Use this power adapter instead transformer X1 and diodes D1, D2.
Price $4.54, free delivery.
View attachment 333938

I did what you suggested, but it didn't yield any results. The current in the circuit is very slow.

 

The voltage at pin 3 is 13 volts

Hi alir,
The LM317 requires at least 3v drop across it to work correctly.

That's why you only measure approx 10V on the output.

Ref Vi - Vo
E

 

Hi alir,
What is the voltage on pin #3?
E

My sim shows you should be getting normal charging current even with a 13V transformer.
What are the voltages at LM317 pin3 and pin 2?

I also used the circuit below instead of the above one, but the charging current is very low, and it charges slowly.

 

hi alir,
Are you still using a13V transformer and not a 15V transformer?
E

 

Hi alir,
The LM317 requires at least 3v drop across it to work correctly.

That's why you only measure approx 10V on the output.

Ref Vi - Vo
E
View attachment 333989

Thank you very much for your guidance.
So, should I supply around 17 volts to the circuit to set 13.8 volts as the cutoff voltage and 12.5 volts as the lower (charging) voltage?

 

Hi,
If you can supply the LM317 input with say 17Vdc and set the LM317 to give 13.8Vout.
Use a variable resistor with the LM317 so that you can control the charge current.

Remember, the LM317 is rated at 1.5A max.
So if you set the charge current to 1Amp, that will cause 1A*3V= 3 Watts heat dissipation, so use a suitable heat sink.
E
Check this PDF

 

hi alir,
Are you still using a13V transformer and not a 15V transformer?
E

I removed diodes 1 and 2 and connected 16 volts to the circuit through a DC power supply.

 

Hi,
If you can supply the LM317 input with say 17Vdc and set the LM317 to give 13.8Vout.
Use a variable resistor with the LM317 so that you can control the charge current.

Remember, the LM317 is rated at 1.5A max.
So if you set the charge current to 1Amp, that will cause 1A*3V= 3 Watts heat dissipation, so use a suitable heat sink.
E
Check this PDF

Your guidance is excellent.
I'll let you know the results.

 

hi alir,
Are you still using a13V transformer and not a 15V transformer?
E

My friend, let me mention that my transformer has two 13-volt outputs and one ground. I am connecting both 13-volt outputs to the circuit.

 

hi alir,
This simulation is a basic representation of your original circuit when using a 13Vrms transformer.
The 3V source represents the Vdrop across the LM317.
As you can see from the images, the Vout is less than the 13.8V voltage at which you want to charge the battery, so the charge current is very low, approx 60mA.

E

Update:
For comparison, I have added a sim, using a 15Vrms transformer as specified in the original circuit.
Note Vout and the possible charge current, but remember the LM317 dissipation, which will be high.
LM317 Vdrop ~4V @ 1.5Amps = 6Watts!!! Suitable Heat sink required

 

hi alir,
This simulation is a basic representation of your original circuit when using a 13Vrms transformer.
The 3V source represents the Vdrop across the LM317.
As you can see from the images, the Vout is less than the 13.8V voltage at which you want to charge the battery, so the charge current is very low, approx 60mA.

E

Update:
For comparison, I have added a sim, using a 15Vrms transformer as specified in the original circuit.
Note Vout and the possible charge current, but remember the LM317 dissipation, which will be high.
LM317 Vdrop ~4V @ 1.5Amps = 6Watts!!! Suitable Heat sink required


View attachment 334074

View attachment 334075

Oh wow! So the circuit is working fine? It's charging at this low current.
I really appreciate your guidance.

 

hi alir,
This simulation is a basic representation of your original circuit when using a 13Vrms transformer.
The 3V source represents the Vdrop across the LM317.
As you can see from the images, the Vout is less than the 13.8V voltage at which you want to charge the battery, so the charge current is very low, approx 60mA.

E

Update:
For comparison, I have added a sim, using a 15Vrms transformer as specified in the original circuit.
Note Vout and the possible charge current, but remember the LM317 dissipation, which will be high.
LM317 Vdrop ~4V @ 1.5Amps = 6Watts!!! Suitable Heat sink required


View attachment 334074

View attachment 334075

"Unfortunately, this circuit isn't suitable for my needs. I'd like to design a circuit using the CN3767 IC to charge a lead-acid battery."

 

Hi alir.
There is no LTSpice for the CN3767.
Check this link.

E

https://www.pinterest.com/pin/594897432054857478/

 

Hi alir.
There is no LTSpice for the CN3767.
Check this link.

E

https://www.pinterest.com/pin/594897432054857478/

Hi
You're a great friend. Thanks so much for your help. I'm going to buy this IC and try it out. I'll let you know how it goes.

Thanks

 

hi alir,
Also check this link for options.
E
https://www.google.com/search?client=firefox-b-d&q=cn3767+schematic

 

hi alir,
Also check this link for options.
E
https://www.google.com/search?client=firefox-b-d&q=cn3767+schematic

Ok
Thanks