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Message |
Verell Boaen (Verell)
Intermediate Member
Username: Verell
Post Number: 1019
Registered: 5-2001
| | Posted on Wednesday, July 30, 2003 - 8:38 am: | |
Hans,
Right on.
You're saying the same thing I posted earlier, If you need to run w/o the ammeter, jumper its leads w/ a heavy gauge wire (no lighter than the leades themselves). Doesn't make any difference whether the shunt is internal or external with 12gsge wires being jumpered. You need the jumper if the shunt is in the meter, don't need it if there's a shunt in the wiring, but the jumper won't hurt anything either.
BTW, IMHO GM used 18gage becaues it's heavy enough to hold up to vibration(& production line handling) & because it's usable with their std crimp connectors.
BTW, The other fuse block thread reminds me of one VERY IMPORTANT SOLDERING STEP OMITTED BELOW:
The joint must be bright metal before you try soldering it.
Solder will NOT cut thru corrosion or form a good bond to it. You must mechanicly clean everything you're going to solder with a small wire brush. The joint must be bright metal before you try soldering it.
I use a stainless steel wooden handled 'toothbrush' style brush. You can easily cut it down to just a few tufts of wire if you need a really small one. For smaller stuff, there are stainles steel Dremel brushes. They can be p ress fitted into a hole drilled in the end of a 1/4" dowel, or used w/the Dremel tool. |
James Selevan (Jselevan)
Member
Username: Jselevan
Post Number: 681
Registered: 6-2002
| | Posted on Wednesday, July 30, 2003 - 7:26 am: | |
Thanks, Verell. Finally answer will come when I can use my Clamp-On amp meter on the shorted wires, and measure the current going through the meter during charging or cranking. Now that I mention it, it is clear that while cranking that the starter current is NOT flowing through the meter, yet the amp meter will measure negative 40 amps or more. Something to think about.
Jim S. |
Hans E. Hansen (4re_gt4)
Intermediate Member
Username: 4re_gt4
Post Number: 1598
Registered: 4-2002
| | Posted on Wednesday, July 30, 2003 - 2:08 am: | |
James: USUALLY, an ammeter is a 'black box' application, in that it has a very low resistance resistor built in. ie - low impedence. The large wires to your meter suggest that this is the case.
My actual car experience is in GM cars circa early 1970's. Some/many/most of these cars had an ammeter that was connected across some external (hidden in the wiring harness) wire/resistor. However, they were characterized by having small wires - perhaps 18 ga or so. It always puzzled me as to "Why 18 ga?" Why not smaller? Does the meter itself carry some significant current? The car had an 80 amp alternator, so the ammeter wiring couldn't possibly be carrying the entire car's power supply. So in this regard, I understand your question/problem.
However...........
Ferrari wiring isn't real sophisticated. I've wanted to get those Fred Flintstone "Dino" sheilds for my car. Real appropriate. I'd really have to guess that your car is wired such that the entire current is directed at the ammeter, meaning that there is an internal shunt resistor. Even if this isn't the case, substituting a simple wire jumper for the meter shouldn't cause much havoc, given that you have 12 ga wire going to the meter. This implies that some serious current must be directed that way, and "shorting it out" shouldn't be a problem. If it were some weanie voltmeter that was sampling a voltage across a hidden resistor in some remote location, they wouldn't have used 12 ga. The meter must be an "ammeter" which implies an internal, *VERY SMALL* built in resistor. Basically a short circuit. A wire jumper would be a functional equivalent.
Verell??????????????????? HELP! |
James Selevan (Jselevan)
Member
Username: Jselevan
Post Number: 680
Registered: 6-2002
| | Posted on Tuesday, July 29, 2003 - 7:09 pm: | |
I'm enjoying this discussion. Thank you Hans. Let's make it simpler. Consider the meter as a black box. Does the black box (meter) in my Boxer have a large input resistance or low input resistance? The answer to this question will answer my question as to whether we are dealing with a (functional) volt meter or amp meter.
As suggested by Verell, however, the original question pertained to the need for the external shunt resistor, and the consequences of shorting across it or leaving the circuit open.
