Transconductance MCQ Quiz - Objective Question with Answer for Transconductance - Download Free PDF

• The drain current (I_D) of a MOSFET in the saturation region is given by the equation:
  
  I_D = (1/2) × μ_n × C_ox × (W/L) × (V_GS - V_TH)²,
  
  where:
  • μ_n: Electron mobility
  • C_ox: Gate-oxide capacitance per unit area
  • W: Width of the MOSFET channel
  • L: Length of the MOSFET channel
  • V_GS: Gate-to-source voltage
  • V_TH: Threshold voltage
• All other parameters (μ_n, C_ox, and L) are constant in this problem.
• Therefore, the drain current is directly proportional to both the width of the MOSFET (W) and the square of the overdrive voltage (V_GS - V_TH).

Solution:

Let the parameters of the second MOSFET (MOSFET 2) be:

• Width: W
• Overdrive voltage: (V_GS - V_TH)
• Drain current: I_D2

For the first MOSFET (MOSFET 1), the parameters are:

• Width: 2W (double the width of MOSFET 2)
• Overdrive voltage: 2(V_GS - V_TH) (double the overdrive voltage of MOSFET 2)
• Drain current: I_D1

The drain current for MOSFET 2 is:

I_D2 = (1/2) × μ_n × C_ox × (W/L) × (V_GS - V_TH)²

The drain current for MOSFET 1 is:

I_D1 = (1/2) × μ_n × C_ox × (2W/L) × [2(V_GS - V_TH)]²

Simplifying I_D1:

I_D1 = (1/2) × μ_n × C_ox × (2W/L) × 4(V_GS - V_TH)²

I_D1 = 4 × (1/2) × μ_n × C_ox × (W/L) × (V_GS - V_TH)²

I_D1 = 8 × I_D2

Conclusion:

The ratio of the drain currents of the two MOSFETs is:

I_D1 : I_D2 = 8 : 1

The correct answer is Option 3.

Concept:

The JFET is a voltage-controlled device so no current flows through the gate, then the source current (I_S) is equal to the drain current (I_D) i.e. I_D = I_S.

$I_D=I_{DSS}{(1-\frac{V_{GS}}{V_P})}^2$

Where;

I_D → Drain current

IDSS → Saturation current

V_GS → Gate to source voltage

V_P → Pinch off voltage

$\mathrm{Transconductance}:\frac{\partial I_D}{\partial V_{gs}}=g_m=\frac{-2I_{DSS}}{V_P}(1-\frac{V_{GS}}{V_P})$

For V_GS = 0 transconductance will be maximum

$g_{m0}=\frac{-2I_{DSS}}{V_P}$

Calculation:

Given;

 IDSS = 1 mA

 Vp = –5 V

Then;

$g_{m0}=\frac{-2I_{DSS}}{V_P}=-\frac{2\times 1\times {10}^{-3}}{-5}=0.4\times {10}^{-3}A/V$

Concept:

The drain current in saturation for a JFET is given by:

$I_D\left(sat\right)=I_{DSS}{(1-\frac{V_{GS}}{V_p})}^2$   ---(1)

IDSS = Saturation Drain Current

Vp = Pinch Off Voltage

The transconductance (gm) is defined as the change in ID with respect to a change in the gate to source voltage (VGS), i.e.

$g_m=\frac{\partial I_D}{\partial V_{gs}}$

Taking the differentiation of Equation (1), we get:

$g_m=\frac{2I_{DSS}}{\left|V_p\right|}(1-\frac{V_{gs}}{V_p})$

The maximum value of gm occurs when VGS = 0

$g_m(max)=\frac{2I_{DSS}}{\left|V_p\right|}$

Calculation:

Given:

IDSS = 1 mA, Vp = -8V

We can write:

Maximum transconductance

gm0 = $\frac{2I_{DSS}}{|V_p|}$

= $\frac{2\times 1}{8}=0.25mS$

= 0.00025 S

Explanation:

For a MOSFET operating in the linear region, the current is given by:

$I_D=\frac{W{\mu }_0{C}_{ox}}{2L}\left[2\left(V_{GS}-V_T\right)\left(V_{DS}\right)-{\left(V_{DS}\right)}^2\right]$  ---(1)

W = Width of the Gate

Cox = Oxide Capacitance

μ = Mobility of the carrier

L = Channel Length

Vth = Threshold voltage

The transconductance is defined as the change in drain current for a given change in Gate-to-source voltage, i.e.

$g_m=\frac{\partial I_D}{\partial V_{GS}}$

Differentiation equation (1) with V_GS we get:

$g_m=\frac{\partial I_D}{\partial V_{GS}}=\frac{W{\mu }_0{C}_{ox}}{2L}\left[2V_{DS}\right]$

as g_m ∝ V_DS

g_m ≈ KV_DS

Hence option (2) is the correct answer.

