know when we're not getting heat. The boss wants a way to tell
immediatly if an element burns out. This isn't really critical
because the mass being heated is large and has "heat inertia" and
won't cool off quickly. Knowing a blown element in the multi-element
application is important though, because I'm using a single controller
and temp pickup, but controlling six elements instead of one.
[/quote]
depending on how fancy you want it:
http://www.tempco.com/Instrumentation/current_indicators.htm
or
http://www.veris.com/pdf/cs/proving/hp/pv-digital/h748_i0b.pdf
-Lasse |
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| Tim Shoppa... |
| Posted: Fri Nov 06, 2009 11:36 am |
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Guest
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On Nov 6, 3:42 pm, Spehro Pefhany <speffS... at (no spam) interlogDOTyou.knowwhat>
wrote:
[quote]On Fri, 6 Nov 2009 12:21:29 -0800 (PST), TimShoppa
sho... at (no spam) trailing-edge.com> wrote:
On Nov 5, 11:27 pm, "lektric.... at (no spam) gmail.com" <lektric.... at (no spam) gmail.com
wrote:
On Nov 5, 7:53 am, TimShoppa<sho... at (no spam) trailing-edge.com> wrote:
What's the current through each element? I'm going to guess 2 amps. If
it's different, change the math below.
I'm sorry, I haven't explained all the details, so please forgive me
when I don't think your idea is practical. We're drawing 11 amps on
the "little" elements, and close to 50 on the big ones. I should have
gone 220 one the big elements. Our "really big" elements in a total
of 3 per application draw about 10KW but run at 440 3 phase.
Your idea *is* useful for lower power applications. I've used
something similar for a launch controller for hobby rockets (mid/high
power).
I think it's just as practical at the higher currents - the shunt
resistors become smaller and smaller as the currents go up. You just
need enough drop across each shunt to light up a flashlight bulb (1.2V
say, although I think some grain-of-wheat bulbs are only 0.6V). At
some point the shunt resistor becomes similar in resistance to a fuse
and that's a good thing.
Tim.
50A at 1.2V is 60W disspation, so it's not going to be a (physically)
small resistor.
[/quote]
Last time I did this I found that fuses designed for 13.8kV and 30kV
work happen to have circa 0.5 to 1.5 V drop across them at the rated
current.
You're right, not physically small at all, they're enormous fuses. And
they aren't even linear resistors (because they heat up a good amount
at rated current). But for "current/no-current" it worked OK.
Tim. |
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| Spehro Pefhany... |
| Posted: Fri Nov 06, 2009 3:42 pm |
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Guest
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On Fri, 6 Nov 2009 12:21:29 -0800 (PST), Tim Shoppa
<shoppa at (no spam) trailing-edge.com> wrote:
[quote]On Nov 5, 11:27 pm, "lektric.... at (no spam) gmail.com" <lektric.... at (no spam) gmail.com
wrote:
On Nov 5, 7:53 am, TimShoppa<sho... at (no spam) trailing-edge.com> wrote:
What's the current through each element? I'm going to guess 2 amps. If
it's different, change the math below.
I'm sorry, I haven't explained all the details, so please forgive me
when I don't think your idea is practical. We're drawing 11 amps on
the "little" elements, and close to 50 on the big ones. I should have
gone 220 one the big elements. Our "really big" elements in a total
of 3 per application draw about 10KW but run at 440 3 phase.
Your idea *is* useful for lower power applications. I've used
something similar for a launch controller for hobby rockets (mid/high
power).
I think it's just as practical at the higher currents - the shunt
resistors become smaller and smaller as the currents go up. You just
need enough drop across each shunt to light up a flashlight bulb (1.2V
say, although I think some grain-of-wheat bulbs are only 0.6V). At
some point the shunt resistor becomes similar in resistance to a fuse
and that's a good thing.
Tim.
