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12/09/2017 3:28 PM

I have a CT of 800:1 and an ammeter with Scale 0-2000 amps AC

It says on its name plate 1MADC which i think means that when it has 1 milliamp dc current it will go to full deflection i measured resistance between its terminals that is coming 172 ohms

Now I took a 12 VDC battery and connected a decade box - a fluke digital multi-meter in ammeter selection in series and this ammeter , when my fluke is telling me that 1 amps is flowing in the circuit this ammeter goes full scale , but i expected it to go full scale at 1ma current , why is this happening

does a ct secondary outputs a dc current or an ac current do i need to add something to make it work

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#1

12/09/2017 3:49 PM

Your meter was designed to work with a scaling SHUNT.

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#2

12/09/2017 4:14 PM

A CT is a current-to-current transformer. 800:1 means if 800 amps are flowing through the primary (1 turn) there will be 1 amp on the output (800 turns). If the primary current is 2000 amps, the output would be 2000/800 or 2.5 amps

If the meter scale says 0-2000 amps, it has a winding resistance of 172 ohms, and 1 ma results in a full-scale deflection, then a shunt resistance of 172 x .001 / 2.5 = .08688 ohms is needed in parallel with the meter.

2000 amps through the CT primary results in 2.5 amps from the CT secondary. 2.5 amps from the secondary produces .172 volts through the .08688 ohms shunt which produces 1 ma through the 172-ohm meter movement causing full-scale deflection.

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#3

12/09/2017 4:48 PM

If the meter was 172 ohms it would drop 172 x 1 = 172 volts at 1 amp. But it would certainly burn-out before the needle got to full scale! There must be a shunt for 1 amp full scale, which "begs the question" how you measured 172 ohms without disconnecting the shunt in parallel with the meter?

Are you sure you are putting in 1 amp - resistance decade boxes do not usually like 1 amp - what resistance is the box set at - current (amp) = 12 volts / resistance - ignoring meter and wire resistance??

A 1 milliamp meter of 172 ohms would drop 172 millivolts at 1 mA and would read full scale 172 millivolts as a voltmeter. A shunt of 172 milliohms would drop 172 millivolts at 1 amp - to be exact it would be slightly higher to allow for the 1 mA current in the meter (the shunt would carry 999 mA with 1 ma meter at full scale).

Current transformers normally have windings on a magnetic core and work only on AC. They are used with AC meters since DC meter would only flicker about zero as it tries to go forward and backward 50/60 times per second.

CTs always have a rated primary current, as well as a ratio - usually they are given as e.g. 2000: 5amp, which tells you the primary is 2000 amp rated (secondary 5 amp) and its ratio is 2000/5 = 400 to 1. A 1 amp secondary is also a standard value.

So I do not think you have a CT and meter which were selected to work together.

But modern CTs may be designed for much less than 1 amp secondary current ( because the power loss and volt drop in long wires to the meter is so much less) and may even have a resistor load built-in to give a voltage output to modern electronics.

A 1 mA DC meter could work with an electronic AC current to DC current converter, called a Transducer, between itself and the AC CT current output.

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#4

12/09/2017 6:42 PM

OK, here is what I think is happening. Your AC ammeter has a full wave bridge built in, plus a parallel shunt. The bridge allows you to use it to measure AC or DC.

If you measured 172 ohms with an ohmmeter, you were actually measuring the voltage/current relationships through the diodes, which are nonlinear. The "resistance" measured depends on the voltage applied and the value measured is irrelevant.

If, as you say, that 1 amp results in full scale deflection, then there is a shunt internal to the meter, approximately 1/1000 the resistance of the meter movement so that 1ma of the 1 amp current flows through the movement.

If the meter is labeled 0-2000 amps, it is meant to be used with a CT of 2000:1, so that 2000 A through the primary (1 turn) results in 1 A in the secondary (1000 turns).

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#5

12/09/2017 10:42 PM

Just as a side note - with some equipment the CT does not encircle all of the conductors - GE 3100 drives were in this category. I found the accuracy of the scheme a little questionable and often needed to calibrate with a shunt.

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#6

12/09/2017 10:42 PM

Let's answer your last question first: All transformers work only with AC input and AC output. This includes Current Transformers. So using a battery as a source is not appropriate for testing this meter.

Any "transformer" that puts out DC voltage/current has more components than just a transformer; at the very least it would have a diode to allow current flow in only one direction.

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#7

12/10/2017 3:44 AM

Minggrabber if the scale say "0 to 2000 AmpAC" why would you measure DC with it? What answer did you expect? It is like using a flashlight to determine distance, if the flash reach the wall it is 50 Meters or whatever.

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#8

12/10/2017 5:03 AM

These meters are rated for 1 amp.

It will have 4 markings or 5 markings.Then markings will be 0.2,0.4,0.6,0.8,1.0

Therefore for 800 /1 Amps CT, the markings to be 160,320,480,640,800.

Best solution would be to rewrite the scale markings accordingly.

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#9

12/10/2017 7:02 AM

i think the current in the secondary of the CT is not like the current of a DC voltage source , during the test i used a battery which was 12 volts , i will perform the test again and record the value of resistance on the decade box this time , also to measure the current flowing through the circuit i have used my Fluke multimeter that has resistance too .

i reversed the polarity of the battery during the test and it was still reading the same so reversing the polarity does not have any effect , i will open this ammeter and look for shunt resistor .

in the old ammeter which i replaced there was a card installed on the ammeter CT wire goes through this card in series , this card has diodes , resistor and a capacitor installed on it i will make a schematic and upload soon

when i installed this card on this new ammeter it does not move at all , when connected directly it reads lesser amps than actual ,

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#10

12/10/2017 9:29 AM

Minggrabber, maybe it is time you tell us what you want to do and give us the information and diagram of what you removed and the reason why you removed it, and also how you got to what you use to replace the original.

It worry me that you are working in an instalation where very high currents are measured and to me it seems that you are busy at a place where you may kill yourself.

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#11

12/10/2017 10:44 AM

"i think the current in the secondary of the CT is not like the current of a DC voltage source"

Exactly! The output of the CT is an AC current! You must use an AC meter to measure it, or else convert it to DC before measuring it.

Do NOT attempt to open the meter itself, unless you mean opening some kind of box that contains the meter. In your original post, you said you measured the resistance of the meter as 172Ω. If the meter contained a shunt, that resistance would be vastly lower.

Now you say you replaced an old ammeter, which included a card that clearly converted the AC to DC, so the old meter was a DC meter, even if the scale indicated AC. I suspect that what failed in the old meter system was in fact something on the card, rather than the meter itself.

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#12

12/10/2017 11:26 AM

Minggrabber, you measured 172 Ohms over the meter so there is no shunt as the shunt will be a fraction of an Ohm and on your Fluke you will think the meter is a short. Your Fluke meter leads are most probably bigger resistance than the shunt. Give us some photos and diagrams that you use to find your way. We may advice you, not having all the information, and you may destroy what you have in front of you. How many Ohms are the old meter? Give us a photo of the PCB.

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#13

12/10/2017 3:10 PM

I think what you have is like Rixter's post #9, with the addition of a shunt resistor from + to - of the meter.

The diodes and shunt resistor are on the card mounted on the back of the meter. There would probably also be a resistor in series with the meter to make its effective full-scale voltage match to an available shunt resistance.

As has been pointed out in posts, the problem may be with the diodes etc - not the meter.

You have not clarified the CT rated primary current - only its ratio.

It might be that the CT is 800 amp to 1 amp, with a 1.00 ohm 1 amp continuous rating shunt [which would give 1 volt at the shunt for 800 amps], but the meter scaled to 2.50 volts to allow for reading short term starting currents. This is workable except that the usual standards for metering CTs only ensure accuracy up to 1.2 times rated current. However, this limit is at the rated load resistance (burden) of the CT and a much lower burden allows more overcurrent with accuracy.

