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Please friends can you help me in telling that how can i get 5 volt constant dc supply from 12v dc supply without using chopper as in chopper the peak voltage remains same and only average voltage changes and henceforth it's costly also.so please help me in getting an idea of the ciruit.0
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Member • Jan 19, 2010
I think you can use buck converters for this purpose.Are you sure? This action cannot be undone. -
Member • Jan 19, 2010
But suppose i want to harge a mobile at three volts and input is 12 volts the the peak input of the output will remain 12 volts only the average dc voltage will change not the full voltage .also it will be costly as size of inductor is very largeAre you sure? This action cannot be undone. -
Member • Jan 19, 2010
For that case you have to opt for Linear regulators.
This is the wiki link-- <a href="https://en.wikipedia.org/wiki/Linear_regulator" target="_blank" rel="nofollow noopener noreferrer">Linear Regulator</a>
This is the inefficient device, where in the excess power is wasted as heat so i think you would need a heat sink.Are you sure? This action cannot be undone. -
Member • Jan 19, 2010
I agree you will need a heat sink. A 12V to 3V linear regulator system will likely get hot at any appreciable current.
You can get high-frequency buck converters that shouldn't cost a lot. Try Linear Technology parts.
- CG
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Member • Jan 28, 2010
THANKS VERY MUCH FOR UR VALUABLE ADBICES😁Are you sure? This action cannot be undone. -
Member • Jan 28, 2010
well i know my answer is not correct, but can't we use resistors to handle these conversions, like we can use a resistor which consumes the voltage of 7v from the 12v power supply nd only 5v will left with us for further use.KAPILBECPlease friends can you help me in telling that how can i get 5 volt constant dc supply from 12v dc supply without using chopper as in chopper the peak voltage remains same and only average voltage changes and henceforth it's costly also.so please help me in getting an idea of the ciruit.
i know it is not correct, but please tell the reason behind it.😕Are you sure? This action cannot be undone. -
Member • Jan 28, 2010
You mean you want to dissipate heat through the resistor and pass the required voltage to the load is it?
It can also be done , but it is very ineffective one!
If you connect a resistor R in series to the voltage and parallel to the load, the resistor consumes the power according to its rating and passes the required power to the load!
In this case we cannot handle the load voltage , I mean if you want to send only voltage above 4 V , without using any semiconductor device you cannot do that thus you are not regulating it at all!!
which doesn't serve the purpose.Are you sure? This action cannot be undone.