transistors
I have a self biased NPN transistor circuit as in the picture below and I want to calculate the base voltage Vb. The most common intuition leads me to apply potential divider and calculate the voltage, which in this case comes to be 3.529V. But whenever measure Vb, i get value near to 3.4V, which is less than what I calculate. Is there a simple explanation?
simulate this circuit – Schematic created using CircuitLab
If you want to conduct up to 150mA of current between node 1 and ground and only drop 1.0V (it's not a perfect switch) then you'll need to assume the current gain is typically 50. If you want a lower saturation voltage then the spec sheets says feed the base with 15mA i.e. the current gain has dropped to only 10 but saturation will only be 0.3V.
So assuming you are happy with 150mA while saturating the transistor to about 1V, you need to push \$\dfrac{150mA}{50} = 3mA\$ into the base.
The base voltage will need about 0.7V so the remainder (1.8V - 0.7V) needs to be across the resistor R1. Ohms law tells us that R = \$\dfrac{1.1V}{3mA} = 366.7\Omega\$.
So choose maybe a 360 ohm resistor.
If this isn't good enough for your needs look for an N channel MOSFET with low \$V_{GS(threshold)}\$ - something like 1V or less.
So as you mentioned it says that the transistor is essentially two diodes.
You should, but may not, know that the typical voltage drop required over a diode to make it conduct is ~0.7V but can vary depending on the diode of course. So if you just 'stick' a voltage across the terminals as when you increase the voltage over the diode current flows:
As the resistance across a diode is very low when this voltage is applied we can work out that the current would be extremely high: I = V/R, simple to see that the lower R is the higher the current, and this can be very damaging to the base terminal, I believe a datasheet of the particular transistor will give you more information on what size current it can take.
What this is saying is you need to have a current limiting resistor in front of the base terminal on the transistor which does exactly what its name describes, limits the current. As the voltage drop across the transistor will remain at 0.6-0.8V we can work out the size resistor we would need quite easily. R = (Vin - Vdrop)/I, 'I' being the base current that it can take, Vdrop being the voltage drop from the base to the emitter and Vin being the supply that is going into base, you also need to look at the hfe of the transistor so see if it will be able to give you the amount of current you need, which coincidentally can be limited, or 'tailored' with a resistor at the emitter pin so the transistor is less reliant on the hfe, but I am sure you will get on to that in the future!
Best Answer
The base takes some current and effectively reduces the value of R2.
The current gain of Q1 (hFE) might be about 100 and this has the effect of taking the emitter resistor (R3), multiplying it by 100 and dropping it in parallel with R2 so, effectively, R2 becomes 12k||200k = 11.3k.
This now reduces the potential divider voltage to about 3.35 volts. Given you measured about 3.4 volts, maybe the hFE of the transistor is a bit higher and maybe you also need to take account of the R1 and R2 resistor tolerances.
For instance, if you used 1% resistors and the hFE of Q1 was infinite, you could still expect to see a voltage that might be as low as 3.46 volts (one resistor at 99% and the other at 101%).
So, no, there isn't a particularly simple explanation but, it depends entirely on what you regard as simple.