Seperating 2 batteries w/o a "smart isolator"

Below is a PM I sent to OldSpark reworded a bit. He suggested I posed it as a thread so others can learn from it.

I have read multiple threads where people were attempting to use a smart isolator between two batteries. In 90% of them OldSpark responds with the same circuit. I understand the circuit (I believe), but am having trouble figuring out the amperage I need the relay to be.

My amp is 5 channels, 50WRMSx4, 150WRMSx1 and my computer has a 320W PSU in it. This adds up to 670WRMS (Granted, I know my amp is set at about 50% on all gains). The amp has 2 40A fuses in it.

By my calculations, I would need a 60A relay @ 12V and 50A @ 14V.
So I should be fine with a 60A? I really appreciate your help.



Im going to let OldSpark reply with what he told me. He really cleared a lot of stuff up. If anyone else has any questions, feel free to ask.

ok travis1581, thanks for the compliment. Here goes....

Funny what a re-read does. I thought you wrote that other people were responding with my circuit as a means of extending or adding to voltage-sensing battery isolators (ie, smart isolators etc).
Now I realise it was I that responded with that circuit, so that's half of my PM taken care of. (D'oh!)
(Who are they? They is me!)


Rewritten offline, I have decided to split my PM reply in two.
This first reply covers the complications of sizing a connection to another battery due to the range of its recharge current, but how under-spec'd components plus reaction times help negate that. It covers some basic stuff about cable * fuse ratings that'll bore the tears out of the more knowledgeable hereon.

The second part gets into the connection dimensioning for loads in general as well as some battery isolator complications. It explains the Div-10 rule (ie, for 12V initial design purposes, input current equals Power/10, eg a 1000W load requires an 1000W/10V = 100A distribution (cable & fuses etc). It explains how Div10 can be a reasonable rule for normal 12V circuit, but it fails miserably for constant-power loads like SMPS PSUs at low voltages.


So, here's some basic complications....

Calculating a battery's relay (isolator) capacity can be somewhat complex though it is helped thanks to double-blind behavior - ie, although people rarely factor in the high recharge current that can initially be drawn by the secondary battery, they tend not to realise the extra current that fuses, cables and relays can handle. IE - fuses and circuit breakers may run at 110% of their rating for hours if not all day. A 120% or 150% overload may take several minutes to blow the fuse.
Likewise cables are rated for an average current capacity. They may take a while to heat up so therefore short-term overloads may not be a problem. Problems occur if it heats too much and then runs at near full capacity and doesn't have a chance to cool down. (How many use room-temperature cable ratings for cables in the engine compartment or in hot boots/trunks?)

By high battery recharge current I mean that a battery can take many tens of Amps upon initial charge. For example, my 40AH AGM cranking battery typically takes 40A (it seems to peak at 45A from a few weekends ago after quite a bit of cranking due to a dead ignition). That current is the same magnitude as the battery's AH rating - ie, 40A into a 40AH battery - and I reckon most will tell you never to recharge at more than 20% of its AH capacity, and generally only 10% of its capacity is recommended - hence 8A or 4A in my case - not 40A!!

The same initial recharge can go to my secondary battery which is the same 40AH AGM (they are Yuasa UXH38-12 UPS batteries; 38AH 12V AGM). (Yes - I use a UIBI to isolate them.)
Yet I have a 60A rated relay with 50A self-resetting circuit breakers at each battery (with more than adequate Telco figure-8 cable to connect both +12V and ground).
If I were to run a 20A load on that (nominally) 50A circuit, that initial 40A battery recharge current could trip the breakers. (Hence why I replaced the original 30A fuse with self-resetting circuit breakers!)

But I'm not aware of the breaker(s) tripping - not that I'd know - I haven't yet installed secondary monitoring (maybe next millennium). (But the warm fridge told me that the previous fuse had blown - probably when I left home 3 hours earlier.)
However, that 40A initial recharge current usually drops to under 10A within 30 seconds. Even if the breakers were to trip and reconnect, each reconnection would see a smaller charge current (unless I had a big load off the second battery).
Since my fridge was about 5A to 8A, a 50A fuse should've been ok. But since IMO it is a "critical" load, it should have self-resetting breakers (like headlights etc).

So why all the verbiage above? I didn't want to leave the empty lines after cutting my whinge about idiots that would use my UIBI in addition to their "smart" isolators. (Mind you, use a voltage sensing isolator in place of the alternator charge-lamp output and it becomes a typical circuit for each battery - the original AND additional batteries - each with its own relay plus fuses or breakers to the primary battery with the relay being controlled by the isolator. And for a BIG relay or multiple batteries using the UIBI, the alternator charge-lamp output should control a smaller 15A or 30A relay which switches +12V to turn on a big relay or many relays - ie, do not overload the alternator charge-lamp output.

Besides, although the above is "basic" knowledge, it may not be obvious to some.
For example, it is important to realise that a fuse (or breaker) will not open immediately for most overloads. A 10A fuse may allow 100A thru for several seconds.
That's another way of looking at equipment ratings and their tolerance for short-term overloading. Overloads may eventually kill them (through thermal fatigue etc), but not immediately.
In addition, their ratings have to be conservative. Manufacturing variations may mean that fuses, relays, cables etc are slightly over-designed so that the worst sample still meets its rating.
(The there is environment. Does a 30A car fuse handle 30A at 25C, or 60C? If it handles 30A at 60C, it might handle 50A at 25C!)


