I am trying to figure out one of two ways to switch off the power in my circuit using an LM2596-5.0 switching regulator.
Option 1: In the datasheet it says you can pull the regulator ON/OFF pin to VCC to turn it off, with a loss of 80uA.
I'm using an SPDT switch, so if I choose the above option, the regular's input capacitor will remain charged, and won't the battery drain?
Option 2 is a more traditional way, leave the regulator ON/OFF pin tied to ground (datasheet says this will turn the regulator on), and the SPDT switches all of the power to the circuit, which charges the capacitor up and I'm sure eventually this might wear out the switch's contacts, despite it being rated for 5 Amps. The circuit consumes less than 50mA average, but I'm assuming that capacitor wants to charge quickly, hence my concern.
In either Option 1 or 2, I'm avoiding a parasitic loss on a voltage divider as shown in my schematic. Option 2 would avoid the 80uA loss if I use the regulator's shutdown feature, but in my application I don't need to worry about 80uA loss.
I see in the datasheet for the LM2596 that if I choose my Option 1 it would truly be a 'soft' start, which would extend the switch life? (not really an issue I'm thinking, due to the switch's rating, but I may be wrong?)
Can someone help me decide what would work best? Thank you.
BVCC is battery voltage which is connected to the common/center pin of the SPDT switch, the top pin of the switch is connected to the common/center pin when the switch actuator points downwards.
1st schematic showing Option 1:
1nd schematic showing Option 2: