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DC DC buck to LDO

Rajinder

Jan 30, 2016
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Hi all,
I have an automotive application (powered from 12V car battery).
I am looking to drop 12V to 5V using a DC DC buck regulator (one of my circuits require a 5V supply, then use this 5V to power a LDO at 3.3V fixed). The current will be 600mA max. Looking at low quiescent current.
I am looking at using a automotive DC DC converter, not sure if it should be a synchronous, asynchronous or a simple switcher, ca anyone assist?

Also are there any issues with me using the DC-DC to power the LDO?

Thanks in advance
 

Harald Kapp

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Nov 17, 2011
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Such modules are readily available at low cost.
I am looking at using a automotive DC DC converter,
The only thing "automotive" here is the input overvoltage protection on the regulator module and the temperature range.
The input overvoltage protection is possibly no issue at all. An often used regulator IC is the LM2596S and these modules are good for up to 35 V input voltage.
The temperature range is another issue. Many commercial modules are rated for 0 °C ... 45 °C only, but e.g. this one is rated -40°C ... +85 °C.
synchronous, asynchronous or a simple switcher
At 5 V × 0.6A = 3 W the topology of the switcher is probably irrelevant,
Looking at low quiescent current.
Connect the step-down module after the main relay so it is powered only when the ignition is on.

Or build your own converter using e.g. this chip (low quiescent current, automotive).

then use this 5V to power a LDO at 3.3V fixed
Why not step down to 3.3 V directly?
 

Rajinder

Jan 30, 2016
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Such modules are readily available at low cost.

The only thing "automotive" here is the input overvoltage protection on the regulator module and the temperature range.
The input overvoltage protection is possibly no issue at all. An often used regulator IC is the LM2596S and these modules are good for up to 35 V input voltage.
The temperature range is another issue. Many commercial modules are rated for 0 °C ... 45 °C only, but e.g. this one is rated -40°C ... +85 °C.

At 5 V × 0.6A = 3 W the topology of the switcher is probably irrelevant,

Connect the step-down module after the main relay so it is powered only when the ignition is on.

Or build your own converter using e.g. this chip (low quiescent current, automotive).


Why not step down to 3.3 V directly?

Hi,
Thanks for your reply. I have one device that powers from 5V hence the DCDC converter, this then will power the LDO (3.3V output) that powers the micro etc.
Is there any consideration i.e. would I need a diode in between the output of the buck regulator and input of the LDO?
I will take a look at the links you sent me.
Thanks in advance
 

Arouse1973

Adam
Dec 18, 2013
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Connecting a LDO to the output of the buck converter is something that is done to reduce noise and improve transient response if required. Because of the improved efficiency of Synchronous mode operation I would choose this in your situation because you are introducing a lower efficiency device (LDO) to drop to 3.3 V. If you are drawing a lot current through the 3.3 V LDO then this will be the device which uses the most power out of the two and you won't get more than 66% efficiency in your example. Thanks Adam
 
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