Hello,

I have this PCB design:
When I solder the LM2596s, the L7805 module is broken(the voltage between Vout and GND is always zero). I don't know the reason. Could you please take a look at my schematic and suggest some problems? Thank you so much.

 

Hi N03,
Welcome to AAC.
Do you have a photo that you could post of the PCB, front and back?
What is the design current in the 3.3v, load.?
E

 

A likely short in the wiring. The schematic does not help since it does not show a short.

 

Hi N03,
Welcome to AAC.
Do you have a photo that you could post of the PCB, front and back?
What is the design current in the 3.3v, load.?
E

Hi Eric, thanks for helping me.

Here is photo of the PCB:
- Front:

- Back:

- The current is 10A from 12V adapter.
I should also note that I soldered the L7805 first and it works fine. After that, I soldered the LM2596S and the L7805 is broken.

 

A likely short in the wiring. The schematic does not help since it does not show a short.

I also think it's short wiring but I don't know why. I posted the PCB front and back photo above, could you please check it?

 

Hi,
As you may know the 7805 can only dissipate about 2.5 Watts without an heat sink, so a drop from 12V to 5V at 10Amps is 7V *10A = 70Watts.
E

 

One possible way to find it is to use an adjustable current limited supply, turn the current up slowly, and see what gets warm.

 

Hi,
As you may know the 7805 can only dissipate about 2.5 Watts without an heat sink, so a drop from 12V to 5V at 10Amps is 7V *10A = 70Watts.
E

I understand but the 7805 only got broken when I solder LM2596s. Before that I test 7805 independently with 12V source and it works normally(though it dissipates lots of heat), so I don't think that's the reason.

 

I understand but the 7805 only got broken when I solder LM2596s. Before that I test 7805 independently with 12V source and it works normally

Hi, N03,
What was the current being drawn by the load when you did that test.?
Also, the LM2596 will initially draw a heavy current during start up, and the 7805 will shut down in current limit.
E

 

Why are you reducing the voltage via an inefficient linear regulator before reducing it further with an efficient switching regulator?

How much current do you need at 5V and at 3.3V?

 

and what is the deal with puny traces? what did you use as guideline for the trace width? specially 5V track need to be meatier.

 

I have used the 7805 for many years. When I first used it, it didn't work. I was told to add a capacitor across the Input - to GND. Sure enough, that fixed it. Turns out, it is also in the datasheet.

"If the device is more than six inches from the input filter capacitors, an input bypass capacitor, 0.1 μF or greater, of any type is needed for stability."

I would also recommend adding one on the output. The datasheet shows a 0.22uF on the Input and a 0.1uF on the Output.

 

Why are you reducing the voltage via an inefficient linear regulator before reducing it further with an efficient switching regulator?

How much current do you need at 5V and at 3.3V?

Yes, I notice that. I didn't design this PCB. The one who designed this PCB told me that he used two regulators to compensate for the excess amount of heat dissipation. He said that it will produce lots of heat to drop from 12V straight to 3.3V so he wanted to drop step by step, from 12V to 5V then from 5V to 3.3V.
I will try to eliminate 7805 and stick with LM2596S. Thanks!

 

I have used the 7805 for many years. When I first used it, it didn't work. I was told to add a capacitor across the Input - to GND. Sure enough, that fixed it. Turns out, it is also in the datasheet.

"If the device is more than six inches from the input filter capacitors, an input bypass capacitor, 0.1 μF or greater, of any type is needed for stability."

I would also recommend adding one on the output. The datasheet shows a 0.22uF on the Input and a 0.1uF on the Output.

Oh nice, I will look over it. Thanks!

 

I think the L7805 is getting almost hot enough to shut down without the LM2596S then shuts down with more heat when the
LM2596S is added.

 

Yes, I notice that. I didn't design this PCB. The one who designed this PCB told me that he used two regulators to compensate for the excess amount of heat dissipation. He said that it will produce lots of heat to drop from 12V straight to 3.3V so he wanted to drop step by step, from 12V to 5V then from 5V to 3.3V.
I will try to eliminate 7805 and stick with LM2596S. Thanks!

The idea is sound, but he got it backwards. The drop from 12 o 5 is where the most power is lost, for two reasons:

1. Both the current used at 5V and that used at 3.3V goes through that regulator.

2. The voltage drop is 7V vs 1.7 from 5 to 3.3.

Using a switching regulator from 12 to 5 followed by a linear one 5 to 3.3 is reasonable. The other way around is silly,

 

Yes, I notice that. I didn't design this PCB. The one who designed this PCB told me that he used two regulators to compensate for the excess amount of heat dissipation. He said that it will produce lots of heat to drop from 12V straight to 3.3V so he wanted to drop step by step.
I will try to eliminate 7805 and stick with LM2596S. Thanks!

The idea is sound, but he got it backwards. The drop from 12 o 5 is where the most power is lost, for two reasons:

1. Both the current used at 5V and that used at 3.3V goes through that regulator.

2. The voltage drop is 7V vs 1.7 from 5 to 3.3.

Using a switching regulator from 12 to 5 followed by a linear one 5 to 3.3 is reasonable. The other way around is silly,

Yes, I understand now. Thank you so much for your great help, guys!