Hi! I am trying to make a dc to dc converter using the 78xx series. I have a dc voltage source of 9V. Can I use a 7805 to bring it down to 5V? If yes, what are the other components which I shall require??

 

Hi! I am trying to make a dc to dc converter using the 78xx series. I have a dc voltage source of 9V. Can I use a 7805 to bring it down to 5V? If yes, what are the other components which I shall require??

The 7805 is a linear regulator not a dc to dc (switching) converter.
It will dissipate power equal to 4V times the output current which may require a heat sink.
All you need to add are the input and output filter capacitors close to the 7805 pins, as shown in the data sheet.

 

Additionally... how clumsy are you? If that 9V is a battery and you ever plug it in backwarsd the 7805 will instantly be toast no comming back, so you probably want to add a series diode to protect from that.

Next, since the input will be open circuited when you turn this off there is a possibility you can reverse voltage this just with the input and output caps. It happens when the input falls faster than the output and too is instantly fatal to the 7805, and is also prevented by a diode from output to input. There should be a schematic of this in every data sheet for the 7805.

 

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Next, since the input will be open circuited when you turn this off there is a possibility you can reverse voltage this just with the input and output caps. It happens when the input falls faster than the output and too is instantly fatal to the 7805, and is also prevented by a diode from output to input. There should be a schematic of this in every data sheet for the 7805.

That problem likely only occurs if the input voltage is forcibly driven to zero by another load at the input or by a switch to ground.
If the input just goes to an open circuit, such as from a series switch at the 9V output, then the input will simply go to zero at the same rate as the output, and a reverse voltage condition will not occur.

 

That problem likely only occurs if the input voltage is forcibly driven to zero by another load at the input or by a switch to ground.
If the input just goes to an open circuit, such as from a series switch at the 9V output, then the input will simply go to zero at the same rate as the output, and a reverse voltage condition will not occur.

If anything else gets connected to the 9V the problem can occur, and things get connected long after the initial circuit with the initial limitations is built. It would be sad for some working circuit to suddenly stop working when some other minor change gets made.

Personally when I've used a 9V battery to get something up I did not bother with the reverse voltage diode either, but when a beginner asks "what are the other components which I shall require" I do believe full disclosure is recommended.

 

The LED and resistor on the output are optional. If the wire to the battery is more than 10 or 20 cm, it would be safest to use a 100 uf capacitor on the input.

 

If you using the 9V cookies battery, because its current is not that much, so you have to reducing the power consumption, you can use LM2576-5 and it will be better than 7805.

 

I would like to thank everyone for their responses to my doubt. By the way, "crutschow" & " ErnieM" , were you referring to use a diode across the terminals 1& 3 of the IC?

 

Sorry, we mean like this:

D1 keeps you safe from putting the battery in backwards and is strongly recommended.

D2 is for keeping the 7805 safe when powering down and Vout >> Vin. It is not essential for a breadboard, I would use it for say a commercial project where I want excellent reliability.

 

I would like to thank everybody for answering my question. My doubt is done.
Thanks
MHDS