Important Points

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

$I_{D\left(sat\right)}=\frac{W{\mu }_x{C}_{ox}}{2L}{\left(V_{GS}-V_{th}\right)}^2$

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. 

$g_m=\frac{\partial I_D}{\partial V_{GS}}$

$g_m=\frac{W{\mu }_x{C}_{ox}}{L}\left(V_{GS}-V_{th}\right)$

Hence in saturation mode, transconductance does not depend on V_DS

Concept:

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

$g_m=\frac{\partial I_D}{\partial V_{GS}}$    

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

$I_{D\left(sat\right)}=\frac{W{\mu }_x{C}_{ox}}{2L}{\left(V_{GS}-V_{th}\right)}^2$

W = Width of the Gate

C_ox = Oxide Capacitance

μ = Mobility of the carrier

L = Channel Length

V_th = Threshold voltage

Application:

The transconductance in saturation is given by differentiating the above w.r.t. V_GS, i.e.

$g_{m\left(sat.\right)}=\frac{\partial I_{D\left(sat.\right)}}{\partial V_{GS}}=\frac{W{\mu }_x{C}_{ox}}{L}{\left(V_{GS}-V_{th}\right)}^{}$

We observe that the g_m(sat.) of a MOSFET is independent of V_DS and depends on W, μ_n, C_ox, L, V_GS, and V_th.

Explanation:

For a MOSFET operating in the linear region, the current is given by:

$I_D=\frac{W{\mu }_0{C}_{ox}}{2L}\left[2\left(V_{GS}-V_T\right)\left(V_{DS}\right)-{\left(V_{DS}\right)}^2\right]$  ---(1)

W = Width of the Gate

Cox = Oxide Capacitance

μ = Mobility of the carrier

L = Channel Length

Vth = Threshold voltage

The transconductance is defined as the change in drain current for a given change in Gate-to-source voltage, i.e.

$g_m=\frac{\partial I_D}{\partial V_{GS}}$

Differentiation equation (1) with V_GS we get:

$g_m=\frac{\partial I_D}{\partial V_{GS}}=\frac{W{\mu }_0{C}_{ox}}{2L}\left[2V_{DS}\right]$

as g_m ∝ V_DS

g_m ≈ KV_DS

Hence option (2) is the correct answer.

Important Points

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

$I_{D\left(sat\right)}=\frac{W{\mu }_x{C}_{ox}}{2L}{\left(V_{GS}-V_{th}\right)}^2$

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. 

$g_m=\frac{\partial I_D}{\partial V_{GS}}$

$g_m=\frac{W{\mu }_x{C}_{ox}}{L}\left(V_{GS}-V_{th}\right)$

Hence in saturation mode, transconductance does not depend on V_DS

The current equation for a depletion type JFET is given by:

$I_D=I_{DSS}{\left(1-\frac{V_{GS}}{V_P}\right)}^2$   ---(1)

IDSS = Saturation Current of the JFET

VGS = Gate to source voltage applied

VP = Pinch off voltage at which ID = IDSS, which is also the voltage at which the channel ceases to exist. It is also denoted by VGS(off).

The transconductance (gm) is defined as the change in ID with respect to a change in the gate to source voltage (VGS), i.e.

$g_m=\frac{\partial I_D}{\partial V_{gs}}$

Taking the differentiation of Equation (1), we get:

$g_m=\frac{2I_{DSS}}{\left|V_p\right|}(1-\frac{V_{gs}}{V_p})$

The maximum value of gm occurs when Vgs = 0

With $g_{m0}=\frac{2I_{DSS}}{\left|V_p\right|}$

$g_m=g_{m0}(1-\frac{V_{gs}}{V_p})$

g_m0 is the maximum transconductance at V_GS = 0

The characteristic graph for a change in V_GS is as shown:

Concept:

The drain current in saturation for a JFET is given by:

$I_D\left(sat\right)=I_{DSS}{(1-\frac{V_{GS}}{V_p})}^2$   ---(1)

IDSS = Saturation Drain Current

Vp = Pinch Off Voltage

The transconductance (gm) is defined as the change in ID with respect to a change in the gate to source voltage (VGS), i.e.

$g_m=\frac{\partial I_D}{\partial V_{gs}}$

Taking the differentiation of Equation (1), we get:

$g_m=\frac{2I_{DSS}}{\left|V_p\right|}(1-\frac{V_{gs}}{V_p})$

The maximum value of gm occurs when VGS = 0

$g_m(max)=\frac{2I_{DSS}}{\left|V_p\right|}$

Calculation:

Given:

IDSS = 1 mA, Vp = -8V

We can write:

Maximum transconductance

gm0 = $\frac{2I_{DSS}}{|V_p|}$

= $\frac{2\times 1}{8}=0.25mS$

= 0.00025 S

Concept:

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

$I_D=\frac{1}{2}{\mu }_n{C}_{ox}\left(\frac{W}{L}\right){(V_{Gs}-V_T)}^2(1+\lambda V_{DS})$

W = Width of the Gate

Cox = Oxide Capacitance

μ = Mobility of the carrier

L = Channel Length

Vth = Threshold voltage

Calculation:

The drain conductance (g_d) is calculated as the rate of change of drain current with respect to the Drain to source voltage, i.e.

