How to Connect DC AMP Meter With Battery

You need a bridge rectifier (or four diodes wired as one):

Make sure the diodes are rated for your highest expected current.

You will lose about one volt, due to the forward voltage drop of the two conducting diodes, whether charging or discharging. Try to select diodes with as small a voltage drop as you can find.

While that circuit will show a reading in amps, and the ammeter will not go below zero thereby achieving one of the original poster's criteria, the way to determine which way the charge is going, which is the original poster's second criterion, would be to measure volts on the battery, volts on the UPS, and to compare the two, perhaps?

Re-reading the OP - I'm not sure that that was a criterion - it's a bit of a language thing, I think. I believe he meant "indicate the magnitude of the current, irrespective of whether it's charging or discharging".

I'd imagine he should be able to deduce current direction from status lights on the UPS.

I agree that the OP appears to not care about identifying battery charge status. A simple differential amplifier circuit across the diode bridge combined with a window detector could easily identify the three battery states; charging, charged, and discharging.

Oh, since the ammeter only reads 40A, then this sets the maximum current required to be handled by the diode bridge.

The 40A meter limit did occur to me - but best to be on the safe side. Then again, depending on the relative cost, it may be better to sacrifice a diode rather than bend the meter .

May get away with LEDs (+ series resistors) across the meter & bridge for indicating current direction - there'd probably be enough voltage dropped across the bridge diodes and meter coil/shunt to light them. Need details of the meter to do the sums.

You can also add a tri-state LED where a forward state condition can give you a green color, a reverse state condition will emit another color, thus satisfying all stated conditions.

I was thinking about those options but realised that to minimize the change in the battery charging voltage and UPS available power the diode bridge and ammeter must drop the least amount of voltage. So I would first choose using Schottky power diodes. But more to the point as far as a directional flow indicator, the voltage drop across the bridge and meter should be less than the voltage drop of even an infra-red LED used in an opto-coupler. This is why I proposed the differential amplifier, window comparator circuit.

Jhon, This is an academic solution, not very practical.

you have 40 amps flowing always thru two diodes,

assuming the drop over one diode is 0.7 V, you have 1.4V drop over initial 12V...12 %, too much, not to talk about heating, 1.4V x 40 amps = 56 watts out of the 500VA ....too much looses just for reading the charging current

I am affraid for that scale any solution will be expensive and not efficient,

I can suggest two relays to revert the polarity of the ammetre connexion when the mains is present and the UPS is charging

Exactly why a Schottky diode like the 65PQ015PbF with its 0.46 V_f @ 65A should be used. Now one can validly argue that this is not a cost effective method in comparison to obtaining an ammeter that can display both polarities as others here have proposed. But I repeat my earlier comment that the OP asked for a method to use the existing unipolar ammeter. Using relays in the current path will mean brief losses of output power. Precisely the task a UPS is supposed to prevent and thus a design destined to fail. IMHO.

Now if you the OP is willing to use something other than this ammeter, I would suggest a calibrated resistive shunt between UPS and battery pack that a differential bipolar LED or LCD panel voltmeter reads the no more than .2V drop @45Amp can then be scaled down to a 45mV difference at the high input impedance terminals of the voltmeter. This will likely be cost effective, very accurate, produce no drop outs and can be well calibrated.

no outage will occur on UPS output

first relay to short circuit the ammeter , second step second relay to change the polarity of the two wires connecting the ammetr, not easy but much better than burning 40 watt just to read the charging current

Ah yes, relay timing circuitry that must always work in the same sequence flawlessly. Zero ohm contact resistance that never oxidize. Relay coils that take no power to operate. Also moving parts that are less fallible than non-moving parts.

Yeah, I am being a little snarky here. Please forgive me.

The advantages and drawbacks of any approach to read the charge/discharge currents should be well known by and accepted by the one designing and building the circuit. This is neither myself or Le_Noble. My preferred choice would still be the shunt and differential voltmeter and neither the diode bridge or relay switch.

redfred, John

The OP is stuck to that meter he already does have and want to use it, he may have his own reasons to start from here .... not to throw it away and put a shunt with zero centered scale, or a DPM as John suggested at 20 STP ( for a 500VA UPS costs may be 70 STP)

any solution to help him save his precious ammeter will be better than burning in best cases (with schotky diodes) 36 watts in the diode bridge, , BTW, such bridge needs a heat sink about 1 pounds of aluminium to cool it down, and with 0.9 volt drop, his UPS will shutdown on low battery signal much much before he get any power from his UPS

John,

sorry if you felt i am underestimating your solution, nothing bad in "academic", I mean by academic too elegant to be applied on such small UPS.

me too if I was going there, I wouldn't start from here ...'

I really do not understand this apparent arrogance. You seem to not accept the idea that there is more than one way to achieve the OP's goal. Or am I misreading how you seem to be taking this.

