| Author | Message | ||
| Jack_In_KC (24.166.166.142) Rating: N/A |
I am wiring up my inverter and generator and realized that separate cabling will be required to install ammeters. Based on a review of the archives, the shunt type are preferred. Under the theory that more is better, I am considering monitoring volts & amps for battery, generator, shore line, and load. Eight gauges, right? Does anyone have suggestions as to specific gauge models or purchase sources? Thanks, Jack In KC | ||
| Jim Ashworth (Jimnh) (172.162.31.54)
Rating: N/A |
You don't need that many meters. The genset, shore and load will share the same meter on the load side of these inputs. This applies for both amps and volts. An additional meter that would be nice is a frequency meter to check on the speed of the genset. If you use the shunt type ammeters, the lines have to be run to the meter and back. With 6ga wire, that's a large bundle. Any ammeter on the 120v lines using the "donut" type of pickup tramsformer for the meter must be placed within a couple of feet from the "donut". More than that requires a "transmitter" to make the gauge accurate. These go for about $175 each and one is required for each AC ammeter. For further info a good site is: http://www.branom.com/literature/actransformer.html As to the DC side for house batteries, an "E-Meter" or Traces "Link" meter is the way to go. Amps, volts, amp-hours and much more info with it. This is just one guy's opinion. I'm sure others are going to chime in on this, too. Jim | ||
| Richard Bowyer (Drivingmisslazy) (24.196.191.70)
Rating: N/A |
I count eleven if you do it your way. In my opinion, separate meters for generator, shore line and load are redundant. Although it is possible, I have never found a need to monitor each one separately. I would recommend the 100-5 donuts with 5 amp full scale meters, calibrated for 0-100 amps. Using proper sized wire (14-16 guage) for the five amps, the meters can be located any reasonable distance away without adverse effects on the reading. Remember that you need two ammeters, one for each hot leg. I would also suggest 0-300 volt voltmeters and I would also recommend including `a frequency meter on the load line. Do not forget the inverter. It needs to be monitored also. If the load shunts are installed in the correct location, they will monitor the inverter AC output voltage and current also. You also need to monitor the DC amps and volts going to and from the inverter, but hopefully your inverter monitor will provide that. Mine did not, so I had to add these meters. I used a center zero, 150 amp ammeter so that it could monitor battery charge or discharge amps depending on which action was happening. Richard | ||
| Scott Whitney (69.35.6.233) Rating: N/A |
Heart sells a shunt that is made for the Link interface panels. I installed a dual shunt rated at 500Amps, I believe it is. I only used the one half, however. Nice to know how much juice is going to and from the batteries. The voltmeter is part of the Link panel too. Might check with Henry to see if his buddy has a shunt. I have the doughnuts for the AC Line 1 & 2, but don't know how they connect to the ammeter. If anyone bought one of the fancy breaker boxes made for Vantare Featherlite from Mallie, and knows how to hook up the doughnuts, I'd be interested to know. Scott | ||
| Sean Welsh (Sean) (64.81.73.194)
Rating: N/A |
I know this ain't the right part of the board for this, but it's on-topic for this thread: I just happen to have two of the 100-5 doughnuts and two 200-amp 50mv shunts that I'm getting rid of -- they came out of my bus from the original conversion and have been made redundant by the Link-10 and SW4024 that I am installing in the re-conversion. If you decide to go with the doughnut and/or shunt ideas, give me a shout and make an offer. I'd just as soon these stay in the bus community, though I was planning on listing them on eBay. -Sean | ||
| Sean Welsh (Sean) (64.81.73.194)
Rating: N/A |
Meant to add this in my last post: Jim - the link is not a Trace product, it's a Heart Interface. Both of those outfits are, of course, now just pieces of the behemoth Xantrex (who also have Pro-Sine and Stat-Power and who knows how many others), but it's an important distinction for anyone running off to find a Link-10 or Link-20. By the way, I bought mine on eBay, and a significant number of sellers still list is as "Heart" instead of "Xantrex." Search on "link battery" and you will find a bunch of them, usually NIB and below retail. Scott -- don't know how the Vantare/Featherlite boxes are wired, but if you can shoot me a picture or a diagram, I can probably tell you how to hook up the doughnuts. -Sean | ||
| Richard Bowyer (Drivingmisslazy) (24.196.191.70)
Rating: N/A |
Scott, the main power lead that you want to measure amps in is where the donuts go. The power lead goes thru the donut. One lead thru each donut. Then the two small wires from the donut go to the two terminals on the ammeter. I am assuming there are two ammeters. Richard | ||
| Scott Whitney (69.35.6.233) Rating: N/A |
Hi Sean & Richard, I know where the doughnuts go, but the strange thing is that the small leads from the doughnuts have no obvious connections to the the ammeter circuit board. (digital) I'll shoot a pic and post it later today. If you can help me figure it out, I know two or three other busnuts with these breaker boxes that will be happy campers too! There is a toggle switch that allows you to monitor either leg A or B on the volt meter and the ammeter. Scott | ||
| Richard Bowyer (Drivingmisslazy) (24.196.191.70)
Rating: N/A |
OK, Scott. Let us have a look. It is fairly common to use one meter that is switched between sources, either for volts or amps. Is there any writing on the donuts? Typically they are like 100-5 or 100-1, depending on the meter movement and the amount of current being monitored. What that means is that for 100 amps of current flowing thru the power lead you would have either 5 amps or 1 amp flowing thru the small leads. Richard | ||
| Stan (24.67.45.35) Rating: N/A |
