Hello,

Maybe is really simple for some but I've never made a circuit without a voltage regulator, I have always used an unregulated voltage regulator to supply to my ciruit 9V, 5V or 3,3V with its capacitor of IN and OUT, even if it wasnt needed I did use it for the stability and protection to the whole circuit.

Currently I am making a battery powered circuit (energy is supplied by two batteries AA in series) as far as all the components work in a range between 1.8V and 3.6V. As I said I dont have any experience making a circuit without a voltage regulator and I dont know if there are measures I could consider to make the circuit stable and secure without the voltage regulator. Note that all IC have decoupling capacitors and the circuit consumption would be around 30mA.

Thank you for reading,

 

I have always used an unregulated voltage regulator

Eh?
Whether or not a regulator is needed depends on what the circuit is for. For example, simply switching an LED on/off is not generally critical, whereas trying to measure signals in the low millivolt range would almost certainly require a regulator.

 

You only need a Voltage Regulator if your circuit needs a specific voltage for it to operate correctly, so if you need a 3.6V supply from a 5v or 9v, then using a deadicated regulator IC is better, like the LM 7xx series, or simple zener and transistor.

 

Since a battery source can't have unanticipated voltage spikes, surges, or output values, a regulator is not needed. Each IC still needs a decoupling capacitor close to its power pins.

ak

 

I understand

I will give it a try, no voltage regulator, direct supply for the AA battery for each IC as far as I wont need measuring or similar application; just powering up the IC and MCU


Best regards and thank you

 

Currently I am making a battery powered circuit (energy is supplied by two batteries AA in series) as far as all the components work in a range between 1.8V and 3.6V.

Certainly neither of those components will likely work properly with the 3V from two AA batteries.

 

Certainly neither of those components will likely work properly with the 3V from two AA batteries.

Could you please explain yourself?

Thank you

 

Could you please explain yourself?

Seems rather clear to me. Those devices need higher voltages than what you propose. An MCU will typically require 5 volts that is pretty stable. Batteries can't guarantee that stability. As voltage drops the MCU will fail to operate correctly. IC's are much the same that they require a higher minimum voltage, though there are plenty that can operate on lower voltages.

To operate an MCU you'll need sufficient voltage and regulation to maintain the needs of the MCU. You may need to reassess your design. Maybe even RE-design your project.

 

Seems rather clear to me. Those devices need higher voltages than what you propose. An MCU will typically require 5 volts that is pretty stable. Batteries can't guarantee that stability. As voltage drops the MCU will fail to operate correctly. IC's are much the same that they require a higher minimum voltage, though there are plenty that can operate on lower voltages.

To operate an MCU you'll need sufficient voltage and regulation to maintain the needs of the MCU. You may need to reassess your design. Maybe even RE-design your project.

Thank you Tonny for the explanation,

I understand what you say about 5V but I dont understand how can I supply 5V to MCU and IC that require a voltage from 1.8V to 3.6V to work. If I supply 5V I will waste the devices, I think?

Two AA batteries in serie will supply 3.2V maximum when fully charged, when discharged they should be around 2.6V-2.8V, from then they will start dropping their voltage until some IC or the MCU fail to work.

I dont have any problem re-designing my project or selecting new components, but I really dont understand how I can supply 5V to a MCU (for example PIC16LF18854) which maximum operating voltage is 3.6V. Or you are recommending me to supply 3.3V through a linear voltage regulator?


Thank you again for reading, and moreover for you explanation

 

As mentioned, there ARE some devices designed to operate on lower voltages. I'm by no means an expert on them. My statement may have been misleading, but my intention was to state a general rule. I'm old school electronics. VERY old school. And there are a lot of devices now on the market that I am absolutely unaware of, let alone have knowledge about. If you have spec sheets on your IC's and they say they operate on those low voltages then by all means don't exceed any of the parameters outlined in the spec sheet.

Again, sorry if I cause any confusion.

 

Could you please explain yourself?

Sorry, I misunderstood your statement about the voltage.
If all the components will work anywhere between 1.8V to 3.6V then yes, you could use 2AA alkaline batteries in series without a regulator.

Just note that if you are using any of them at near their performance limits, then the voltage change as the batteries discharge may also affect the circuit operation.
Since I know nothing else about your circuit function I can't advise you further.

