Variable power supply using 9V batteries in series

[awa123]

Hi,

I am trying to create a power supply that outputs 0-35V using 9V batteries.
I need to be able to vary the voltage while keeping the current as low as possible.

I was thinking a potential divider would work, but I am not sure. Please help, thanks.

 

[tpetar]

Hi,

I am trying to create a power supply that outputs 0-35V using 9V batteries.
I need to be able to vary the voltage while keeping the current as low as possible.

I was thinking a potential divider would work, but I am not sure. Please help, thanks.

2x 9V is 18V (two 9V batterie in serie)

Voltage divider will make lower voltage, first you need to make higher voltage, with some step-up circuit, from 18V to 35V, then regulation in rage 0-35V.

 

[sgreen]

Hi,

I am trying to create a power supply that outputs 0-35V using 9V batteries.
I need to be able to vary the voltage while keeping the current as low as possible.

I was thinking a potential divider would work, but I am not sure. Please help, thanks.

After raising to 36 volt u can use LM317 Regulator ic. Using this u can vary the voltage from 1.2v to 36V.See the datasheet where u can find the typical application.

 

[awa123]

2x 9V is 18V (two 9V batterie in serie)

Voltage divider will make lower voltage, first you need to make higher voltage, with some step-up circuit, from 18V to 35V, then regulation in rage 0-35V.

Hi thanks for your reply.

I only need the power supply for short bursts as I need to apply voltage across a component in a piece of equipment I'm using, and I don't really want any active components. Couldnt I use 5 9v batteries in series to give me 45V and then use a potential divider using a potentiometer to vary between 0-35V.

 

[tpetar]

If you want to use 35V I think is better to use LM317 for High Voltage (LM317HV), because standard LM317 have limit to 37V its on edge.

What current you plan to drain from thise two batteries, maybe switcher solution is better for that.

- - - Updated - - -

Hi thanks for your reply.

I only need the power supply for short bursts as I need to apply voltage across a component in a piece of equipment I'm using, and I don't really want any active components. Couldnt I use 5 9v batteries in series to give me 45V and then use a potential divider using a potentiometer to vary between 0-35V.

You can use four 9V batteries to get 36V, and use some switcher regulator to regulate voltage.

What is minimum needed voltage for sure ? 0 or can be 1,25V?

Battery power supply can be turned off completely with switch, and turned on only when is needed with some switch ?

 

[awa123]

If you want to use 35V I think is better to use LM317 for High Voltage (LM317HV), because standard LM317 have limit to 37V its on edge.

What current you plan to drain from thise two batteries, maybe switcher solution is better for that.

- - - Updated - - -



You can use four 9V batteries to get 36V, and use some switcher regulator to regulate voltage.

What is minimum needed voltage for sure ? 0 or can be 1,25V?

Battery power supply can be turned off completely with switch, and turned on only when is needed with some switch ?

The current is another issue, I need it to be very low micro amps, so it doesn't damage the equipment. The main use is to apply a bias to a small instrument in the machine I am using.

The voltage does not have to be exactly on zero. In fact I need a negative voltage. And yes a switch would be useful as i don't need the supply to be one for more than a few minutes every time.

 

[Audioguru]

A little 9V alkaline battery voltage quickly drops to 7V then slowly drops lower when it has a low current load.
Then 5 of them in series produce 45V when they are new then the voltage drops to 30V or less without using a voltage regulator.

 

[BradtheRad]

This is a simulation of what you are asking about:



You will need to install 2 switches, if you don't want to waste any juice.

Since you say you only need a few uA, I gave the load high resistance.

I also put one battery below zero ground, because you say you need negative polarity as well.

- - - Updated - - -

In case you'd rather build a voltage multiplier, and use just one 9V battery...

This screenshot shows how to do it:



You can install a potentiometer at the final output (similar to the previous schematic).

 

[awa123]

This is a simulation of what you are asking about:



You will need to install 2 switches, if you don't want to waste any juice.

Since you say you only need a few uA, I gave the load high resistance.

I also put one battery below zero ground, because you say you need negative polarity as well.

- - - Updated - - -

In case you'd rather build a voltage multiplier, and use just one 9V battery...

This screenshot shows how to do it:



You can install a potentiometer at the final output (similar to the previous schematic).

Thanks a lot for the help.
If I use the circuit in the first screen shot, will I be able to vary my output voltage from say -10V to -36V in steps of 1V ? It seems as if the 1M pot will only vary the current and not the voltage. Correct me if I'm wrong.

Also I only need a negative voltage, so one switch should be sufficient right?

 

[BradtheRad]

Thanks a lot for the help.
If I use the circuit in the first screen shot, will I be able to vary my output voltage from say -10V to -36V in steps of 1V ? It seems as if the 1M pot will only vary the current and not the voltage. Correct me if I'm wrong.

Also I only need a negative voltage, so one switch should be sufficient right?

Yes, since you only need a negative voltage, then all the batteries can be at negative polarity. One switch is sufficient.

Here is a revised schematic.



The potentiometer acts as a voltage divider. It will allow you to make infinitely fine adjustment between 0 and -36V. The scope traces portray the full range.

With a 1 megohm pot, only a little juice is wasted. However you may find you need a 100k pot, or 10k pot, etc. It depends on how much current your load requires.

Stacking several 9V batteries is a simple and easy method, but it requires several dollars worth of batteries.
Although you say you don't want active components involved, it would not be too difficult, nor expensive, to construct a -40 V source that is powered from a single 1.5V battery. It uses a blocking oscillator built around a small transformer.

 

[awa123]

Although you say you don't want active components involved, it would not be too difficult, nor expensive, to construct a -40 V source that is powered from a single 1.5V battery. It uses a blocking oscillator built around a small transformer.

Hi, how difficult would it be using a blocking oscillator and transformer, and would it be better for me to use some sort of negative voltage regulator. Stacking 9V batteries is taking up too much space and too much heat.

- - - Updated - - -

2x 9V is 18V (two 9V batterie in serie)

Voltage divider will make lower voltage, first you need to make higher voltage, with some step-up circuit, from 18V to 35V, then regulation in rage 0-35V.

Hi tpetar,

Do you have any suggestions what kind of voltage regulator I would need for negative voltage up to -35V

 

[BradtheRad]

Hi, how difficult would it be using a blocking oscillator and transformer, and would it be better for me to use some sort of negative voltage regulator. Stacking 9V batteries is taking up too much space and too much heat.

One blocking oscillator that has become popular is called the joule thief. It will power a bright white LED from a half-used AAA battery.

The primary and secondary windings each consist of a couple turns of thin wire around a small ferrite core. An internet search will turn up instructions. (Someone ought to find a suitable core of everyday material instead of needing to obtain a ferrite.)

To create a constant DC source you'll add a diode, zener, and smoothing capacitor.

Since you want negative polarity you will configure thus: