# Converting 1.5V from AAA cell to 0.8V output

#### Winters893

Joined Jun 29, 2021
8
Hi all,

I'm designing a circuit that needs to take 1.5V from two AAA cells in parallel and outputs 0.8V to a DPDT toggle switch (on-off-on) that connects the load to +0.8V, 0V, and -0.8V depending upon the switch position. The output current would ideally be able to reach up to 1.5A for safety reasons but will average under 100mA (about 30-60mA). The circuit should technically be able to work with one AAA cell. I want two to extend battery life.

I've attached my schematic. I have the circuit set up on a solderless breadboard using thru-hole components. I'm having the problem of the AAA cells not working to power the circuit; the output voltage is only about 15mV when I use them. I'm using an LT1307 as a 5V regulator to provide the voltage to Vbias on the LP38851 which should output the 0.8V (the LP38851 is the same as the LP38852, which is what's shown in the schematic, but handles less current). The LT1307 is able to generate the 5V when it's not connected to the LP38851, but its voltage drops to 1.2V (the voltage of its feedback pin) when it's connected to the LP38851's Vbias pin. A 3.3V microcontroller and a 3.7V Li-ion battery both work to power the circuit: the LT1307 provides 5V to LP38851's Vbias and the output is the appropriate 800mV. My initial assumption was simply that the circuit requires a higher voltage source, however, both chips should be able to operate just fine with 1.5V according to their datasheets.
Note: I do not yet have my DPDT toggle switch, so I have been directly connecting my load to the output and swapping the leads when I want the negative voltage.

I also tried rewiring the circuit to have the AAA power only the LT1307, and the LT1307 then provides 5V to both the Vbias and Vin for the LP38851. But that didn't work either. The voltage still dropped to 1.2V when everything was connected. Perhaps it has to do with how the grounds are connected? I'm currently going through and testing the circuit while powering one regulator with the AAA and the other with the Li-ion battery as a "controlled power supply" that I know works.

Am I going about this the wrong way with two voltage regulators? Are there better regulators to be using (if so, what are they)? For all I know, I may need to completely scrap this design. My background is in systems engineering not electrical, so I would appreciate any and all assistance or advice. I also want to make this into a PCB once I'm able to test and confirm the circuit works on a breadboard. I did some cursory research on rail splitters, but I couldn't find any information for splitting 1.5V since everything was for a higher voltage. Technically a simple voltage divider works, but it's inefficient and I'd like my batteries to last a good while.

At the end of the day, I just want to find the most efficient way to have a AAA battery provide 0.8V (0.7V would be even better) for my load. I was also hoping to be able to design something a little cheaper; the thru-hole LT1307 costs over $5 and the LP38851 over$2 on Digikey.

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#### KeithWalker

Joined Jul 10, 2017
2,055
What is the voltage across the battery when you connect the load? I suspect that the AAA cells are not able to supply enough power.

#### Winters893

Joined Jun 29, 2021
8
What is the voltage across the battery when you connect the load? I suspect that the AAA cells are not able to supply enough power.
The voltage is 1.296V across the batteries when I connect the load. The load is getting 10mV.

#### Papabravo

Joined Feb 24, 2006
17,318
Egads. Can I ask why you have a boost converter from 1.5V to +5V and then what looks like a linear regulator to 0.8V? This does not conform with my understanding of efficient. On top of that, building an SMPS (Switch Mode Power Supply) on a solderless breadboard is not a recommended method of implementation.

#### BobaMosfet

Joined Jul 1, 2009
1,900
@Winters893 What calcs did you do for BUCK/BOOST conversion? You do realize that voltage and current work inversely with one another, right? BOOST converters take a lot of current to raise voltage... and can only output within a designed window of current- they don't output what you have available on the input.

As an example- if we take the ubiquitous MC34063A BUCK/BOOST converter and try to make it convert 2V input to 5V output at 1.5A Max, it has input over 10A from the input source to make that conversion happen- at 21W through the IC- so it also needs significant heat dissipation.

There is a lot more you need to understand about BUCK/BOOST regulation than most of the videos showing basics will tell you.

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#### Winters893

Joined Jun 29, 2021
8
Egads. Can I ask why you have a boost converter from 1.5V to +5V and then what looks like a linear regulator to 0.8V? This does not conform with my understanding of efficient. On top of that, building an SMPS (Switch Mode Power Supply) on a solderless breadboard is not a recommended method of implementation.
It's much less than ideal... This was a result of my limited knowledge and attempt to wade through Digikey's endless sea of components.

#### Winters893

Joined Jun 29, 2021
8
@Winters893 What calcs did you do for BUCK/BOOST conversion? You do realize that voltage and current work inversely with one another, right? BOOST converters take a lot of current to raise voltage... and can only output within a designed window of current- they don't output what you have available on the input.
I do understand the inverse relationship between current and voltage. I only need a few milliamps on the output of the 5V boost for the Vbias pin on the LP38851. Sounds like I'm very much on the wrong path with these components... What would be the superior alternative?