$g_d=\frac{\partial {\mathrm{I}}_{\mathrm{D}}}{\partial {\mathrm{V}}_{\mathrm{D}\mathrm{S}}}$

$g_d=\frac{1}{2}{\mu }_n{C}_{ox}\left(\frac{W}{L}\right){(V_{Gs}-V_T)}^2\lambda$

Putting on the respective values, we get:

$g_d=\frac{1}{2}\times 70\times {10}^{-6}\left(4\right){\left(1.8-0.3\right)}^2\times 0.09$

g_d = 28.35 μ Seimens

Concept:

Whenever gate and drain are shorted of MOSFET, we can replace directly MOSFET by resistance which is equal to

Analysis:

Since the gate and drain of MOSFET 1 are short-circuited.

It will work like a resistor.

→ drawing AC - equivalent model of MOSFET - 2

$V_{g{s}_2}=V_{in}$

$V_0=-g_{m_2}{V}_{g{s}_2}\left\{r_{0_2}||r_{0_1}||\frac{1}{g_{m_1}}\right\}$

$\frac{V_0}{V_{in}}=-g_{m_2}\left\{r_{0_2}||r_{0_1}||\frac{1}{g_{m_1}}\right\}$

Important Points

When gate and source of MOSFET are short-circuited:

Given: |V_tp| = V_tn = 1 V

r_ds = 6 MΩ

For NMOS:

μ_nC_ox = 60 μA/V²

${\left(\frac{W}{L}\right)}_{nmos}=5$

For PMOS:

μ_pC_ox = 30 μA/V²

${\left(\frac{w}{L}\right)}_{pmos}=10$

Current through PMOS M1 and PMOS M2 will be same because they are connected in series.

Both the transistors will be in saturation because the drain is connected to the gate.

$V_{SG\left(M1\right)}=V_{SG\left(M2\right)}=\frac{V_{DD}}{2}$

$V_{SG\left(M1\right)}=\frac{4}{2}=2V$

For MOS M3 and M1, the source voltages are the same and the gates are shorted, which gives:

V_SG(M3) = V_SG(M1) = 2 Volts

$I_2=\frac{1}{2}{\mu }_p{C}_{ox}{\left(\frac{W}{L}\right)}_p{\left[V_{SG\left(M3\right)}-\left|V_{tp}\right|\right]}^2$

$I_2=\frac{1}{2}\times 30\times {10}^{-3}\times 10{\left(2-1\right)}^2$

I₂ = 0.15 mA

I₂ = 150 uA

The same current I₂ will pass through M4

The transconductance of M4 can be given as:

$g_{m4}=\frac{\partial I_{DS}}{\partial V_{GS}}={K_N}^{\prime }\left(\frac{W}{L}\right)\left(V_{GS}-V_{tn}\right)\dots 1)$

where K_n’ = u_nC_ox = 60 μA/V²

For NMOS (M4)

$I_{DS4}=\frac{1}{2}{u}_n{C}_{ox}\left(\frac{W}{L}\right){\left(V_{gs}-V_{tn}\right)}^2$

$V_{GS}-V_{tn}=\sqrt{\frac{2I_{DS4}}{{K_N}^{\prime }\left(\frac{W}{L}\right)}}$

Substituting this in (1), we get:

$g_{m4}=\sqrt{2I_{DS4}}\sqrt{{K_N}^{\prime }\left(\frac{W}{L}\right)}$

I_DS4 = I₂ = 150 uA

$g_{m4}=\sqrt{2\times 150}\times \sqrt{60\times 5}$

g_m4 = 300 uA/V

Drawing the small-signal model of m4.

