voltage drop problem, what battery to buy?

tommy... wrote: Odyssey and Northstar are a few brands a buddy uses for his competitions(group 31's...) I got a few for him this past year...Couldnt even find them for awhile...They said miltary was buying them up...! What size is the stock battery in those(group #)...There really is no upgrade for the alternator in those ...huh...? What about a "factory" upgrade alternator...(just throwing it out there) On a side note...Did you fuse the 8 gauge extension seperately(from the dist block)...?

I will look into these brands. A factory upgrade alternator isn't really going to work, considering that if they took the alternator apart and increased the number of windings, I wouldn't get that much more considering that the alternator already has no room barely for any additional windings (just in stock configuration). I don't know the stock battery size at the moment. Also, the distribution block for power is fused, both the 8 and the 0 guage are on separate fuses as part of the fused distro block.

michigan_tech wrote: Can anyone else verify this?

I hope so! It's called Ohms Law. But I'm talking about the voltage drop after the battery - ie, the 1.5V - not the hold-up voltage available. A battery will not reduce this 1.5V drop. Sure, a bigger battery will stay closer to 12.7V for longer, and AGMs like Odyssey etc have lower internal resistance hence a higher output voltage, but I figure it's cheaper upgrading the big-3 than an expensive AGM battery. (Isn't cost an issue here?) Or if it is the distribution block that is causing the drop, how much are magnetic breakers compared to an Odyssey battery? Else what is the total load Amperage - maybe it's better sizing the main fuse for the load rather the cable, thereby possibly avoiding the smaller branch fuses. (Unless you require discrimination - but usually all or nothing should be fine for car power-amplifier systems. No??)

oldspark wrote: michigan_tech wrote: Can anyone else verify this? I hope so! It's called Ohms Law. But I'm talking about the voltage drop after the battery - ie, the 1.5V - not the hold-up voltage available. A battery will not reduce this 1.5V drop. Sure, a bigger battery will stay closer to 12.7V for longer, and AGMs like Odyssey etc have lower internal resistance hence a higher output voltage, but I figure it's cheaper upgrading the big-3 than an expensive AGM battery. (Isn't cost an issue here?) Or if it is the distribution block that is causing the drop, how much are magnetic breakers compared to an Odyssey battery? Else what is the total load Amperage - maybe it's better sizing the main fuse for the load rather the cable, thereby possibly avoiding the smaller branch fuses. (Unless you require discrimination - but usually all or nothing should be fine for car power-amplifier systems. No??)

Well if I already had upgraded the charging wire to 0 gauge, what more can I do to upgrade the big 3? The SAZ 1500D has 4x40 A fuses The SAZ 50.4 has 2x25 A fus I have a distroblock that has a 250 A fuse between each one. 0 gauge in, 250 amp fuse to 0 gauge out (1500D)             250 amp fuse to 8 guage out (50.4) Hopefully this can help some. What is a magnetizing breaker by the way? And what did you mean by sizing the main fuse for the load rather the cable? The load is two Dual 3 ohm voice coils wired parellel down to .75 ohms. They can handle 2000WRMS all day.

The big 3 - like I said - duplicate what you have. But you need to determine where the biggest voltage drop is occurring - ie, battery to chassis, chassis to amp, battery to distribution block, the distribution block, or d-block to amps etc. If for example your drop is 1V in the ground and to the d-block, and say 0.5V across the d-block, then duplicating your battery to chassis and battery to d-block (2 of the big 3) will HALVE that 1V drop. IE - two parallel 0 gauge cables will halve the resistance of a single 0 gauge cable (of the same length). A magnetic or Hall-Effect ammeter or circuit breaker is one that senses current WITHOUT using a series impedance. Traditional breakers and ammeters use at least a shunt resistor - eg, say 1 milli-Ohm for 10-500A, or a 0.25V drop for a 250A amp load (ie, 0.25V & 62.5 Watts lost). Magnetics don't - they sense the e-mag field - not the voltage drop or heat given off. Usually to overcome the long cables from the alternator and main battery, a second battery is located next to the load (amps etc), or the main battery is moved there. But you say you do not have the space. (And if you can fit a capacitor, use a small AGM/VRLA battery instead - a small 12V 4AH or more common 7AH (usually the same price or cheaper than a 4AH) battery provides at least 1,000 times the storage of a more expensive 1F or similar cap.) Fuse sizing: You have 2 series 250A fuses. (That is strange - normally it'd be say 250 followed by 125A or smaller, or 500 by 250 etc.) If 250A is enough to protect the 8G cable, then just use a single 250A fuse at the input to the 0G cable. That halves that resistive loss. But say your load is 4x40 + 2x25 = 160 + 50 = 210 Amps total. A 200A fuse is probably ample, but the 250A fuse will certainly handle the max load with a bit of headroom. 8G is about 0.000641 Ohms per foot, & 0G about 0.000100 (nominal ratings of about 45A & 170A depending on application). Or 8G has over 6x the resistance of 0G. Or at 250A, 8G drops .16V per foot and 0G drops .025V per foot etc (V=IR). So keep your 8G as short as possible (unless you split the 0G into at least 6 8G cables - ie, 8G has 6x the resistance....). And if the 8G is short, you do NOT need fuses to protect it - the theory being there is little chance of a short to ground (aka physical security - no fuse type protection needed). And if you have an 8G short to ground, although it is rated for a nominal 50A, it should carry enough current long enough to blow or trip a 250A fuse/breaker. EG - assume a 300A current - that's .000641 x 300 x 300 = just under 60 Watts of heat per foot (P=RxIxI) - which shouldn't generate too much heat. So hence why not use a single 250A fuse rather than two? (That halves fuse losses.) And double your existing big 3 (or 2/3rds anyhow) since that will halve ground losses to the amp/s, and power losses to the distribution block. Forgive my long examples, but it shows how a physical layout can be manipulated for best outcome - in this case, minimal voltage drop along the conducting path. Most wiring protection (fusing etc) is designed to protect downstream wiring. But sometimes the wiring isn't worth protecting. [Or the wire itself is the protection - remember those primitive "wire" fusible links? (Not that that is a good example...)] And all those calcs are just Ohm's Law V=IR and Power P = VI, and substitutions to get P=VV/R = IIR. Easy isn't it?

