Your description of the relay is unclear. Usually we speak of a pull-in and a drop-out voltage of a monostable relay. Both are considerably different due to the properties of the magnetic circuit. If I understand right, you are talking about contact closure with decreaing coil voltage, which would refer to a normally closed contact.
The purpose of the shown circuit is also unclear, because it just powers the coil permanently. I think, you should try to describe more clearly, what you want to achieve. Also a specification of the relay in terms of pull-in and drop-out voltage would be helpful, or a reference to a manufacturer and specific relay type.
You have asked a strange question, I understand your question as follows:
"How to energize the relay , (which needs more than 1,6 V to be energized), using the voltage less than 1,6 V".
May be it would be good if you would describe the task you want to realize.
Important is, if there are a stable power source ( say 5V)
or the 1,6V +/- delta V is the only voltage available.
Thanks for your responses and sorry for the late reply.
I have attached the block diagram of my application.
Initially the battery voltage will be 1.5V. Both the relays are open and no current flows.
When a magnet is placed closer to the reed relay, the reed relay will on which turns on the RELAY and current flows through the load (here i have shown resistor as load) and battery voltage decreases gradually. When ever the Battery voltage reaches to 1V i need to cut off the Relay.
The power supply available is Battery supply only.
I am looking for the circuit to cut off the Relay. i tried to design the circuit with the NPN transistor but no success.