Richard: I think the ouput of a current donut is a voltage, just like the output of a shunt is usually 50 milivolts. Ain't old age hell when we forget things like that? I say things every day that I would have laughed at twenty years ago. | ||
| Gary McFarland (Gearheadgary) (209.128.79.46)
Rating: N/A |
Uh....What else could the output be? Gary | ||
| Sean Welsh (Sean) (64.81.73.194)
Rating: N/A |
The output is a current, and it's read with an ammeter. Typically, CT-driven ammeters are 5-amp full scale, with the scale calibrated according to the CT ratio. If you leave the terminals (or wire leads, as the case may be) unconnected, then, yes, you will have a "voltage" (potential difference) between the terminals. You will also likely fry the CT (in high-current applications) because the EMF can be very high indeed -- high enough to arc. That's why you should short the CT terminals if no ammeter is connected (so, if, for example, you use a switch to toggle the meter between two CT's, the switch should short the unmetered CT). This is the reason the doughnuts are called "current transformers" and not "voltage tranformers". A current of X amps in the primary induces a current of rX amps in the secondary, where "r" is the current ratio as stamped on the device. -Sean | ||
| Gary McFarland (Gearheadgary) (209.128.79.46)
Rating: N/A |
Nonsense, there is no such thing as an "output of current". The only measurement of electrical supply is voltage. Current is a measurement of demand. What you have is a simple Torous coil, that induces a voltage when current passes through it, it is an inductive process. The voltage present in the leads on the torous coil is directly proportionate to the amount of current that passes through the toroid. Your Ammeter is actually a voltmeter. scaled to display the present voltage as a function of the current passing through the toroid. I have never heard the term "Current Transformer" before, you can only "Transform" voltage. Also, there is no such thing as a "DC Transformer". Gary | ||
| Sean Welsh (Sean) (64.81.73.194)
Rating: N/A |
Gary, You're not measuring an electrical supply -- you're measuring a flow. And, look it up -- the doughnuts are called "Current Transformers" and are usually abbreviated CT. ALL meters meassure current, and only current. The way a Voltmeter works is to draw a miniscule amount of current through a known resistance (extremely high). Ammeters work directly, except that the mechanism (whether analog or digital) is sensitive to micro-currents (just as the mechanism in a voltmeter), so to measure large currents, the bulk of the current must be shunted around the meter, leaving only a tiny percentage of the current for the meter to measure. To put this another way, a VM is a high-impedance device. An AM is a low-impedance device. You need the latter to get a correct reading from a CT. You connect an Ammeter in-ine, directly and with no other resitance or load, and it measures the current flowing around the CT (same current as would flow if the CT leads were shorted). If you put a voltmeter in there, the impedance of the VM itself will change the amount of current flowing around the coil. -Sean | ||
| Sean Welsh (Sean) (64.81.73.194)
Rating: N/A |
Follow up... Gary, Just for you (well, OK, also for everyone else who's interested), I hunted around and found this nice page on how AC meters work: http://www.allaboutcircuits.com/vol_2/chpt_12/1.html The part closest to the bottom of the page discusses CT's and shows a diagram of how they go into the circuit. -Sean | ||
| Gary McFarland (Gearheadgary) (209.128.79.46)
Rating: N/A |
OK, I have just never run across the term, but it does not "Transform Current" it is basically half a transforer converting current to voltage. You cannot "Supply Current", it's a contradiction in terms. All Meters Measure Voltage, not current. The shunt "Conducts" Current across itself, it is not a high resistance, but rather a very low resistance. the miniscule IR Drop, (I=Current, R=Resistance) is Voltage. (V=IR AKA Ohm's Law) A meter is nothing more than a sprung electro-magnet that pivots and has a needle attached to it. An electro-magnets intensity is controlled by the amount of voltage applied to it. If you connect an ammeter to an energized toroidal coil, you have a short. Gary | ||
| Tony (64.215.196.148) Rating: N/A |
you can always tell when someone really knows what they are talking about. | ||
| Gary McFarland (Gearheadgary) (209.128.79.46)
Rating: N/A |
I don't see anything on the site to support what you are saying. This is a common misconception, about current. Let me say this another way. There is no such thing as current, without voltage and resistance/conductivity. Current is only the reaction of voltage flowing across a conductor. This is basic stuff, but it is worth learning. Once you really understand what electricity is, it all gets real simple. Gary | ||
| Richard Bowyer (Drivingmisslazy) (24.196.191.70)
Rating: N/A |
Gary, before you go any further and dig yourself into a bigger hole, go find an ammeter in a catalog. You will find that it is a current measuring device. Typically five amps full scale, and you can put a clamp on ammeter on one of the leads and measure the current. And you definitely must use a make before break switch when switching from one CT to another. Your switch must also maintain a short across the ammeter leads that were disconnected. Trust me!! Richard Also for who said there was no such thing as a DC transformer. You are also wrong. A hall effect curent transformer will measure DC amps. I personally own a Fluke DC ammeter which is a clamp on type. You clamp it around the conductor amd measure DC amps, just like you measure AC amps with a AC clamp on ammeter. Richard | ||
| Sean Welsh (Sean) (64.81.73.194)
Rating: N/A |
Gary, Please go back to school and redo Electromagnetism 101. You flunk. Sorry. I can not spend any more time trying to educate you -- this stuff is available on the internet. You might start at the beginning of the site for which I posted the link. For anyone still reading this who intends to use a doughnut (CT): You MUST connect an ammeter (low-impedance, nearly a dead short) to the leads, or short them together, or you risk frying the CT. -Sean | ||
