Voltage Regulator Chips

is i/p meaning current and power?

if you connet 24v at input, the power certainly changs. no problem.

the current depends on lyou r load. if load constant, the current dont change.

i= v/r

as your voltage increase, the power dissipate increase. your chip iwll be hotter than front.

I mean to say:

If 7805 is used to provide regulated +5v to one ciruit, with +12v at its input terminal in one case and with +24v at its input terminal in another case with the same ciruit(i.e. load does not get changed but only input voltage to the 7805 get doubled.)

As fas as i think as per your comment 7805 will dissipate more heat with +24 v at its input terinal than +12 v input.

Is it right?

As other said the power dissipation goes up with the input voltage. The power dissipation of t he 7805 is (Vin - Vout) * Iout. The efficiency of the power conversion is Pout / Pin or Vin / Vout is a very close approximation for linear regulation. If you can afford the power dissipation and want to keep the 7805 cool you can pre-regulate with an LM137, a zener in series or just a properley sized resistor in series. Either way you will probably need a heat sink. If you want to save some power and their is a few extra bucks in the budget try a switching regulator.

Well , your warm Hi to all is quite apropriate.

Using your example , the 7805 is a 3-pin +5V 1A voltage reulator.

To calculate the power dissipated in the regulator , you subtract the output voltage ( 5V in tis case ) from the supply voltage ( say 12 V) and multiply by the load current.

The load current wil vary somewhat, but lets assume full load of 1A.

then the power dissipated is (12-5)X 1 = 7 Watts.

If the supply vlotage was 24v then the power dissipated would have been 19 Watts.

Your heatsink wil have to be sized to dissipate the heat from the regulator.

Hai everybody....

Actually the 7805 Regulator is a +5V regulator. When it is fed with +12V it will regulate the o/p voltage to +5V consuming some current internally... When the 7805 is connected to +24V input the o/p anyway will be regulated to +5V but the current taken internally to regulate is more than compared with the +12V input..

The load current depends upon the load. The total current is the summation of the Load current & the device internal consumption current.

Thanks & Plz. reply I am wrong.....

Thank you very much Everyone here to contribute your valued experience.

My concept gets cleared regarding regulator chip.

Does anyone know what the maximum input voltage is for the 7805?? I don't have a book handy, but I would check that parameter before running it from 24 volts.

Bill

Yup!

definately i knew it!

Max. i/p to 7805 is 35v