5 amp fuse keeps blowing?

[Hoohaa]

Originally Posted by smac999

your instrument fuse should be large. say 15amp. I normally run 30a breakers for nav gear. then run cable to bus bar, then each unit has it's proper size fuse from bar based on the manual. or buy a blue sea fuse block to feed from the switch. https://www.bluesea.com/products/502..._Bus_and_Cover

Sounds like a plan, Thanks..
I will figure out the gouge of the wire I have used and will probably use a 10 amp and make shore the in-line fuses are installed.
Thanks.

[TurninTurtle]

Originally Posted by smac999

your instrument fuse should be large. say 15amp. I normally run 30a breakers for nav gear. then run cable to bus bar, then each unit has it's proper size fuse from bar based on the manual. or buy a blue sea fuse block to feed from the switch. https://www.bluesea.com/products/502..._Bus_and_Cover

Even better than the 10 amp feeder line idea... Leaves room for adding another device (with its own fuse)

****************

I required the guy installing my dashcam and rear camera in my truck (I didn't want to take the dash apart myself) to put them each on their own fuses and the 3.5 inch rear camera monitor is on its own fuse.

The cell phone chargers that plug into the cig lighter ports... have a fuse in them.

If you get a "splitter" to plug 4 devices to the cig lighter, it has a fuse and then each device has its own fuse.

This is ALL done for a reason.

[dlymn]

separate fuses might be ok as long as you remember where they are. If all of the components go back to the original fuse it will still kepp on blowing. the ratings you are quoting might be average use and don't consider the occasional surge such as what happens when you turn a device on or the dvice loads up to operate a relay or other piece of equipment. So go with a small panel of fuses(like your car) just to keep them together but put the lot through a larger circuit breaker (aka fuse) to cope with total load

[cal40john]

There are lots of different fuses with different time/current ratings. One example is old style panels quite often had AGC fuses in them. This fuse is called a fast acting fuse. From page OC83 of this link - http://www.cooperindustries.com/cont...ne_Catalog.pdf
- we see that a 5 amp AGC fuse will take 100 seconds to blow with 6 amps of current, 1 second with 10 amps, 0.1 seconds with 23 amps

[reed1v]

Fuses need spare capacity. Breakers already have that built into them. Rated amps per appliance is an average, not a constant. Start up amps usually more than operating amps. And yes, not just an ideal, but a wise move to have everything separately fused, ideally with breakers not fuses. Easier to track down the short, isolates the problem so other systems can still operate, and easier to replace.

[CLady]

What size wire on the power side?

[Cadence]

skipmac summed it up correctly. I think the first thing is to identify the wire gauge to and from the switch and fuse. It could be all you need is a larger fuse because you have increase the load. Just don't fry an inadequate wire size.

[MarkSF]

Originally Posted by smac999

your instrument fuse should be large. say 15amp. I normally run 30a breakers for nav gear. then run cable to bus bar, then each unit has it's proper size fuse from bar based on the manual. or buy a blue sea fuse block to feed from the switch. https://www.bluesea.com/products/502..._Bus_and_Cover

That's what I used when installing new instruments. Works great.

The wire from the breaker to the fuse block should be rated for more than the breaker, and also for less than 3% voltage drop with all the instruments on.

There's a good voltage drop calculator, here :

Voltage Drop Calculator Genuinedealz.com

For example, although 18 gauge will handle the current, you need 14 gauge to stay under 3% over a 10ft length at 5A.

Don't forget that the voltage drop from breaker to fuse block, and from block to instrument, is accumulative, and the total should stay under 3%.

I've always got a good length of 14 gauge two core boat cable, as it's so useful - a great general cable to have around.

[smac999]

Originally Posted by Cadence

skipmac summed it up correctly. I think the first thing is to identify the wire gauge to and from the switch and fuse. It could be all you need is a larger fuse because you have increase the load. Just don't fry an inadequate wire size.

each of the devices will have inadequate wire size. that is why they each need a proper sized fuse. among other reasons. you can't just put one big one for everything.

[ramblinrod]

Originally Posted by Hoohaa

I have recently installed a new AIS and splitter. We wired in a bus, which takes its power from the instrument switch on the pannel. If everything is running, the fuse will blow after a minute or so. I suspect we are overloading the 5 amp fuse but would like some input on what size I should be running. Note that the fuse does not blow if the AIS is switched off.
Here is a list of the equipment currently using that fuse and the manufacturers power consumption specs.

Class B Icom AIS MA-500TR, with its own gps antenna - Uses TX 1.5 amps & RX 0.7 amps (could someone explain what TX & RX mean?).

GME AIS Splitter (AISS120) - uses 1.15 amps.

Garmin GPS map 550S plotter with a Transducer - uses 1.25 amps.

If this data is correct then its a total of 4.6 amps..

Thanks for your help.
Jeff

Refer to ABYC E11 - AC and DC electrical systems on vessels.

The panel current limiting device should be sized to protect the wire, not the devices connected to it. Each instrument should be protected by a current limiting device that will not disable other devices connected to the branch circuit.

Fortunately, based on your description, the wiring errors you have introduced into your system are not likely to start a boat fire (like many of the DIY wIring mods I see). When it comes to vessel wiring, don't guess; follow the standards or hire a pro, your life or the life of loved ones may depend on it.

[reed1v]

Still not clear why fuses and not circuit breakers? Is there some advantage to having fuses?

[EzzyD]

There could be another issue... What is the Voltage that the stuff is running on?
As your voltage drops ( due to use of battery ) the current could be more then specified, (as the Power requirements are a function of Voltage: P=VI).

What is the state of your battery health? Does this happen while hooked to shore power?

[ramblinrod]

Originally Posted by EzzyD

There could be another issue... What is the Voltage that the stuff is running on?
As your voltage drops ( due to use of battery ) the current could be more then specified, (as the Power requirements are a function of Voltage: P=VI).

What is the state of your battery health? Does this happen while hooked to shore power?

While the formula you referenced is correct, the application may not be.

Reducing applied voltage usually reduces current drawn.

Most devices represent a fixed impedance regardless of the voltage applied.

In that case P = E^2/R

e.g. A device rated 60 Watts at 12 Volts represents a 2.4 Ohm load and draws 5A.

If the voltage is reduced to 10 Vdc, P = E^2/R = 100/2.4 = 41.7 W. The current is actually I=E/R = 10/2.4 = 4.17 A.

[cal40john]

Originally Posted by ramblinrod

While the formula you referenced is correct, the application may not be.

Reducing applied voltage usually reduces current drawn.

Most devices represent a fixed impedance regardless of the voltage applied.

In that case P = E^2/R

e.g. A device rated 60 Watts at 12 Volts represents a 2.4 Ohm load and draws 5A.

If the voltage is reduced to 10 Vdc, P = E^2/R = 100/2.4 = 41.7 W. The current is actually I=E/R = 10/2.4 = 4.17 A.

This is correct except for devices with switching power supplies built in.

[TurninTurtle]

Originally Posted by ramblinrod

While the formula you referenced is correct, the application may not be.

Reducing applied voltage usually reduces current drawn.

Most devices represent a fixed impedance regardless of the voltage applied.

In that case P = E^2/R

e.g. A device rated 60 Watts at 12 Volts represents a 2.4 Ohm load and draws 5A.

If the voltage is reduced to 10 Vdc, P = E^2/R = 100/2.4 = 41.7 W. The current is actually I=E/R = 10/2.4 = 4.17 A.

Works for a resistance based incandescent light bulb... not for most electronics.

Generally its a bad idea to feed electronics the wrong voltage. While most "12 v" devices can tolerate 10.5 to 14.0 v (some a little wider range) due to expected output range of automotive alternators and batteries, some can't tolerate being outside 11.0 to 13.0 v.

Electric motors will attempt to deliver the power demanded by the load. Reduced voltage applied invariably results in increased current demand. Reduced voltage results in lower operating RPM which means lower "counter-electromotive force" which is the main "resistance" to current flow while a motor is spinning.

Actually the "fixed impedance" is mainly seen in lights (and not LED or fluorescent lights...)