| Sean Welsh (Sean) (64.81.73.194)
Rating: N/A |
Follow-up/Apology Well, after re-reading my last post in the calm light of day, I would like to apologize to Gary and to everyone else for the rather childish first paragraph. I think my hackles were up after the "This is basic stuff, but it is worth learning" comment. Of course it is worth learning, Gary, and I commend you on a good start. Clearly, you have a good handle on Ohm's Law. Ohm's Law is indeed a necessary part of understanding the concepts involved in AC and DC electrical circuits, but it is by no means the only part. For Gary and anyone else interested in a more in-depth understanding of how meters work, I suggest you start at this page: http://www.allaboutcircuits.com/vol_1/chpt_8/4.html and go on through to the next lesson as well, along with the AC page I posted earlier. And just to be clear: No, this is not how I learned this stuff. Before there was an internet, I did spend four years in engineering school, and I have a graduate degree in computer engineering. We went just a bit further than Ohm's Law :-) -Sean | ||
| Sam Sperbeck (206.230.105.230) Rating: N/A |
Hi Sean, I think this thread demonstrates that a little knowledge is a dangerous thing, especially in a little mind. Thanks, Sam Sperbeck La Crescent, MN | ||
| Gary McFarland (Gearheadgary) (64.134.135.51)
Rating: N/A |
OK, you guys. I'm trying to help here. I have been working with Electricity as long as I can remember. I grew up in an apartment that was attached to my dad's electrical repair shop. I learned Ohm's law at like 9. Please consider that I just might know what I'm talking about, just for a minute--Humor me. look at your website--the first drawing See you have a meter, with resistance to the left, a Current at FS= 0.1 mA. referring to our venerable Ohm's law, if you have known resistance, and a known current, what is left? What I didn't mention above, and they don't directly mention in your example website, but they allude to, is that the ammeter has an internal shunt. All ammeters do. "....extend the usable range of the movement are connected in parallel...." Parallel with what? Parallel with the internal resistance of the meter. Again the Meter is measuring the voltage drop across the meter's internal resistance. Again, all meters, the movements themselves, work from voltage. In your first AC site you posted the two D'Arsonval (AC one is only a rectified version of the other) meter movements and the Iron-Vane movement use an electro-magnets to deflect the needle. Electro-magnets vary their intensity based on application of voltage. Again, using conventional technology, you can only measure current as a function of a voltage drop. Richard, If you take apart your meter, you will find one form of a shunt, the meter measures the voltage drop across the shunt. Your Fluke--Mine is an Amprobe, uses an inductive clamp that induces a voltage at the meter. the meter itself, the movement, measures voltage. Again, all Ammeters are actually voltmeters that measure a voltage drop across a shunt. Including the ones you are installing in your coach. Your hall-effect yoke is exactly what your doughnut is. it takes a current passing through the doughnut and creates a voltage at the ends of the coil. Again--look at your terminology. Our original discussion--what is coming out of the doughnut--you said current. you are dead wrong, and giving away a limited understanding if fundamental electircal theory. You are measuring a potential. You cannot have "Potential Current" Now, if we're comparing pedigrees--I grew up in an electrical repair shop, as I mentiond, was doing electrical circuit design (Professionally) at age 17, Apprenticed as a Millwright-Electrician until the wood products industry failed. after several years in Steel fab and mechanical work, I went back to college and graduateed third in my class as EETII. (The other two had a calculus bacground, which was weak for me). I now am a network operations director for a tier 1 automotive supplier. this discussion is at Kindergarten-level, I recommend you do some review before you post a condecending reply. Nice to see you too sam. Gary PS: I mentioned that there is no such thing as a DC Transformer, I can't remember what I was getting at, but there isn't such a thing. There are devices that will alter DC voltages, but they're not transformers. | ||
| Scott Whitney (66.82.9.28) Rating: N/A |
And as it all relates to bus wiring. . . there are two items that one may encounter that might need more clarification from the above discussion. That is 1) the doughnuts that go around the main AC legs to measure AC current . . . and 2) a big honking brass shunt that 0000 battery cable bolts to in order to measure DC current for the Link panel. My understanding of the brass shunt is that it has a very precise, very small, but known resistance and so one can therefore measure the slight voltage drop across it and use Ohms law to calculate the current passing thru it. This is all done transparently to the user inside the Link panel. Will post pics of my set-up soon. But I am afraid they might not help much. All you can really see in them is the ammeter circuit board and a whole lot of AC wires for the loads etc. Scott | ||
| Gary McFarland (Gearheadgary) (64.134.135.51)
Rating: N/A |
Simplified, both the the doughnuts and the brass shunt are like "Sening Units" to the actual gauge. Your description of the shunt is exactly correct. both devices produce a voltage that is proportionate to the current draw. Gary | ||
| Scott Whitney (66.82.9.14) Rating: N/A |
Anyway, here are the pics: http://www.dustyfoot.com/Electrical.html Hard to see what is what in the pics. It's not much better in person. : ^ | But other than not being able to get the ammeter function hooked up, it is a great breaker box. : ^ ) Scott | ||
| Sean Welsh (Sean) (64.81.73.194)
Rating: N/A |
