lektric....@gmail.com wrote: > On Nov 3, 11:48 pm, ehsjr <eh...@nospamverizon.net> wrote:
>>I had the same situation. CTs, as others have mentioned (with >>a burden resistor, a current limiting resistor and a LED for >>each CT) were the correct solution. A glance tells you instantly >>if all elements are heating, or, if not, which element(s) is/are not >>heating, whether due to an open element, blown fuse, failed >>contactor, whatever. Mine lit the LEDs when working properly, >>so it required no power source - the CT's powered the LEDs. Yours >>will require power to run the inverter circuits, if you must >>have led on = failure. Simpler to have led on = working.
>>Ed
> 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.
On Nov 3, 1:16 pm, "lektric....@gmail.com" <lektric....@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?
What's the current through each element? I'm going to guess 2 amps. If it's different, change the math below.
Take a miniature flashlight bulb, 1.5V. To get 1.5V across a resistor with 2 amps, you need 0.75 ohms. Put a 0.75 ohm resistor in parallel with the miniature flashlight bulb. Then put this in series with each heating element.
When the heating element is on, lamp is on. If it goes open circuit, lamp goes out. If it goes short circuit, you burn up that 0.75 ohm resistor real fast. In fact in some schemes the 0.75 ohm resistor *is* a fuse.
I did not invent this scheme.... it is identical to that used half a century ago in electric ovens. Because the lamp socket can be hot with AC line voltage I think maybe they stopped using it. Or maybe elements became reliable enough that front-of-oven indicators weren't really necessary anymore.
<sho...@trailing-edge.com> wrote: >On Nov 3, 1:16 pm, "lektric....@gmail.com" <lektric....@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?
>What's the current through each element? I'm going to guess 2 amps. If >it's different, change the math below.
>Take a miniature flashlight bulb, 1.5V. To get 1.5V across a resistor >with 2 amps, you need 0.75 ohms. Put a 0.75 ohm resistor in parallel >with the miniature flashlight bulb. Then put this in series with each >heating element.
>When the heating element is on, lamp is on. If it goes open circuit, >lamp goes out. If it goes short circuit, you burn up that 0.75 ohm >resistor real fast. In fact in some schemes the 0.75 ohm resistor *is* >a fuse.
>I did not invent this scheme.... it is identical to that used half a >century ago in electric ovens. Because the lamp socket can be hot with >AC line voltage I think maybe they stopped using it. Or maybe elements >became reliable enough that front-of-oven indicators weren't really >necessary anymore.
> 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?
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:P...@Hovnanian.com ------------------------------------------------------------------ Opinions stated herein are the sole property of the author. Standard disclaimers apply. All rights reserved. No user serviceable components inside. Contents under pressure; do not incinerate. Always wear adequate eye protection. Do not mold, findle or sputilate.
> > 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.
On Nov 5, 7:53 am, Tim Shoppa <sho...@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).
On Nov 5, 8:20 am, Spehro Pefhany <speffS...@interlogDOTyou.knowwhat> wrote:
> The elements glowing red are their own indicator.
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.
> 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, 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.
: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.
> On Nov 5, 12:04 am, ehsjr <eh...@nospamverizon.net> wrote: > > lektric....@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.
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.
>>>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.
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.
> 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.
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.
On Nov 5, 11:27 pm, "lektric....@gmail.com" <lektric....@gmail.com> wrote:
> On Nov 5, 7:53 am, TimShoppa<sho...@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.
<sho...@trailing-edge.com> wrote: >On Nov 5, 11:27 pm, "lektric....@gmail.com" <lektric....@gmail.com> >wrote: >> On Nov 5, 7:53 am, TimShoppa<sho...@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.
> > > 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.
> On Fri, 6 Nov 2009 12:21:29 -0800 (PST), TimShoppa
> <sho...@trailing-edge.com> wrote: > >On Nov 5, 11:27 pm, "lektric....@gmail.com" <lektric....@gmail.com> > >wrote: > >> On Nov 5, 7:53 am, TimShoppa<sho...@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.
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.
On a sunny day (Fri, 6 Nov 2009 13:36:30 -0800 (PST)) it happened Tim Shoppa <sho...@trailing-edge.com> wrote in <2ba24dee-7bfd-44a4-b884-9772cac6f...@v25g2000yqk.googlegroups.com>:
>On Nov 6, 3:42 pm, Spehro Pefhany <speffS...@interlogDOTyou.knowwhat> >wrote: >> On Fri, 6 Nov 2009 12:21:29 -0800 (PST), TimShoppa
>> <sho...@trailing-edge.com> wrote: >> >On Nov 5, 11:27 pm, "lektric....@gmail.com" <lektric....@gmail.com> >> >wrote: >> >> On Nov 5, 7:53 am, TimShoppa<sho...@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.
> 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.
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.
<pNaonStpealm...@yahoo.com> wrote: >On a sunny day (Fri, 6 Nov 2009 13:36:30 -0800 (PST)) it happened Tim Shoppa ><sho...@trailing-edge.com> wrote in ><2ba24dee-7bfd-44a4-b884-9772cac6f...@v25g2000yqk.googlegroups.com>:
>>On Nov 6, 3:42 pm, Spehro Pefhany <speffS...@interlogDOTyou.knowwhat> >>wrote: >>> On Fri, 6 Nov 2009 12:21:29 -0800 (PST), TimShoppa
>>> <sho...@trailing-edge.com> wrote: >>> >On Nov 5, 11:27 pm, "lektric....@gmail.com" <lektric....@gmail.com> >>> >wrote: >>> >> On Nov 5, 7:53 am, TimShoppa<sho...@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.
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..... 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.
>>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..... >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...
Jan Panteltje wrote: > On a sunny day (Sat, 07 Nov 2009 08:38:49 -0500) it happened Dave M > <masondg4...@comcast.net> wrote in > <l0uaf55drljlg6gconmv7q0ddnuod1n...@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..... >>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...
On Nov 3, 2:05 pm, "lektric....@gmail.com" <lektric....@gmail.com> wrote:
> On Nov 3, 2:05 pm, "amdx" <a...@knology.net> wrote:
> > Don't miss Spehro's point, he's suggesting putting a current transformer > > in the circuit. When ac current flows it develops an ac voltage >... The pieces are in the range of $10 > each though.
For that kind of money, you'd get something that has a good output waveform and takes calibration. Neither quality is required. You could put ten turns of wire around a soft steel washer for simple detection.
It's bothers me when only the COTS parts are considered. Heck, a common coathanger is a perfectly good 10W resistor, you don't always have to buy something with "10W resistor" printed next to it in a catalog.
>>Not "automated" at all. A light that comes on to indicate when an >>element is burned out is enough.
>>> Another approach is to pass a small current through the heaters in the >>> 'off' state, but it's hard to detect bad heaters when others are in >>> parallel.
>>And there lies the problem. If this were a DC circuit, it would be a >>LOT easier because I could isolate each element with a diode and do a >>simple current/no current measurement.
>>Thanks for the reply.
>Don't miss Spehro's point, he's suggesting putting a current transformer >in the circuit. When ac current flows it develops an ac voltage on the >output, >you can use that to drive an LED.
>"- One CT* and one LED per element (each LED should light whenever the > related controller is calling for power). "
> I like this idea, (3 in one"direction" and 3 in the other), but I need to >think >out the phase situation. But it is a neat idea.
>" One CT and one LED with 6 wires running through the core (3 in one > "direction" and 3 in the other) .. if the controller is calling for > power and any one element is burned out then the LED lights. ;-) )
>* current transformer-- look it up.
> Mike
Oh yeah, Speff is onto something alright. Use current transformers, set the current ratio to provide about 5 mA into an antiparrallel pair of LEDs. Light == current to load. Cheap simple and works nice.
On Tue, 3 Nov 2009 18:24:57 -0600, "amdx" <a...@knology.net> wrote:
><lektric....@gmail.com> wrote in message >news:b5641ed1-fad9-4af6-988d-a698582b2ad9@j24g2000yqa.googlegroups.com... >On Nov 3, 2:05 pm, "amdx" <a...@knology.net> wrote: >> Don't miss Spehro's point, he's suggesting putting a current transformer >> in the circuit. When ac current flows it develops an ac voltage on the >> output, >> you can use that to drive an LED.
>>>I see his point, and agree. This is probably the easiest and most >>>straightforward way to do this. The pieces are in the range of $10 >>>each though.
> Ya, I agree CTs can be expensive, What is the current your heating >elements >use?
> Mike
If you do not know what you are doing (with current transformers) you can get skinned real easily. So i will give some clues:
The core only needs to be sized to support the output load power (usually milliwatts). For an example take apart a GFCI outlet, see just how big those cores are, yet they detect and trip on 5 mA out of a possible 15 A load.
A transformer is a transformer, the windings ratings are properties of the copper wire size, the insulation withstanding voltage, and the expected temperature rise. To a large extent the core as well. The canonical load of a current transformer is nearly a short circuit.
Depending on volume is may be cheaper to just hand wind a few (dozen or hundred).
Provided that the core and windings can take the currents and heat surprisingly small audio transformers can be used. Much the same with hand wound, see GFCI above.
Pulse systems have very different requirements than steady state line power systems.
That is about as much as i can say without getting into the actual application.
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