C1=100 uF 20V
C2=300uF 20V
D1=FAST RECOVERY Schottky diode 30V 3A
R2=3K 1W
R1= 1K 1W
L1=100 uH
Heat dissipation:
Pd=5Vx2A=10W
Pd=10W/77%(0.77)
Pd=13W~
Pd=13W-10W= 3W
Junction to case=2C/W
3Wx2C/W=25+6=31C°
Max temp. of the regulator should be 31C° If my calculus is right.
Q2=They have small PCB mounted transformers.
This is very small like 2x2cm.

C2=300uF 20V
D1=FAST RECOVERY Schottky diode 30V 3A
R2=3K 1W
R1= 1K 1W
L1=100 uH
Heat dissipation:
Pd=5Vx2A=10W
Pd=10W/77%(0.77)
Pd=13W~
Pd=13W-10W= 3W
Junction to case=2C/W
3Wx2C/W=25+6=31C°
Max temp. of the regulator should be 31C° If my calculus is right.
Q1=LM2576 will deliver 2A of current incase you draw more the voltage will drop and regulator overheat unless it has overload protection.Q1=I'd like to add, how will I know what LM2576 delivers?
Q2=But how do they make them so small?, is my matchbox sized nokia charger got a transformer in it?
Q2=They have small PCB mounted transformers.
This is very small like 2x2cm.

Last edited:
