so confused about fusing battery to amp

I wasn't confused at all, until I did a search and read the masters' replies. Now I've been searching a reading for the last two hours...I just need an answer at this point. Here's what I (now) understand: Since I have chosen to use 4 gauge wire to run my Massive Audio NX5 and JBL MS-8 combo, I (now) know that I will put a fuse on the battery to amp/distro block not to exceed 150A to protect the wire and vehicle. The MA NX5 does not come with a fuse, but rather a "protection circuit" inside the unit. I have figured the most it will draw, in watts, is 1360w. Now, I think that I multiply that number by 60% (?why), then divide by 14.4, which gives me ~ 56A. This seems way too low to me, so my math is probably junk. But let's say it is...using that plus the 25W fuse that's in the MS-8 as my additional info, I would add those together to get ~ 80A fuse. Ok, that's all fine and good, but here's where nothing makes sense to me, please explain: 1) the fuse is there to protect the wire, not the equipment; I've read this multiple times (and adamently stated, I might add!). So why do equipment manuals always say to put an xx sized fuse on the line (within 18" of the battery, etc.)? The manual doesn't know what size wire I'm using, and usually doesn't recommend one. My MS-8 says to throw a 25w fuse on it...the same as the equipment uses. 2) once I get from the battery to the distro block safely (with my 150A in place), do I then need to put an "appropriately sized" fuse on each wire leaving the block going to each piece of audio equipment, fan, capacitor (NO, I don't use them), etc? If the answer to that is yes, God forbid, then how safe is my BMW since it has the battery in the trunk, but the fuse box in the glove compartment? Meaning, those wires are very unprotected over that distance from the battery, no? If you got through this post, "thanks!" If you decide to reply, "THANKS!"RadarContact

As you mentioned, the fuse is there to protect the wire, not the amp. If the book that came with an amp tells you to put an inline fuse then they are most likely telling you to put a fuse at the amp and they assume you are using sufficient wiring. If they are referring to the battery than again they assume the installer is using the proper size wire. That fuse rating can be used at the battery as well (and added to fuse requirements for all other amps). As you mentioned, at the battery, with 4awg, you should NOT put anything larger then 150A and it is beneficial to put the smallest fuse in the fuse holder that won't blow under normal operating conditions (for example, if your amps only draw 80A a 100A fuse would be more then sufficient and would provide better protection then a 150A fuse). To rough guess current consumption take total output watts (1360) and multiply by battery voltage (14.4) and then multiply by 1.10 (110%) for a class D amp (assuming 90% efficiency). This gives you 104A. If you are using a class AB amp you would multiply by 1.50 (150%). If your sub amp needs 104A and your smaller amp is 25A then 150A should be a good fuse to use!Kevin Pierson

Kevin, thank you. I will use that formula from now on. My particular amp is both D and A/B, so I'll get to try it twice today! Could you, or someone else, please address my question #2? This has me concerned, especially the fuse for every piece of equipment's power wire part of it. I am in the middle of an install today and I want to do things right, and safe! Thanks again!RadarContact

I just looked in the om and it says: BEST EFFICIENCY AT 4OHM -- 63% (A/B) 82% (D) The sub section will not be used at 4ohms, but at 2ohms, however there's nothing doc'd for 2 ohm use. So, knowing that, would I substitute 1.37 (for the 1.5) and 1.18 (for the 1.10)?RadarContact

Remember, we are just rough guessing here - the formula I gave is a ball park. You are correct, if you have published efficiencies for your amp use those. I would guess that efficiency will only go down as ohm load goes down but that is just a guess. To answer question two I would say yes. Think about the first 150A fuse - it is sized based on the maximum current that a 4ga wire can safely handle. So, if you fuse a 150A 4ga feed to a distribution block and branch off of it with a bunch of 8ga wires you fuse won't protect the 8ga wires. I would guess your BMW has a fuse right at the battery somewhere to protect the main feed going up to the the in cabin fuse block.Kevin Pierson

