# Battery backed up power supply using LDO

I was looking for a battery(12V, 11.2Ah lead acid) backed up power supply solution for my ESP microcontroller(3.3V) along with few relays(3V) and came across some reference design in the following LDO data-sheet which seems to be simple and effective. It seems to charge the battery and provide power for the ESP when there is input supply and switch to battery during an outage.

Data-sheet link - https://in.element14.com/on-semiconductor/ncp1117st33t3g/ic-linear-voltage-regulator/dp/1652366 (Refer data-sheet Figure 31)

Below is the modified version to provide 3.3V output. I have few queries as below.

1. What must be the values or specification of those diodes D4 and D5
2. Will a 10 Ohm resistor that is in series with the ground pin of the upper LDO ensures that the lower LDO is off until the input source is removed (approx 0.1V difference).
3. What must be value for the resistor R1?
4. Is this a stable solution that can be used in production systems ? simulate this circuit – Schematic created using CircuitLab

Note - The power supply is rated 15V,2A and battery is 12V, 11.2Ah. The ESP along with the relays would only take a peak current of 850mA. A cutoff for charging would be added for battery safety. I also understand that a buck converter would be much efficient, however I was looking for a solution with least and low-cost components and prefer cleaner output without adding too many capacitors.

• 15V in, --> 3.3V out is a LOT to ask of for an LDO. – Ron Beyer Mar 10 at 20:59
• As Ron alluded to, you haven't considered the power dissipation. The LDO needs to dissipate ~8 W, but according to figure 21/22, the part is only good for 1.4 W or so. You can sometimes resolve this with resistor in series with the LDO input, but I think that you are too far off. – Mattman944 Mar 10 at 21:54
• @Mattman944 850mA is the peak current, the normal current consumption will be around 250mA or less. How did you calculate 8W? – Zac Mar 11 at 6:57

What must be the values or specification of those diodes D4 and D5

It depends on the voltage and current you want to use to charge the battery.You should choose a diode and a current limiter resistor according to the following:

$$R_1= \frac{15V - V_{charge} - V_{D,FWD}}{I_{charge}}$$

where $$V_{D,FWD}$$

is the forward voltage of the diode you are going to select,

$$I_{charge}$$ is the charging current of the battery, and

$$V_{charge}$$ is the charging voltage of the battery.

The lead acid battery in question has a capacity of 11.2Ah. Considering 2.4V per cell and a very low charging rate of 0.01C, the battery's charging voltage and current are given by:

$$V_{charge} \approx 14V$$

$$I_{charge} \approx 112mA$$

The low charging rate was selected due to the following:

• A backup system should have lower priority in comparison to the main power source
• It should not disrupt the normal operation of the main power source
• In case a large charging current was used, it could lead to a temporary input voltage drop, since the battery would need to charge first, before resuming normal operation
• It alleviates the required specs of the rectifier

Since there's limited headroom between the supply voltage and the battery's voltage, a schottky diode with a low forward voltage is preferred. The device MBR120VLSF MBR120VLSF should be alright, by dropping the supply voltage only 280mV for the required charging current. Now you can calculate the resistor:

$$R_1= \frac{15V - 14V - 280mV}{112mA} = 6.4 \Omega \approx 6.2\Omega$$

and its power rating:

$$P_{R_1} \ge 280mA^2\cdot 6.2\Omega = 1/2W$$

Will a 10 Ohm resistor that is in series with the ground pin of the upper LDO ensures that the lower LDO is off until the input source is removed (approx 0.1V difference).

a 10Ohm resistor will probably create a ground offset voltage of approx. 50mV ~ 60mV.

What must be value for the resistor R1?

See answer above. It depends on the battery information.

Is this a stable solution that can be used in production systems ?

Normally such backup systems consist of two power sources (PS) connected to the output / load via diodes, in order to prevent current from one PS from flowing into the other PS (Back Feeding). In the datasheet it is not specified (at least I did not find) the internal architecture of this LDO, and as such it is not so trivial to say whether it is going to be a robust solution. But nonetheless, the same reference circuit is shown in the datasheet of this IC, so I assume it should be alright.

Advantages in having two power supplies for such backup systems?

I can think of a few reasons:

1. Data Logging: Let's say you are collecting data via a sensor over time in batches, and you have to guarantee that every batch of information is completed and not corrupted. In this case the backup system would make sure that you don't run into such problems.
2. Power Sequencing: Some fancy / state-of-the-art SoC require some proper power management, consisting of powering up and down in a proper sequence, otherwise, it can damage the chip. In this case the backup power source makes sure that the SoC has enough time for a proper shutdown.
3. Buffer: Let's say you are connecting your device to an external battery, and it is running low. You want to replace the battery but do not want to shut down the circuit. In this case the back up power source would provide you with some extra time.

For points 1 and 2 it is necessary that the load or SoC monitor the power source, in order to know when to initiate the safety procedure.

As already mentioned by the other members, your circuit will suffer a lot from the high power loss (large voltage drop across the LDO). You might consider other solutions, such as the use of a SMPS - Buck converter.

I hope it helps.

• The battery I am referring to is a Sealed Lead Acid battery (12V, 11.2 Ah). A sample is amazon.com/Mighty-Max-Battery-VT750C2-Product/dp/B00YNOFZ1S – Zac Mar 11 at 2:39
• Also are there any advantages in having two power supplies for such backup systems? – Zac Mar 12 at 5:38
• I will update my answer. – vtolentino Mar 12 at 7:21
• By power supply, do you mean two power sources? i.e one from battery and one from the external power supply? Isn't that this circuit is doing! Sorry I got confused. – Zac Mar 12 at 10:09
• I removed the power supply term and replaced it according to: Supply Voltage = Input voltage of the circuit Power Source = LDOs you are using – vtolentino Mar 12 at 11:30