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7.4V and 9V on same line

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tiwari.sachin

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I have a circuit that provides 7.4V (from battery) and when DC jack plugged in provide 9V. Basically its a power path control circuit.

The load connected needs 7.4V at max.

I am just not able to come up with a concept of how I can possibly use it.

Can anyone show me the direction to proceed with the design like this.

Below is the simple image explaining my requirement.

 

The load connected needs 7.4V at max.

More important is the min. voltage which the load needs. This decides between a very simple solution with a few diodes, or a more complex solution using a buck- or buck-boost converter.
 

DC to DC is a costly option and with input varying from 7.4 to 9 V and getting 7.4V using a DC to DC. I am not sure on this...

Using a diode in series would mean drop with battery voltage too which i donot want
 

DC to DC is a costly option and with input varying from 7.4 to 9 V and getting 7.4V using a DC to DC. I am not sure on this...

Using a diode in series would mean drop with battery voltage too which i donot want

I would use a SEPIC regulator, which is ideal for the listed requirements.

Having said that.............Engineering is the science of understanding design constraints. And then choosing a solution which impacts the least the less important parameters, while still meeting your most important requirements.

So you have to list YOUR requirements in order of importance.
For example:

Maximum and minimum input voltage
Maximum and minimum output voltage
Cost
Size
Component availability
Efficiency
Complexity
Noise
Shutdown capabilities
other...
other....
other....

Once that you have listed those constraints, we can provide suitable suggestions.
Something we cannot do, is to know what you don't want or can't do if you don't tell us. And without that piece of information, we'll be providing useless recommendations.
 

For a long time the safe method to do what you want has been a switched coax DC power jack. It has 3 terminals. Inside is a switch which connects and disconnects two terminals, when you insert and remove a coax plug.

Thus you can draw battery power, or supply power from an external adapter. It does not cause any voltage drop.

You must make sure it is a break-circuit type, with 3 terminals. It must not be 2-conductor.



The power plug can be two conductors. It must be the correct diameter, outer and inner.

- - - Updated - - -

To clarify... the 2-conductor plug makes contact with 2 surfaces inside the jack, so in that sense it is a 2-conductor system.

You'll have 3 wires going to the jack: (a) circuitry, (b) battery, (c) ground.
 

I have my board working on 7.4V and was trying to add power switching, over voltage and reverse polarity protection.

Was thinking of using something simple like the below.



I did find a turn around where I first use the option above for OVP and reverse voltage and probably something else for power switching.

Hope it works fine... waiting for some components to test the same.
 

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