There is also a scaling to consider - the DC meter will read the mean value of the rectified AC current - not the r.m.s. value used for AC current. This may be no problem if the AC current measured by the CT is actually turned into DC by a power rectifier (DC gear is usually rated by mean current).

The ratio mean current/r.m.s. current is 0.900 for sinusoidal current and multimeters which are not "True rms" apply this "fix" in their scaling for AC amps and volts.

If the diode current is going up to 2.5 amps or more, this may be too much, too long a time, for the usual 1N540* wire end "3 amp" rectifier - one or more of the four in the bridge may have shorted (If they have used 1N400* type "1 amp" diodes, I suggest you replace with 1N5407 type). In usual rectifier operation diodes "fuse open-circuit" after they short, but since the CT gives a limited current this is unlikely.

So, over to Minggrabber for diode & resistor measurements and more information......

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#14

12/11/2017 2:23 AM

Be careful..Do not energise a CT if the secondary 1amp terminals do not have a burden or are open circuit ..assume it is a low voltage system...

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#15

12/11/2017 3:19 AM

Clue: it is not possible for a <...12 VDC battery...> to supply <...1 amp...> into <...172 ohms...> without some form of voltage multiplier circuit.

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#16

12/11/2017 5:37 AM

"It says on its name plate 1MADC"

Does it say that? Or does it say 1mADC?

m is for milli; M is for Mega: that's 1 billion times as much.

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#19

12/11/2017 7:38 AM

<...that's 1 billion times as much...> only on the other side of the pond...

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#31

12/11/2017 4:51 PM

Why do you Brits keep mixing up million (106) and billion (109)? I laugh every time we go over this. You guys keep acting like you invented the language or something.

Oh hell, you did, didn't you! LOL!

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#21

12/11/2017 10:46 AM

To elucidate PWSlack's comment:

In the United States one Billion equals one Thousand Million (1,000,000,000)

In Europe and South America, and I presume most of the rest of the world, one billion equals one Million Million (1.000.000.000.000).

After all, the prefix "bi" means two (as in Million Million), so the US interpretation is less logical.

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#33

12/12/2017 9:34 AM

I have distinctly heard Brits using the word "billion" when they were referring to 106.

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#45

12/12/2017 2:38 PM

I frequently misunderstand British accents too...

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#46

12/12/2017 2:43 PM

Nah, not you!

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#51

12/12/2017 4:12 PM

Even though both sides of my ancestry are from Great Britain!

You just might remember "BabyBear" here on CR4 a few years back. (I suspect that he has since passed on...) I had zero problems conversing with him in text, but then I made a few FaceTime calls to him, and really had to work at it to understand. I like FaceTime, as I have used lip movement as cues to understanding speech for many years.

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#48

12/12/2017 3:42 PM

In fairness to USA, scientific and technical folks worldwide decided to go kilo, mega, Giga and Tera, logical 103 steps - so USA was ahead of the pack with thousands, millions & billions!

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#49

12/12/2017 4:00 PM

I clearly agree with kilo, Mega, Giga, Tera, etc. and their corresponding powers of 10.

The question remains: does 'Billion' correspond to Giga (109)or to Tera (1012)?

I was always taught that a billion corresponded to (109), while my late father-in-law (from Chile) always insisted that a billion corresponded to (1012). I don't believe I ever asked how much was a trillion...

Please: will a few of our European friends tell me what 'billion' and 'trillion' mean to you?

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#52

12/12/2017 4:33 PM

One billion is a book of one thousand pages each with one thousand rows of one thousand columns of 0's or 1's in each location. Hint- the print is pretty tiny.

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#53

12/12/2017 5:31 PM

When I was a boy in England, billion was 1012, trillion was 1018 , mv was microvolts and mils were 1/1000 inch, milk came in pints that were 8 to a "10 pounds of water" gallon, pints were "20 fluid ounces" and a "hundredweight - cwt" was 8 stone or 112 pounds and a ton was 20 cwt. But things got rationalised and metricated over time.

