An interesting question, Tony!
I think your 2nd to last statement (Is the power drop of the fuse/diode negligible, due to the short length) is correct & having the fuse close to the battery is mandatory. Telecommunications exchanges' batteries where the load could be over 200A, always had the fuses (generally HRC type) close to the batteries.

I think of it was a unit - "battery, fuse & feed" to the load via cables or busbars. I couldn't guess what a "6sqmm fuse" would be rated at and it certainly wouldn't protect the cable + equipment hanging of it.

Your diodes on the solar panel fall into the same system - the heavy 6mm feed back to the battery(ies) prevent further voltage loss.

My thoughts only - left most of that stuff behind 12.5 years ago now.

Thanks for your input, Warren and Jahhn
It was appreciated

It does make it a little clearer to me now

I probably tripped over this answer twice, while researching it, but it got lost (to me) in the technical jargon I was reading

Short length of thin (fuse/diode) cable/wire, gives minimal and acceptable loss, but long length of too thin cable/wire, gives enough loss, that we now do not have enough power, to charge battery/run appliances 

Tony Bev wrote:

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Can someone explain, in layman speak, so that I can  understand it, please

[1]We are informed that 12VDC cable should be a minimum of six square millimetres area, and that larger square area is better, for less voltage loss

[2]A normal 12 volt 30 amp blade fuse has a much lower, (never measured it), square area than six millimetres
Normally a fuse is close to the batteries, so would incoming/outgoing power, be affected?

[3]A diode of 12/15/20 amps has a wire with a diameter of 1.2 to 1.3 mm, (1.13 to 1.327 square mm)
In a normal situation, the diodes are at/near the solar panels.
How does this affect the power from the solar panels?

[4]Is the power drop of the fuse/diode negligible, due to the short length?

Edit to say that it is not my intention to start any flame wars, also when I am absolutely sure, that I know the answer, I do not ask the question

 

-- Edited by Tony Bev on Tuesday 27th of November 2018 02:39:49 PM

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 Hi Tony

A little more to chew on.

[1]  To minimise voltage drop in any cable to acceptable levels for the particular application , a number of factors have to be consider  :

[a] max current through the cable. 

the total length of cable[ both pos & neg], 

From that  ,the required cable size to limit the voltage drop to acceptable limits can be calculated 

In some cases such as AS3001 "Transportable Structures "a minimum cable  cross sectional area & number of strands may be specified. This is more for mechanical reasons cable movement  [flexing in normal use] just as with extension leads.

[2] A thermal fuse or thermal circuit  breaker must generate some heat in them to operate,

If you look at most 12V blade fuses you will see there is a small section of reduced section in the middle of the fuse That is the section calibrated to blow. It does limit the voltage drop across the fuse quite a bit compared to the small 12V glass fuses .

The  length of reduced section when blowing must be long enough gap  for the arc to  self extinguish .That is a critical condition for all thermal fuses [the higher the voltage the longer the reduced section must be]

The fusing function is not dependent on where it is located in the circuit . It's location  should always be as close as possible to the starting point of the positive line it is protecting. & should never have a higher rating than the current rating of the cable  it is protecting. Also note All fuses are not the same ,  they can be fast blow or ,slow blow. even with the small  12V glass fuses depending on what protection is required & the actual load type .

[3]Others have answered the diode question very well so I do not need to comment on that ,except to say the actual  size  ,cross sectional area ,of the pigtail must be of sufficient current carrying capacity to suit the diode rating  .It also serves as a heat transfer for  the actual diode junction.

[4] The voltage drop is dependent on the following factor :

The actual current " Amps"

Cross sectional area of the  actual conductor .

The total length of the conductor both pos & neg ,or with AC ,Active & neutral

The actual conductor material. [for normal wiring ,usually  copper ,but can be aluminium.[with  a  higher voltage drop for the same current & conductor cross sectional area]

It should not start  a flame war unless some who does not understand tries to tell you different,

Just one other comment on fuses

They are not all the same even though they may have the same current & voltage rating , the actual  operational requirements will determine the fuse type required but that is not really relevant to the 12V wiring in a van., 

-- Edited by oldtrack123 on Wednesday 28th of November 2018 02:11:34 AM

The formulae for calculating voltage drop in copper cables is .
VOTLAGE DROP = [cable length (in metres) X current (in amps) X 0.0164] divided by cable cross-section in mm.sq.

Cheers,
Peter

Tony Bev wrote:

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Normally a fuse is close to the batteries, so would incoming/outgoing power, be affected?

snip

Is the power drop of the fuse/diode negligible, due to the short length?

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 In any circuit where the conductor varies in size the the voltage drop in volts per millimeter will be related to the size and resistance of the conductor. In radio equipment rooms we needed to run long leads between devices. Often the length of cable means it has to be a large one to limit the voltage drop to an allowable value. The large cable size will not fit the terminals on the battery charger or the device being supplied. In these cases a short piece of smaller wire is used between the heavy one and the terminal. If this piece of wire is short enough then there is only a very small voltage drop along its length and this will not affect the service by a significant amount.

