pvb
Well-Known Member
But almost all meters and monitors are built for 50mV shunts.
Cheap make before break SPDT switches?
- Thread starter PaulMcC
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But almost all meters and monitors are built for 50mV shunts.
JumbleDuck
Well-Known Member
I know. But they could be built for 0 - 1V pretty easily too. My point, such at it was, is that HE sensors are pretty easy to use, as all the driving and signal conditioning are built in, They also require no high-current connections and are completely immune to damage from overcurrents.
jfm
Well-Known Member
Yes, we're violently agreeing as they say. A shunt is contraption where you create two parallel routes for the current, in a known ratio eg 99:1 or 99,999:1 or whatever. You then know that the current in the high resistance leg is 1/100,000th (say) of the current in the other leg. You take your measurements on the high resistance leg (by measuring volt drop) but you print the card in the voltmeter to read what you deduce are the amps in the main leg. Yes JD, the high resistance leg is actually the meter itself; I forgot that, but it doesn't change the principle nor does it change what I'm saying which is that Vic's 20amps mathematics is wrong and there is no energy consumed in a shunt
jfm
Well-Known Member
See above. All shunts work on the principle of creating two parallel legs, of known relative resistance, but as JD corrected me one of those legs is the meter coil so you don't "see it". There should be no 20Amps factor when VicS was calculating energy consumption, because the 20Amps is the current passing thru the low resistance leg and shouldn't therefore be in the calculation. A shunt doesn't consume energyI don't recognise your description of a shunt. All the shunts I've ever seen on boats have all the current passing through, and don't have a "high resistance leg". The voltage drop is masured across the whole shunt, not just part of it.
VicS
Well-Known Member
I knew should not have started this!
First point though you are right that with no load and therefore no current there is no energy consumption. That's why I said, " at the full 20 amps"
Two ways of looking at the way an ammeter shunt works.
One is that most of the current flows through the shunt while only a small current flows though the higher resistance of the meter The proportions depend on the relative resistances of the two. To do any calculations you need to know the FSD in mA and the resistance of the meter.
The other is to look at it as you have.
The current flowing through the shunt produces a small potential difference across it proportional to the current. You measure this PD with a voltmeter which has a high resistance and through which therefore only a very small current passes.
Which ever way you look at it you have the very low resistance of the shunt in parallel with the higher resistance of the meter. The result is a very small "bottle neck" caused by the resistance of the shunt. The resistance of the meter is many times that of the shunt so the two in parallel have to all intents and purposes the same resistance as the shunt
The formula for the combined resistance R of two resistances r1 and r2 in parallel is 1/R = 1/r1 + 1/r2
so if the shunt has a resistance of 0.0025 ohms and the meter has a resistance of 50 ohms
1/R = 1/0.0025 + 1/50 = 400 + 0.02 = 400.02
if 1/R =400.02, R= 0.0024999, which, as near as makes no odds, is 0.0025 ie the resistance of the shunt on its own.
The figures BTW would apply to a 20A, 50mV shunt and a moving coil meter with a FSD of 1mA and a resistance of 50 ohms. looked at as a voltmeter it would have a FSD of 50 mV
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pvb
Well-Known Member
The maths is correct. The shunt has a small resistance; current flowing through it causes a voltage drop across it; this causes a very small power loss, manifested as a miniscule rise in temperature. In just the same way, high current flowing in an undersized wire causes a volt drop which makes the wire heat up.
jfm
Well-Known Member
In reply to both the above posts you are assuming the X section/resistance of the shunt is smaller than the resistance of the 2 inches of wire you've chopped out to insert the shunt. I don't know that it is generally, and it need not be. If it isn't, there isn't any bottleneck, and therefore as I said there is zero energy consumption/conversion to heat. If there is a bottleneck it's because you chose too small a shunt; you cannot blame that energy loss intrinsically on using a shunt. In other words the 1 watt heat conversion above would happen with the same 2 inches of wire if you never inserted the shunt, so is not attributable to the shunt, if you choose a non bottlenecking shunt.
