Now coming to your requirement of 9 V battery. You would need a voltage converter (ideally a Buck regulator) which can convert 9 V to 5 V at 200 mA. Choose a converter for about 500 mA for getting the max efficiency. Take 90 % efficiency for this conversion. So, about 0.6 ~ 0.7 (750 mAh) battery would do.
About this item 【Safety and Performance】9V Li-ion Rechargeable Batteries there are five protections for this battery, with intelligent control of the entire charging process, built-in protection device to prevent overcharge, overvoltage, overcurrent, over discharge,can protect the battery very good. My take is that the multiplier in the formula is incorrect. For 1.5 volt alkaline batteries it is (voltage-1)*200. For 9 volt alkaline batteries it is (voltage-6)*33.3. A 1.5V battery is exhausted at 1V and a 9V battery is exhausted at 6V. A 1.5V battery has .5V of life and a 9V battery has 3V of capacity. The formula governing the output voltage is. Vout = 1.25 ( 1 + R2 / ( R4 + R5 ) ) + IAdjR2. Using this formula the output voltage is set to 12.8V which is the charging voltage for 9V NiMH batteries. LM338 is capable of delivering up to 5A to the load. It is essential to use heat sink with LM338 voltage regulator. Hi everyone, I bought some rechargeable 9V NIMH batteries, only problem is my charger can only charge AA and AAA. They are 200mAh and list the standard charge as 20mA for 16 hours. Do you reckon I could charge these using a lab power supply with current limiting?Or we could calculate it from the battery energy and solenoid power: 18000W s 1 1 225 W = 80s 18000 W s 1 1 225 W = 80 s. But maybe it's enough. The question is how to do it efficiently. Electrical power in a resistance is given by: P = I2R P = I 2 R. The internal resistance of a 9V battery is maybe 1.5Ω 1.5 Ω, when fresh.
jlgD.