In this circuit, the charging and USB port part is still the same, but I add in S2 and R3 at the LED side. So, the S1 and R2 will still function like the previous one, that by switching on S1, you will get the 1KOhm resistor brightness. By adding S2 and R3, it means that if you switch on S2, you will get 330Ohm brightness, which is brighter than 1KOhm brightness, because the resistor that limiting the LED current is now smaller and higher current on the LED means higher brightness. I say you will have 3 level brightness, so, the third level brightness is by switching ON S1 and S2 at the same time.
When S1 and S2 is being switch ON, the R2 and R3 form a parallel pattern, and you need some calculation to get the total resistance for the White LED.
The total resistance from the parallel of R2 and R3:
1/R = 1/R2 + 1/R3
R = 1 / ( 1/R2 +1/R3 )
R = 248.12Ohm
So, the total resistance at the White LED if you turn ON both S1 and S2 is 248.12Ohm which is 3rd level of brightness for the White LED.
- 1st level (S1 ON) – 1KOhm Brightness
- 2nd level (S2 ON) – 330Ohm Brightness
- 3rd level (S1 and S2 ON) – 248.12Ohm Brightness