 

Sorry, I misunderstood your statement about the voltage.
If all the components will work anywhere between 1.8V to 3.6V then yes, you could use 2AA alkaline batteries in series without a regulator.

Just note that if you are using any of them at near their performance limits, then the voltage change as the batteries discharge may also affect the circuit operation.
Since I know nothing else about your circuit function I can't advise you further.

Thank you crutschow to clear out,

As I commented before I count on the discharge of the batteries and the decrease of the voltage drop when they are discharged. From what I read, I could state that two AA batteries in series will be discharged when the total voltage is around 2.6V-2.8V, which is even sufficient for a MCU that works from 1.8V, even with the drop voltage due to a silicium diode in VDD; that considered I think everything should work fine while the batteries are charged. My worry is about the stability and "cleaness" of the energy of the two AA batteries, as others commented before, I shouldnt worry about this fact.

Again, thank you

 

I could state that two AA batteries in series will be discharged when the total voltage is around 2.6V-2.8V,

That's a reasonable point, but I generally run alkaline batteries to below 1.2V if I can, such as by using them in my battery powered clocks after other higher power devices that have used them quit at a higher voltage.
.
I have a wireless mouse that will operate down to about 0.9V on a single AA before it quits.
Don't know how they get the LEDs to operate that low.
Must have some kind of boost converter.

 

That's a reasonable point, but I generally run alkaline batteries to below 1.2V if I can, such as by using them in my battery powered clocks after other higher power devices that have used them quit at a higher voltage.
.
I have a wireless mouse that will operate down to about 0.9V on a single AA before it quits.
Don't know how they get the LEDs to operate that low.
Must have some kind of boost converter.

Exactly, they can keep working when the more demanding devices cant work anymore, but the voltage drop will be fast depending on how much energy demands the device (a clock is always a good option as you say). The mouse... well I dont really know how that is possible, because as you said is wireless, its crazy for me to imagine how that is possible, as you said a boost converter would be needed.

About the LEDs, I just know leds that can operate from 1,7V, I havent seen them with lower voltage and only red, green, orange and yellow colors if I remember well


Regards,

 

I´ve been digging about this topic of filtering and supplying stable energy through AA batteries. I have to say that I dont find much about the topic, but I got to find something that I represented in a schem in the attached picture.

I understand that the diode should be to avoid damage in case of inverting the position of the battery? But the rest I understand the porpose


Thanks for reading,

 

I understand that the diode should be to avoid damage in case of inverting the position of the battery?

Doesn't look like battery reverse polarity protection to me...

But the rest I understand the porpose

Then, kindly explain the purpose to me. There are no values for the capacitors or inductors and the schematic is poorly drawn to boot. Anyone who isn't consistent in their use of connection dots should be ignored.

Tell me where you found it so I can make sure to avoid that website.

 

Doesn't look like battery reverse polarity protection to me...

It shorts the battery if it's reversed connected.
Not sure if that would be considered protection.

 

Doesn't look like battery reverse polarity protection to me...

Then, kindly explain the purpose to me. There are no values for the capacitors or inductors and the schematic is poorly drawn to boot. Anyone who isn't consistent in their use of connection dots should be ignored.

Tell me where you found it so I can make sure to avoid that website.

Sorry I wanted to say I dont understand haha, I really do not understand

 

On the question of voltage regulator, the 3-terminal linear voltage regulator serves two purposes.

Firstly, it is common to supply your circuit from AC mains into a step-down transformer and a full-wave or bridge rectifier and a reservoir capacitor. Under loaded conditions, this unregulated DC supply will have 100Hz or 120Hz ripple. The function of the regulator device will remove this ripple voltage.

Secondly, the output voltage of the unregulated DC supply will vary with load conditions. Again the regulator device will hold the output voltage constant over the entire range of spec'd load current.

If you can match the battery voltage to the spec'd operating range of all the devices in your circuit you may avoid having to use a voltage regulator device.

 

Sorry I wanted to say I dont understand haha, I really do not understand

Where did you find that nonsense?

If we assume that the inductors are being used as chokes, you wouldn't need the one on the negative battery terminal. You also wouldn't need the one on the positive terminal. If there was any noise, preventing it from affecting the power source would be the least of your concerns.

If we assume that the parallel capacitors were for decoupling, they would offer better suppression if they were placed near components that were either generating noise, or could be affected by it.

The next time you see such a poorly drawn schematic, make a note of the source, ignore it, and never use information from that source again.