Now if we're going to continue this, I'd like a little parity in your proposal. For the reasons you've stated, you dislike the .9V drop across two diodes that a bridge will produce. You prefer relay contacts for they will have a lower impedance drop. Well at 0.9V/40A=0.0225Ω the total effective series resistance of the two diodes combined seems to me to be a reasonably low impedance. To match this impedance your relay contacts that are capable of handling 40A of current must each be less than 0.01125Ω. I've provided a manufacturer's data-sheet for one of many available Schottky diodes capable of handling this bridge circuit. Will you provide a manufacturer's form C relay data-sheet capable of handling 40A that has less than 11 mΩ of contact resistance. Remember you will have to find a form C or B contact to not include the power dissipated by the coil(s) to the added battery power dissipation calculation.

Don't forget since it won't be known what the minimum UPS current draw will be, the relay contacts may also have to provide a good contact with only 100mA being drawn through the contacts. An easy feat for a diode bridge even though the voltage drop will be similar.

You know, if I was actually being paid for this little foray I could probably configure a switched MOSFET bridge that had less than a 0.1V drop at 40A. But I'm not, so I won't.

And I really do not understand how you see an apparent arrogance, ( even I tried my best to keep it hidden :) )

...yes you misread it, I think the language difference is the cause, I am not english native speaking, Sorry for this misunderstanding,

we are debating on an issue that will never happen in professional word. Just to keep our guest happy with his ammeter.

As I said,as this will not happen in professional world, both solutions are not ideal, the ideal is a shunt with zero centered scale or digital DPM. [p]

and as John DG said, I would not start from here

AS for your challenge, it is easy to find a 4 poles 10 amps form C relays, contact resistance is typical 20-30 milli ohm, with the 4 in parallel you will have 5 to 8 m-ohm contact resistance, Chinese suppliers can offer even more ampicity, not sure about NEC and UL types

again, sorry for the misunderstanding

Le Onble, have you heard the old, old joke? - this version's set in Ireland in my part of the world, but I assume that versions exist everywhere.

"A traveller, somewhat lost, stops to ask directions from an old man, leaning on a farm gate. 'Which is the way to Dublin?' he asks the old man. 'Well, Sur, if I was going there, I wouldn't start from here ...' "

There are many ways to skin a cat. If you're spending out on relays+bases+wiring, you may as well say "BUGGER IT - I'll do it another way" - and throw the meter away.

How about a cheap'n'cheerful DVM module (this one costs about £20), with a 5mR sampling resistor? The power loss would be 40 X 40 X 0.005 = 8W at full current (plus a few μW for the meter supply).

As we have no details of the ammeter being used, we can't comment about power lost or saved.

I assure you that there was nothing "academic" about my initial solution - it was absolutely practical, and was the easiest way to achieve the OP's requirement. To reiterate, there are many ways to skin a cat.

Would you build this circuit to charge your batteries ?

The point is you didn't understand [or read] the question [on other threads too]

The OP already has a charger & isn't looking for exact readings, just some indication of the magnitude of the current flow, without having to buy more than a minimum of components

Before you go on a tirade like this you really should carefully read the OP's question and others already posted answers. The OP asked how to use the ammeter that he already owns to measure the current flow in both directions, not what extra equipment should be purchased to replace the ammeter. As far as the complication of the voltage drop of the diodes I want to point out three things. First, the lower voltage drop of Schottky diodes has already been suggested. Second we do not know the charging circuitry topology. So diode voltage drops may not be a real problem, calm down. Third, a review of the Shockley diode equation

shows that a small current will still go forward even when not at the traditional turn ON voltage. So the sealed lead acid batteries will still get a trickle charge.

Lastly, engineering is not all about you demonstrating superior technical knowledge to a customer or an international group of bloggers. A good engineer should always keep a customer's financial and power budget in mind when trying to solve a problem, particularly when the customer specifically asks for one to be conserved. This will occasionally mean that a potential customer may ask for something that cannot be done in your opinion. But we only have here the piece of this puzzle that the OP has given us. To claim that anyone's approach cannot work from such a small amount of information just seems a little pushy to me.

If the batteries in the circuit you have supplied are to be correctly charged this circuit will not work.

As you say Diodes have a "volt drop."

You will constantly undercharge the batteries by 1.2 v dc with silicon type and 0.4 v dc with germanium type.

Fitting very large condensers instead of batteries would work with this circuit though.

This circuit would destroy the batteries.

They would not charge properly.

once again read the question

This is an absolutely useless circuit. What are you trying to achieve with this? Do you know anything about electronics or the charging of batteries. Disgraceful design and Admin, if they actually knew anything, would remove this before some one gets hurt.

Did you read and understand the original post?

If so, you would not need to ask what I am trying to achieve.

Please explain - in the light of your understanding - how the circuit in post #1 could harm someone.

You need a Zero Center ammeter.

This type of ammeter can measure current in both directions. From what you are saying, the ammeter tells you the charge current. However, when your UPS draws power from the batteries, the current reverses. This is why the meter you have "goes in reverse direction". With a Zero Center ammeter you will be able to read both the charge current and the discharge current.

Quite true but the OP wishes to use the ammeter he/she presently has.

You can use the analog ampere gauge installed in cars, with the following specs: 0 to -/+ 50 Amps. with the 0 at the center of the gauge. (+) readings indicate when battery is charging and alternator is working. (-) reading, indicate not charging condition or defective alternator, and (0) when engine is off. Meter range goes up to 50 or 60 DC Amps.

Thanks to posters. It means i have to spend some more money to buy a relay etc. to achieve my goal.

A zero centered meter, better with shunt, not direct connection , it is cheaper than a DPM .

Do not take my word as the truth, see other posters ideas

Good luck

You may have to spend some gas money going to automotive junk yard to get hold of an amp gauge.

An Amp meter needs a "shunt" or "hall effect device" to work.

Technically an ammeter requires neither of these, yet nearly all tangible ammeters do use these current to voltage phenomena to indirectly measure the current. This is a subtle but important distinction to keep in mind when one does circuit simulations.

Most ampere meters work on these principles.

How would you measure current ??

I know redfred is fully capable of answering for himself, but as I'm here ...

"Traditional" methods of measuring current (moving coil or moving iron meters) require that the current being measured passes through a coil of wire - which has a finite resistance. Methods employing a voltage-measuring device need a voltage (proportional to the current) to measure - this to be generated across some form of resistor.

In either of these cases, the act of measuring the current introduces resistance into the circuit - which must be taken into consideration when simulating or modelling the circuit, if the model is accurately to reflect the circuit being modelled.

I think redfred was just pointing this out.

[Aside - current measurements using Hall effect transducers step around this problem to a large extent.]

Exactly John. Any measuring device will perturb the system it is trying to measure, ammeters are no exception. In contrast, a simulator display of current is an unperturbed value of the simulation current. Now indirect measurements of current by measuring some other phenomena can be extrapolated back to the actual current. This is what I was trying to convey.

All this diagram displays is that you recommend fitting 2 diodes between battery and charger which no matter elegantly represented with PCA software is still going to not charge the batteries correctly without fitting such devices as a sense wire.

I know plenty of chargers which do provide such a facility so why not just include it in the diagram or advice.

Regards.

Philip.

WHEN WILL YOU GET IT INTO YOUR ***** (adjective deleted to pre-empt Admin intervention) HEAD - IT IS AN UNINTERRUPTIBLE POWER SUPPLY NOT A ******* (adjective deleted to pre-empt Admin intervention) BATTERY CHARGER.

STOP MAKING DEROGATORY COMMENTS AND TELL US ALL WHAT DEVICE YOU WOULD FIT IN ANSWER TO THE ORIGINAL POSTER'S QUESTION. IF YOU CANNOT DO ANY BETTER, SHUT UP AND GO AWAY.

AND YES, I AM SHOUTING! I'VE TRIED BEING REASONABLE, AND IT CLEARLY JUST DOES NOT WORK. I AM NOW ANGRY.

Calma Calma......

I said already I would fit a Hall effect device and associated circuitry and then you can even tell if it charging or discharging....which would be a proper engineering solution and not a bodge.

Sorry I do not understand these "***" things.

I am in Portugal and we use some letters with accents but I have never seen these words before.

If you are worried about : pre-empt Admin intervention please e-mail me at my inbox thingy then I can reply to your queries about battery chargers and UPS's as they are in fact the same thing just with a mains fail detection circuit Incorporated into the UPS where upon the batteries will then discharge into the circuit for which it is intended.

The clue to this device is it is a UPS...i.e. If the mains supply fails the power will still be available...hense the battery it is like a back-up affair and if it has been poorly charged...for instance through 2 unnecessary diodes then the system will fail premiturly due to this fact....that is all.

Uniteruptable Power Supply I beleve they are properly named.

"I said already I would fit a Hall effect device and associated circuitry ...".

In your statement you have already failed to answer the question posed^[1], as you have included "associated circuitry".

How are you intending to connect the 0-40A ammeter to the Hall effect device?

Please do tell, as this is the really interesting bit ...

I assume you're going to add an external 0-40A power supply, modulated from the output of the Hall effect device using your "associated circuitry" so that the meter will read the correct magnitude of the current? Will this not reduce the efficiency somewhat?

I think you must include this external power supply as part of your "associated circuitry", as the meter would not operate without it.