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#54

12/12/2017 8:41 PM

Thanks.

...so I gather that a billion is now accepted as 109 and trillion is now 1012 on both sides of the pond?

We still use mils =1/1000 inch. Plastic sheeting is commonly specified in mils of thickness. For metal foils (such as used to fabricate the honeycomb of my avatar), we normally specify the thickness in 'thou', which also means 1/1000 of an inch. I continue to be amazed that the aerospace industry here still uses the inch as the standard measurement unit of distance.

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#17

12/11/2017 6:09 AM

Minggrabber, when you supplied 12V to your meter it was passing 14mA through a coil that you say yourself according to the markings need only 1mA for full scale. You did get full scale with 14mA, what do you want. We can not help you if you just keep following your own head. Give us all the answers on the questions we asked you and some photos of the PCB and the old meter and the resistance of the old meter. Tell us why you are doing these tests. Is it to fix something broken or what?

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#18

12/11/2017 6:53 AM

Seems your meter needs callabrating;some i,e low entry do yourself others high entry have to be sent away.

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#20

12/11/2017 8:01 AM

If it was me, with my next to zero experience of this meter, I would inclined to set the decade box to it's highest resistance (with highest safe voltage available)-(the simplest way to get a constant current circuit) with 1 mA flowing in it. Then measure the current with your meter.

Give or take a few uA's (depending on the actual internal resistance of the meter) the current through your meter will be 1 mA.

The actual reading should help work out what is going on with your meter.

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#22

12/11/2017 11:58 AM

Dear All ,

Today i repeated the test with a 12.99 volts battery (measured on my fluke) and 13k resistance on the decade box , this will allow exactly 1milliamp current to flow in the circuit from V=IR ohms law .

observation - meter is deflected full scale . just touches the 2000 A point on its scale . so the meter is 1milliamp full scale meter

last time i have added my digital fluke multimeter in series that was making all the readings wrong

https://www.dropbox.com/sh/4nbj8x9lv6e8m40/AABmvghUIt6b_Sha1emR5Qi9a?dl=0

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#23

12/11/2017 12:17 PM

Minggrabber, do the test again with the Fluke on mA scale in series and you will find the same result. The resistance of the Fluke can be ignored and will change the 172 Ohms with only fractions. You will find that you did something wrong.

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#24

12/11/2017 12:49 PM

"so the meter is 1milliamp full scale meter"

And furthermore, it is a DC meter mechanism, since the battery did make it deflect. This means it MUST be used with some form of AC>DC converter (like the card used with the original meter).

When a DC meter is connected directly to a sinusoidal (or other symmetrical wave shape) AC source, it should indicate zero. Thus it is to be expected that this meter should not deflect when connected to the output of a Current Transformer.

Note that zero deflection does NOT necessarily indicate zero current; it simply indicates zero average current. The average value of an AC sine wave is zero, since the signal is positive half the time and negative by the same amount the other half of the time. It is quite possible to destroy a DC meter by connecting it to an AC source, while the meter always indicates zero.

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#66

12/14/2017 9:45 AM

Torreg
Connect the two terminals of the voltage source to either side of the 1K ohm resistor.
Connect the 2 terminals of the ammeter Across the resistor or in parallel, this will allow the current flowing in the resistor to be determined.
Switch on the voltage supply and set it to 1 V.
Calculate the expected value of current using Ohm's law Ohm's Law states v = IR where V is the voltage I is the current and R is the resistance in this case, the expected current is I=V/R and is equal to 1 mA. Compare this with the measured value shown on the ammeter. If the values are different, adjust the calibration knob (sometimes a sunken screw) on the ammeter to match 1 mA.

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#68

12/14/2017 1:14 PM

NO!

If you connect a meter across the 1kΩ resistor to determine (calculate) the current flowing in that resistor, the meter must be a voltmeter.