Jim S. |
Verell Boaen (Verell)
Intermediate Member
Username: Verell
Post Number: 1016
Registered: 5-2001
| | Posted on Tuesday, July 29, 2003 - 12:35 pm: | |
Hans,
You're right on with one tiny quibble. It's irrelevant to James's concerns tho:
True Ammeters do exist, & they're fairly common. They're usually cheap & somewhat inaccurate tho. See the last paragraph in my post # 966 below. I've got one in a battery charger. |
Hans E. Hansen (4re_gt4)
Intermediate Member
Username: 4re_gt4
Post Number: 1590
Registered: 4-2002
| | Posted on Monday, July 28, 2003 - 11:18 pm: | |
James: I think you are way overanalyzing this. ie - 'forest for the trees' syndrome.
First, lets set down a couple of definitions.
Voltmeter: In conventional usage, a voltmeter measures some typically meaningful voltage - in automotive terms, 0 - 15 volts usually suffices. Note that in a laboratory, micro volts and micro amps are measured. But lets stay with real world typical household/automotive values here.
Ammeter: Measures some 'reasonable' amount of current. If we are talking about the total car's system - alternator output, etc. - we're talking 10's of amps, probably a lot more on newer cars. If we're talking about some of the car's subsystems, we're still in the high milliamp territory, probably more like amps.
Now consider: The typical meter movement has a high resistance. So it takes little voltage or current (remember ohms law) to move it to full scale.
A 'True Ammeter' probably doesn't exist - by this I mean a meter movement that directly carries the large currents typically encountered in everyday life w/o supplemental parallel resistors. (I have seen some galvanometers that directly take big current.)
A 'True Voltmeter' probably doesn't exist - by this I mean a meter movement that directly measures the large voltages typically encountered in everyday life w/o supplemental series resistors. If you go to Radio Shack and buy a 15V panel meter, I guarantee that it has a resistor hidden somewhere inside.
Something labeled an 'Ammeter' has some sort of resistor bypassing the meter movement - but this resistor is built in. It doesn't appear externally. All you see is some large terminals designed to carry large current. It's a "black box". Internally, it's a resistor, and a low current (and, remembering ohms law, a low voltage) meter.
Perhaps an example: As a poor starving EE student, I needed a VOM (couldn't afford a VTVM). Radio Shack had a decent one for $39, but the same meter was available as a kit for $29. (BTW, I still use this meter.) The *actual meter* had a 50K ohm resistance, and would do full scale at .25V - which is 5 microamp. Quite remarkable for a cheap meter. Now consider: This meter will measure a large range of voltages and currents (as well as resistance). How? Resistors that are switched in as necessary.
How does this relate to an automotive ammeter? Well, this VOM had two terminals that the test leads plugged into. These sufficed for all measurements - except one. There was a 3rd terminal. Internally, this 3rd terminal had a short, large wire connected directly to the ground terminal. It was the 10 amp measurement terminal. Damn near a short circuit. For a 10 amp full scale, this 'shunt resistor' has to be 50 MICRO-OHMS.
So the ammeter that you are fiddling with is probably similar, altho the meter movement itself is more likely to be around 1 milliamp. This would mean an *internal - built in* shunt resistance of perhaps 20 MICRO-OHMS to measure 50 amps full scale. Hmmm..... Upon reflection, the meter movement probably isn't quite that sensitive. But I'd doubt that the *internal - built in* shunt resistance is more than 1 milli-ohm.
For all practical purposes, the ammeter is a short circuit. It's presence in the circuit is largely irrelevant. You can substitute a wire jumper (large) in it's place w/o effect. If you disconnect it and leave it out, you will have an open circuit.
However....... Some automotive systems will measure amperage via some leads from some remote (small) resistance. I doubt this is the case for you, as 12 ga wire wouldn't be needed if this were the case. 30 ga wire would suffice - altho typically they use larger wire in autos simply for strength/durability considerations. |
Verell Boaen (Verell)
Intermediate Member
Username: Verell
Post Number: 1013
Registered: 5-2001
| | Posted on Monday, July 28, 2003 - 9:46 pm: | |
Sorry, sure don't.
Best guess is to ask around the speedometer shops.
