Fuse Box Question....

[Diesel]

If this fusebox......

http://cgi.ebay.com.au/ws/eBayISAPI.dll?ViewItem&item=221128206855&fromMakeTrack=true&ssPageName=VIP:watchlink:top:en#ht_574wt_1397

says that it is rated at 20 amps max per fuse - then should I run a 30A constant power cable to it from my battery with a 20A in line fuse?

i.e. the multiple circuits on thhis line can't overload a 30A wire can they?             

Cheers, Diesel

[Brock]

If you use a 30Amp supply line, then the in line fuse should be 30A to protect the wire. Each fuse in the blockl can be anything up to 20 A, if you have a 3Amp item, then a 5 A fuse would be fine. Not many items on the bike would need 20 amps.

remember, the total that the alternator can supply is 40 Amps, any draw more than this will start to drain the battery which means no start, and the bike will eventually stop.

[Diesel]

Thanks Brock - here is some text I nicked from Whizz when commenting on Shiney's power socket installation:

Ohms Law says that Power equals Voltage multiplied by current, or P=V x I. This transposes to I=P/V or 60/12, so the headlight draws 5A when on high beam (4.58A on low beam). This means that the maximum current you can draw from your new accessories block, with the headlight on is 5A, not as you might think 10A (from the size of the fuse).

So you need to be a little careful what you plug into these sockets because the  headlight is on all the time, day and night! Not forgetting that the current draw is cumulative, i.e. you add together all the currents required by all the accessories that you plug in at any one time to get the total current requirement. Under these circumstances it is really quite easy to pop a 10A fuse, especially as there is a significant surge current requirement when you first turn on your ignition. This means that when you turn the key the headlight appears as a dead short and draws a hefty spike of current to make the lamp glow. If you also have other stuff plugged in which also has surge currents then that is the moment when the fuse is most likely to blow. There is an old adage that says that a light bulb only blows when you turn it on and this is simply because of the surge current required to make an incandescent bulb glow white hot.

[Diesel]

So - to use this fuse block.....

http://www.ebay.com.au/itm/221128206706?ssPageName=STRK:MEWAX:IT&_trksid=p3984.m1438.l2649&autorefresh=true#ht_574wt_1398

I should go:

Battery ---->  30A wire  --------> 30A Relay   -------->   Fuse Block -------->    acessories (load)


Yes? - Do I really need the relay in there?

Cheers, Diesel

[Brock]

The relay is only needed if the devices you want to power are only used when the ignition is on. If you want to power items when the bike is off, no.

The relay could be energised (turned on ) by the tail light, it will add no load to speak of to the tail light.

[saaz]

As Brock says. Depends what devices you are planning to power.  The total output of the fusebox should not exceed the wire and fuse powering it, otherwise the fuse going to the fusebox will blow to protect the wire.  The fuse for each device does not have to be 20amp, just what is required to protect the device/wiring to it.

[DaleMcLean]

Thanks Brock - here is some text I nicked from Whizz when commenting on Shiney's power socket installation:

Ohms Law says that Power equals Voltage multiplied by current, or P=V x I. This transposes to I=P/V or 60/12, so the headlight draws 5A when on high beam (4.58A on low beam). This means that the maximum current you can draw from your new accessories block, with the headlight on is 5A, not as you might think 10A (from the size of the fuse).

So you need to be a little careful what you plug into these sockets because the  headlight is on all the time, day and night! Not forgetting that the current draw is cumulative, i.e. you add together all the currents required by all the accessories that you plug in at any one time to get the total current requirement. Under these circumstances it is really quite easy to pop a 10A fuse, especially as there is a significant surge current requirement when you first turn on your ignition. This means that when you turn the key the headlight appears as a dead short and draws a hefty spike of current to make the lamp glow. If you also have other stuff plugged in which also has surge currents then that is the moment when the fuse is most likely to blow. There is an old adage that says that a light bulb only blows when you turn it on and this is simply because of the surge current required to make an incandescent bulb glow white hot.