[/quote]
50A at 1.2V is 60W disspation, so it's not going to be a (physically)
small resistor. |
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| Jan Panteltje... |
| Posted: Fri Nov 06, 2009 6:06 pm |
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Guest
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On a sunny day (Fri, 6 Nov 2009 13:36:30 -0800 (PST)) it happened Tim Shoppa
<shoppa at (no spam) trailing-edge.com> wrote in
<2ba24dee-7bfd-44a4-b884-9772cac6f116 at (no spam) v25g2000yqk.googlegroups.com>:
[quote]On Nov 6, 3:42 pm, Spehro Pefhany <speffS... at (no spam) interlogDOTyou.knowwhat
wrote:
On Fri, 6 Nov 2009 12:21:29 -0800 (PST), TimShoppa
sho... at (no spam) trailing-edge.com> wrote:
On Nov 5, 11:27 pm, "lektric.... at (no spam) gmail.com" <lektric.... at (no spam) gmail.com
wrote:
On Nov 5, 7:53 am, TimShoppa<sho... at (no spam) trailing-edge.com> wrote:
What's the current through each element? I'm going to guess 2 amps. =
If
it's different, change the math below.
I'm sorry, I haven't explained all the details, so please forgive me
when I don't think your idea is practical. We're drawing 11 amps on
the "little" elements, and close to 50 on the big ones. I should ha=
ve
gone 220 one the big elements. Our "really big" elements in a total
of 3 per application draw about 10KW but run at 440 3 phase.
Your idea *is* useful for lower power applications. I've used
something similar for a launch controller for hobby rockets (mid/high
power).
I think it's just as practical at the higher currents - the shunt
resistors become smaller and smaller as the currents go up. You just
need enough drop across each shunt to light up a flashlight bulb (1.2V
say, although I think some grain-of-wheat bulbs are only 0.6V). At
some point the shunt resistor becomes similar in resistance to a fuse
and that's a good thing.
Tim.
50A at 1.2V is 60W disspation, so it's not going to be a (physically)
small resistor.
Last time I did this I found that fuses designed for 13.8kV and 30kV
work happen to have circa 0.5 to 1.5 V drop across them at the rated
current.
You're right, not physically small at all, they're enormous fuses. And
they aren't even linear resistors (because they heat up a good amount
at rated current). But for "current/no-current" it worked OK.
Tim.
[/quote]
It is all very simple,
just put old tape recorder playback heads next to the power carrying leads.
ftp://panteltje.com/pub/play_back_head_current_sensor_img_1153.jpg
Use some opamps to amplify and detect the 50 Hz or 60 Hz signal.
A PIC can then collect the data and control LED indicators,
or send serial info to a PC, or over the internet for example.
No power losses, no expensive current transformers.
The playback heads only need to be close to the wires, no need to strip
isolation or even interrupt the circuit, or change any wiring.
Cheap too. |
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| Paul_P... |
| Posted: Fri Nov 06, 2009 11:24 pm |
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Guest
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[quote]Is it possible to add another heating element or two? If so, you can
switch on a spare, until you can shout it down for repairs. Just turn
off the control line to the relay on the failed heater, and turn on a
spare.
[/quote]
Add a back up or secondary controller that kick in when more heat is needed
due to an element failure. And sound an alarm or indicator.
-or-
Can you crimp a durable heat tolerant tap to the resistive element as a
voltage divider circuit and drive a voltage sense circuit. There should be
a way. They had to attach a lead wire or connector to the beginning and
end.
[ tap element length / total element length ] times applied voltage = sample
voltage
If the tap opens you see applied volts. If the element opens any were else
you would see zero volts.
