Definitely a diode, not a resistor. However as @allObjects says, at small power draws diodes won't reliably drop 0.7v (it can be a bit less). While personally I'd say a diode was fine to power a Puck in this case (I've personally used one a few times), ideally you'd use a proper voltage regulator - and as you say the voltage reading won't be great.
What I'd do instead for the voltage is to use a potential divider (in this case, just two equal value resistors) to halve the voltage from the battery, which you can then feed into an analog input on the Puck.
The lower value the resistors the more accurate the reading but the more power draw - however you can also put a small capacitor between GND and the analog input, and then you can use high-value resistors (1 MOhm?) which will draw tiny amounts of power from the LiPo battery.
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Definitely a diode, not a resistor. However as @allObjects says, at small power draws diodes won't reliably drop 0.7v (it can be a bit less). While personally I'd say a diode was fine to power a Puck in this case (I've personally used one a few times), ideally you'd use a proper voltage regulator - and as you say the voltage reading won't be great.
What I'd do instead for the voltage is to use a potential divider (in this case, just two equal value resistors) to halve the voltage from the battery, which you can then feed into an analog input on the Puck.
The lower value the resistors the more accurate the reading but the more power draw - however you can also put a small capacitor between GND and the analog input, and then you can use high-value resistors (1 MOhm?) which will draw tiny amounts of power from the LiPo battery.
So: