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.
Distribution dimensioning using Div10; battery isolator issues.
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...)