Board1 Injector Simulator.

 Image. Wiring diagram.

Red LED and  Green LED were chosen for this circuit. According to the LED datasheet forward current for RED LED 10ma<I<30mA
GREEN LED 10mA<I<25mA
Let'assume that working current for both LED's is 20mA.

Vdrled- voltage drop for Red LED is 1.785 V
 Vdgrled-voltage drop for Green LED is 1.755 V
 Transistor BC547A is chosen for this circuit. From this transistor datasheet:
  Vbe=700 mV
  Vce=90 mV
  Thus: R14=(Vcc-Vdrled-Vce)/Ic=(12V-1.785V-0.09V)/0.02A=506 Ohms
From resistor value table choose the resistor with 5% tolerance. This is E24 series.
Therefore, R14=510 Ohms
R15=(Vcc-Vdgrled-Vce)/Ic=(12V-1.755V-0.09V)/0.02A=507.75 Ohms
For the same reason as it is mentioned above in R14 calculation, choose resistor 510 Ohms > R15=510 Ohms
At the next stage let's calculate Ib=Ic/b, where b=Hfe- current gain
b varies from 110 to 800 for this type of transistors. To be on the safe part of the road choose b=110. That means transistor will work under minimum conditions. If b is higher the transistor will work for sure.
Ib=Tc/b=0.02A/110=0.18 mA
However, returning back to the datasheet, we can see Ib=0.18 mA is not enough to get the transistor hard saturated. According to the datasheet for BC547A Ib must be 5mA.
Thus, R13=R16=(Vpwm-Vbe)/Ib=(5V-0.7V)/0.005 A= 860 Ohms
Choose R13=R16=910 Ohms from the resistor value table
Limiting resistors which were calculated will provide saturation mode.
The results are :  R13=R16=910 Ohms + - 5%
                          R14=R15=510 Ohms + - 5%
                          T1=T2= BC547A


Board for this circuit is on the photo :
This is a working board examle. Soldering has revealed that power supply wire was covered by oxide . Ideally, all these wire must have been treated with flux in order to remove oxidation before soldering.
Other components were soldered with common bar solder.
The measurements on the diagramm show volatge drop accross each device.
It is obvious that voltage drop accross current limiting resistors provides accord to the calculation above with reference to the components tolerance. Both transistors are saturated. That's why for the transistor T1 Vbe=0.817V and Vce=0.041V. For the transistor T2 Vbe=0.818V and Vce=0.042V. Voltage drop accross LED1=2.129V and LED2=2.136V what allows them to glow brightly.

 

Reflection: In this particular case the elements layout is a bit irrational. If I could rebuild it I would do it in more compact way. There is no high demand for heat exchange because the transistors work in the saturation mode.