Improved DECT Battery Charger Schematic Circuit Diagram
After buying a number of DECT telephones, we noticed that these became quite warm while charging. That surprised us, because the manufacturer wrote in the manual that the batteries had to be charged for 14 hours. That would lead you to conclude that the batteries are charged at 1/10th of the nominal capacity. But we had the feeling that the batteries were getting rather warm for such a small charging current. That is why we quickly reached for a screwdriver and explored the innards of the charging station. The accompanying schematic reveals what is going on. The batteries are ‘simply’ charged from a 9-V mains adapter. In series with the output are a diode and a resistor.
A quick calculation shows that a current of about 160 mA will flow and this was indeed the case when measured with a multimeter. That means that, for the AAA cells used here, rated at 650 mAh, the charging current isn’t 1/10 C, but 1/4 C! This is rather high and certainly not good for the life expectancy of the batteries. The remedy is simple: increase the value of the 25 Ω series resistor so that the charging current will be less. We chose a value of 68 Ω, resulting in a charging current of about 60 mA. You may ask what the purpose of the remainder of the circuit it. This is all required to turn on the LED, which indicates when charging is taking place. During charging, there is a voltage drop of about 4 V across R1.
T1 will then receive base current via R2 and the LED turns on. The resistor in series with the LED limits the LED current. The fact that the batteries are charged with an unregulated supply is not such a problem. The mains voltage will vary somewhat, but with a maintenance charge such as in this application it is not necessary to charge with an accurate constant current, provided that the current is not too high. For the curious, here is the calculation for the charging resistor: three cells are being charged. The charging voltage of a NiMH cell is about 1.4 V. The voltage drop across D1 is about 0.7 V. That leaves a voltage across the resistor of: 9mains adapter – 4.2batteries – 0.7diode = 4.1 V Therefore, there flows a current of 4.1/25 = 164 mA.