I use a LM317 voltage regulator as a power source for a battery charger because we have frequent power interruptions. The problem arises when the electricity grid cuts out and I have a sealed lead acid battery connected to the output of the LM317. The input of the regulator goes to something like 11V, obviously coming from the battery even though there is no input voltage from the electricity grid. Obviously this is not a good or preferable state of affairs during power cuts. What efficient methods are there to deal with the LM317 during power cuts? I have checked many, many spec sheets and application notes and I have not seen this problem discussed. Any suggestions will be highly appreciated. There are ways to solve the symptoms like putting a diode between the LM317 and the battery. But this represents merely avoiding the symptom, not designing in a feature to fundamentally deal with misusing the LM317.

 

hi w,
Welcome to AAC.
This image shows one method of protecting the LM317.
E

 

Thank you for a rapid response!
The LM317 does not need D2 because it is built into the chip these days.
D1 protects against shortcircuiting the input. In the case of a power outage the output of the battery is fed through to the input, minus the voltage drop over D1. In fact that is exactly what I have. Effectively a resistor is put in parallel with C1, representing the (relatively large) resistance of the power supply to ground when the power supply has no AC input.

 

I have a few battery chargers using an LM338 (Similar to the LM317 but higher current rating.) I us a set of contacts on a relay to connect the battery only when the coil is energised from the output of the bridge rectifier in the charger. You also have to add a diode between the bridge rectifier and the LM317 input so the back feed through the regulator does not hold the relay on. You could use the same idea but use a MOSFET in place of the relay. An even simpler method is just to put a diode between the LM317 output and the battery. With that method you would have to adjust the voltage at the output of the LM317 to compensate for the 0.7V (Or so.) voltage drop across the diode.

Les.

 

hi w,
D1 is not for input protection of the Lm317, its for reverse voltage protection, which covers the requirement of your failing mains supply.

E

 

Without your actual schematic it is difficult to say. As shown on the LM317 datasheet, there is an overvoltage protection 6 V zener diode and 50 ohm resistor inside the 317 that is activated when the main regulator circuit is not energized. The zener prevents reverse breakdown of internal transistors. This, plus your external resistances, form at least part of the reverse DC path from the output to GND.

When the grid cuts out, is the input to the 317 left floating or is it pulled to GND by something (not counting any input capacitors)? If it is pulled to GND, D1 in the above schematic is strongly recommended to discharge the output filter capacitor C3, but it increases the battery discharge current during a blackout.

Nothing will be more simple-and-efficient than a 3 A Shottkey diode in series with the output. An SPST relay driven by the AC line would be my second choice for you. Another option is a MOSFET switch driven by an optocoupler (basically a low energy solid state relay), but that is a more complex circuit that has low voltage electronic components connected directly to the AC power line.

Can you measure the current through the 317 when the grid is down?

ak

 

ak,
The current is around 50 mA. That is because the 11V on the LM317 input still drives (without any voltage regulation or battery charging) the other electronics and LCD display. Without going in to much details this is an Arduino-driven constant-voltage charger that controls the charge voltage via a PWM that is Sallen-filtered and, via an op amp, gives a voltage that is applied to the adjust pin of the LM317 in order to control the charge voltage.

To ALL of you who interacted,
I find it amazing to have so many informed opinions within a few hours of posting the question. I really appreciate your time with this. I hope that, at some time, I will be able to return some of your goodwill to this forum.
Kind regards,
willem