Also ask any exotic car restoration shops in the area. |
Dr. I. M. Ibrahim (Coachi)
Member
Username: Coachi
Post Number: 329
Registered: 5-2002
| | Posted on Monday, July 28, 2003 - 8:13 am: | |
Verell, or anyone please, does anyone know of a shop who rebuilds ammeters? I think the problem can be repaired with an experienced technician...thanks for any input. The internet search yielded little. |
Verell Boaen (Verell)
Intermediate Member
Username: Verell
Post Number: 1008
Registered: 5-2001
| | Posted on Sunday, July 27, 2003 - 9:16 pm: | |
Coachi,
Sure sounds like you cooked the shunt resistor & it's resistance has increased enough to let enough current flo thru the meter so that it's pegging the needle.
Good news is that the needle mechanism still seems to be working.
If you can disassemble the meter & figure out a replacement for the shunt you may be able to recover w/o buying a new meter.
|
James Selevan (Jselevan)
Member
Username: Jselevan
Post Number: 679
Registered: 6-2002
| | Posted on Sunday, July 27, 2003 - 1:13 pm: | |
Coachi - there is a technical term in electrical engineering that is used in this situation: "You toasted it."
Jim S. |
Dr. I. M. Ibrahim (Coachi)
Member
Username: Coachi
Post Number: 326
Registered: 5-2002
| | Posted on Sunday, July 27, 2003 - 11:19 am: | |
speaking of ammeters, I recently had an accident with my boxer's ammeter. While trying to reinstall the ammeter, I accidentally shorted it..I had forgotten to disconnect the battery. Now it reads zero when the ignition is not on, the gives a high amps reading when the ingition is turned on and the car is started. It never goes back down again as the battery is charging. It stays off the gauge. Any ideas what happened and how to correct it? |
James Selevan (Jselevan)
Member
Username: Jselevan
Post Number: 677
Registered: 6-2002
| | Posted on Sunday, July 27, 2003 - 9:09 am: | |
Verell - thanks for the input. I am unable to scan the schematic at this time, but can in 2 weeks. Sorry. Perhaps someone else has a schematic.
Jim S. |
Verell Boaen (Verell)
Member
Username: Verell
Post Number: 998
Registered: 5-2001
| | Posted on Saturday, July 26, 2003 - 9:37 pm: | |
Any chance you could scan a section of the schematic around the meter in question & post it.
An shunt is wire-wound ceramic resistor, or even just a strip of steel or wire across the meter terminals.Odds are that the shunt is built inside the meter, but I have seen them externally mounted.
While the shunt is carrying a fairly heavy current, it's dissipating very little power because it's resistance is so low.
BTW, all this is academic. You still jumper the leads together. If the shunt is externaly mounted away from the meter (IMHO highly unlikely), then the jumpering is unnecessary, as the shunt would have carried the current. If it's inside of the meter, then the jumper will replace the shunt.
And technically, all moving coil 'voltmeters' are really current meters, typically with a 1mA full scale. Additional resistance is inserted series with the meter to limit the current to 1mA at the max voltage the meter is calibrated for.
Digital & other electronic voltmeters are a different story. There the input signal is handled by being either amplified, or digitized depending on the specific meter type.
re:"Input from electrical engineers appreciated"
BTW, BSEE '69, 8 years as a radio/tv shop tech prior to that, self-taught thru correspondence courses while in high school. Designed computers including power supplies for DEC for another 10 years, then made mistake of going into management. No fun there. Oh, yes, Ham license N1OJP. |
James Selevan (Jselevan)
Member
Username: Jselevan
Post Number: 675
Registered: 6-2002
| | Posted on Saturday, July 26, 2003 - 4:19 pm: | |
Steve M. - I do not know where the shunt resistor is physically located in the Boxer. I could measure the resistance of the meter. If low, presumably it is an amp meter. If high, then a voltmeter.
I find this thread particulary interesting in that surely someone in the past has questioned whether the charging amp measuring device is an amp meter or a voltmeter. At first thought, it would seem to be an amp meter as the 12 gauge leads suggest. It is the presence of the shunt resistor noted on the schematic that raised the issue in my mind.
I appreciate everyone's input. I am an EE by training (undergraduate and graduate), and this seems to be a rather simple exercise, but I never paid attention to this issue before.