Paul P. |
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| Dave M... |
| Posted: Sat Nov 07, 2009 8:38 am |
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Guest
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On Fri, 06 Nov 2009 23:06:43 GMT, Jan Panteltje
<pNaonStpealmtje at (no spam) yahoo.com> wrote:
[quote]On a sunny day (Fri, 6 Nov 2009 13:36:30 -0800 (PST)) it happened Tim Shoppa
shoppa at (no spam) trailing-edge.com> wrote in
2ba24dee-7bfd-44a4-b884-9772cac6f116 at (no spam) v25g2000yqk.googlegroups.com>:
On Nov 6, 3:42 pm, Spehro Pefhany <speffS... at (no spam) interlogDOTyou.knowwhat
wrote:
On Fri, 6 Nov 2009 12:21:29 -0800 (PST), TimShoppa
sho... at (no spam) trailing-edge.com> wrote:
On Nov 5, 11:27 pm, "lektric.... at (no spam) gmail.com" <lektric.... at (no spam) gmail.com
wrote:
On Nov 5, 7:53 am, TimShoppa<sho... at (no spam) trailing-edge.com> wrote:
What's the current through each element? I'm going to guess 2 amps. =
If
it's different, change the math below.
I'm sorry, I haven't explained all the details, so please forgive me
when I don't think your idea is practical. We're drawing 11 amps on
the "little" elements, and close to 50 on the big ones. I should ha=
ve
gone 220 one the big elements. Our "really big" elements in a total
of 3 per application draw about 10KW but run at 440 3 phase.
Your idea *is* useful for lower power applications. I've used
something similar for a launch controller for hobby rockets (mid/high
power).
I think it's just as practical at the higher currents - the shunt
resistors become smaller and smaller as the currents go up. You just
need enough drop across each shunt to light up a flashlight bulb (1.2V
say, although I think some grain-of-wheat bulbs are only 0.6V). At
some point the shunt resistor becomes similar in resistance to a fuse
and that's a good thing.
Tim.
50A at 1.2V is 60W disspation, so it's not going to be a (physically)
small resistor.
Last time I did this I found that fuses designed for 13.8kV and 30kV
work happen to have circa 0.5 to 1.5 V drop across them at the rated
current.
You're right, not physically small at all, they're enormous fuses. And
they aren't even linear resistors (because they heat up a good amount
at rated current). But for "current/no-current" it worked OK.
Tim.
It is all very simple,
just put old tape recorder playback heads next to the power carrying leads.
ftp://panteltje.com/pub/play_back_head_current_sensor_img_1153.jpg
Use some opamps to amplify and detect the 50 Hz or 60 Hz signal.
A PIC can then collect the data and control LED indicators,
or send serial info to a PC, or over the internet for example.
No power losses, no expensive current transformers.
The playback heads only need to be close to the wires, no need to strip
isolation or even interrupt the circuit, or change any wiring.
Cheap too.
[/quote]
The mention of using a playback head to sense current reminded me of a
similar solution at
http://www.discovercircuits.com/circuit-solutions/Pump-Motor-Monitor.html.
This approach uses a common inductor as the sensor, which should be
considerably cheaper than a tape head if you need to buy it.
You could make the detection circuit a bit simpler by using a quad
comparator IC such as the LM339. Sensitivity could be made adjustable
to handle different current levels.
David
masondg44 at comcast dot net |
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| Jan Panteltje... |
| Posted: Sat Nov 07, 2009 9:34 am |
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Guest
|
On a sunny day (Sat, 07 Nov 2009 08:38:49 -0500) it happened Dave M
<masondg4499 at (no spam) comcast.net> wrote in
<l0uaf55drljlg6gconmv7q0ddnuod1no0o at (no spam) 4ax.com>:
[quote]It is all very simple,
just put old tape recorder playback heads next to the power carrying leads.
ftp://panteltje.com/pub/play_back_head_current_sensor_img_1153.jpg
Use some opamps to amplify and detect the 50 Hz or 60 Hz signal.
A PIC can then collect the data and control LED indicators,
or send serial info to a PC, or over the internet for example.
No power losses, no expensive current transformers.
