Power Supply Clarification

[electronicsLearner77]

I want to design a solution where 2 possible sources of power (12V) one using a Power Jack and other using connector are possible, which IC shall i use as switch? Please advise. Two input voltages at the same time may not be possible, but even if it is given i need to protect. Please advise.

 

[Martaine2005]

Why an IC? Why not a switch with a diode on each power supply?


Martin

 

[electronicsLearner77]

Ok I will try with diode, I don't know how to do it so I was thinking if there will any such IC.

 

[John Canon]

How to do it ...

 

[electronicsLearner77]

Thank you for the diagram, what power rating of the diode shall I choose, any normal diode would do? Also a drop of 0.7V across diode will have any impact?

 

[John Canon]

This circuit allows Supply A, supply B, or both to provide power to the output, automatically isolating the 2 supplies from affecting each other. If you want to control which supply is being used while both are connected, then a SPDT switch is all you need, and no diodes. You haven't told us the current draw of your output device, and you haven't told us the need for auto or manual control. Are both supplies always connected and you want to switch between them, or do you want the device to accept power from whichever one is connected ?

 

[electronicsLearner77]

Accept power from whichever is connected. Both supplies are not always connected, only one of them will be connected. As a part of safety in case if both are connected it shall take power from one of them power jack or PS1. This shall be automatically done without manual intervention. The current draw I am not 100% sure but I can say around 3A, it is a guess, I will find out and update

 

[electronicsLearner77]

If both power supplies are connected I can decide a strategy for easy solution.

 

[John Canon]

Good. The diode circuit above will allow power draw from one or both. The 0.7 volt drop is nominal and should cause no problem unless your device is medical equipment or satellite transmitter. hehe. Choose rectifier diodes with minimum 5 Amp and 50 volt ratings

 

[crutschow]

You can do it with a mechanical interlock if you use a Jack with three-terminal connections, such as below:
The contact between pin 2 and 3 is normally closed, but is mechanically opened when a Plug it placed into the Jack.

The Jack's power supply circuit common is pin 2, and the Connector's power supply common is connected to pin 3.
Both power supplies 12V are connected to pin 1.

If a Plug is not connected, the common circuit is completed for the Connector power supply through the contacts.
But if the Jack's power supply is plugged in, the common for the Connector's power supply is open and it is disconnected from the circuit.

If you prefer to break the 12V instead of common, you can make pin 1 the common for both supplies, pin 2 the 12V circuit connection, and pin 3 the 12V connection from the connector.
Either way works the same from the circuit point-of-view.

Using that connector means you aren't buying two large diodes needing possible heat sinks, and avoid the voltage drop they subtract from the circuit voltage (which may or may not be a problem in the circuit).

 

[electronicsLearner77]

I think this is how it looks like

 

[crutschow]

I think this is how it looks like

No, you did not connect it as I stated.
Read that more carefully.
There is an electrical contact connection between pins 2 and 3, not 1 and 2.

 

[Martaine2005]

As most items are ‘centre positive’ pin 1, it may be better to do it that way. This diagram has ‘negative centre’ pin 1. The battery will be your main power supply.



Martin