So you're saying I need to double the zero gauge from my alternator to my battery, from my battery ground to chassis, and engine ground to chassis? Also, if I change the 8 gauge on the 4 channel to a 4 guage on the 4 channel, does that help or hurt me? The 4 gauge will have less resistance in theory I believe, but won't it allow it to draw more current and hence suck more life from the battery? Also, how do I hook that up on the distribution block? My block is set up like this:       ANLFUSE1 ---> (0/4/8) 0 --> Dist.BLK       ANLFUSE2 ---> (0/4/8) You said to use a single 250A fuse. How do I do this given the way the block is? Also, do you recommend running a direct ground to the battery for the sub amp? I thought you weren't supposed to have long grounds on sub amps, even if its 0 gauge. By further upgrading the big 3 and changing this fusing/grounding, do you think I can achieve stable voltage?

Cables don't drain batteries - it's the overall load that drains a battery. Ideally cables would have zero resistance (aka super-conductors) so there is no power loss. Using the water analogy, whether you water your garden with a 10' or 1/2" diameter hose doesn't make much difference to the flow - it's the nozzle at then end that determines that. Fattening your pipe (cable) reduces your transmission resistance hence reducing voltage/power losses, hence your load (amplifiers) get more voltage/power. As to "stable voltage" - this merely reduces voltage drops. With a fixed resistance (ie, 1, 2 or 10 parallel 0G cables etc), the voltage drops vary with the load current. So is that stable? The load voltage will still vary, but with lower transmission losses, its variation will less. The battery will dip. A large AGM battery might have 5mR internal resistance, hence it drops 200A x 0.005R = 1V. A larger (Odyssey 75/86-PC1230) is 2.5mR hence a 0.5V drop at 200A. Then there is the batteries capability... For now, keep it simple and try to minimise your losses.    Later you can decide if are you concerned with voltage losses (resistance), or voltage dips (from varying current with power output)? And if concerned about dips, why? (Keeping in mind you are already down at least 1.5V for high power outputs because your alternator isn't big enough.) As to running a "direct ground to the battery for the sub amp" - only if you have a high resistance chassis or connections. But I would be surprised (read: worried) if a cable had a lower resistance than a car chassis. But all those voltages are easy to measure with a voltmeter anyway. And there is unlikely to be much difference in length between chassis and cable anyhow (who cares? and why?) Normally I'd say the more parallel cabling the better, but not for a metal chassis (unless cable-connector-chassis joints have a higher resistance than end-to-end cable).     For more info though, I suggest experienced gurus. Whilst I keep hearing things that seem strange to me (like single cables not paralleled, or ground loops, or antennae planes etc), I don't have the practice. And they can advise of the best equipment and fusing etc. I'd run the fattest cable as far as possible. And if lower gauge fanouts are short and secure, they need not be fused.    The rest of the crap below is optional detail. Feel free to ignore it! EG: Lets assume you measure a 2V drop with the amps using 200A - ie, the battery as 12V; the amp as 10V; 12-10 = 2V. Your Amp is consuming 10V x 200A = 2,000W = 2kW (P=VI). Your cable is dropping 2V @ 200A = 2x200 = 400W (20% of 2kW the same as its 2V drop is 20% of 10V.) What resistance is the cable (including fuses etc)? V=IR hence R=V/I = 2V/200A = 0.01 Ohm = 10 milli-Ohm. That's the equivalent of 100' of 0G cable (from 0G = 0.0001 Ohms per foot). I'll use "R" to represent "Ohms". You might then find that the distribution block is dropping 1.5V and the cable 0.5V. So the d-block is 1.5V/200A = .0075R = 7.5mR. The cable is 0.5V/200 = .0025R = 2.5mR. The total "transmission" resistance is cable + d-block = 7.5mR + 2.5mR = 10mR (same 10mR as original (1.5V + 0.5V) 2V at 200A. Say if you halve the transmission resistance. Instead of 0.01R we have 0.005R. The voltage drop is V=IR = 200A x 0.005R(ohms) = 1.0V. That's half of the original 2.0V drop. (As we'd expect - we halved the resistance & since V=IR and we made 1/2 x R, that means 1/2 x V where V is the voltage (drop) across the Resistance R with current I running through it. So now you have one extra volt for your amps. But that's looking at it in simple terms. The lowered cable & d-block resistance means more current will flow - eg V=IR hence I=V/R = 12V/(50+5)mR = 218Amps (where 50mR is the effective amplifier(s) resistance - ie, 10V/200A) so hence you now lose 1.1V across the cable & d-block, and the amps get 10.9V or 0.9V (~10%) more than it did. (1.1+10.9 = 12V = battery voltage.) In practice, the battery voltage will drop a bit due to the increased current, and amplifiers may not behave like a resistor.... But the main point is that by reducing your transmission losses aka resistance, the amps get more power.