The playback heads only need to be close to the wires, no need to strip
isolation or even interrupt the circuit, or change any wiring.
Cheap too.
The mention of using a playback head to sense current reminded me of a
similar solution at
http://www.discovercircuits.com/circuit-solutions/Pump-Motor-Monitor.html.
This approach uses a common inductor as the sensor, which should be
considerably cheaper than a tape head if you need to buy it.
You could make the detection circuit a bit simpler by using a quad
comparator IC such as the LM339. Sensitivity could be made adjustable
to handle different current levels.
David
masondg44 at comcast dot net
[/quote]
Yes, that method should also work.
The playback-heads are so small that you can just glue those against the wire.
After all the talk here about glue, finding a suitable glue should be easy  .
Maybe if you have more then one, as each head has about 10 cm of shielded wire,
connect those to one big shielded multi core cable, and do the electronics is a different more convenient place.
After all the talk here about soldering that should be easy too .
As to the cost, I think if you shop around, less then a dollar a piece, for the playback-heads,
but that may well go up over time as that article is getting scarce...
Strip an old walkman perhaps.
Somebody here in this group mentioned he had a whole lot of those in a box, so maybe he wants to sell some?
I use a 1/4 LM324 on 5 V as preamp, and 1/4 as peak detector, so 1 chip can do 2 PB heads.
Since the output is an analog voltage, I can actually measure current, and use it to monitor
the electric heater here, the PC calculates the kWh and electricity cost from that:
ftp://panteltje.com/pub/xhcs.jpg
System has been working OK for many years... |
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| Baron... |
| Posted: Sat Nov 07, 2009 10:56 am |
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Guest
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Jan Panteltje wrote:
[quote]On a sunny day (Sat, 07 Nov 2009 08:38:49 -0500) it happened Dave M
masondg4499 at (no spam) comcast.net> wrote in
l0uaf55drljlg6gconmv7q0ddnuod1no0o at (no spam) 4ax.com>:
It is all very simple,
just put old tape recorder playback heads next to the power carrying
leads.
ftp://panteltje.com/pub/play_back_head_current_sensor_img_1153.jpg
Use some opamps to amplify and detect the 50 Hz or 60 Hz signal.
A PIC can then collect the data and control LED indicators,
or send serial info to a PC, or over the internet for example.
No power losses, no expensive current transformers.
The playback heads only need to be close to the wires, no need to
strip isolation or even interrupt the circuit, or change any wiring.
Cheap too.
The mention of using a playback head to sense current reminded me of a
similar solution at
http://www.discovercircuits.com/circuit-solutions/Pump-Motor-Monitor.html.
This approach uses a common inductor as the sensor, which should be
considerably cheaper than a tape head if you need to buy it.
You could make the detection circuit a bit simpler by using a quad
comparator IC such as the LM339. Sensitivity could be made adjustable
to handle different current levels.
David
masondg44 at comcast dot net
Yes, that method should also work.
The playback-heads are so small that you can just glue those against
the wire. After all the talk here about glue, finding a suitable glue
should be easy . Maybe if you have more then one, as each head has
about 10 cm of shielded wire, connect those to one big shielded multi
core cable, and do the electronics is a different more convenient
place. After all the talk here about soldering that should be easy too
. As to the cost, I think if you shop around, less then a dollar a
piece, for the playback-heads, but that may well go up over time as
that article is getting scarce... Strip an old walkman perhaps.
Somebody here in this group mentioned he had a whole lot of those in a
box, so maybe he wants to sell some? I use a 1/4 LM324 on 5 V as
preamp, and 1/4 as peak detector, so 1 chip can do 2 PB heads. Since
the output is an analog voltage, I can actually measure current, and
use it to monitor the electric heater here, the PC calculates the kWh
and electricity cost from that:
ftp://panteltje.com/pub/xhcs.jpg
System has been working OK for many years...
[/quote]
I've a few if needed.
--
Best Regards:
Baron. |
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