To get you started:
- The current drawn by the relay is still unknown
- The green LED has no current limiting resistor. It will burn quickly as it is directly connected between 5 V and gnd by T2.
R3 can be 10 kΩ as it is. It is only required to draw the base of T2towards 5 V, thus making Vbe(T2) ~ 0 V. This is good enough to turn T2 off.
- Add a series resistor to the green LED. I think you may have meant R6 to be this current limiter, but both sides of R6 are tied to gnd, thus rendering R6 useless.
- Calculate the total worst case collector current through T2 (sum of relay current and current through green LED.
- calculate the worst case base current for T2 by dividing max. collector current by the min. DC gain (as this is a slow on/off switching circuit DC gain is relevant, not AC gain).
- From the base current and the voltage drops Vce(T1) and Vbe(T2) calculate the minimum required base resistor R4. If you want to add a safety margin, reduce that value by e.g. 10 % (increase current by 10%) and select the next standard value for R4.
- Calculate the collector current of T1 (use the currents through R3 and R4) and perform the same steps as above to get a value for the T1's base resistor R1.
The same goes for R2. However, R2 will contribute to the current through R1 as I(R2) = Vbe(T1)/R2 needs to be taken into account. If you don't want this to happen, move R2 to the left of R1 where it will still work as pull-down. Whether 10 Ω is sufficient for R2 depends on the impedance of the output stage driving into the input (R1) of your circuit). R2 needs to fulfill these conditions:
- If the output of the driver is active High (I assume 5 V as the rest of your circuit), the driver needs to be able to supply current both into the base of T1 and through R2. For a typical µC output capable of driving a few mA this is no problem.
- If th eoutput of the driver is high impedance, the value of R2 needs to be low enough such that Vbe(T1) is much less than 0.6 V (typical value for an NPN) to turn T1 securely off. Design for 0.1 V max. and you're save. If, for example, the driver's output would deliver 1 µA (just a ballpark figure as I don't know what's driving into R1), this 1 µA would drop 10 mV across R2, definitely hood enough to turn T1 off.
I have calculated the following:
Current through T2 and Green LED is 72mA
Base current for T2 = 72mA/120 = 600uA
Base resistor for R4 = T1 VCE 5V - VBE T2 0.7V
4.3/600uA = 7.1K
They have this as a 1K.
T1 Ic = 72mA + (5/10K)= 72.5mA
T1 base current is 72.5mA/200=0.362mA
T1 base resistor is 3V (GPIO) - 0.6/0.326mA=7.3K
Can I add a saturation factor of X3
Therefore 3V-0.6/(0.326mA x 3)=1K
Is my value of R2 too big? As the transistor T1 switches on at 1.1V?
Have I got anything wrong?