Charging a 1.2 rechargeable battery

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SUMMARY

The discussion centers on the charging mechanism of a cordless phone handset that operates with two 1.2V rechargeable batteries while using a 9V DC output charger. The 9V output is not directly applied to the batteries; instead, the internal circuitry of the charger steps down the voltage to the required 2.4V for charging. This design choice is attributed to cost-effectiveness in mass production, as 9V chargers are cheaper to manufacture and can efficiently power various devices by drawing only the necessary wattage.

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mendes
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I have a simple question I hope):

My cordless phone handset works with two 1.2V rechargeable batteries, but the DC converter that comes with it and that is supposed to recharge the batteries has a 9V DC output ! How come ? Is it applying 9 V to recharge two 1.2V rechargeable batteries ?! Or perhaps the 9V go only on the charger (where sits the headset to be charged) and the charger applies only the appropriate voltage to the batteries, that is around 2.4 V. But then why so much voltage (9 V) for the charger ?!
 
Last edited:
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To power the transceiver functions of the base
 
somecreepyold said:
To power the transceiver functions of the base

The base is a separate unit (from the handset charger) and has a separate power supply.
 
Then its most likely because its cheaper to produce a bazillion 9-volt chargers than it is to make separate ones for each unit.

The internal circuitry will step down the voltage to what is needed and draw only as much power as needed.

Stepping down DC voltage is much much much easier (less waste) than going from AC to DC, hence the preference for mass production of something that can properly power a wide range of devices.
 
you mean the device will draw only 1.2 v even if we supply 9v to it?
 
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Well a better way to put it would be its going to draw as many watts as it needs to charge
 

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