Gary, Well, I guess I just wasted an engineering education. The mysterious force that makes an analog movement work is magnetism. The magnetic field strength is proportional to the current passing through the coil -- voltage, or EMF, has nothing to do with it. (This is first-year electromagnetism, folks.) There will, of course, be a voltage drop across the coil, due to the coil's internal resistance as well as its reluctance. ANALOG METER MOVEMENTS ARE QUANTIFIED BY AMPERES OF FULL SCALE DEFLECTION, period. Note I am talking about the movement itself, without any shunt or multiplier involved. If you connect a movement directly to a voltage, it will go to full scale (and probably fry). (N.B.: Microvoltage is a different story, pricipally because, at extremely low voltage, the internal resistance of the wire in the movement's coil, as well as the coil's reluctance, becomes significant, acting, as it were, as an internal multiplier.) Again, I see you have Ohm's law well in hand. When you can explain Kirchoff's law to me, or show me how to determine a Thevenin Equivalent Circuit, then I will let you expound to everyone here on how meters work with impunity. Until then, frankly, it is just too dangerous to let this sort of misinformation get posted without rebuttal. I will say again, for everyone's benefit: The current (NOT voltage) induced in a CT is the exact multiple of the current passing through the primary as determined by the winding's ratio. This assumes that the CT is a CLOSED LOOP. The loop is closed with an AMMETER, having a very low impedance -- so low as to be a negligible "load" on the transformer. IF you foolishly choose to connect, instead, a high-impedance voltmeter to the leads or terminals of your CT, you will: 1. Not get a meaningful reading of any sort, unless you know exactly the internal impedance of the meter. 2. Likely fry the CT. Of course, I've always liked my doughnuts fried.... To use a CT, you need an AMMETER, which presents very low impedance to the loop, and reads the CURRENT flowing in the CT from 0 to the full-scale value of the highest measurable current in the primary. -Sean | ||
| Sean Welsh (Sean) (64.81.73.194)
Rating: N/A |
Scott, Can't quite make anything out in the photo you posted. However, the meter board appears to be an "Intellitronix" part number 570. Intellitronix is now "Nordskog Performance Products" (Oxnard, CA), phone 805-483-2411. You might give them a call and see if they still have a circuit diagram of that board in their files. If they do, and the current inputs are not immediately obvious, I can probably look at the drawing and noodle it out. -Sean | ||
| Sean Welsh (Sean) (64.81.73.194)
Rating: N/A |
Follow-up: Scott, Also found this number for Intellitronix in Lakewood, OH (might be out of date, though): 216-521-9796. That might be a better bet, since the board is marked "Cleveland." -Sean | ||
| Sam Sperbeck (206.230.105.230) Rating: N/A |
Hi Sean, I think this thread also demonstrates that for ignorance there is a cure, it's called education. If one refuses to learn it is called stupidity, for which there is no cure. Thanks, Sam Sperbeck La Crescent, MN | ||
| Gary McFarland (Gearheadgary) (209.128.79.46)
Rating: N/A |
Well, we've gone far afield on this one and it's long since departed from a useful discussion. The original point, is you cannot have an "output that is a current". It's a contradiction in terms. Current varies based on the conductor it is flowing across, but Voltage is always what it is. I don't have time to continue this, I'm installing a LP-Gas Backup Generator today, and I'm behind schedule. Gary | ||
| Chris 85 RTS (66.194.150.45)
Rating: N/A |
Personally I think they are both right and are just agruing different angles of the same picture. V=IR is fact. Everything has R is a fact. A shunt has an extremely low R, thus a large I will only produce a small V, which is safe to supply to the coil of a meter, which has an R itself, which produces an I, which causes the needle to deflect. Think of it this way. The source wire is actually one side of the transformer. Transformers are usually specified by the ratio of the primary and secondary windings. In an ideal sense the Power into a transformer is equal to the power out of the transformer. P=VI. If the CT is a 100:1 CT, then the coil likely has 100 windings, cause the wire being measured is considered 1 winding. Thus, there is a 100:1 reduction in current, but a 100:1 increase in voltage. Thus, if you connect the CT to a high impeadance voltmeter, fry! When you connect it to a low impeadance ammeter, it works. An ammeter does measure current. It does so by producing a low voltage across a low R shunt, which induces a small current in a coil, which creates a magnetic field which causes a needle to deflect. Of course you can't just use a 100 winding coil as there are losses in every part, so the coil is optimized for the correct reading. | ||
| Gary McFarland (Gearheadgary) (209.128.79.46)
Rating: N/A |
The measurement of supply is always Voltage. Ok let me get this straight. I can connect my VOM to a 480 supply, and read an output. No problem. But if I connect it to the output of the CT something will fry? What fries, the meter or the coil? of course an ammeter measures current. but the meter movement is actually a milivolt meter across an internal shunt, measuring a voltage drop. Of course, when you place a Voltage across a conductor, you have a current, that causes EM to deflect the movement. You cannot have a supply of Current. Plain and simple. If you have a supply, it is a potenial. Now what is the other word for potential? OK, I'm going to hookup that genset now... Gary | ||
| Chris 85 RTS (66.194.150.45)
Rating: N/A |