Copy all. Again, thanks...I feel much better now (although my install costs just went up another $86 in fuses, wires, and distro blocks!). (found some great prices, though, at Darvex.com...can I say that here?)RadarContact

Good to see some REAL formulae - ie efficiencies etc. As KP said, fusing at the battery is to protected the downline cable (and battery) to prevent smoke & fire (and battery explosions). The fuse should be the smallest link - ie, lowest rated current - until the next downstream fuse takes over. (EG - 100A fuse for 100A or 150A cable until a 50A fuse for 50A cable etc.) Cable aka distribution fusing is different to equipment fusing in that equip fusing is to protect the equipment and is usually part of and supplied with the equipment. IMO it must be supplied else defined by the equipment manufacturer/supplier because only they know what profile is required - ie, what gives first in their design. But many have fuses internal to the equipment - ie, "non-user serviceable" else inconsiderate or money making manufacture. As to "rules" - I used the common "divide by ten" rule - ie, current = power / 10, so (800W + 240W =)1240/10 = 124A. So a 125A fuse => 130 or 150A. That div-10 is simpler than 12 or 14.4 etc and tends to allow for efficiency (ie, converts output power to input power) or voltages lower than optimum - ie, a 1,000W INPUT single channel amp is 1000/14.4V = 70A (@14.4V) or 1000/10 = 100A @ 10V (assuming the amp has a proper dc-dc convertor). But that rule is a ball-park and only for initial guesstimation purposes. Furthermore the general rule that fuses and protection (and cables?) should not generally be run at more than 70-80% of rating. IE - a 100A load should use at least a 100/0.8 = 125A fuse and cable. Both the above combine to enable selection of a "reasonable" sized cable and fuse etc. Reasonable includes practical considerations - eg, maybe 200A is required, but a 200A fuse is a different class or price, and 150A should be fine since (1) it protects the cable, & (2) we designed for MAX power output which won't be run for more than a few seconds if at all. (And since a spare fuse is always carried...) And sometimes circuit breakers are cheaper. (And sometimes better.) Dang - I really just wanted to know how close the div-10 estimate came to the above efficiency etc calcs - especially after selecting the next or closest "real" fuse etc.

Lol, that's okay, I am always willing to learn! Thanks for taking the time! In truth, I found that the Watts x voltage x % didn't work for me, I kept coming up with numbers in the tens of thousands. Kevin, I'm sure I was doing something wrong. However, I had a formula written down from the hours of reading the night before and I got it to work for me. It's: watts / efficiency / volts (example 560 / 63% / 14.4). I ran this separately for the A/B and the D since the efficiencies were so different. I used the 82% for the sub section even though I'll be using it at 2ohms versus the 4 ohms the efficiency rating is based on. It came out to a total of 129A. Pretty damn close if I would have used the tens method, 1360 / 10 = 136. In my head, even (on a good day! Lol). Add that to the 25A required by the MS-8 and I'm maxed out for a 4 ga (150A) wire. I'm never gonna be pushing a solid 1360w though, so the 4ga is fine. Branching at a distro block going to a 4ga fused at 125A for the NX5 and an 8ga fused at 30A for the MS-8. That should cover it, right?RadarContact

Not Watts times voltage... it's watts DIVIDED by voltage! Two simple formulas that cover just about everything on this forum: V = IR === Voltage (across a resistance) equals the Current (I) thru it (the resistance R); in Volts, Amps, Ohms (or Volts, milli-Amap & kilo-Ohms). aka Ohm's Law. P = VI === Power equals Voltage times Current (Watts, Volts, Amps). (P in something = voltage across it x current thru it) From the above 2 formulae and basic algabraic rules (commutative, substitution etc) come others like: I = P/V P = VxV/R = IxIxR (= V^2/R = R*I^2 depending on your preferred nomenclature).

Yeah...see..well...I'm gonna go play with my Lego blocks now...RadarContact