Possibly you can do it by making a simple voltage doubler.

However Pls. check below info, before any attempt.

Considering you have an 2 Volt AC Output.

İf you directly rectify this voltage, you can obtain 2.82 Volt DC. (2 X 1.41 the peak value of 2 V Sinuswave = 2.82) Certainly, this is not your solution.

If you rectify 2V AC by using a simple voltage doubler, you can obtain twice of 2.82 VDC and this equals 5.65 VDC . You will need two units of 1000 microfarad 16 Volt electrolyt capacitors.

Now that you have 5.65 V Direct current, this is a higher voltage value for your purpose. You have a 1.95 Volt of excess to match 3.7 Volts.

Now what you will need is to get a bunch of diodes and make a chain by connecting them in serial. (You can use 1N4005 or 1N4007 diodes for an example.)

You know Diodes are need to hold a certain amount of voltage across them to start conductivity. It depends the character of the diode in use.

Try a bunch of diodes connecting them in serial to form a diode chain, for the purpose to hold a 1.95 Volt of difference across the total chain.

Make trials by measuring the output voltage by Voltmeter to determine the number of diodes to be used in the chain configuration. (Theoretically the voltage across the diode is not dependent with the load in normal limits.)

make a few units of serial diods at the output of 5.65 Volt to obtain 3.7 Volt.

Wishing you success

If your input is DC, you should consider a simple boost converter design. See here for design ideas and calculations.

http://www.maxim-ic.com/appnotes.cfm/an_pk/2031

Many OTHER IC's are available from other manufacturers. If you are in the US, suggest you search Digikey for best price and availability. Look for "Voltage Regulators - Switching DC DC Converters & Controllers"

http://www.digikey.com

Good luck!

Or:
Shopping results for 3.7V and 350mA. power supply

thanks for the reply, i will try it out

You can level-switch using a comparator at 3.7V set your threshold at 1V if this is a digital signal.

It would really help if you would give a bit more detail. Starting from scratch, the simplest way to power this circuit is a single Li cell, but we don't know whether this is part of a portable or fixed appliance and whether it is for intermittent or continuous use. The next simple suggestion is to use a 5V source and a Zener diode. How is it that you only have a 2V source and where does it come from? Peter Harris