PS dont forget within tha calculation that the "V" actually when running is about 14  so the "I" may well be larger and the "P" therefore too, as ohm law dictates P=I^2 * R and the only constant really is the "R" of the lamp, so just be cautious when calculating total load of alternators 

[Dennisgw]

Pardon me if I am wrong Diesel,.. I agree with the formula and calculation of current based on W=ExI (W = Watts, E = Volts and I = Amps), using a 60W globe in a 12V system hence 5A. If you're talking headlamps, don't forget to multiply by 2, which gives a total current draw on hi-beam of 10A based on your figures.

Relays are required to enable switching high current with a light (small)  switch. It seems that you are just using the heavy cable as a feed to the fuse box and hence would not necessarily need a relay. That said, Brock is correct if you do want to switch power off when the bike isn't running. In that case you would need a minimum of a 35A relay.

The size fuse that you would put on the 30A cable depends on the overall anticipated current draw on the cable. If all devices fed by the cable have a combined max. current of 20A then a 25A fuse should be suitable. Your maximum constant current for that cable is obviously 30A in which case you go up to a 35A fuse to protect that cable if need be. Don't fit a fuse size that equals the anticipated maximum current draw as you may find yourself continually replacing fuses.

I don't think I can agree totally with Whizz's comments about surge currents etc. Headlights never equal a dead short as the electrical resistance always exists in the globe filament. The resistance may increase ever so slightly once the globe is illuminated but it is always there. A nominated Voltage can only push a limited amount of current through a set resistance. In this case, using nominal figures. a 12V globe pulling 5 Amp would have a resistance of around 2.4 Ohms by my electrical theory and calculations.

I am curious to know why Brock suggests that the relay should be powered from the tail light circuit. I would have thought any live, switched feed accessible would be acceptable.

Good advice too from Brock about checking on total capacity of your alternator v's total current required by accessories and the bike itself.

Cheers
Dennis

[Brock]

why Brock suggests that the relay should be powered from the tail light circuit

I didnt say should, but could. The tail lights are easy to get at to pick up an ignition on detection. but any source is good. The P models are easy, as there are a couple of points in the wire cluster that are ready to go.

[Dennisgw]

Thanks Brock - here is some text I nicked from Whizz when commenting on Shiney's power socket installation:

Ohms Law says that Power equals Voltage multiplied by current, or P=V x I. This transposes to I=P/V or 60/12, so the headlight draws 5A when on high beam (4.58A on low beam). This means that the maximum current you can draw from your new accessories block, with the headlight on is 5A, not as you might think 10A (from the size of the fuse).

So you need to be a little careful what you plug into these sockets because the  headlight is on all the time, day and night! Not forgetting that the current draw is cumulative, i.e. you add together all the currents required by all the accessories that you plug in at any one time to get the total current requirement. Under these circumstances it is really quite easy to pop a 10A fuse, especially as there is a significant surge current requirement when you first turn on your ignition. This means that when you turn the key the headlight appears as a dead short and draws a hefty spike of current to make the lamp glow. If you also have other stuff plugged in which also has surge currents then that is the moment when the fuse is most likely to blow. There is an old adage that says that a light bulb only blows when you turn it on and this is simply because of the surge current required to make an incandescent bulb glow white hot.

PS dont forget within tha calculation that the "V" actually when running is about 14  so the "I" may well be larger and the "P" therefore too, as ohm law dictates P=I^2 * R and the only constant really is the "R" of the lamp, so just be cautious when calculating total load of alternators 

Buggar!!!,

Got my brain working now. When I was a jolly apprentice, Ohm's Law stated "The current flowing in any electrical circuit is directly proportional to the Voltage applied across that circuit and indirectly proportional to the resistance contained within the circuit". The basis is that an electrical circuit with 1 Ohm resistance and with 1 Volt applied will have a current of 1 amp flowing. Therefore if you increase the Voltage to 2 Volts, then 2 amps will flow. Conversely, if the resistance is doubled to 2 Ohm, then the current flow will halved to 0.5amp. (I=E/R)

Dale is right, if you work on the nominal voltage of 12 in your calcs, you will get one result, ie 5A for each headlamp on high beam. If you work with the maximum voltage for an alternator, potentially up to 14.6V then the current will be closer to 6.1A. Power will also increase proportionally though for the purposes of this discussion, current flow (I) is the important and overriding factor. You also need to consider that 12V is a nominal voltage only. A normal lead/acid automotive battery, fully charged, has a voltage closer to 13.2V hence an alternators voltage regulator will be set to 14 - 14.5V. Any higher and there is a risk of overheating the battery.