Can we agree that putting the coil around the wire is essentially a transformer? If so, can we then agree that with a primary winding of 1 and a secondary winding of 100, you will get a 1:100 reduction in current and a 100:1 increase in voltage? Pin = Pout , Vin * Iin = Vout * Iout If so, if you have a 110V line at 50A in, you will see a 11,000V at .5A on the output. Putting a high impeadance voltmeter will read 11,000 volts, if it is designed to do so, otherwise it will fry. Putting a low impeadance (ie shunt) ammeter will read 0.5A, which will represent the 50A of current in the source. The method by which the ammeter reads the 0.5A seems to be what is debated here. I think we can agree that an ammeter is designed to measure the current present in a wire, independent of the source voltage, thus no connection to ground, does so by passing said current through a shunt, or a very low resistance. In a digital meter, you would read the voltage(V) across the shunt(R), and convert that via a I=V/R. In an analog meter you would connect a coil across the shunt, thus creating a current in the coil proportionate to the voltage in the shunt, the current in the coil will produce a magnetic field, which will move a magnetic needle, and calibarate the needle movement to read out the current in the shunt as desired. | ||
| Sean Welsh (Sean) (64.81.73.194)
Rating: N/A |
Gary, Chris' math is correct. Also, just to point out, that not all ammeters have shunts in them. In an analog meter, the meter movement itself, which works strictly by electromagnetism, measures current (not voltage, as you continue to claim) directly, albeit only very small currents. It is not uncommon for a D'Arsonval movement for example, to be designed to respond to currents in the range of 0-0.5a. Again, this is the movement itself, with no internal shunt. And while the 120v*50a=>12,000v is a bit simplistic (because the actual potential through the transformer primary is probably different than the supply voltage, as we are reading a conductor going to a load), it is reflective of the general magnitude of these values. That's why I keep saying you can fry the CT: if you leave the leads completely unconnected, the potential difference (or EMF, or Voltage, whatever your preferred term is) can be high enough to cause arcing between the turns of the CT, thus damaging it. Simply shorting the leads together (presumably BEFORE the potential builds to that level) safely drains this potential through a very low-current connection. The ammeter, being a very low-impedance device, also safely completes this circuit. In practice, the CT's we are probably using for our applications are more like 20:1, and we will seldom see a full 50A of current flowing in the primary, so the probability of CT arcing is low. But it does exist, and I would be remiss in failing to recommend, as I have now several times, to you and everyone else on the board that either an Ammeter or a direct short be maintained across the CT leads/terminals at all times. Specifically, if people intend to use electromechanical switches to switch one ammeter between two or more CT's (doughnuts), that switch must: 1. Short the leads of all unused CT's (to themselves, not each other) and 2. Do so in a make-before-break manner. Lastly, I need to point out that E=IR (Ohm's Law) is sufficient only for purely resistive circuits. In order to apply Ohm's law to non-resistive loads, one needs first to derive what is known as an Equivalent Circuit, which is a subject that we probably do not need to go into on this board. Good luck with your genset install. -Sean | ||
| CaSteve (208.19.54.101) Rating: N/A |
SSSOOOOOO,,,, HOW 'BOUT THEM CUBS? | ||
| Sean Welsh (Sean) (64.81.73.194)
Rating: N/A |
Uhh, is it baseball season already? Seriously, Steve, I appreciate the sentiment, and I'm sure a number of people (myself included) are fed up with this discussion. However, I think it's an important dialog to have -- many people read this board for advice on how to do things, and folks who have neither the desire, nor really the need, to know the underlying physics of this thing at least need to know the right way to hook up their meters. -Sean | ||
| Sean Welsh (Sean) (64.81.73.194)
Rating: N/A |
Hey Scott -- Were you able to get anything out of Intellitronix? Let us know! -Sean | ||
| Scott Whitney (67.234.91.245) Rating: N/A |
I called the number (Thx!), and found the new number. Intellitronix was bought bought out. So I called the new company and spoke with a very nice person. He checked out my pics and printed them out. He said he was going to check with a couple people in the company and failing any luck with that, he would contact the EE guy who sold Intellitronix. So. . . if he follows thru, I hope to have an answer. I also found out that the tech guy I talked to, as a side business, buys sandstone from near me here in the Tulsa area and sells it to his clients in the Santa Barbara area. SO. . . I may just look him up over Xmas as I pass thru SB on my way to SD. . . and see if he had any luck. Will report any findings. The only thing I can figure is that there are some pins on the PCB that the doughnut leads are supposed to be soldered to. Either that or I am blind in overlooking the obvious location to connect the CTs. It will probably seem obvious once I find out what it is. Will be cool to get the digital ammeter function working. Regarding this whole ammeter topic, shouldn't it be clarified about the difference btwn a DC shunt, which is utilized by measuring the voltage drop across it, to an AC CT (doughnut)which, as I have think I have learned, is like a single winding transformer and actually inducing a current in the coil, which is measured by the ammeter circuit? Seems to me like the two devices are operating on a different principle that has not been made very clear in all of the theoretical and mathematical discussion so far. The whole DC vs. AC issue might be where some of the misunderstandings lie? Scott | ||
| Mallie (65.38.24.87)
Rating: N/A |
Hey Scott Whitney,, for one cool nickel I will tell you how wire it up the meter. Mallie | ||
| Richard Bowyer (Drivingmisslazy) (24.196.191.70)
Rating: N/A |
Scott, I believe the current from the ct goes thru an internal resistor in the meter that we never see in a schematic. The current thru the resistor develops a voltage, proportional to the current flow, and the meter is actually reading this voltage. This internal configuration of the meter is never shown in a schematic. At least this is my opinion FWIW. Richard | ||