Please supply a suggested circuit diagram, indicating all your "associated circuitry". Failing this, please describe how you intend to implement your proposed solution. A list of the items required would be very illuminative^[2].

[1] From the Original Post: "Can any body help me out how to use this ampere meter so that whether current follow from battery or to the battery, AMP meter reading should correct? What device should I connect?"

[2] My items list: 1 bridge rectifier.

.....[2] My items list: 1 bridge rectifier.....which would not charge the batteries properly.

So it is not a good circuit.

In reply to :

How are you intending to connect the 0-40A ammeter to the Hall effect device?

Not directly but with associated circuitry as I say.

I answered the bit which said :

"Can any body help me out how to use this ampere meter so that whether current follow from battery or to the battery, AMP meter reading should correct? What device should I connect?"

And I still would fit a Hall effect device with associated circuitry so that the batteries would be charged correctly and that the ammeter would give a correct reading.

It is not so difficult to use an hall effect device.

In reply to : A list of the items required would be very illuminative^[2]

A cheap and cheerful clamp meter would do...they are Hall effect principle devices.....so we have :

Plastic case.

PCA.

LCD display.

power supply (PP3 maybe.)

some wires (conductors covered with insulator.)

A hall effect circuit.

on/off switch.( A conductor which you can make and break.)

brand symbol.(optional.)

Bin to put ammeter in and a warning sign to the effect of :

DO NOT FIT A RECTIFIER BETWEEN BATTERY AND CHARGER.

This sign might have to flash on and off in a highly visible fashion to try & prevent non competent persons but then depending on display type may require more associated circuitry....shall I proceed in more explanation or is that enough ?

Your Sincerely.

Philip.

This is getting very tedious.

There's so much crap here, I'm not sure where to start - anyway, life isn't long enough.

Let's try this approach - which returns to the point you miss time after time after time. You have not answered it yet , so let's try to narrow it down and get this one cracked, so that you can prove what a wonderful electronics designer you are.

-----------------------

1) You are given an ammeter, which shows a full-scale reading when passing 40A between the positive and negative terminals. This is stated in the Original Post. Do you agree?

2) It is required (by the Original Poster) to use this meter to indicate the magnitude of the current into or out of the uninterruptible power supply (UPS) battery. Do you agree?

-----------------------
Hint: if you disagree with either of the above, your entire premise is unfounded.
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3) To use this meter to indicate the current accurately requires that the current passing between the battery and the rest of the UPS circuitry is equal to the current passing through the ammeter. Do you agree?

4) You propose (in your post above) to power the ammeter and "associated circuitry" using a battery. Let us simplify the question by ignoring the power consumption of your Hall effect device and all its associated circuitry. Concentrate on the current required to move the needle on the 0-40A ammeter. Do you agree that this a fair simplification?

5) You suggest (in your post above) to use a PP3 size battery. This typically has a capacity of, shall we say, 1200mAh (using the maximum figure suggested here, relating to a Lithium type PP3). Do you agree?

6) The probable maximum charge rate for the 120Ah lead-acid battery (using the "C/8" rule) is 15A. Do you agree?

7) The 0-40A meter would need 15A flowing through it, supplied by your PP3 battery, in order to correctly indicate the charging current. Do you agree?

8) The maximum possible time for which a 1.2Ah (PP3) battery could sustain a 15A output would be 1.2 ÷ 15 = 0.08 hours (4.8 minutes). In practice, the time would be rather less. Do you agree?

9) A circuit implementation requiring battery replacement after about 5 minutes is crap. Do you agree?

-----------------------

Oh, and while you're at it, please give an approximate cost for your suggested solution.

And please explain what a "PCA" is, and why you need one.

http://www.choiceful.com/choiceful-id-103932-Skytronic-Digital-Clamp-Meter.html

A PCA is a basic standard acronym used in electronics.

This post was edited because it was an attack on another user. Please review the CR4 Site FAQ and the CR4 Rules of Conduct.

I've avoided this irritating farce since you started the other thread and I lost control there. Now that John has also lost control with you, (I'll grant you John you held your tongue better than I did but I did get 4 GAs before the admin correctly removed my comment.) I would like a simple answer that must be known to say that the addition of two forward conduction diodes will be such a large percentage of the charge and discharge voltages to render this a bad design. How many nominal 2.0 volt lead acid cells are in series in this UPS? You must know this portugalphilip to be able to say that this is a bad design. Now a qualified and experienced engineer would know that assuming anything as simple as operating voltages would be bad engineering. So tell us the operating voltage of this battery pack this guest has. While you're at it, maybe you could tell us the model number of this UPS.

The emoticon I posted was to show my appreciation for Red Fred's comment

fairly straightforward

care to explain your reply?

or why you continue to post insults in portuguese?

or even to actually answer John's questions

PCA as far as I can tell is the name of a company certainly not a common acronym in the rest of the world

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