The brain hurts now.

Cheers
Dennis

[Tipsy]

Thanks Brock - here is some text I nicked from Whizz when commenting on Shiney's power socket installation:

Ohms Law says that Power equals Voltage multiplied by current, or P=V x I. This transposes to I=P/V or 60/12, so the headlight draws 5A when on high beam (4.58A on low beam). This means that the maximum current you can draw from your new accessories block, with the headlight on is 5A, not as you might think 10A (from the size of the fuse).

So you need to be a little careful what you plug into these sockets because the  headlight is on all the time, day and night! Not forgetting that the current draw is cumulative, i.e. you add together all the currents required by all the accessories that you plug in at any one time to get the total current requirement. Under these circumstances it is really quite easy to pop a 10A fuse, especially as there is a significant surge current requirement when you first turn on your ignition. This means that when you turn the key the headlight appears as a dead short and draws a hefty spike of current to make the lamp glow. If you also have other stuff plugged in which also has surge currents then that is the moment when the fuse is most likely to blow. There is an old adage that says that a light bulb only blows when you turn it on and this is simply because of the surge current required to make an incandescent bulb glow white hot.

PS dont forget within tha calculation that the "V" actually when running is about 14  so the "I" may well be larger and the "P" therefore too, as ohm law dictates P=I^2 * R and the only constant really is the "R" of the lamp, so just be cautious when calculating total load of alternators 

Buggar!!!,

Got my brain working now. When I was a jolly apprentice, Ohm's Law stated "The current flowing in any electrical circuit is directly proportional to the Voltage applied across that circuit and indirectly proportional to the resistance contained within the circuit". The basis is that an electrical circuit with 1 Ohm resistance and with 1 Volt applied will have a current of 1 amp flowing. Therefore if you increase the Voltage to 2 Volts, then 2 amps will flow. Conversely, if the resistance is doubled to 2 Ohm, then the current flow will halved to 0.5amp. (I=E/R)

Dale is right, if you work on the nominal voltage of 12 in your calcs, you will get one result, ie 5A for each headlamp on high beam. If you work with the maximum voltage for an alternator, potentially up to 14.6V then the current will be closer to 6.1A. Power will also increase proportionally though for the purposes of this discussion, current flow is the important and overriding factor. You also need to consider that 12V is a nominal voltage only. A normal lead/acid automotive battery, fully charged, has a voltage closer to 13.2V hence an alternators voltage regulator will be 14 - 14.5V. Any higher and there is a risk of overheating the battery.

The brain hurts now.

Cheers
Dennis

Mine just flew a buse reading that
Tipsy

[Dennisgw]

Sorry Tipsy. Couldn't help myself!
Cheers

[Diesel]

Dale is right, if you work on the nominal voltage of 12 in your calcs, you will get one result, ie 5A for each headlamp on high beam. If you work with the maximum voltage for an alternator, potentially up to 14.6V then the current will be closer to 6.1A.

Hi Dennis - is the above why you'd give the system a little scope by putting, say, an 8A fuse in this example's circuit? (To allow for a little fluctuation)?

Cheers, Diesel

[saaz]

Being in the dark about what is being powered by this circuit, if a HID light is in the circuit just remember that the maximum power it draws is much more on start up than on constant power (it spikes up when starting up, then goes back to a more consistent power use), so you have to allow for the start up spike in power draw in the wire and fuse requirement.

 Allow for something like up to 30 to 50% extra amps required on power up compared to normal power on.  If you are just powering halogen headlights or other power sources there is no such spike in power requirement.  The ballasts should have some specifications on peak power draw and saturation power (ie when on all the time)

[Biggles]

That's the good thing about the HIDs, of course.  One they're over the spike, their current draw is lower than their brightness suggests.
I noticed when I flicked my two 100w halogens on tonight the engine quit.   

