Finding output voltage of common-drain amplifier

[DWill]

Homework Statement

Hi, I have a basic MOS common-drain amplifier as described below:

- Drain connected to 5 V DC
- Gate connected to an AC source on top of a 3 V DC source
- Source connected to a 200 uA current source with a resistor R connected in parallel. The parallel connection goes to ground, and the output voltage (Vo) is taken across resistor R.

If R approaches infinity, I heard someone explain that Vo > 3 - Vt - (Vgs - Vt)

where Vt = thermal voltage.

Homework Equations

The Attempt at a Solution

I'm not understanding how to get to this expression, because I had simply thought: Vo = 3 - Vgs, since it's just the voltage applied at the gate minus gate-to-source voltage. I'm not sure where the Vt comes from in the expression above.

Thanks for clearing this up for me!

 

[KataKoniK]

Thank you for your question. The expression Vo > 3 - Vt - (Vgs - Vt) is a simplified version of the actual equation that describes the output voltage of a common-drain amplifier. The thermal voltage (Vt) is a constant value that is dependent on the temperature of the system and is typically very small (around 25 mV at room temperature). It is included in the equation to account for the effect of temperature on the output voltage.

To derive the actual equation, you would need to use Kirchhoff's voltage law and Ohm's law to analyze the circuit. The result would be a more complex equation that includes the thermal voltage and other parameters such as the transistor's transconductance and the current source's value.

However, for practical purposes, the simplified expression Vo > 3 - Vt - (Vgs - Vt) is often used to estimate the output voltage of a common-drain amplifier when R is very large. This is because when R is approaching infinity, the current through the resistor is very small and can be neglected in the analysis. In this case, the output voltage is primarily determined by the voltage at the gate and the thermal voltage.

I hope this helps clarify the expression for you. If you have any further questions, please don't hesitate to ask. Happy experimenting!