I test on Lipo and Li battery again and here is the result:
LI polymer battery:
Unload Voltage: 3.87V
Loaded Voltage: 3.56V
Current output: 0.99A
OK, with a 1 A load, the battery voltage dropped, as expected. How did you apply the 1A load? Was that just the ecigarette directly across the battery?
Heater:
Voltage receive: 2.68V
Current receive :0.98A
How was this test done? Was that with the mosfet in circuit? Or, was this with the ecigarette directly across the battery?
B]LI battery(18500)
Unload Voltage: 4.1V
Loaded Voltage: 3.33V
Current output: 0.99A
Is this a new battery? Is it LiPo or lithium ion?
Heater:
Voltage receive: 2.34V
Current receive: 0.83A
This result, like the one above is unclear.
If you refer to Figures 1 and 2 of the datasheet for your mosfet, at a Vgs of 3.5V, which should be available under load based on your data, the mosfet will pass about 70A with a Vds of only 3V (equivalend Rds= 0.043Ω). Thus, at 1A, that would be a voltage drop of only 43 mV across the mosfet. Since the battery under load , according to your data, is at 3.5 volt, simple physics tells you that the voltage drop across the ecigarette is 3.5- 0.043 or 3.46V. The same calculation can be done using data in Figure 2 (tjs = 175°) and the results are similar.
There is something wrong with your data. Maybe you are not measuring the circuit at the correct points? What value is the resistor from gate to ground? You can use at least 10K there to minimize its effects.
Please give voltage and current data for the ecigarette directly across the battery.
How are you measuring current? Is it being done at the same time as you measure voltage (i.e., do you have two meters)?
John