| Gary McFarland (Gearheadgary) (64.134.135.51)
Rating: N/A |
Good lord. Isn't that what I said in the first place? Here is what I said: "What you have is a simple Torous coil, that induces a voltage when current passes through it, it is an inductive process. The voltage present in the leads on the torous coil is directly proportionate to the amount of current that passes through the toroid. Your Ammeter is actually a voltmeter. scaled to display the present voltage as a function of the current passing through the toroid" What is the difference, except for the fact that I gave more detail? Gary | ||
| Sean Welsh (Sean) (64.81.73.194)
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Yes Gary, that's what you said, except that it's still incorrect. Yes, you are right in the sense that for real-world meters, we can't measure a current without inserting some type of resistance into the line, just as, in the real world, we can't measure voltage without drawing off some current. Think instead in terms of "ideal" meters. An ideal Ammeter would be able to read current without presenting any resistance (meaning there would be NO, zero, nada, voltage drop across the meter). An ideal Voltmeter, on the other hand, presents infinite resistance -- this is so much of an important issue, in fact, that voltmeters are qualified by how many "ohms per volt" they present, more being better. Let's put this another way -- if you take the leads of the CT and short them (which, as I keep saying, you MUST do if they are not connected to anything), there will STILL be a current in the CT that is directly proportional to the current in the primary -- for example, if you have a 100:5 CT and you have 40 amps running through the primary, you will have 2 amps of current circulating in the CT. This is just a fact of physics. Of course, you can't see the current, because you left out the meter. The VOLTAGE between the leads of the CT (where you proposed to put your voltmeter) is now Zero, because the leads are shorted together. On the other hand, if you leave the meter out all together, and do not short the leads, the VOLTAGE between the leads will be on the rough order of 3kv. In other words, the voltage reading between the leads depends strictly on the value of the internal resistance of the voltmeter, because that value will affect the current transiting across the meter. The value you are trying to read is the current induced in the CT --- you do this by inserting as little resistance as possible into the circuit. While, yes, you can conceptualize this process as inserting a miniscule resistance into the circuit (a shunt) and then measuring the voltage drop across that with a very sensitive millivoltmeter (which, incidentally, is how some digital ammeters work), you are still measuring current, not voltage. And, just for further education, while constant voltage (or relatively constant, anyway) sources such as batteries and most commercial power supplies are certainly more commom, there are, indeed, such things as constant-current sources. As a further note following up on the suggestion that a distinction be drawn between shunt-based and CT-based current measurements: whoever said this raises an excellent point. In fact, one could measure AC current with a shunt (and the associated millivoltmeter) just as we do for DC. The reason this is seldom done is that shunt-based current measurements dissipate more power from the circuit under test, and for AC current, the CT-based method is cleaner. DC, of course, does not lend itself to this type of non-invasive measurement. There are devices (known as Hall-effect transducers) that can do this, but they are more expensive, more finicky, and require an external power supply, making them less attractive overall. Both shunt- and CT- based measurement are self-powered (at least with analog magnetic movements). Whew. I'm tired of typing about this.... maybe I should do a seminar in Arizona. -Sean | ||
| Stan (24.67.45.35) Rating: N/A |
Sean: "The VOLTAGE between the leads of the CT (where you proposed to put your voltmeter) is now Zero, because the leads are shorted together." Under these conditions, the lead wires are the shunt and to measure the voltage across the shunt you have to put the meter at the source (that is at the ends of the coil). | ||
| Sean Welsh (Sean) (64.81.73.194)
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Umm, Stan, if you short the lead wires to each other: 1. The volatge "between" them must be zero because they are connected to the same point. Any single point in a circuit can be at only one potential (Kirchoff's Law). If you want to call that a shunt, fine -- it's a perfect shunt (shunts 100% of the current). 2. There are no "ends" to the coil -- it's now a continuous loop. That's what's meant by connecting the two leads together. Look, folks, for anyone who still does not believe this: Do not believe me, and do not research this on the internet. Instead, go down to your local community college, and find a copy of the first year physics text book. "Halliday and Resnick" is the canonical text that has been in use for longer than I've been alive (now in its umpteenth edition). Turn to the section on Electricity and Magnetism, and bone up on this for yourself. For anyone who doesn't care to take the time to do this, I suggest you follow my earlier advice and: 1. Connect an AMMETER, (not a voltmeter) between your CT leads. 2. Short the leads of your CT when an ammeter is not connected. 3. Make sure the CT leads are shorted before disconnecting the ammeter (use a make-before-break switch). -Sean | ||
| Richard Bowyer (Drivingmisslazy) (24.196.191.70)