If you connect an ammeter as described, you have that meter connected directly across the two supply terminals, and you will simply be measuring the output current capability of the voltage supply at a setting of 1 Volt, or, more likely, have just blown out your ammeter. Since an ammeter always has a very low resistance, that 1kΩ in parallel with it has almost no effect.

If you are able to set the voltage source to 1 Volt, then that voltage source must already include a voltmeter for setting the voltage, so the second meter is of no use, except to verify the reading of the built-in meter.

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#25

12/11/2017 1:05 PM

i will make a schematic of the card and upload it here

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#26

12/11/2017 2:06 PM

A photo or two of the card and how it connects to the meter would also be good.

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#27

12/11/2017 2:46 PM

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#28

12/11/2017 3:02 PM

Very good. I only saw a single photo in each dropbox folder in your previous post.

I'll look again when I have time to analyze...

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#29

12/11/2017 3:37 PM

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#30

12/11/2017 4:48 PM

It is happening because you have DC source connected, and the back emf drops off, so the forward current pegs out. DOH!

Go back and read the manual this time, and stop mixing carrots with oranges in the cart.

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#32

12/11/2017 10:31 PM

All you have to is look at the photos. There are three main components in this circuit: the CT, the metering circuit board, and the meter.

The 600:1 CT produces 1Amp AC when 600Amps AC is flowing in the conductor it is surrounding.

The 1Amp AC is applied to the circuit board directly to R1, the 0.5Ω 10Watt wirewound resistor, which provides the burden for the CT circuit. The voltage drop across R1 is then fed to the 4 diodes to produce a full-wave pulsating DC output that is smoothed by C1. This DC voltage is then applied to the series-parallel combination of R2, R3, R4, and the meter resistance. Those resistor values are calculated to provide 1mA DC to the meter coil when the Full Scale current is applied to R1.

Your original meter required 1mA DC to produce a reading of 2,000Amp AC; i.e., the CT secondary current would have to be 2,000/600 = 3.3Amp presented to R1. Your new meter requires 1,500/600 = 2.5A to produce a full scale reading; therefore, you will need to change the values of R2, R3, and R4 so that 1ma will flow through the meter coil.

This ends the FREE portion of this session, you will need to use Ohm's Law to calculate the values, or provide your PAYPAL account for further information.

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#34

12/12/2017 9:37 AM

There were photos? I missed part of the lecture. Where should I direct payment?

Just include your routing and acct number, and I will be happy to un-munerate you.

(this was spoken in jest, as I would never actually steal from you, or anyone else on here, well, OK, maybe Lyn).

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#35

12/12/2017 9:46 AM

Incorrect.

Old CT 600:1/new CT 800:1 is 600/800 = 0.75

Old 1 mA meter scale 1500A/ new 1 mA meter scale 2000A = 1500/2000 = 0.75

Conclusion - nothing needs adjusting - unless high accuracy is required & precision testgear is available to re-calibrate with new CT and meter errors.

The volt drop across R1 at 1 amp rms is 0.9 x 0.5 volts = 450 mV mean DC. (0.707V peak).

Note 0.5V to 1.25V [800 - 2000A] is inadequate for linearity with two diodes of bridge dropping 0.5V each. The resistor R1 must be in the DC circuit of the bridge.

The following is the circuit got from the photographs posted.

The capacitor is on the AC side.

The R2 in parallel R3 combination plus the meter resistance must drop 0.450 V x 1500/600 = 1.125 V at meter full scale. Since meter is 1 mA, that requires 1125 ohms.

With meter resistance of 165 ohms, R2//R3 would be 1125 - 165 =960 ohms [ignoring R4, for which the purpose & connection can only be guessed at present].

I cannot read value of R2 with certainty, but with R3 at 20 kilohm it is probably 1k, or near, precision type.

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#36

12/12/2017 10:16 AM

Why is the capacitor not down stream of the bridge? I don't see this correctly?