Jim S. |
Steve Magnusson (91tr)
Intermediate Member
Username: 91tr
Post Number: 1975
Registered: 1-2001
| | Posted on Saturday, July 26, 2003 - 10:32 am: | |
James -- Can you measure the resistance of the external shunt resistor and the resistance of the "voltmeter" separately to compare? I think you're right to be concerned about what's what if the "voltmeter" leads seem designed to carry a significant current. Could there be an unseen second internal shunt resistor in parallel with the external one? |
Mike Florio (Mike_in_nevada)
New member
Username: Mike_in_nevada
Post Number: 19
Registered: 6-2003
| | Posted on Saturday, July 26, 2003 - 10:23 am: | |
Just a thought - are there any other leads connected to the same post as the 12GA wire? I have vehicles where the wire to the Ammeter is also used to feed other dashboard and lighting circuits. If any of them are drawing too much current it drops the voltage at the meter, so it reads negative charge. That might be why the input lead is 12 GA. |
James Selevan (Jselevan)
Member
Username: Jselevan
Post Number: 674
Registered: 6-2002
| | Posted on Saturday, July 26, 2003 - 10:19 am: | |
Hans and Craig - let's cut to the chase. If the device reading charging amps on the dash board is a voltmeter, then minimal current will flow through the wires connecting it to either end of the shunt resistor. A volt meter will have, for all practical purposes, infinite input resistance.
If the device reading charging amps is an amp meter, then it will have, for all practical purposes, zero resistance, and ALL the charging current will pass through it.
The only time the resistance of the amp meter is of consequence is when, as Verell suggested, the resistance of the shunt resistor and the input resistance of amp meter are comparable, in which case the amp meter / shunt resistor will act as a voltage divider, and this parallel resistor must be accounted for.
In either case, if one removes the measuring device from the circuit, as when servicing it, the circuit is open. If it was an amp meter, then all the current will run through the shunt resistor, which I assume was not designed for the full load. On the other hand, if it is a voltmeter, and removed from the circuit, then one would NOT want to short the leads for fear of too large a current running through the circuit. I have done both with my meter out without untoward sequelae.
Still not convinced that I know what this meter reads. Thanks for everyone's input. Keep it coming.
Jim S. |
Hans E. Hansen (4re_gt4)
Intermediate Member
Username: 4re_gt4
Post Number: 1584
Registered: 4-2002
| | Posted on Friday, July 25, 2003 - 3:21 pm: | |
James: For a slight clarification - The large wires are needed, not for the voltmeter-calibrated-in-amps, but for the shunt resistor. The wires are feeding the shunt resistor. The voltmeter happens to be hooked across it, but it isn't drawing much current. The shunt passes 99.99+% of it. |
Craig Dewey (Craigfl)
Member
Username: Craigfl
Post Number: 653
Registered: 1-2001
| | Posted on Friday, July 25, 2003 - 6:51 am: | |
"So the ammeter is set up internally to read about 0.4V for max scale with this example."
Actually I meant the ammeter. The ammeter electrical system consists of two parts -- the Shunt and the meter that reads the voltage across the shunt.
Since an ammeter is a series device, all the current in the circuit must pass thru it, therefore the wire size must be large enough to handle the maximum system current.
A voltmeter only measures the "potential difference" so very little current flows thru it and only very small wires are required. |
James Selevan (Jselevan)
Member
Username: Jselevan
Post Number: 673
Registered: 6-2002
| | Posted on Friday, July 25, 2003 - 3:33 am: | |
Craig - you mean the voltmeter. But why such heavy leads?
Jim S. |
Craig Dewey (Craigfl)
Member
Username: Craigfl
Post Number: 652
Registered: 1-2001
| | Posted on Thursday, July 24, 2003 - 12:59 pm: | |
James..
Power = I**2 * R (current squared times the resistance)
For a 12 volt circuit, 40 amps, the resistance may be less than 0.01 Ohm which means only about 1.6 watts of power are dissipated.
If you use ohms law, E=I*R, the voltage drop across the shunt would be 0.4V which is reasonable.
So the ammeter is set up internally to read about 0.4V for max scale with this example. |
James Selevan (Jselevan)
Member
Username: Jselevan
Post Number: 672
Registered: 6-2002
| | Posted on Thursday, July 24, 2003 - 12:02 pm: | |
Verell - thus, the "amp meter" is really a volt meter measuring voltage across a known resistor, and calibrated in amps to do the calculation of volts to amps.