In the case of fuses they are built material of higher resistance. This allows the fuses to build up enough heat to be blown is a short length. It does not matter where the fuse is placed in the circuit, it's effect on voltage drop will be the same. (The importance of a fuse being near a battery is purely to limit the possibility of a short occurring between the fuse and the battery.)

This dissertation was given to demonstrate the answer to your last question in the piece I quoted from your question. You can see how much voltage drop there is across the fuse from the following tables. You can use these fuses to measure the current flowing them. Just use a good digital volt meter, connect the leads to the little pads on the top of the fuse. Cross reference the voltage reading to the table and you can get a reasonably accurate reading of the current flowing through the fuse.

 

With a diode, the junction inside the diode causes more voltage drop that the short piece in and out of the diode. If the circuit doesn't need them than the whole cable run will more efficient without them, but as has been mentioned if they are necessary then it is just one of the losses we have to put up with and add into the calculation. Small conductor size does not stop current flow as such, but the resistance does create heat and that uses up both amps and volts that can not be regained so lost from the total electrical supplied into the cable.
This heat generation is most obvious where blade fuses are used, it is not uncommon to see a fuse rated at greater than 20 amps with the plastic bit actually melted and often the blades welded into the fuse holder. The small piece of actual fusible material is not the cause but rather the poor design causing a high resistance joint between the fuse blades and the fuse holder.

T1 Terry

PeterD wrote:

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Tony Bev wrote:

---

 

Normally a fuse is close to the batteries, so would incoming/outgoing power, be affected?

snip

Is the power drop of the fuse/diode negligible, due to the short length?

---

 In any circuit where the conductor varies in size the the voltage drop in volts per millimeter will be related to the size and resistance of the conductor. In radio equipment rooms we needed to run long leads between devices. Often the length of cable means it has to be a large one to limit the voltage drop to an allowable value. The large cable size will not fit the terminals on the battery charger or the device being supplied. In these cases a short piece of smaller wire is used between the heavy one and the terminal. If this piece of wire is short enough then there is only a very small voltage drop along its length and this will not affect the service by a significant amount.

In the case of fuses they are built material of higher resistance. This allows the fuses to build up enough heat to be blown is a short length. It does not matter where the fuse is placed in the circuit, it's effect on voltage drop will be the same. (The importance of a fuse being near a battery is purely to limit the possibility of a short occurring between the fuse and the battery.)

This dissertation was given to demonstrate the answer to your last question in the piece I quoted from your question. You can see how much voltage drop there is across the fuse from the following tables. You can use these fuses to measure the current flowing them. Just use a good digital volt meter, connect the leads to the little pads on the top of the fuse. Cross reference the voltage reading to the table and you can get a reasonably accurate reading of the current flowing through the fuse.

 

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 Thanks for your input Peter, it was appreciated

Sorry that I did not reply to, Peter n Margaret (Peter), earlier

As I now understand what his formula means

 

This link may help members to understand that a fuse does not blow " @ it's rated current.

The speed of blowing is dependent on the actual excess current & fuse type.The same applies to thermal circuit breakers

All have a overload to blow time curve. That does vary with actual type  ie if standard ,fast blow or slow blow

https://www.google.com.au/search?q=Time+current+curve+to+blowr+typical+12V+auto+fuses&tbm=isch&source=iu&ictx=1&fir=k7B4W7UtuGznqM%253A%252C8unE8TdxA1UDPM%252C_&usg=AI4_-kRS4_5Ng7ZP0NZAvApPN-f5v0adDA&sa=X&ved=2ahUKEwjp7-322fveAhXKEnAKHdN2DX0Q9QEwAHoECAAQBA#imgrc=k7B4W7UtuGznqM:

-- Edited by oldtrack123 on Friday 30th of November 2018 11:24:55 PM

Tony Bev wrote:

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Very interesting, Ralph

It just goes to show, that I never knew, what I did not know, so thanks for that

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 Hi Tony 

To quote a now famous persons speech "There are things we know that we know. There are known unknowns. That is to say there are things that we now know we don't know. But there are also unknown unknowns. There are things we do not know we don't know." 

Precisely 

Jaahn

oldtrack123 wrote:

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Jaahn wrote:

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Tony Bev wrote:

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Very interesting, Ralph

It just goes to show, that I never knew, what I did not know, so thanks for that

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 Hi Tony To quote a now famous persons speech "There are things we know that we know. There are known unknowns. That is to say there are things that we now know we don't know. But there are also unknown unknowns. There are things we do not know we don't know." 

Precisely 

Jaahn

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 I just had to laugh, when I saw the (clap gif) and then realised why it was there

I can assure everyone, that I had no intention of trying to be funny, with my reply

It is good to see, that no matter how serious we can get on this forum, our sense of humour can still shine through

Hopefully no offence was given, as certainly none was taken, by me

 

Great answer Iza. Something to be very careful of is misleading advertising regarding cable size. 6 B&S cable is a min of 13.5mm sq, yet we had one supplier try to pass off 10mm sq conductor cable as 6 B&S. There are quite a few sellers peddling cable that is a mix of aluminium and copper, for the same voltage drop per mtr the cable would need to be a much larger cross sectional area if the conductor wasn't high grade copper.

T1 Terry