Current draw is based on the circuitry, what's in the box and what it want's to eat; Circuit capacity in amps is 30A, based on the wire and relays and such you used, as you want to drive a fuel pump and stuff, but each separate bit draws what it personally needs.
The total needs to be less than that; I use a "60% of max rating" number for designing stuff like that; I like headroom better than redoing it if I need to add "one more thing".
A fuel pump is a good example of a "dynamic electrical load", because it changes how much current it draws, depending on what's happening.
You'd think a 255 walbro would draw the same ~8A across the spectrum of drag racing, but it can be much higher, and then there's the pump-boosting boxes and all that try to fix it.
The big difference is that the 255lph of fuel are Way harder to drive into the pipe if the car is moving (pipe is moving) relative to the fuel.
At ~2G's, the power should be ~4x, ignoring losses.
That doesn't last for long, maybe a second or less, but if the pump stalls from undercurrent, it takes seconds to recover.
I used this exact circuit to keep car eecs alive while I did stereo work on them, WBW. Plugged it into the cigarette lighter, until they started turning them on with transistors, then it has to be clipped in to the fusebox with clips.
I never blew one up.
You only need one diode for a keep alive, as long as the load you're hooking it to is not greater than ~20mA.
You only care about keeping the eec alive; if there are parasitic loads connected to it, I'd move them to a different switch terminal, on a dpdt or 3pdt switch.
The use of two diodes, and isolating the pin, is something I'd have to look into, because some things don't like that.
In my personal opinion, the keep alive only needs to keep everything over 8V; that's the point where the internal regulator has started to droop, and they want a 3V dropout voltage on the 5V (lm7805 family) regulator in the eec. This is years before LDO voltage regulators... but I digress.
If you hook it up as I said above, it can't blow up unless the diode is bad, and/or it's carrying the load into the pin fully powered up.
EHH: I said:
As was said, the thing should go in the line between the connector and the eec; + terminal of battery to non-banded end of diode, band on the diode toward the eec.
Ground to ground.
"IN" is not correct there, "TO" is what I should have written. There's only one wire to that pin 1 circuit, the way I've done this; the banded end of the diode. The other terminal of the battery is Ground. Diode only hooks to + battery terminal, and pin1.
As long as it's over 8.4V, the 9v batt is disconnected, with the 1 point hookup.
The problem with 2 diodes, an Isolation Circuit, is that That circuit, pin 1, might draw 10A when everything's else is running at 12v; it likely draws less power when the other circuits are powered down; EEC-V does this.
That makes things complicated; you need to measure the input to that pin under different conditions, to prove the other parts of the circuit.
If it tries to draw 1A when everything's off, it also needs to be rethunk, a 9v won't work for that; and even a 1n5400 used for isolation would blow if pin 1 draws 10A powered up. (IDK, I think EEC KAL is a 7A fuse in my car)
Isolating the pin with diodes is a different problem, and will require some testing; it might work fine, but it might not. I've never tried that, so I can't speak to that.
But I know from experience that if you install this device and the battery gets warm, the battery isn't going to last long.
I use 9V batteries for rocketry, to deploy parachutes using electric matches, and they draw ~3A for .5-2 sec, which is a lot for a tiny battery like that.
I'm not trying to blow up your car, that would be tragic.
I just needed to give better detail.
EDIT: I see the graphic now; I didn't see it before.
I wouldn't use the 30A diode you show, but fuck it, it should work fine.
It is rated for the full circuit load, so it should be fine.
These are cheap: https://www.mpja.com/Stud-Diode-85Am...tinfo/34996+DI
If the dark ends are the bands, both bands should be toward each other; the EEC is the load, and "bands (kathodes) go toward the load". (In a negative ground system)