Hi,
I need help in making a circuit for automatically switching between the power supply for my door bell.

The bell uses 2 AAA batteries rated 1.5v each ( 3v total). I want to also supply it by a 3v ac-dc adapter.

I need a circuit to switch the power source from the adapter to batteries when the adapter is off and use only the power from adapter when it is on.
My problem is similar to this one
@danadak provided this circuit in the above thread

Will these diodes ( D1N4148) in this circuit work for 3v as well ?
PS: I think the bell allows 3.4 v as well since the AAA batteries I have even though rated at 1.5 v give about 1.65v each.

 

Use 1N4004 diodes as they have a 1A current rating. The 1N4148 diodes would be running close to their max ratings. Not a good idea for long life.

 

Use 1N4004 diodes as they have a 1A current rating. The 1N4148 diodes would be running close to their max ratings. Not a good idea for long life.

Alright. Also I'd like to ask if there is a resistor in the circuit or it just denotes the device ?
And wherever it shows as grounded , should I connect all of these to the negative wire from the adapter ?

 

Hello,

When you want a lower voltage drop, use a shottky diode.

Bertus

 

Hello,

When you want a lower voltage drop, use a shottky diode.

Bertus

Will a voltage drop affect the door bell ?
Sorry I am completely new to this field so i know nothing about all this.

 

Hello,

When the powersupply is giving 3 volts and the diode drops 0.7 volts, the bell will get 2.3 volts.

Bertus

 

Alright. Also I'd like to ask if there is a resistor in the circuit or it just denotes the device ?
And wherever it shows as grounded , should I connect all of these to the negative wire from the adapter ?

No resistor. That is the bell.
And there is probably no ground. This is a shortcut in drawing. It makes the circuit easier to read as there are not lines everywhere. Yes, all the "ground" points are the supply - line.
Oh, just a thought. Have you checked that the bell is running on a 3VDC plug pack? Often the supply for doorbells can be an AC supply. That may change things a bit.

 

No resistor. That is the bell.
And there is probably no ground. This is a shortcut in drawing. It makes the circuit easier to read as there are not lines everywhere. Yes, all the "ground" points are the supply - line.
Oh, just a thought. Have you checked that the bell is running on a 3VDC plug pack? Often the supply for doorbells can be an AC supply. That may change things a bit.

No, it runs on 2 AAA batteries only, I plan to use a spare mobile charger as as the main power source and the batteries as a backup.

 

Hello,

When the powersupply is giving 3 volts and the diode drops 0.7 volts, the bell will get 2.3 volts.

Bertus

I have another adapter rated 5.7v and 800mA, how should I reduce the voltage just enough such that the diode reduces it to around 3v ?

 

I'd like to ask if there is a resistor in the circuit or it just denotes the device ?
And wherever it shows as grounded , should I connect all of these to the negative wire from the adapter ?

1. The resistor is there to simulate your load so the simulation software can see some current.

2. Yes, all ground symbol points connect together. As noted, this point does not have to be connected to an actual ground, such as a cold water pipe. Again, many simulation programs will not run correctly unless something in the circuit is "grounded".

ak

 

I have another adapter rated 5.7v and 800mA, how should I reduce the voltage just enough such that the diode reduces it to around 3v ?

Diodes are cheap. 3 or 4 of them in series are ugly, but still cheap.

BUT ...

Check that adapter carefully. If it is a switching type, the output will be regulated reasonably tightly. But if it is the old transformer-and-rectifier type, the output could be way higher than the rating when it is lightly loaded or unloaded. With no load, measure the output with the meter in both AC and DC modes. Or, if you have a scope . . .

Please post photos of the adapter, with a close-up of the label.

Where are you located?

ak

 

Diodes are cheap. 3 or 4 of them in series are ugly, but still cheap.

BUT ...

Check that adapter carefully. If it is a switching type, the output will be regulated reasonably tightly. But if it is the old transformer-and-rectifier type, the output could be way higher than the rating when it is lightly loaded or unloaded. With no load, measure the output with the meter in both AC and DC modes. Or, if you have a scope . . .

Please post photos of the adapter, with a close-up of the label.

Where are you located?

ak

I tried simulating the circuit and also used just the diodes, but it supplies 3.98V when the bell is idle.
I don't know if it would be able to handle that much. ( It's rated for 3v but also works with about 3.36v too)
I'll attach the images of the pcb of the bell, maybe someone can tell if it works fine with 4v.
I'm located in India btw