Well, not quite, but it certainly noticed a bit more work was required.

[Dennisgw]

Dale is right, if you work on the nominal voltage of 12 in your calcs, you will get one result, ie 5A for each headlamp on high beam. If you work with the maximum voltage for an alternator, potentially up to 14.6V then the current will be closer to 6.1A.

Hi Dennis - is the above why you'd give the system a little scope by putting, say, an 8A fuse in this example's circuit? (To allow for a little fluctuation)?

Thats right Diesel, Allow a little more that what the constant current draw will be. If you expect a constant current draw of 6, then an 8, or 10, amp fuse would be OK.

Remember that the fuse is there to protect the circuit in the event of damage, shorting and so forth and does nothing as long as all is working OK. A lead/acid battery can punch out 400 plus amps when there is little resistance so don't be too concerned about giving yourself some leeway. If anything goes wrong it won't take much to get 10amps flowing to cause the fuse to blow.

Another point to remember is to place the fuse as close as possible to the power source, ie the battery and make sure that the cable from the battery to the fuse is well protected.

Welcome to Tuesday
Dennis

[Diesel]

Thanks gentlemen - I was wondering how to run numerous 'switched' circuits if I pinched in to the tail light circuit. I thought that I'd overload the tail light circuit - but you've showed me that putting in a relay will draw next to no current, supply sufficient power for additional circuits, and give me the 'switched' power source I am after.

Thanks to all for a most helful and informative set of posts.       

Cheers, Diesel

[Biggles]

Another happy customer!

What's obvious to the tech-heads is a revelation to the uninitiated.  So everyone does their bit here.

[Dennisgw]

No Worries Diesel,
Hope it all works out for you.

Cheers
Dennis

[Diesel]

Here is my wiring diag for my switched fuse block set-up.....

Main relay is activated by switched power from tail light circuit (lower right hand corner of diagram).....



This is what the sounds of angle grinders, crimpers, and soldering has been caused by that has been emanating from my garage this week.     

Quick math would suggest I need 15A fuses for the Halogens and HIDs - I haven't gotten that far yet with detail.

Cheers, Diesel

[saaz]

Good job there    The HID harnesses usually come with a 30amp fuse in them, but a 20amp should be fine, based on the maximum power draw during startup.

Very similar to what I have done. An extra I have is that the switched circuit - tail lights (I assume that is a relay?) can be turned off, so that the lights etc don't have to be going while the bike ignition is on.

[Whizz]

Guys, I'm not sure I understand the layout in the diagram supplied by Diesel. According to the diagram the 20A fuse feeds a fuse box via the tail light relay, two 10A fuses then feed a 40A relay each via a switch, and the relays then power either the headlights or the HID's. In this configuration the power for the headlights and the HID's is sourced from the fusebox, so exactly what are the two 40A relays doing? because as far as I can see the answer is nothing at all.

[Brock]

The driving lights are no longer fed from the switched fuse box

[Diesel]

... According to the diagram the 20A fuse feeds a fuse box via the tail light relay, two 10A fuses then feed a 40A relay each via a switch, and the relays then power either the headlights or the HID's. In this configuration the power for the headlights and the HID's is sourced from the fusebox, so exactly what are the two 40A relays doing? because as far as I can see the answer is nothing at all.

My goal was to have (essentially) a 'switched' fuseable bus bar that could run a variety of items, and be neat, tidy and easily accessible. The HIDs have a built in relay (so nothing I can do there), and I am under the impression of not having a switch wired directly to the main load (Halogens), so I left the Relay that comes with the kit in line.

I am open to suggestions, but if I run the spotlights straight from the battery (through a fuse), a have to pinch in to a switched circuit anyway (the headlight hi-beam circuit) so I have drawn the diag a little wrong. I am trying to keep the wiring loom tidier through the bike frame and avoid having a spider web of earths and +ives going everywhere.

Is this better.....



All suggestions welcome.

Cheers, Diesel

[Brock]

A slightly modified diagram.