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#38

12/12/2017 12:56 PM

the capacitor is in series 500--l l---501

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#39

12/12/2017 1:24 PM
1. Cannot pick out terminals or wires 500/501, but connection of multiple devices in series in CT circuit, so each receives CT current, is frequently done.
2. Is it correct to suppose only 4 connections used on meter PCB, so R4 is not used?
3. Any comment on my post #35 with circuit diagram?
4. Have you measured diodes etc on PCB to see if any faulty/short/open circuit parts causing your problems?
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#40

12/12/2017 2:09 PM

James Stewart,

By my reading of image IMG3895 of PCB, C1 is connected to cathode of D2 and anode of D4. The view of back of board shows other end of C1 connected to D1 cathode, D3 anode.

So the capacitor is across the bridge AC side. You do not need a capacitor on DC side for a rectifier to work and I would not put one there in this circuit because it would push towards peak reading rather than mean/average & cause error. The DC meter does the averaging in a mechanical way.

I guess the capacitor is to avoid a high impedance when current is almost zero and no diodes conduct. With the diodes alone I would expect almost a step change in voltage as current changes polarity - diodes can generate RFI by charge recovery.

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#41

12/12/2017 2:14 PM

Where are you getting this image, I guess that memo never made it to my desk.

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#43

12/12/2017 2:23 PM

Post #27 by minggrabber, the OP.

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#42

12/12/2017 2:20 PM

SO, the capacitor may be there to prevent RF feedback into the CT? Or is just there to remove noise in the local instrument environment?

You seem to indicate that the C value in parallel with the meter would result in V drift? Why wouldn't the meter coil bleed that off?

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#47

12/12/2017 3:08 PM

The CT will not care a damn about RFI. I have personal experience of having to fit RC filter at input of electronic frequency indicator on generator - reading of which was jittering about without it. Spikes near the zero crossing can mess up frequency/volt converters. The OP's wiring diagram shows a volt & frequency control box - maybe motor drive and the meter CT current also goes into that box.

The meter will bleed off charge from capacitor, but charge/second is current and current moves the meter needle. The capacitor steals current from the meter when the voltage is high and gives out, but with a lower voltage [and hence current], later. I would be surprised if the give and the take are the same.

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#44

12/12/2017 2:26 PM

I agree that the Capacitor is on the AC side, but I don't quite agree with your diagram.

Here's mine:

Likewise, I'm not sure about R2. After enlarging it considerably, I see it as a 10MΩ 10% value, but I could easily be mistaken. Then why a 10MΩ in parallel with 20kΩ?

I have the impression that R4 can either be used or not, depending on the meter movement. I've drawn the circuit to roughly correspond to the board terminal locations. None of the terminals connect to the D3,D4 junction, as your drawing indicates.

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#50

12/12/2017 4:10 PM

You are right, no terminal for D3, D4 junction - as you spotted, the terminal holes could be made to fit the + - studs on two physical sizes of meter, with different resistance requirements.

The OP has not sent a picture with the board bolted on the back of the meter. I agree it is very difficult to read R2, my mind fogs up on the colour code for precision resistors. It just occured to me that R2 may have been "cooked", changing the colours and its resistance.

I asked "minggrabber" to check the diodes and other PCB parts for faults, so we await an answer.

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#60

12/13/2017 2:19 PM

Yes, I mixed the values for old and new. A CT ratio 1/3 higher will produce a full scale reading 1/3 higher, but since the meter reads 1/3 higher all is good.

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#37

12/12/2017 12:52 PM

the card (pc12) is connected like this in series , only one wire of ct goes through this card the other goes direct to module

there are three CTs but this is only connected on one of the phase for measurement and then the singnal goes generator ac control module which controls voltage and frequency

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#55

12/12/2017 11:13 PM

The upper two holes are bolted on the ammeter

the lower two holes are not used

the two holes on the right side of the card were connected to the CT as i have marked on the card , CT has two wires only one wire comes to this card and then goes to a generator control module

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#56

12/13/2017 8:02 AM

MingGrabber,

You have shown the meter is 1 mA by test. Some measurements are needed on the PCB to prove all the parts are good. Also it is clear that R4 is unused in your application.

