## Q. 2.2

Consider the transistor shown in Fig. 2.18, where the total width of the four parallel transistors is 80λ, its length is 2λ, and λ = 0.2 μm. Assuming node 2 is the source, node 1 is the drain, and the device is in the active region, find the source-bulk and drain-bulk capacitances given the parameters $C_j=0.24 \mathrm{fF} / \mu \mathrm{m}^2 \text { and } C_{j-\mathrm{sw}}=0.2 \mathrm{fF} / \mu \mathrm{m}$. Also find the equivalent capacitances if the transistor were realized as a single device with source and drain contacts still evenly placed. ## Verified Solution

Starting with node 1, the drain, we find that the areas of the junctions are equal to

$\mathrm{A}_{\mathrm{J} 2}=\mathrm{A}_{\mathrm{J} 4}=6 \lambda \times 20 \lambda=120 \lambda^2=4.8 \mu \mathrm{m}^2$

Ignoring the gate side, the perimeters are given by

$\mathrm{P}_{\mathrm{J} 2}=\mathrm{P}_{\mathrm{J} 4}=6 \lambda+6 \lambda=12 \lambda=2.4 \mu \mathrm{m}$

As a result, $\mathrm{C}_{\mathrm{db}}$ can be estimated to be

$\mathrm{C}_{\mathrm{db}}=2\left(\mathrm{~A}_{\mathrm{J} 2} \mathrm{C}_{\mathrm{j}}+\mathrm{P}_{\mathrm{J} 2} \mathrm{C}_{\mathrm{j}-\mathrm{sw}}\right)=3.3 \mathrm{fF}$

For node 2, the source, we have

$\mathrm{A}_{\mathrm{J} 1}=\mathrm{A}_{\mathrm{J} 5}=5 \lambda \times 20 \lambda=100 \lambda^2=4 \mu \mathrm{m}^2$

and

$\mathrm{A}_{\mathrm{J} 3}=\mathrm{A}_{\mathrm{J} 2}=4.8 \mu \mathrm{m}^2$

The perimeters are found to be

$\mathrm{P}_{\mathrm{J} 1}=\mathrm{P}_{\mathrm{J} 5}=5 \lambda+5 \lambda+20 \lambda=30 \lambda=6 \mu \mathrm{m}$

and

$P_{\mathrm{J} 3}=P_{\mathrm{J} 2}=2.4 \mu \mathrm{m}$

resulting in an estimate for $\mathrm{C}_{\mathrm{sb}}$ of

$C_{s b}=\left(A_{J 1}+A_{J 3}+A_{J 5}+W L\right) C_j+\left(P_{J 1}+P_{J 3}+P_{J 5}\right) C_{j-s w}$

$=\left(19.2 \mu \mathrm{m}^2\right) 0.24 \mathrm{fF} / \mu \mathrm{m}^2+(14.4 \mu \mathrm{m}) 0.2 \mathrm{fF} / \mu \mathrm{m}$

$=7.5 \mathrm{fF}$

It should be noted that, even without the additional capacitance due to the WL gate area, node 1 has less capacitance than node 2 since it has less area and perimeter.

In the case where the transistor is a single wide device, rather than four transistors in parallel, we find

$A_J=5 \lambda \times 80 \lambda=400 \lambda^2=16 \mu \mathrm{m}^2$

and

$P_{\mathrm{J}}=5 \lambda+5 \lambda+80 \lambda=90 \lambda=18 \mu \mathrm{m}$

resulting in $\mathrm{C}_{\mathrm{db}}=7.4 \mathrm{fF} \text { and } \mathrm{C}_{\mathrm{sb}}=9.0 \mathrm{fF} \text {. }$ Note that in this case, $C_{\mathrm{db}}$ is nearly twice what it is when four parallel transistors are used.