Note: MOS – 3 is replaced by its output drain resistance r_d3 = 6mΩ

$\frac{V_o}{V_{in}}=\frac{\left(-g_m{V}_{gs}\right)(r_{d4}||r{d}_3)}{V_{gs}}$

= -g_m(rd₄ || rd₃)

= -300 × 10^-6 (6 × 10⁶ || 6 × 10⁶)

= -300 × 10^-6(3 × 10⁶)

= -900

magnitude wise

$\left|\frac{V_o}{V_{in}}\right|=900$

Concept:

For VDS < VGS – VT, n-channel enhancement type MOSFET will operate in triode region:

$I_D=\frac{\mathrm{W}{\mu }_{\mathrm{n}}{C}_{ox}}{2L}\left[2\left(V_{GS}-V_T\right)V_{DS}-V_{DS}^2\right]$

For V_DS ≥ V_GS – V_T, the MOSFET will be in saturation with current given by:

$I_D=\frac{\mathrm{W}{\mu }_{\mathrm{n}}{C}_{ox}}{2L}\left[{\left(V_{GS}-V_T\right)}^2\right]$

Also, the transconductance is defined as:

$g_m=\frac{\partial I_D}{\partial V_{GS}}$

Calculation:

Given,

V_DS = 0.1 V

V_GS = 0.7 V

V_th = 0.3 V

V_GS – V_th = 0.4

V_DS = 0.1 < 0.4;

So, the MOSFET is in the triode region:

$g_m=\frac{d{i}_D}{d{V}_{GS}}=\frac{\mathrm{W}\mu C_{ox}}{2L}.2V_{DS}=\frac{\mathrm{W}\mu C_{ox}{V}_{DS}}{L}$

Putting values, we get,

g_m = 50 × 100μ × 0.1

Concept:

The drain current when the MOSFET is in saturation is given by:

$I_D=\frac{1}{2}{\mu }_n{C}_{ox}\left(\frac{W}{L}\right)\times {\left(V_{GS}-V_T\right)}^2$

VT  = Threshold

The transconductance (gm) is defined as the change in the output current with a change in the Gate to source voltage, i.e.

$g_m=\frac{\partial I_D}{\partial v_{GS}}$

$g_m={\mu }_n{C}_{ox}\left(\frac{W}{L}\right)\left(V_{GS}-V_T\right)$

The output resistance (r0) is given by:

$r_o=\frac{V_A}{I_D}$

$V_A=\frac{1}{\lambda }$

λ = Channel length parameter

Calculation:

Given: ID = 1 mA

Channel length modulation coefficient = 0.05 V-1

$V_A=\frac{1}{\lambda }=20$               

$r_o=\frac{V_A}{I_D}=\frac{20}{1mA}=20k\mathrm{\Omega }$

Hence the value output impedance is 20 kohm. 

Concept:

For a field-effect transistor (FET) under certain operating conditions, the resistance of the drain-source channel is a function of the gate-source voltage alone and the JFET will behave as an almost pure ohmic resistor

For a MOSFET in the saturation region:

VGS > Vth and

VDS > VGS – Vth

The current equation for a MOSFET in the saturation region is given by:

$I_D=\mu C_{ox}\frac{W}{2L}\left\{{(V_{GS}-V_{th})}^2\right\}$

Taking the derivative of the above w.r.t. V_DS, we get:

$\frac{\partial I_D}{\partial V_{DS}}=\mu C_{ox}\left(\frac{W}{L}\right)\left(V_{GS}-V_{TH}\right)$

Voltage-controlled Resistor (r_DS) is defined as:

$r_{DS}=\frac{V_{DS}}{I_{DS}}=\frac{1}{{\mu }_n{C}_{ox}\left(\frac{W}{L}\right)\left(V_{GS}-V_{TH}\right)}$

Calculation:

Given:

V_GS = 2.0 V,

V_TH = -0.5 V,

$\frac{W}{L}=100$

C_ox = 10^-8 f/cm²

μ_n = 800 cm²/V-s,

V_DS = 5 V

Putting on the respective values in Equation (1), we get:

$r_{DS}=\frac{1}{800\times {10}^{-8}\times 100\times \left(2-\left(-0.5\right)\right)}$

$r_{DS}=500\mathrm{\Omega }$

Concept:

Condition of Operation of NMOS in linear region:

V_DS < [V_DS - V_T]

‘OR’

V_DS < V_GS

And the current equation is:

 I_D = K_n’ [V_GS - V_T] V_DS

Calculation:

Given:

g_m = 0.5 μA/V for V_ds = 50 mV, V_GS = 2V,

g_d = 8 μA/V for V_GS = 2V, V_DS = 0 V

Since,

V_GS > V_DS, MOSFET is in linear operation,

So, I_D = K_n’ (V_GS - V_T) V_DS

$\frac{\partial I_D}{\partial V_{GS}}=K_n^{\prime }{V}_{DS}$ 

$g_m=K_n^{\prime }{V}_{DS}$                 

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

So, .5 × 10^-6 = K_n’ [50 × 10^-3]

⇒ K_n’ = 10^-5

Now,

$\frac{\partial I_D}{\partial V_{DS}}=K_n^{\prime }\left[V_{GS}-V_T\right]$ 

$g_d=K_n^{\prime }\left[V_{GS}-V_T\right]$ 

$\therefore g_d=\frac{\partial I_D}{\partial V_{DS}}$  (Given)

So, 8 × 10^-6 = 10^-5 [2 - V_T]

V_T = 1.2 V