Hi,

I am trying to replace a battery powered circuit with a AC to DC adaptor circuit. I have a 5V adaptor but the device requirement is 1.5V. So, what is best way to reduce this voltage? Is using a resistor good enough? If so can you help me in understanding how to choose the value of resistor?

Regards

 

A resistor is unlikely to produce acceptable results under any circumstances. Why do you think that it would?
The best method for your circumstances depends on how much current the load requires. Can you estimate or measure what that might be?
Failing that you can experiment with a purchased 1.5V battery eliminator.
https://www.amazon.com/s?k=1.5v+battery+eliminator&crid=121J4KNH6S6YA&sprefix=1.5v+battery+eliminator,aps,801&ref=nb_sb_noss_1

 

So, what is best way to reduce this voltage?

See:
https://forum.allaboutcircuits.com/...r-in-my-hacked-dmm-fs8233.190182/post-1779154

 

Hi,

I think i can simply use a voltage divider with appropriate resistors right? Please check the attached image.

 

I am sure that my load doesn’t consume more than what my adaptor can supply. It is very simple device when very low current consumption.

 

You can if the current drawn is always the same and doesn't vary when the device operates. You will then only need one resistor in line to drop 5V - 3.5V = 1.5V.

If the current drawn by the device varies you will need an active regulator of some kind.

What is the current drawn by the device and does it vary with operation? Actually - what is the device?

Depending on device behaviour and tolerances, anything from a simple resistor to a zener diode to a regulator chip may be required to meet specs.

 

How about this?

ADDED capacitor C1.

 

The device is actually a simple wall clock. Are you saying that we can just put one resistor to drop that additional 3.5V and make this happen instead of going for a resistor divider ?

 

Do you have a way to measure the current your clock draws, or does the power requirement or current appear in your clock? Without that information all you will get is a good guess and probably have to do some experimentation.

 

A wall clock is pretty tolerant of voltage as it will run on a cell from new to nearly flat. The circuit by Danko in post #7 will be all you need, except the resistor should be more like 10k or something. You could get away with just a carefully calculated single resistor but the diodes are a clamp that makes it more tolerant. The 100u capacitor is probably not needed - maybe a 1u if your supply is poor quality.

Knowing the current drawn by the clock would help calculate the resistor.

 

Heres someone whos done it:
https://www.instructables.com/Quartz-Clock-Power-Supply-Hack-AA-battery-to-AC-p/

It seems that about 2200 ohms for the resistor would be about right. The design in the link uses 1000 and three diodes. Close enough.

 

Put two red LEDs in series with it, or one white one.

 

Wait a minute, one red LED should be about right. White LEDs run 3 ro 4 volts.

I would go with @Danko 's circuit in post #11 since it has the best chance of working.

 

If it's a "simple wall clock" then it runs for a year on a battery, so draws very little current. The forward voltage drop of the LEDs would be at the lower end of the range.
At that level of current, a micropower three-terminal regulator would be the best option. It might work of a 1.25V version.

The battery is going to be the cheapest method of powering it.

 

(Some text removed for clarity)
The battery is going to be the cheapest method of powering it.

Hahaha (or as they say in Asia in and around China: 555) !

Very good point that in the end a battery may be the best (though less interesting) solution.

 

A linear voltage regulator will likely be the easiest solution. A buck converter would also work, and are typically more efficient than linear regulators (longer battery life), but at the power levels required by a wall clock maybe this isn't an issue?

Here are some linear options

Here are some switcher options

 

If an 1.5Ah AA cell lasts a year, then the current consumption is 170uA.
You would need a switcher with a very low quiescent current to compete with a linear regulator.

 

Unless you know the current for the load a resistive divider will not give you the result you expect.

 

hi srih,
One option is the LM317L set for a 1.5V output.
E

 

A battery operated clock draws current in a burst once every second. I would go with a 3 terminal regulator since the current is nowhere near constant.