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dc adapter with battery backup

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Advanced Member level 6
Jan 5, 2008
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I have found this circuit which comes from elektor
**broken link removed**

I was wondering what is the purpose of R2 ??

Also can I use 1.2v ni-cd batteries in series, will they be charged and discharged ok, instead of a big 12v one?

Also I see no reason why I could not substitute the IC with a 7805 to get 5v instead of a 7809, is this ok?

Yes, you can use 1.2V batteries in series, a 12V nicad battery is just ten 1.2V cells in series.

No problem with switching out the 7809 to a 7805. It will just get hot if the load get too high. That can be helped by scaling down the voltages by using a 9V adapter and 7 battery cells for 8.4V into the regulator.

As for R2, I can't see of a reason for it but that doesn't mean there isn't one.


Input Capacitor:
Most of the time, the voltage specification on transformers and power adapters refers to the rms voltage of the output. If this is the case then 12 volts rms is 17 volts peak. The input cap in this example is 16 volts. This would be bad in that exceeding the voltage spec for a cap will cause the cap to fail. I would recommend using a 35 volt cap on the input.

A 9 volt adapter would have a 12.7 volt peak voltage. This may work better for a 5 volt output 7805 version. The minimum input to the regulator is the output voltage + dropout voltage + 2x diode Vf or 5 + 2 +1 = 8 volts. Assuming 0.5 volt schottky diodes are used.

Using 7 NiCad cells the valid battery voltage range is 7.7 to 10.5.
There is nothing to prevent over discharging of the battery down to 0 volts. This can damage the battery. The minimum battery voltage is therefore 0 volts.

Max charging current = battery Amp hour rating /10
Max charging current for a 1100mahr battery is .11amps
R1 = (Max input voltage – Minimum battery voltage)/Max charging current.
R1 = 10.5/.11 = 95.5ohms
Since the battery will be continuously charging as long as power is applied I would keep the 180 ohm resistor. For max charging current of 10.5/180 = 58milliamps
Minimum charging current is 12.7-10.5-.5 /180 = 1.7/180 = 9.4milliamps

I can not explain R2. The current in R2 is 25uamps in parallel with the 300milliamps of the regulator. I think that R2 can be removed.

If you use a 9 volt adapter you will not need a heat sink at room temp.
0.3 output current
5 output voltage
12.72792 input voltage
7.727922 differiental voltage
2.318377 watts
degC/watt TO-220 thermal resistance junction to ambient
115.9188 delta T = 50 * 2.318
34.08117 max ambient temp deg C = 150 junction -115.9

If you use a 12 volt adapter you will need a heat sink to operate at 25deg C.

Thank you all very much!

I was wondering if there is such a simple schematic for NI-MH batteries as well.

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