# Output voltage in a BS270 circuit

I am reading through Exploring Raspberry Pi and in a section on FETs, there is an example circuit. I understand how the controlled current is calculated $$I = \frac{V}R = (5V)(100\Omega) = 50mA)$$, but how is the voltage of 0.17V calculated?

It's calculated by looking at the BS270 datasheet:

There, above, you can see that the absolute maximum drain-source resistance is $$\3.5\:\Omega\$$. So if you multiply that resistance by the current, $$\50\:\text{mA}\$$, you get $$\3.5\:\Omega\cdot 50\:\text{mA}=175 \:\text{mV}\$$. That is close to what they wrote.

They were being conservative here. This is at a high temperature for the device. You can pretty much plan that the resistance will be less in most common circumstances. But it is good to plan for worst-case.

I think they use the maximum Rdson value :

In practice, for Vgs=3.3V at 25°C, it's not the correct value.

The current values are rough approximations just to give a clue what is happening in the circuit, they are not calculated from exact values.

As the FET is not an ideal short circuit, there would not be exactly 5V over the resistor, so there would not be exactly 50mA current flowing through it.

Since in that circuit there is 4.83V over the resistor, there is 48.3 mA flowing. The FET has approximately 3.5 ohms of resiatance then.

In real life your resistor would not be exactly 100 ohms either, it would have some tolerance, and for general purpose a 10% or 5% resistors are cheap and common so the resistor can have more variation in the resistance (e.g. 95 to 105 ohms) than what the FET adds.