Hi David,
That¡¯s great news, well done.
The supply current seems higher than I would expect.
Components:
Specified nominal coil resistance = 720 ?
D3 diode forward voltage at 0.5 A = 0.9 V
Therefore:
Relay current = (Vsupply ¨C D3 diode drop) /720 x 15 = 252 mA
Assuming 18mA for the PIC and OLED display, Total = 270 mA
I think your test LEDs may be the culprits. Looking at the photo, the LED has a 470? current limit resistor, which probably means the current is 20mA/LED, 300 mA total.
That would give a total of 270 + 300 = 570 mA, which looks close to what your meter is showing.
I would consider reducing the current for the LEDs. I would expect a few mA would be more than adequate, and it is only for testing.
73, Dave