(deleted)
OK
start at the beginning. Ohms law.
V=IR. or re-arranging, R = V/I
Note I use 1mA == 1/1000 ampere. By convention.
THE LEDS:
You have 2 leds in series, 1.5 v each = 3v.
Saturation voltage of transistor probably 0.3v .
Total 3 + 0.3 = 3.3v.
the voltage across the resistor is thus : 5 (your supply) - 3.3 = 1.7v
The leds would work well with 10mA ie 1/100A or 0.01A
So R2 should be 1.7/0.01 = 170 ohms. Use 150 or 180 as commonly available values.
THE BASE DRIVE:
You normally reckon a base current of 1/20 collector current to ensure a bipolar transistor is turned hard on. so 10mA/20 or 0.5mA (0.0005A) should be ok.
The base/emitter voltage of a bipolar transistor is about 0.7v. You have 3.3v to drive, so the resistor R1 needs to drop about 3.3 - 0.7 = 2.6 v.
So R1 should be about 2.6/0.0005 or 5200 ohms . Use a 4700 or 5600.
regards
Allan
ps Red LEDS have a voltage drop of about 1.5 - 1.6 v. Blue ones about 3.1v. Other colours in between.
Nice to see Planck's and Einstein's work accurately reproduced in real things you can buy for pennies !
The semiconductor bandgap ( in volts ) and the photon energy in electron-volts all tie up nicely.
Except I reckoned on 10mA led current.. for 50mA re-do the sums and get R2 = 34 ohms, and R1 becomes 1040 ohms. Practically 33 and 1k respectively.
ie divide all values by 5..
Sorry - I missed the 50mA..
Allan
(deleted)
Don't forget to modify R1 => 1k as well - more collector current needs more base current to drive it.
The values aren't critical , and components are probably +/- 10% tolerance anyway.
And no - at these currents the collector saturation voltage won't change much.
Allan
(deleted)