65.1.6. Input Capacitance Calculation

So, to get the input capacitance C which is C + C (1 ‒ whatever the gain we do have from here to here which is let you call this is A ) and what is the A ? A = ‒ g v1 v1 v1 m1 multiplied by r in parallel with the equivalent resistance coming from this circuit.

This chunk introduces the formula to calculate the input capacitance (C). The total input capacitance combines the gate-source capacitance (Cgs1) and the gate-drain capacitance (Cgd1) modified by the gain (A). The gain is defined as the product of transconductance () and the load resistance (r) in parallel with the equivalent resistance of the circuit. Understanding this formula is crucial for assessing how capacitance affects circuit performance.

Think of a water pipeline where the capacitance is like the amount of water the pipe can hold before it overflows (input capacitance). The resistance is the size of the pipe (load resistance). A larger pipe allows more water to flow through, similar to how a higher gain in an electrical circuit can lead to a larger effective input capacitance.

And we made the calculation if this is 5 MΩ, if this is 5 MΩ then the this resistance it is 50 kΩ. So, that gives us these gain of ‒ g is 2 × 10‒3 and then we do have 50 k and 50 k. So, that is and so, these 2 are getting cancel, this is also getting cancel that is the giving us ‒ 50 gain.

In this segment, we calculate the gain based on the resistance values provided (5 MΩ). Given the circuit conditions, we find that the transconductance () is 2 mA/V and the load resistances are each 50 kΩ. The calculations demonstrate how the gain contributes to the modified capacitance value. The gain of -50 indicates that the output signal is inverted and amplified. Understanding how to calculate gain helps in designing efficient circuits.

Imagine you are amplifying the sound from a microphone. If the microphone picks up a sound softly (like your voice), the amplifier boosts that sound so that it can be heard clearly across a large space. The gain represents how much louder the sound gets, and similar principles apply in electronic circuits, where a small input can lead to a significantly larger output.

So, the gain from this point to this point it is ‒50 and hence the input capacitance with this value of this A it is C it is 10, C it is 5 and then we do have (1 + 50) here and that gives us 265 pF yes.

Here we put the calculated gain into the overall input capacitance formula to get the final result. The input capacitance is the sum of contributions from both capacitances, scaled by the gain. A final value of 265 pF illustrates how the gain can effectively increase the input capacitance, which can be significant in high-frequency applications. Knowing how to compute this helps engineers design circuits with desired performance criteria.

Consider a sponge soaking up water. The initial water content (10 pF and 5 pF) represents the basic capacitance, and the additional water (gain factor of 50) shows how much extra the sponge can hold. The sponge’s total soaking capability is akin to our final capacitance value of 265 pF, demonstrating how gains can significantly enhance system characteristics.

So, it is increasing the input capacitance, but probably still it is not so, alarmingly high, maintain this factor it is so, high, but yes depending on the situation we may or may not be able to accept that.

This conclusion discusses the ramifications of the computed input capacitance value. While it is increased due to the circuit design, the figure of 265 pF is not excessively high in most situations and can be acceptable. However, the designer should evaluate whether such an increase aligns with circuit requirements, particularly in high-speed applications where capacitance can affect timing and frequency response.

No real-life example available.