Let's imagine that you knew that 2 inches of wire had a resistance 100,000th of your voltmeter using the specs above. You could press the voltmeter probes thru the power wire insulation, 2 inches apart, and with the right printed amps card in the meter you would get a reading of amps flowing thru the power wire. That is, as I know you know, a shunt. No way would that consume any energy or make more heat. It would actually reduce the circuit's total resistance by a very small amount indeed, so saving you energy consumption
Let's imagine that you knew that 2 inches of wire had a resistance 100,000th of your voltmeter using the specs above. You could press the voltmeter probes thru the power wire insulation, 2 inches apart, and with the right printed amps card in the meter you would get a reading of amps flowing thru the power wire. That is, as I know you know, a shunt. No way would that consume any energy or make more heat. It would actually reduce the circuit's total resistance by a very small amount indeed, so saving you energy consumption
pvb
Well-Known Member
You're now trying to wriggle! In post #17, you said " So the shunt (if the main leg of the shunt is correctly sized so as not to be a bottleneck) itself adds no resistance and consumes no energy at all." That's incorrect. In post #23, you said "there is no energy consumed in a shunt". Again, that's incorrect. In post #27, you've introduced a different argument, which is that the shunt doesn't necessarily use any more energy than a similar length of wire. This is correct, but is rather different from your previous claims. Good wriggle, though!
JumbleDuck
Well-Known Member
There most certainly is. A shunt only works by resistance - a superconducting one would be no use at all - and where a current flows through a resistance there is a power loss. VicS is quite right - a 20A/50mV (ie drops 50mV when 20A is flowing through it) shunt will dissipate 1W in heat.
Incidentally, traditional mechanical meters almost invariable measure amps, not volts. I have an mechanical electrostatic voltmeter here, but the finest scale is 0 - 10kV and it weighs around 20kg, so not wholly practical for boat use. Electronic meters measure volts rather than amps.
JumbleDuck
Well-Known Member
There is conversion, just (maybe) not additional conversion.
Adding any parallel component will always increase overall power consumption, just as adding any series component will always reduce it. The overall heat generated in the circuit is V2/R: adding the voltmeter as you suggested reduces R and therefore increase V2.
Appeals to authority are a pain, of course, but I might mention here that around half my hands-on research career was in measuring electrical power.
VicS
Well-Known Member
Ok continuing with the model I was using.
If the distace from battery to ditrubution panel is not huge, say between 3 and 4 metres as the cable runs, AWG 8 might be the wire size used.
AWG 8 has a resistance of 2.2 ohm per 1000 metres. Therefore the resistance of 50 mm is 0.00011 ohms That makes the resistance of a 20A,50mV shunt about 22 times greater.
OR
Crunching the numbers differently, if you used the wire itself instead of a shunt the connections would have to be about 1.14 metres apart. Possible to do that I guess but not the way it would normally be done.
You'd be saving a max volts drop of 50mV and reducing the resistance of the circuit by 0.0025 ohms. The effect of connecting a meter across just over a metre of 8 AWG would be infinitesimal. Ive already done the arithmetic above. It would reduce 0.0025 ohms to 0.0024999.
jfm
Well-Known Member
Yup all looks good. We all understand the maths and on that we're violently agreeing. You have a kit of parts there that results in (tiny) energy used by the shunt which is fine but nevertheless that consumption is a result of the compromised design, not of shunts intrinsically. Such a terribly wasteful (!!) shunt might be all that is available to Joe Average boat owner and if so then on that level I'm happy to concede 
halcyon
Well-Known Member
Cable resistance changes by temperature, air temp, current flow in cable and number of cables in bundle, so reading is not accurate, that's why they developed the special material to give a stable reading,
Brian
halcyon
Well-Known Member
JumbleDuck
Well-Known Member
Indeed, and if I had remembered your name ("that bloke in Kidderminster who makes VSRs" seemed a bit vague) I'd have said so. So ... I can't see why most of the marine electronics industry uses such old technology.
BERT T
Well-Known Member
Two relays and a delay off timer with a DPDT switch will do what you want.
http://www.ebay.co.uk/itm/4-Pin-12v...arts_Vehicles_CarParts_SM&hash=item43b41fabf8
and http://www.ebay.co.uk/itm/DC-12V-LE...200?pt=LH_DefaultDomain_3&hash=item5d4b248968