Can I decrease voltage in a battery to get higher amp hour rating?
Assuming that the motor draws 1 Amp, my question is will I will still get 1 Amp for only 1 hour from the output of the converter or I will get 1 Amp for two hours ??


You'll get 1A for 2 hours (approximately).


  The total watt hour for my battery is as I know 7.4 watt hours (V*Ah) but my motor draws 1 Amp at 3.7v so watt hours of 3.7 watt, so If I divided the watt hour of my battery by the watts my motors draw I get 2 hours


Correct, but you have to substract losses in the DC-DC converter. If your converter is 90% efficient and uses a 7.4Wh battery as power source, 7.4*90% will be available on the output, or 6.66 Wh. Basically, conservation of energy (input energy = output energy + losses).

Note that you don't need a DC-DC converter. You can use a MOSFET instead:



A permanent DC motor does not require accurate voltage regulation, and it is quite inductive so PWM works very well. If the question was about using a 110V rated motor on 230V we should definitely check if the wire insulation would withstand the extra voltage, but for 7.4V I don't think this would be a problem.

This simple schematic uses PWM to apply 7.4V to your 3.6V motor with 50% duty cycle, so the average voltage on the motor is 50% of supply voltage, or 3.6V. This saves a buck converter.

Efficiency calculation: suppose the PWM is fast enough so we can neglect the current variations in the motor, it will be an average of 1A. Since DC motors have quite large inductance it will smooth the current. 

Duty cycle 50%, so D=0.5

When mosfet conducts: it has resistance RdsON, and dissipates \$ R_{dsON} I^2 \$

When diode conducts, it has forward voltage Vf, and dissipates \$ V_f I \$

Note that mosfet conducts with duty cycle D and diode conducts with duty (1-D), so:

Total losses: \$ D R_{dsON} I^2 + (1-D) V_f I \$

You can add switching losses which are estimated by (Switching time) * Frequency * I*V but at motor PWM frequency they will be pretty small.

With realistic values RdsON = 0.02 ohm, Vf=0.5V using a schottky, D=0.5, Losses=0.26W, Motor power=VI=3.6W, efficiency=93%.

Another option is to wire the battery in accordance with the motor voltage. Since your battery is made of two 3.6V elements in series, if you wire them in parallel instead, you got 3.6V 2000mAh instead of 7.4V 1000 mAh (the energy capacity in Wh stays constant of course). But you may need a new charger, protection pcb, etc, so that may not be an option.