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Sean, I do not think anyone is disagreeing with you on the above items. Although I am not disagreeing, I really do not recall what happens inside the ammeter after you connect the CT to it. In other words, the meter guts itself. It is overall an ammeter designed to read amps. Is the internal guts a low resistance shunt that converts the 0-5amps to 0-1mv meter movement, or is there another internal transformer that converts the 0-5 amps to 0-1 ma? I am really looking to understand the meter movement itself, not what the overall device is designed to do. Somehow the amps must be converted to mechanical movement. Can you help? It has been way over 50 years since I studied meter movements and quite truthfully, I am a little rusty. Richard | ||
| Stan (24.67.45.35) Rating: N/A |
Sean: Where can I buy this zero resistance wire to make meter leads? The winding is a coil within the magnetic field. The leads are external to the coil and form a shunt across it. You have spent a lot time trying to prove that you can have current flow without any voltage. Millions may believe you but there are still some non-believers. | ||
| Chris 85 RTS (65.35.68.187)
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I am thinking that a current, the movement of electrons from one atom to another can be induced by two methods, first a voltage potential, and second a changing magnetic field. In fact I believe that if you have a coil, made from a superconductor, a near zero resistance wire, you can in fact setup a current, which would theoretically last forever. SInce the coil is a closed loop, the current exists with no measureable potential present. The electrons are forced to move by the magnetic field, not nay potential present in the circuit. However, any interaction with a magnetic field would steal power from that coil and the current would eventually be lost. | ||
| Sean Welsh (Sean) (64.81.73.194)
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OK, three posts to answer: Richard: The simplest form of (real-world) meter is the analog, magnetic type. If you go up to near the beginning of this thread, and follow the links I posted, there are some good diagrams. In any case, the way these work is that the wire goes into the movement and forms a coil and then comes back out (note: no shunt or other parallel connections at all inside the movement). The coil is attached to the needle and the base contains a permanent magnet. As current flows through the coil, a magnetic field forms, moving the needle. The more current, the stronger the field, the more the needle moves. The sensitivity of an analog movement and the amount of current required to deflect the needle to full scale depends on a number of factors, any of which can be adjusted by the designer to achieve the type of movement he wants. These factors include the strength of the permanent magnet, the number of windings in the coil. In general, these movements are fairly sensitive and fairly delicate, so common full-scale deflections for movements alone run anywhere from just a couple of milliamps to an amp or two. Above that, it becomes necessary to "shunt" some of the current around the movement, to protect the movement and allow it to provide a reading. The movement's internal resistance to current flow is not constant, but movements are designed so that the resistance will vary only a small amount within the movement's operating range (movements, for example, designed to measure current in amps are not acurate in the micro-amp range). Since this resistance is known, the amount of current to be "shunted" can be controlled by precisely controlling the resistance of the bypassing pathway. This is whence comes the name "shunt," and why shunts are precisely calibrated resistance values. Note that both analog ammeters and analog voltmeters are made using the same basic movement. The difference comes in how those movements are connected to the circuit under test. A voltmeter, by definition, is intended to measure a potential difference (voltage) without affecting said difference unduly, thus it must be extremely high impedance relative to the circuit under test. Because the movement itself, being just a wire coil, has very low impedance, a very high resistance value is placed in series with the movement. Of course, since the movement itself measures current and not voltage, the meter necessarily draws current through both this resistance (called a Multiplier) and the movement. An ammeter, by contrast, is a device to measure current. Again, ideally, the current would be measured without impeding it in any way. As a result, ammeters by definition are extremely low-impedance devices. Last note on this subject: The above movement description applies to DC meters. AC meters can, of course, be built exactly the same way simply by rectifying the source so that all the current flows in the same direction. More practically, though, due to the electromagnetic charactaristics of a reversing current, it is easier to build the movement with a fixed electromagnet and a ferromagnetic vane attached to the needle. DMMs, by the way, are a whole different subject unto themselves. OK, next question, from Stan: Yes, one can have current flow without any voltage. In fact, this is precisely what happens in a CT (doughnut). You are correct, though, in your understanding that even wire has resistance, so, in fact, there is no free lunch: the "ideal" current which would be induced in the CT if it were a perfect conductor, which, by the way, is what we are really after, is the exact multiple of the current in the primary. If you had a "perfect" voltmeter (infinite resistance -- no effect on circuit under test), and attached it to any two points in the coil (say, for example, two points as far from each other as possible), it would read 0, because the perfect conductor would ensure that all points were at the same potential. Nevertheless, a current would be flowing, and that current would be directly proportional to the current in the primary (the wire going through the center of the doughnut). You could read this current, of course, only with a perfect ammeter (one which has zero resistance). OK, back to the real world: This unimpeded current which would be running in a perfect conductor through an ideal ammeter is what we are trying to read, because that (and only