Are there any solid state options available for a simple relay for a 9v battery backup?

Power to the device will be DC +5V. If this voltage drops below a specified level, say 2.5V, a relay should switch to a 9V battery backup (with a voltage regulator, of course).

I've done some searching but I'm having trouble coming up with something like a part number or datasheet.

Thanks!

 

hi,
Look at using blocking diodes when connecting the two voltage sources in parallel.

Do you follow OK.?

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Eric " Good enough is Perfect "
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So a blocking diode has a higher current rating than a normal diode and will block current flow when one side has a higher voltage than the other?

So if I have 5 volts on both sides (one Vcc and one the regulated 9V), and I put a blocking diode in series with the backup and then connect the two sources in parallel, the 5V backup source will be enabled when Vcc drops below 5V?

 

hi,
Assume that the current required by the load is less than 1A.
By connecting a 1A diode [1N400x] from the +5V to the load will give approx 4.7V at the load.
Do the same for the output of the regulated +5V, that is connect a 1N400x diode in series with the 5Vreg out and the load.

Post a sketch showing the basics on how the 5V and the 5Vreg are going to be used and I will draw in the diodes..

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Eric " Good enough is Perfect "
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I've got more than a sketch

 

hi,
Look at this cct.

This is the BAT54
STMICROELECTRONICS | BAT54 | Diodes | Semiconductors - Discretes | Farnell UK

Attached Thumbnails

 

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Eric " Good enough is Perfect "
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Ok, so in your circuit, when Vdc is operative at anything above 3.9V the diode will be forward biased and out goes the voltage.

When it drops below 3.9Volts, the BAT54 becomes reverse biased and 4.3 volts comes down from the regulated source?

 

No, the 5V will supply the load until the cct voltage drops to 4.3V, then the +5Vreg [9v] will supply the current to the cct, the voltage being 4.3V.

The 1N4001 is reversed biassed when the cct voltage is greater than 4.3V.

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Eric " Good enough is Perfect "
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A 9V battery cannot properly power a 7805 regulator. The regulator needs a minimum input of 7.5V but the voltage from a 9V battery quickly drops to 7.2V then drops slower to 6V.
Use a low-dropout regulator that will still provide a regulated 5V when its input is as low as 5.5V.

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Uncle $crooge

 

A 9v battery should work fine. I have built many test modules using a 7805 regulator with no problems.

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t the datasheet for a 9V alkaline battery. Its voltage quickly drops to 7.2V.

Look at the datasheet for a 7805. it is spec'd with an input of at least 8V. Its dropout voltage is typically 2V but could be higher. When it has dropped out then it is not regulating.

Attached Thumbnails

 

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Uncle $crooge

 

An LM2940-5 is a low dropout replacement for the 7805

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A relay with a 5v coil will probably drop out at 3v. If this is close enough to 2.5v, and relays with 5v coils can handle your current draw requirements then hook up the relay contacts so the 9v batt/reg arrangement is switched in when the relay drops out.

 

Thanks for all the suggestions guys. Analog electronics aren't exactly my strong point so I appreciate the help!

 

I understand that and I'm sure what you are saying is correct. I'm only speaking from my own experience with these devices.

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