Hey Everyone,
Just want to make sure I'm doing this right.
Suppose I have a voltage divider like in the first attachement. You'll see that R1 = 680Ω and R2 = 20k giving me an output voltage of 8.7V [Vout = R2/(R1+R2)]. Now, to calculate the output current of the voltage divider, you do the following: Iout = Vin/(R1+R2). So my current output is approximately 435μA. Let's say I want to have 2.5μA for my output, I simply add a resistor like in the second attachment. What would the output current be now? Also, wouldn't there be a voltage drop across R3 so I would have to make the voltage a bit larger to compensate for that drop? Ignoring the voltage drop, this is what I did: 435μA - 2.5μA = 437.5μA. So, 8.7V / 437.5μA = 19885.71429Ω ≈ 20k.
Please let me know if I'm doing this right! This is to get the correct base current and voltage to an NPN transistor.