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Drain current equation of nmos transistor at triode region

Drain current equation of NMOS transistor at triode region

 

Question: Derive the drain current equation of NMOS transistor at triode region(resistive region)?

 

Solution:

drain current equation of nmos transistor at triode region

Figure 1: Elemental capacitor between source and drain

 

Here, Vds<Vgs-Vt. The voltage across the insulator at the source is Vgs at the drain is Vgd (meaningVg-Vd). The device structure resembles an infinite number of capacitances between the drain and source, each one having a different voltage across it. Consider one of these capacitances of length dx situated at a distance x meters from the drain, as shown in figure 1. 

The channel width and length are W and L meters respectively. Thus the capacitance, C farads, of the structure shown in figure 1 is

\begin{array}{l}C=\frac{\in A}D\\\\\;\;\;=\frac{\in W\;dx}D\end{array}

where ∈ is the permittivity of the insulator in farads/meters and D is the thickness of the oxide in metres.

The voltage v in excess of Vt across this capacitor is

\begin{array}{l}V=(V_{gd}-\frac xLV_{ds}-V_t)\\\\\;\;\;=V_{gs}-V_{ds}\;+\frac xLV_{ds}-V_t\end{array} 

Thus the charge q coulombs,induced on this capacitor is

\begin{array}{l}q=CV\\\\\;\;=\frac{W\in dx}D(\begin{array}{l}V_{gs}-V_{ds}\;+\frac xLV_{ds}-V_t)\end{array}\end{array}

The total charge Q induced in the channel is

\begin{array}{l}\\Q=\boldsymbol\int_{\mathbf0}^{\mathbf L}\frac{W\in}D(\begin{array}{l}V_{gs}-V_{ds}\;+\frac xLV_{ds}-V_t)\end{array}dx\\\\\;\;=\frac{\in WL}D\lbrack(\begin{array}{l}V_{gs}-V_t)\end{array}-\frac{V_{ds}}2\rbrack\\\\Now\\\;\;\;\;\;\;\;\;\;\;Q=tI_{ds}\end{array}

Where t is the time in seconds for an electron to move across the channel and

t=\frac{Channel\;length\;(L)}{Electron\;velocity}

μn is the electron velocity per unit electric field (measured in metre2/volt-second) and is called the ‘electron mobility’.

Thus the electron velocity in the channel is

\frac{\mu_nV_{ds}}L


Hence the current Ids in amps is

\begin{array}{l}I_{ds}=\frac Qt\\\\\;\;\;=\frac{\in W\;\mu_n}{LD}\lbrack(\begin{array}{l}V_{gs}-V_t)V_{ds}\end{array}-\frac{V_{ds}^2}2\rbrack\\\\Since\;drain-source\;(V_{ds})\;voltages\;are\;very\;small\;compared\;with\;V_{gs}-V_t\\\\{\boldsymbol I}_{\mathbf d\mathbf s}\boldsymbol=\frac{\boldsymbol\in\mathbf W\mathbf L}{\mathbf D}\boldsymbol\lbrack\boldsymbol(\begin{array}{l}{\mathbf V}_{\mathbf g\mathbf s}\boldsymbol-{\mathbf V}_{\mathbf t}\boldsymbol)\end{array}{\boldsymbol V}_{\mathbf d\mathbf s}\boldsymbol\rbrack\end{array}

Drain current equation of NMOS transistor at triode region

Read: Different operating regions of CMOS inverter

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