Electronics 2000

Hi,

I am having a bit of a problem with a design I am working on and I would appreciate some advice. I need to produce a circuit that will accept an input voltage of 12 - 14V and will output 5 volts at a continues load of 0.9A. My initial design was to use a single L78S05CV as that is rated up to 2A. However, when I did the maths, I realised that the heat sink that I would need was impractical for the box it has to fit in. After hitting Google I found an article that said that I could put 2 of these regulators in parallel as long as I managed the current and it provided the following circuit, using 2 regulators, some diodes and a some caps which it claimed would do the job. The circuit shows 2 x 12V regulators, but I cannot see why it could not be implemented with 3 x 5V regulators?

I do have room on my circuitboard for 3 regulators each with a small heat sink (I have a max height limit of 10MM) which should be OK as I think each reg would be producing less than 3W which should not be too difficult to dissipate using a 24C/W heat sink.

If I understand the circuit correctly, the voltage drop across D1 and D2 is offset by D so the output voltage is still 5V?

Any comments, advice, alternative solutions would be much appreciated. To be a viable solution it really needs to cost no more than £3.00 to implement and it must be reliable.

Regards

Tim

To get the heat out of the 7805 regulator you can put a resistor in series with the power source. To get the value start with the lowest input voltage (12V), subtract the output voltage (5V) and the 7805 dropout voltage (2V), result is 5 volts. Divide by the maximum current ( 0.9A) for a resistance of 5.5 ohms.

With the resistor in place the 7805 will see 3.6 watts at full input voltage and maximum current , without it it would see 8.1 watts. Of course the missing 4.5 watts of heat will now be dissipated in the resistor.