So are you saying I SHOULD use the 4 gauge on the 4 channel? Also, how do I hook up a 4 gauge input with a 0 gauge input and only use one fuse? Remember, this is how it looks in my vehicle right now:       ANLFUSE1 ---> (0/4/8) 0 --> Dist.BLK       ANLFUSE2 ---> (0/4/8)

No - I'm saying that to reduce your voltage drops, you need reduce you resistances. How you best do it is a design issue. Are you designing for bling and hence want the best looking system, or maybe the loudest system? (Both IMHO&E usually resulting in poor sound quality.) Maybe you might want minumum voltage drops hence 0G (0.1mR/ft) splitting into two equal loads with 2 x 3G (0.2mR/ft), or into 2G & 4G (.16 & .25 mR/ft) for a 40/60 load split. (But why not 0G splitting into 0G? See below...) But if I understood you correctly, you have a single 250A fuse on your 0G (presumably near the battery!) which then feeds 2 downstream 250A fuses at each split? If that is correct, then why have the downstream fuses - they are meaningless. (Did I miss the obvious or misunderstand?). IOW - remove them. Get another disto block or bypass them or short them with cable. If I had such a system, I'd have the 250A fuse near the battery (obviously) into 0G going as far as possible - right to the loads if they handle it. For the split to each load I'd use some heavy secure cable joiner or clamp (twin screws; pressure spreaders; insulation clamps etc). Or for Mr non-Bling Cheaps8t me, maybe at least two hose clamps or U-bolts - with maximum cable-cable copper interweaving - in an insulated box or surrounded by copious insulation. If my splits are long runs and their gauge was too unsafe at the "master fuse" rating (250A), and assuming I couldn't split into 0G and then the smaller G at the last few inches, I'd install fuses at the splits as per normal practice, else guarantee physical security. [FYI - These audio systems are similar to traditional telephone exchanges us DC power at HUGE currents. They use copper busbars and NO fuses - instead relying on physical security and maintenance procedures. If a fault happens, they burn to the ground (as did some New York exchange a few decades ago). But they have no choice - you cannot safely break (fuse) DC loads of several thousand Amps. And the practice works - despite the odd short circuit that usually vaporises whatever metal ladder or crowbar or nose-ring that somehow managed to touch both busbars, rarely has an exchange burnt due to shorts.] But as I said, you need a designer. If you understand what I have said, you should be able to do it yourself. (Hence my sample calcs & analysis.) It's Ohm's Law (V=IR) for given cable sizes & lengths and the "combining resistor in parallel formula". But the latter is simply "half the resistance every time you parallel the same sized resistor" (eg if a cable is 12R, two in parallel is 6R, two of those = 4 in parallel = 3R etc). There is a bit more circuit theory (sum of load voltages = source voltage; sum of parallel currents = source current; and the same total current flows thru a circuit, and at a given connection point, all voltages are the same)... But all the above are really just interrelations of the same V=IR Ohm's Law. But instead you could just keep improving your system. If you find your biggest single voltage drop is from battery negative to to chassis, double the cabling. That will halve the resistance and hence halve that voltage drop. If it's still the highest, double again, or add as much as you can - the fatter the pipe, the lower its resistance. If it's the distribution block, then overcome it somehow. If you can't, its loss has to be accepted. (Not that anything is ever acceptable!) Eventually though you hit a limit - you can't attach a 1' diameter cable to a battery etc. Or it is cheaper buying a vehicle that can accommodate battery(s) next to the loads, etc etc. Or cheaper using a AC system.... That's all design & installation stuff which is why we have talented installers. I'm not one of those (nor talented). I "merely" expanded the simple Big-3 principle in reaction to what I considered unacceptable voltage drops. My experience does however include people that install expensive batteries for no significant gain, or even install capacitors to (quote) "increase alternator output voltages". (I think they need a true-RMS voltmeter if that's what they really did measure!) And lots of people that think having a HUGE battery can overload their vehicle, or that a HUGE alternator can blow their battery etc. (Noting that eg batteries may have charge-rate limitations, but that's usually not the issue....)

Brilliant! I used to work for Allen Bradley - I wonder if they still discount? And that dude (Carnes?) explains it so well. If only I could do the same with my plasmic recycler.