how to convert 12V dc to 5V DC?

[pratik_parmar]

hy...
I want to convert 12V DC to 5V DC. But at i/p end current is 1.2A and at o/p end i need only 70mA .Will any body help to make such converter with 7805.?
please post your answer with circuit diagram.
Thank you.

 

[oyvdahl]

If you don't need to limit the current you can do something like this:

https://www.google.no/imgres?imgurl...lO5UZGuKMjKswakm4HYBA&ved=0CDUQ9QEwAg&dur=470

 

[nitishn5]

How is it that you are taking in an
Input power of Pin = 12V x 1.2A = 14.4W
but your output power is only Pout = 5V x 70mA = 0.35W
This means a power loss of 14.05W ie an efficiency of 2.4%. This could possibly be the worst power supply ever!!

Even by just putting a 7805 like oyvdahl has suggested, it would give:
Pin = 12 x 70mA = 0.84W
Pout = 5 x 70mA = 0.35W
Efficiency = 41.66%

 

[pratik_parmar]

How is it that you are taking in an
Input power of Pin = 12V x 1.2A = 14.4W
but your output power is only Pout = 5V x 70mA = 0.35W
This means a power loss of 14.05W ie an efficiency of 2.4%. This could possibly be the worst power supply ever!!

Even by just putting a 7805 like oyvdahl has suggested, it would give:
Pin = 12 x 70mA = 0.84W
Pout = 5 x 70mA = 0.35W
Efficiency = 41.66%

first of all thans for your valuable suggession.
Actual the problem is that i have battery of rating of 12v 1.2A and my controller need only 70mA curretnt and 5V. So, if i not make such kind if converter it is not allowing me to load program to controller.I also need 12V and high current to run my motor simulteniously. If i not make such converter either thing will put me at trouble. I know that this is waste of power but i got trapped between both requirements on same board.
Better solution is welcomed.

 

[nitishn5]

I think that you have not understood the concept of battery rating and battery load.

A rating of 12V, 1.2A means that the battery can support a max of 1.2A current load at 12V. It can also supply any current below this too.

It doesn't mean that the battery will always supply 1.2A only. Simply putting a 12kOhm resistor across it would only deliver 1mA of current by Ohms Law and not 1.2A.

You should use the 7805 solution for the 5V. and simultaneously use the battery for your motor.

 

[tpetar]

I will say 12V and 1,2A is rather 1,2Ah (check Ah mark this is not A).

Ah tell us capacity of battery. Maximum draining current is limited with internal resistance of battery and battery chemistry and battery model possibility. Check manufacturer datasheet for that battery.


7805 (1A), 78L05 (100mA for you Ok) or 78M05 (0,5A) voltage regulator is linear regulator and will make higher losses, maybe look to use switcher regulator. With linear you will waste 0,5W, its too much on longer usage, but if load is used for shorter time intervals and you need simple solution then try linear.


Best regards,
Peter

:wink:

 

[Ogu Reginald]

The best thing for you to do is to convert the 12V 1.2A to 5V 1A and then use a current limiter or there is a circuit in the datasheet of LM78XX that incorporates output current control.
Check the datasheet.

 

[oyvdahl]

I don't think you need a current limiter. I would go for nitishn5's suggestion. Just use a 7805 for 5V and the battery directly for the motor.