#### BobTPH

Joined Jun 5, 2013
4,056
What is the load?

Bob

#### Papabravo

Joined Feb 24, 2006
17,318
It's much less than ideal... This was a result of my limited knowledge and attempt to wade through Digikey's endless sea of components.
I would not imagine that it was possible to learn circuit design by wading through a parts catalog. Did you even consider that you could apply the batteries directly to the LP38852. Why do you need the Boost converter? All this information is readily available in the datasheet, but you have to learn how to read them, then read them carefully, then read them again just to make sure.

#### Winters893

Joined Jun 29, 2021
8
I would not imagine that it was possible to learn circuit design by wading through a parts catalog. Did you even consider that you could apply the batteries directly to the LP38852. Why do you need the Boost converter? All this information is readily available in the datasheet, but you have to learn how to read them, then read them carefully, then read them again just to make sure.
The LP38852 requires 3V - 5.5V on its bias pin, hence me getting a boost converter.

#### Winters893

Joined Jun 29, 2021
8
I appreciate everyone's comments, and it seems we can all agree the current design is bad. But my current conundrum remains: taking a AAA battery and getting 0.7 - 0.8V.

#### BobaMosfet

Joined Jul 1, 2009
1,900
I do understand the inverse relationship between current and voltage. I only need a few milliamps on the output of the 5V boost for the Vbias pin on the LP38851. Sounds like I'm very much on the wrong path with these components... What would be the superior alternative?
Correct me if I'm wrong, but I think you also mentioned 100mA... so here is the setup I'd use to get that:

MC34063A (DIP-8) from Mouser: P/N: 595-MC34063AP (approx \$0.30 ea as I recall)

Now, as for going from 1.5V to 0.8V.... use a signal diode that drops 0.7 to go through it.

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#### Papabravo

Joined Feb 24, 2006
17,318
The LP38852 requires 3V - 5.5V on its bias pin, hence me getting a boost converter.
I do not think that is correct

#### ElectricSpidey

Joined Dec 2, 2017
1,972
A simple common collector circuit might get you want you need.

Maybe just a diode and resistor...depending on load.

#### BobTPH

Joined Jun 5, 2013
4,056
I do not think that is correct
Actually, from reading the datasheet it is. It uses an NFET as a high side switch and needs it for the gate voltage.

The question is why you would choose this unusual part, especially since he expects to draw 1.5A from a regulator limited to 800 mA.

And we still don't know the characteristics of the load. Why are we bothering to continue helping when we are blind.

Bob

#### Papabravo

Joined Feb 24, 2006
17,318
Actually, from reading the datasheet it is. It uses an NFET as a high side switch and needs it for the gate voltage.

The question is why you would choose this unusual part, especially since he expects to draw 1.5A from a regulator limited to 800 mA.

And we still don't know the characteristics of the load. Why are we bothering to continue helping when we are blind.

Bob
You're right. I misread the input voltage range did not realize there would need to be a boost converter/charge pump for the internal switch.

#### Winters893

Joined Jun 29, 2021
8
Actually, from reading the datasheet it is. It uses an NFET as a high side switch and needs it for the gate voltage.

The question is why you would choose this unusual part, especially since he expects to draw 1.5A from a regulator limited to 800 mA.

And we still don't know the characteristics of the load. Why are we bothering to continue helping when we are blind.

Bob
I don't currently have the LP38852 on hand (which is rated for the 1.5A), so I'm using the LP38851 in the meantime as 800mA should be fine for my current setup.

The load's similar to an auto-dimming mirror that I'm trying to design a simple battery power supply for. Like I mentioned, 1.5A is for ensurance, and the current should be below (+/-)100mA outside of transients. The current decreases over time.

#### Winters893

Joined Jun 29, 2021
8
A simple common collector circuit might get you want you need.

Maybe just a diode and resistor...depending on load.
I've tried a voltage divider with just resistors and about 25% of power is lost as heat. Please excuse my limited knowledge, would a diode be much different? I'm also unfamiliar with common collector circuits outside of knowing they use BJTs, would it be more efficient than a diode and resistor combo?

#### Papabravo

Joined Feb 24, 2006
17,318
I appreciate everyone's comments, and it seems we can all agree the current design is bad. But my current conundrum remains: taking a AAA battery and getting 0.7 - 0.8V.
It is very difficult to get a regulated voltage on the order of one diode drop. Semiconductors do not work well or efficiently in that regime. This may be easier to do on a silicon die but difficulties persist even there. Standard logic chips only go down to 1.65V for a reason.