In this case, any voltmeter will have "infinite" resistance, thus by shorting the leads one should do bad things. Normally, in a voltmeter, very little current flows. There is something wrong in Peoria, as the leads to the amp meter are 12 gauge, implying large current flow.
Something does not smell right for this to be a voltage meter interpreting amps by passing current through a known resistor. Furthermore, passing 40 amps through a known resistor will develop enough heat to toast your bread as you drink your coffee between shifts
Jim S. |
Verell Boaen (Verell)
Member
Username: Verell
Post Number: 966
Registered: 5-2001
| | Posted on Wednesday, July 16, 2003 - 1:00 am: | |
James,
Yes, it's an ammeter. So short the wires to it together to temporarily work around your ammeter while it's out.
You described it correctly, the shunt is a known resistance, & the meter measures the voltage across it. (Well, technically, the current is divided between the shunt resistor & the much higher resistance of the meter coil as a function of the ratios of the two resistors.)
Hans, Many automotive ammeters just used a magnetic needle, and ran a single turn of the input lead in a little housing on the back of the meter. The lead's magnetic field acted directly on the meter pointer! Cheap, but good enough for battery chargers & auto dashboard ammeters.
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Hans E. Hansen (4re_gt4)
Intermediate Member
Username: 4re_gt4
Post Number: 1563
Registered: 4-2002
| | Posted on Tuesday, July 15, 2003 - 11:53 pm: | |
Oh, PS: If you see what appears to be a "true amp meter", it just has an internal shunt resistor built in. I've never heard of a meter that had actual coil windings that would take several amps of current. Most meters from the 1940's to the late 1960's were basically made to work on 1 millamp, give or take. Voltmeters simply had a series resistor built in to make the 1 ma current give whatever full scale reading it was calibrated for. Ammeters were the same. Shunt resistor calibrated to bypass all but 1 ma to register the necessary current printed on the scale.
The ammeter on my old WWII jeep had a copper bar across the terminals. With that huge shunt, I would have never thought that enough current could reach the meter, but it apparently worked! |
Hans E. Hansen (4re_gt4)
Intermediate Member
Username: 4re_gt4
Post Number: 1559
Registered: 4-2002
| | Posted on Tuesday, July 15, 2003 - 11:27 pm: | |
You had it right, in that amp meters (for the most part) don't measure the full current, but rather measure the current across a shunt resistor. The shunt resistor carries the vast majority of the current. If you short out the meter, you simply bypass the shunt resistor, and nothing gets measured. The current simply goes around it. Regardless what you do with the meter, you need to either leave the shunt resistor in the circuit (especially if you still want the meter to function), or short out the shunt resistor (which will make it impossible for the meter to function). Shorting out the meter effectively shorts out the resistor. Leaving the meter open will have no effect, other than the meter not working.
Oh, BTW, in these old circuits, the meter itself won't have infinite (or even very high) resistance. Probably on the order of 1000 ohms or so.
Hans. (who used to tinker endlessly with really old radio transmitters)
PS - If you short out the meter with the resistor still in the circuit, all the current will go thru whatever you shorted the meter with. So make sure that you use reasonable sized wire. |
James Selevan (Jselevan)
Member
Username: Jselevan
Post Number: 671
Registered: 6-2002
| | Posted on Tuesday, July 15, 2003 - 9:13 pm: | |
Thought about my question a bit more, and realized the answer is staring me in the face. The leads to the amp meter must be 12 gauge, implying significant current flow. A voltmeter would require 16 gauge. End of story. It is truely an amp meter. But why the shunt resistor?
Jim S. |
James Selevan (Jselevan)
Member
Username: Jselevan
Post Number: 670
Registered: 6-2002
| | Posted on Tuesday, July 15, 2003 - 8:54 pm: | |
Removed my instruments and dash to re-leather. Studied schematic diagram carefully, and note that there is a "shunt resistor" across the amp meter. I am studying the issue vis-a-vis leaving the amp meter leads "open" or tapeing them together.
Based on the schematic, it appears that the amp meter is really a voltmeter measuring voltage across a known resistor, and reading this voltage as amps ( amps = volts/resistance).
The important question is: If this is an amp meter, then its impedance is zero, and one can tie the leads together. If this is a voltmeter, then the impedance is infinite, and the leads should be left apart.
Input from electrical engineers appreciated.
Thank you.
Jim S. |
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