Analog Circuits Test 3

Concept:

Transconductance indicates the amount of control the gate has on the drain current.

Mathematically, the transconductance (gm) is defined as:

\( g_m=\frac{\partial I_{D}}{\partial V_{GS}}\)

It is given the name transconductance because it gives the relationship between the input voltage and the output current.

Analysis:

The current for a MOSFET in saturation is given by:

\({I_D} = K_n{\left( {{V_{GS}} - {V_{th}}} \right)^2}\)

VGS = Gate to source voltage

Vth = Threshold Voltage

Taking the square root of the above equation, we get:

\(√{I_D} = √{K_n}{\left( {{V_{GS}} - {V_{th}}} \right)}\)   ---(1)

Now, the transconductance is calculated as:

\( g_m=\frac{\partial I_{D}}{\partial V_{GS}}=2K_n{\left( {{V_{GS}} - {V_{th}}} \right)}\)   ---(2)

Evaluating (V_GS - V_th) from Equation (1) and putting it in Equation (2), we get:

\( g_m=2K_n.\frac{√{I_D}}{√{K_n}}\)

\(g_m=2√{K_n.I_D}\)

So \(g_m\propto\sqrt{I_D}\)

The transconductance of a MOSFET in saturation is directly proportional to the square root of I_D.

Concept:

For a MOSFET in saturation, the current is given by:

\({I_{D\left( {sat} \right)}} = \frac{{W{μ _x}{C_{ox}}}}{{2L}}{\left( {{V_{GS}} - {V_{th}}} \right)^2}\)

W = Width of the Gate

Cox = Oxide Capacitance

μ = Mobility of the carrier

L = Channel Length

Vth = Threshold voltage

The transconductance of a MOSFET is defined as the change in drain current(ID) with respect to the corresponding change in gate voltage (VGS), i.e. 

\({g_m} = \frac{{\partial {I_D}}}{{\partial {V_{GS}}}}\)

\(g_m = \frac{{W{μ _x}{C_{ox}}}}{{L}}{\left( {{V_{GS}} - {V_{th}}} \right)}\)

Application:

Given:

\(g_{m1} = (\mu_x C_{ox})(\frac{{W}}{{L}})_1{\left( {{V_{GS}} - {V_{th}}} \right)}_1\)

Since the transistor current for both the cases is the same:

I_D(sat)1 = I_D(sat)2

\(\frac{{μ _x}{C_{ox}}}{2}(\frac{{W}}{{L}})_1{\left( {{V_{GS}} - {V_{th}}} \right)_1^2}=\frac{{μ _x}{C_{ox}}}{2}(\frac{{W}}{{L}})_2{\left( {{V_{GS}} - {V_{th}}} \right)_2^2}\)

Given:

\((\frac{W}{L})_2=2(\frac{W}{L})_1\)   ---(1)

\(\frac{{μ _x}{C_{ox}}}{2}(\frac{{W}}{{L}})_1{\left( {{V_{GS}} - {V_{th}}} \right)_1^2}=\frac{{μ _x}{C_{ox}}}{2}(\frac{{2W}}{{L}})_1{\left( {{V_{GS}} - {V_{th}}} \right)_2^2}\)

\((V_{GS}-V_{th})_2=\frac{(V_{GS}-V_{th})_1}{√2}\)

Now, the new transconductance will be:

\(g_{m2} = (\mu_x C_{ox})(\frac{{W}}{{L}})_2{\left( {{V_{GS}} - {V_{th}}} \right)}_2\)

\(g_{m2} = (\mu_x C_{ox})(\frac{2{W}}{{L}})_1\frac{{\left( {{V_{GS}} - {V_{th}}} \right)}_1}{√2}\)   ---(2)

Using Equation (1) and (2), we can write:

\(g_{m2} = \sqrt2(\mu_x C_{ox})(\frac{{W}}{{L}})_1{\left( {{V_{GS}} - {V_{th}}} \right)}_1\)

∴ g_m2 = √2 g_m1

Concept:

An n-channel MOSFET biased in the saturation region will result in a current, given by:

\({I_D} = K_n{\left( {{V_{GS}} - {V_{th}}} \right)^2}\)

Where K_n is the conduction parameter defined as:

\(K_n=\frac{W\mu _nC_{0x}}{2L}\)

And the transconductance is defined as:

\({g_m} = \frac{{\partial {I_D}}}{{\partial {V_{GS}}}}\)

\( {g_m} = 2K_n\left( {{V_{GS}} - {V_{th}}} \right)\)

Also, the unity-gain bandwidth of MOSFET is given by:

\({f_T} = \frac{{{g_m}}}{{2\pi \left( {{C_{gd}} + {C_{gs}}} \right)}}\)

Application:

The small-signal Transconductance will be:

g_m = 2K_n(V_GSQ - V_TN)

= 2(0.25) (3 - 1) mA/v

= 1 mA/v

The unity-gain bandwidth of MOSFET will now be:

\({f_T} = \frac{{{g_m}}}{{2\pi \left( {{C_{gd}} + {C_{gs}}} \right)}} = \frac{{1 \times {{10}^{ - 3}}}}{{2\pi \left( {0.03 + 0.3} \right) \times {{10}^{ - 12}}}}\)

f_T = 4.823 × 10⁸ Hz

f_T = 482.3 MHz