I have an application with two AA/AAA batteries (so total of 3V or less) powering a unit with a micro controller. I want to design a low cost switch over circuit that can disconnect the battery supply when an external supply is connected with 3.3V. I also want to be able to detect with a micro controller when the external cable is connected supplying 3.3V. I am aware that this can be done with some source selector switch OR controller ICs like LTC4413. But these chips are relatively very expensive and the minimum input volt is only about 2.5V. I would need to go down as much as about 1.8V, as battery can go down to that level. Looking for some ideas how this can be done, thank you.
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\$\begingroup\$ Many of the common coaxial power connectors include suitable switches. \$\endgroup\$– Peter BennettFeb 25, 2016 at 16:47
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\$\begingroup\$ Thank you, I am planning to use just a two/four pin snap in connector to connect this cable between two units - one will supply 3.3V to a second battery powered unit. So, not planning to use any power connector jack with switch at the moment. \$\endgroup\$– user101095Feb 25, 2016 at 17:06
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1 Answer
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The most simple solution is to simply Diode-OR both supplies like this:
simulate this circuit – Schematic created using CircuitLab
The downside is that you'll drop 0.7V across a regular silicon diode and about 0.15-0.2V across a Schottky, which may or may not be acceptable to you.
If you want something a little more complex, but will actually turn off any supply from the battery (and probably have less voltage drop when the battery is connected):
The FET should be some sort of low-RDSon FET. Either way you will still need some sort of power conditioning circuit, like a buck-boost so that you can accept voltages higher and lower than your required output voltage. These circuits take care of the power source arbitration.
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\$\begingroup\$ In the second example, keep in mind that the resistors will dissipate nearly 0.25 watts as heat when the power supply is connected, which is not the most efficient solution. Also, why are you using 1N4148s as power diodes? \$\endgroup\$– 3871968Feb 25, 2016 at 17:16
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\$\begingroup\$ The 10 ohm was supposed to be a 10k, the diodes were picked at random to demonstrate the concept, not an actual part selection. \$\endgroup\$ Feb 25, 2016 at 17:19
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\$\begingroup\$ I think I wouldn't even need that diode as when cable is not connected, the 3.3V line would be floating and when connected, it would always be higher that the battery volt. Isn't it? \$\endgroup\$ Feb 25, 2016 at 17:26
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\$\begingroup\$ I would make the pull down higher, like 100K to reduce current through the external line. So, can I make R1 as 10K and R2 as 100K or so? And have V1 as 3.3V instead of 5V? As gate would be higher than source volt, I think it should be ok. \$\endgroup\$ Feb 25, 2016 at 17:33
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1\$\begingroup\$ Ok, consider this: V1 is gone, and D1 is replaced by a short. Lets treat the battery as a step input: At t=0, the gate voltage is 0, and thus, RDS is close to 0 ohms. (the FET is conducting) At t>0, Current flows through M1, and a voltage develops across the load. This voltage is now connected by a voltage divider (R1 and R2) and begins driving the gate voltage higher and higher. As the gate voltage increases, M1 conducts less and less, until steady state is reached. This is sort of like negative feedback. If we add the diode, then the gate voltage will remain at 0V even when M1 is conducting. \$\endgroup\$ Feb 25, 2016 at 18:52