that) is the exact representation of the current in the primary. Soooo... we use an ammeter which, while not perfect, has as little resistance as possible, and we build the coil to have as little resistance as possible. The we cheat: we put an extra winding or two on the coil to make up for the losses in the coil wire and the ammeter itself. Last thing for Stan: Contrary to your assertion, I have spent no time at all trying to prove any of this. I have only spent time explaining it. The proof would take far too long to type, but rest assured that I have done it (as has every student who made it through engineering physics 101 and 102). If it's proof you seek, then follow my earlier suggestion and get a copy of Halliday and Resnick or some other first-year Physics text and look it up. It's all there in black and white. Moreover, if you get a good text, it will walk you through the experiments, many of which can be improvised at home, which will let you see these principles in action for yourself. OK, last post was from Chris: Chris, you get an A+ for research. Yes, that's exactly right, there are indeed superconducting experiments that have been performed where a current has been induced in a coil and then sustained itself for a good long time with no potential whatsoever. Even in our real-world CT's, since they are closed loops (at least, if you hook them up as intended) there is very little measurable potential difference anywhere in the loop. -Sean | ||
| Sean Welsh (Sean) (64.81.73.194)
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Source of confusion! Whoops... I just discovered that I goofed, way back at the top of this thread, when I cut-and-pasted the link to the pages on how meters work. The correct link is http://www.allaboutcircuits.com/vol_1/chpt_8/1.html which will start you at the beginning. The link I actually posted earlier puts you two pages later in the course, at Ammeter design. I think this may be why some folks seem to believe that current can not be measured without a shunt. If you start at the correct page, you will see how current is measured directly using the principles of electromagnetism. Shunts come into the picture only when current gets too high for the delicate electromagnetic mechanism. -Sean | ||
| Richard Bowyer (Drivingmisslazy) (24.196.191.70)
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Sean, thank you very much for the detailed explaination. I know it took quite a bit of time to type all this and I certainly appreciate your efforts and your willingness to try and put it in laymens terms so we could all understand it. One of the odd things that I never thought about before is the secondary current in the CT. I knew that you had to keep them shorted, either directly or thru the ammeter, but I never really thought about the fact that with the leads shorted there was a current flowing, but nowhere to measure an actual voltage causing this current flow. Magic, huh? Unfortunately my PHD* came from the SHK** University, and it did not really adequately cover this subject. LOL * Post Hole Digger **School of Hard Knocks Again my sincere thanks for taking the time. Richard | ||
| Chris 85 RTS (66.194.150.45)
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Thanks Sean. It really is all in the explaining of things. I still think that this thread was so difficult due to the very different backgrounds of the people trying to approach the same problem. | ||
| busjock (4.41.168.36)
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I want to touch on another aspect: Sometimes you may want to measure a current where there is no standard readily available CT. I wanted a 75 Amp meter for a particular job. It is easy to change a CT ratio. For the standard 100:5 units it is obvious that the torroid has 20 turns on it. You can add or subtract turns externally. The magnetic field doesn't care. for a 75 amp meter you need 15 turns right.. Just add 5 turns evenly spaced around the toroid. Connect them in "BUCKING" and you have your 15 effective turns. Accuracy will not be compromised if the turns are evenly spaced and of reasonable wire size. I want to thank Sean for taking the time to clarify a lot of serious misconceptions. I think a lot of engineers were very uncomfortable with the considerable amount of grossly bad information in this therad. Jock Fugitt | ||
| Sean Welsh (Sean) (64.81.73.194)
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Jock, You're correct. I just want to add an important clarification, though, for all our listeners out there in BBS-land: When Jock says "a 75 amp meter" in this context, what he means is "a 5-amp meter with a scale calibrated from 0-75amps." Most ammeters made for use with CT's to measure high currents are made as 5-amp full-scall meters, and you select one whose dial face is calibrated to read the actual current you are measuring. Then you get the right ratio CT to achieve a 5-amp CT current at the upper-end reading of the scale. I'm a cheap SOB, so I bought a "10 amp AC ammeter," which actually was intended as a direct-read (in-line in the circuit) meter for low-current applications. (I assume it has an internal shunt or an internal CT, as I'm sure the movement itself can't handle 10 amps!) Since I have a 100:5 CT, I just run the primary through the CT twice (so there's a loop in the primary -- one wire passes through the doughnut twice) and then put a little sticker on the meter that says "X 10" just like on your tachometer. -Sean | ||
| BrianMCI96A3 (69.34.171.174) Rating: N/A |
Thank you Sean, I think your typing fingers have to be sore, about now. I have to admit that while I understand the general idea, and have a working knowledge of electricity it has always been a bit of a boogie man... I just try real hard not to let the smoke out. (grin) Brian | ||
| Chris 85 RTS (66.194.150.45)
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Now Brian that is funny. It is not commonly known that smoke it what makes electronic circuits work at all. Afterall, once you let the smoke out, it no longer works. Makes sense to me. |