Incidentally, I'd separate PC supply from the audio supply, but that should not effect the inter-battery isolation - unless you use separate batteries for each (which is a good idea - I would not run a PC off an audio-thumping battery - the cranker would be better; however how do you want to protect the cranker form PC discharge - a low voltage disconnect or a separate battery again?).


BTW - any battery system should have a low-voltage warning or disconnect to prevent battery damage. PC OpSystems usually have UPS software to enable graceful shutdown given a low battery warning before the supply is cut. But that's another story. A reasonable "simple" LV disconnect is the cig-plugged MW-728 (~$20) that handles 10A and connects above 12.5V and disconnects below 11.2V. IOW it is probably the cheapest "smart" isolator on the market, though it has no switching delays of any significance. Its voltage thresholds can be increased with some trick circuitry (by adding diodes, and maybe a relay if the two threshold voltages (connect and disconnect) are increased by differing amounts) and an added relay - the relay of which can be sized to extend the MW-728's 10A current capacity to 30A, 60A, or 400A etc. But that too is another story (though previously mentioned or described on mp3car).


Now for the remaining pasted part from the PM: calculating distribution capacity (ie, cables, relay, fuses).
But this is where I'll split into the next reply....

Continuing on for distribution dimensioning in general, and some battery isolator issues...


For 12V loads I use the simple "divide by 10" rule (Div10). This is close enough for most applications and it factors in that the voltage may drop as low as 10V. EG - for a 100W headlamp, assume 100W/10V = 10A. Hence 10A cabling and a 10A fuse. The 200A inrush current is for well under 1 second so that shouldn't be a problem (with regards to distribution - not other circuits).

But amplifiers, power supplies, inverters etc are usually rated in OUTPUT power. Their input power is higher due to their inefficiency - typically about 20-30% - ie, efficiencies are typically 80% or 70%.

That "divide by 10" rule is a good way of factoring in that inefficiency.
EG - we could calculate that a 1,000W amplifier or inverter with 80% efficiency has and input power of 1,250W (ie, 1,000W/0.8). Therefore at 12.5V that is 1,250W/12.5V = 100A.
Or we could just use the Div10 rule - output power is 1,000W therefore input current is roughly 1,000W/10V = 100A.

That's good enough to get an idea of what we need.
It also may oversize the fuse so that it runs at its (generally) recommended "no more than 70-80% of rating (on average)". They will handle 100% and even 110%, but they won't last as long. (And then there is fuse de-rating due to ambient temperature...)


But an important point - that 100A is based on 12.5V which is probably the highest battery voltage for such a big load.
It is more likely to be 11V - hence 1250W/11V = 114A, that's 14% higher than our original 100A @ 12.5V. (Yes - 11V is 14% lower than 12.5V, ie, 12.5V - 11V = 1.5V; 1.5/12.5 = 0.12 = 12%. ok - so there is 2% rounding error.)

But if the car is charging, then it might be 14.4V, hence 1250W/14.4V = 87A. That's lower, but we don't care about that wrt cabling etc - we want the worst case which is the highest current.

Example: If using an SMPS like an inverter, or say a dc-dc converter similar to the M2 or M4, it might be (say) 120W output, and it might provide that as low as (say) 6V.
So 120W @ 12V = 120/12 = 10A.
But 120W @ 6V = 20A. That's twice the current at 12V!
So if we used 120/10 = 12A, we'd be far short of 20A. (When charging, 120W/14.4V => 8.3A)
Consider too that the 120W is output. If it's 80% efficient, that means 120W/.8 = 150W input, or 150W/6V = 25A at 6V.

In your case, you have a total of output of 200W + 150W + 320W = 670W for the amps and PC PSU. (That's RMS of course!!)
670W/10V = 67A, hence an 80A relay should handle that. (Round up - not down!) Or a 60A relay if you do not draw full power - at least not for long periods.

But that Div by 10 rule is an estimate.

If you also have a secondary battery, you need to factor its maximum charge current (which will probably drop to under 2A once full) and add that to the total load. So if its 40A like mine, 67A + 40A = 107A. Hence a 120A relay, though maybe a 100 relay with 100A breakers (self resetting?) or fuses might generally do.


Many may think "But if the voltage drops below (say) 12.5V, then the isolator opens so that the inter-battery circuit does not handle the higher load currents at at 11V or 10V etc.".
Well the UIBI will stay connected as long as the alternator thinks it is charging - which it will do even at very low voltages.
And although that is one possible advantage of a voltage-sensing or smart isolator (however I argue not except for systems where the alternator cannot normally supply that total load so you want to prevent the main cranking battery from flattening; though even then I prefer a dash voltmeter or a delayed alarm and a manual isolate switch), some isolators may delay up to 60 seconds before isolating, and all but truly intelligent designed units (are there any?) will go into a cycling mode where they continuously reconnect with a cycle time equal to isolate delay plus (re)connect delay. For simple units, that may be seconds, but most will cycle at 15 second to 1 minute intervals. Some may be as infrequent as 2 minutes.
(Remember - remove the heavy load that pulls down the alternator voltage and the alternator voltage goes up, hence reconnecting the not so smart isolator.]


And there are a few things I've missed - I might back-edit later. (I forget what things I was to remember to add...)