How can I calculate the minimum switch time between main battery and back-up battery for my board?
When the battery is disconnected, the load will have to only rely on the capacitors on the board. The voltage will drop at a rate depending on the total capacitance available.   

If allowed drop on voltage is dV, then time will be
Time  = dV* C/I
Where I is the load current.   

Share the actual values of load current, allowed dip and capacitance available, so that we can see how much will be the time.   



Edit 2:    

Example:    

Load current - \$800 mA\$
Change in voltage allowed = \$5 V\$
Capacitance available:  4\$70 mF\$  

calculated time will be  = \$293.75 ms\$