First you need to measure the resistance of R2/R3. As calculated, this is expected to be 960 ohm - it could be more/less to allow for R1 tolerance, which could be +/- 5% around 0.5 ohm.

To check out the diode bridge, R1 and R2/R3, it would help to mount the board back on the meter.

You will need a source of DC current at about 1 amp. This could be a variable bench supply 12V DC with a 10 to 12 ohm 12W ceramic wirewound resistor, or better several 10W resistors in series.

High power resistors can get very hot, maybe 250'C at maximum rating. Their temperature coefficient of resistance is not negligible e.g. 300 ppm x 250 = 75000 ppm = 7.5% resistance change. If I have a choice, I use the "finned" gold ones designed to mount on a heatsink. This minimises temperature rise and drift during measurements.

If a bench supply is not available, a 12V car battery can be used.

Put an ammeter on as a load (commonly, they will actually drop 0.4V, even 1V on 1 amp range) and check you are getting about 1 amp.

Next feed the "1 amp", resistor, meter source into the AC (CT) terminals of the PCB. Noting that the total resistance R2/R3 and meter is 1.125 kohm, full scale voltage 1.125V. if the current is 1.000 amp, the voltage across R1 would be 0.500V and the reading on the 1 mA meter would be 2000 amps x 0.5/1.125 = 800 amps [ ideally if R1 were 0.500 ohm, meter + R2/R3 = 1125 ohm, meter fullscale 1.000 mA].

Two diodes e.g. D1 and D4 should be forward biased (black bar, cathode, negative) with a volt drop about 1 volt +/- 0.1 volt. A shorted diode would have much lower volt drop.

Reverse the polarity of connection to the CT terminals, the 1 mA meter should read the same (or in proportion to source current, if that has changed on ammeter). The other two diodes should now be forward biased by 1V +/-.

This test should indiicate if the circuit is working correctly.

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#57

12/13/2017 9:19 AM

R2 could be a 1070 1% br bk v bn bn. Make the combination 952 Ohm. Why would the firs brown ring be brown but the other 2 faded?

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#58

12/13/2017 12:27 PM

Maybe that resistor is faulty, bad connection under end cap. About 1 volt & 1 mA is too little power to do damage. But an overvoltage fault? And whatever happened burnt-out the meter coil?

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#61

12/13/2017 2:43 PM

"...bad connection under end cap." I thought the same, but the blue body color appears the same at both ends, so I don't think that is the problem. This is a 30-year-old circuit board. Perhaps the paint/ink used to mark those last two bands was different, and simply faded with age.

Here are similar resistors of 1.00kΩ and 10.0kΩ, probably of similar vintage, and similarly enlarged, which have never been used:

All of their browns are very similar... Of course the 4th color on the left resistor (10kΩ) is red, although not a very bright red. I observed many years ago that bright reds rarely keep their brightness. Look at old cars and fire hydrants...

And here is my latest circuit diagram:

I've removed the values from R2 and the meter, since I'm not sure about either, and added the control module. I don't know whether the CT was actually connected to the upper or lower terminal, but it should make no difference.

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#62

12/13/2017 3:23 PM

You have pictured resistors in the five band code, 4th is multiplier and fifth is tolerance.

Thus, you have shown on left, 10K 1%, and on the right 1K 1%.

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#63

12/13/2017 3:31 PM

Correct, as indicated in the text below the resistor photo. I did list them in the opposite porder

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#59

12/13/2017 1:19 PM

That R2 is a real dilemma!

Fortunately, the original images in DropBox were very good quality, allowing considerable enlargement. Obviously, the colors will be seen somewhat differently on different computer screens, so doubt will remain.

Once enlarged to this level, there is little doubt that R3 is 20kΩ and the unused R4 is 180Ω, both 5% tolerance.

R2 definitely starts with Brown, Black, so 10... I think I see the third color as Blue, or possibly violet, so 6 or 7. Those last two bands look like Grey or Silver, but Grey is not used for multiplier or tolerance on 5-band resistors, so I have to interpret them as Silver (unless they are faded, as you said). Silver in the 4th band means a multiplier of 0.01, and in the 5th band means 10%, Which would make it 1.06 or 1.07 Ω, but a 20kΩ in parallel with it wouldn't even change the third digit, so that does not make sense. Furthermore, a 1Ω resistor in series with a 165Ω meter would be a very small correction!

Thus I am led to accept your idea of those last bands as faded Browns and the value of 1070Ω ±1%. I get 1070 in parallel with 20k to be 1016Ω.

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#64

12/14/2017 1:37 AM

dkwarner, I just did a "you need 20x 20k in parallel to make a 1k therefore a 20k in parallel with 20x 20k resistors is 21x 20k resistors in parallel and that is 20k/21 is 952 Ohms". Your answer is more accurate.

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#65

12/14/2017 7:55 AM

Some good detective work on R2!

Taking my design centre value of 1125 ohm for meter circuit and estimating the tolerances effect on 1125 total resistance.....

Shunt R1 +/- 5%, 56 ohm

Current Transformer +/- 3%, 33 ohm

meter resistance 165ohm +/- 10%, say 15 ohm

Meter full scale mA +/- 1%, say 11 ohm

Total 115 ohm, add to R2/R3 combination 960 ohm [1125 minus 165 ohm in meter] gives 1075 ohm for R2. The meter nominal resistance might be 150 ohm.

So 1070 ohms seems a good value to allow a calibration by varying R3.

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#67

12/14/2017 12:54 PM

Sorry! I don't follow that... R1 is 0.5Ω; where do you get 56Ω?

Here is the circuit with the terminals and R4 removed, followed by an equivalent circuit considering the diodes as shorts.

1 Amp from the CT will produce just under 0.5 V across R1 and across the M1/R2/R3 group. 1 mA through the 165Ω of the meter will produce a 0.165 V drop, leaving ≈0.335 V across the R2/R3 pair. The same current that passes through the meter must pass through the R2/R3 pair, which tells me that the pair should have a resistance of ≈335Ω. There is no way I can see R2's colors indicating anywhere near that! Where did I go wrong?

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#69

12/14/2017 4:00 PM

My second sentence of post had "estimating the tolerances effect on 1125 total resistance.....".

So 5% to allow for a +5% value of R1 is 1125 x 0.05 = 56.25 ohms.

You have supposed that 1 amp/0.500 volt r.m.s. corresponds to full scale of the meter (1 mA).

For the old scheme, CT is 600:1 and the meter is scaled 0 to 1500A - a 2.5 times over-range on CT. The new scheme has 800:1 CT and meter scale 0 to 2000A, same over-range, 2.5 :1.

The meter full scale must be 2.5 amps r.m.s. - but, since the mean rectified D.C. value of a sine voltage is 0.900 of the r.m.s., meter full scale must be 0.5 ohm x 2.5 amps x 0.9 = 1.125 volts DC, requiring 1125 ohms total nominal, including meter at 1 mA.

The value of R2 must allow for the maximum resistance required for the "worst case" effect of component resistance variation.

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#71

12/17/2017 3:27 PM

Got it! (I think...) Thanks!

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#70

12/14/2017 4:44 PM

Apologies, minggrabber!

My arithmetic is wrong in post #56.....

2000 x 0.5/1.125 = 889 amps, not 800 amps.

This is why engineers check & double check before finalising a design or test schedule!

Fortuitously, dkwarner "rattling my birdcage", made me realise that 1 amp DC fed into the PCB must take account of the mean/rms = 0.9 factor and could not give a reading of 800A equal to CT primary rating.

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