Math Problem Statement

For two matched MOS transistors with W/L=12, k'n=390µA/V^2, and V_m=0.5V, design the circuit to provide Io=90µA. Given that R=9Ω, find V_DD and the minimum value Vo can have while Q2 still operates in saturation.

Solution

The problem involves two matched MOS transistors with the following parameters:

  • WL=12\frac{W}{L} = 12
  • kn=390μA/V2k_n' = 390 \mu A/V^2
  • Vm=0.5VV_m = 0.5 V
  • Desired output current: IO=90μAI_O = 90 \mu A
  • R=9ΩR = 9 \, \Omega

We are tasked with:

  1. Designing the circuit to provide IO=90μAI_O = 90 \mu A.
  2. Calculating VDDV_{DD}.
  3. Finding the minimum value of VOV_O while transistor Q2Q_2 is still in saturation.

1. Design the Circuit to Provide IO=90μAI_O = 90 \mu A

For a MOS transistor in saturation, the drain current IDI_D is given by: ID=kn(W/L)2(VGSVT)2I_D = \frac{k_n' (W/L)}{2} (V_{GS} - V_T)^2 Where:

  • ID=IO=90μAI_D = I_O = 90 \mu A
  • kn=390μA/V2k_n' = 390 \mu A/V^2
  • WL=12\frac{W}{L} = 12
  • VmV_m is often treated as VT=0.5VV_T = 0.5V.

Substituting the known values, we have: 90μA=390μA/V2×122(VGS0.5)290 \mu A = \frac{390 \mu A/V^2 \times 12}{2} (V_{GS} - 0.5)^2

Simplifying: 90μA=2340μA/V2(VGS0.5)290 \mu A = 2340 \mu A/V^2 (V_{GS} - 0.5)^2 90μA2340μA/V2=(VGS0.5)2\frac{90 \mu A}{2340 \mu A/V^2} = (V_{GS} - 0.5)^2 0.03846=(VGS0.5)20.03846 = (V_{GS} - 0.5)^2 VGS0.5=0.03846V_{GS} - 0.5 = \sqrt{0.03846} VGS0.5=0.1961V_{GS} - 0.5 = 0.1961 VGS=0.6961VV_{GS} = 0.6961 V

Thus, VGS=0.6961VV_{GS} = 0.6961 V.

2. Calculate VDDV_{DD}

The voltage across RR (denoted as VRV_R) is given by Ohm's Law: VR=IO×R=90μA×9Ω=0.81VV_R = I_O \times R = 90 \mu A \times 9 \Omega = 0.81 V

The total supply voltage VDDV_{DD} is the sum of VGSV_{GS} and VRV_R: VDD=VR+VGS=0.81V+0.6961V=1.5061VV_{DD} = V_R + V_{GS} = 0.81 V + 0.6961 V = 1.5061 V

3. Minimum Value of VOV_O for Q2Q_2 in Saturation

For transistor Q2Q_2 to remain in saturation, the condition is: VDS2VGS2VTV_{DS2} \geq V_{GS2} - V_T Since VGS2=VGS1V_{GS2} = V_{GS1} (because the transistors are matched and in the same configuration): VDS2=VOVGS2VT=0.6961V0.5V=0.1961VV_{DS2} = V_O \geq V_{GS2} - V_T = 0.6961 V - 0.5 V = 0.1961 V

Thus, the minimum value of VOV_O for Q2Q_2 to remain in saturation is VO=0.1961VV_O = 0.1961 V.

Summary of Answers:

  1. Circuit design provides IO=90μAI_O = 90 \mu A.
  2. VDD=1.5061VV_{DD} = 1.5061 V.
  3. The minimum value of VOV_O for Q2Q_2 to remain in saturation is 0.1961V0.1961 V.

Do you want more details or clarifications?

Related Questions:

  1. How does the W/LW/L ratio affect the operation of MOS transistors in this circuit?
  2. Can you derive the expression for IDI_D in the linear region of operation?
  3. What happens if VDSV_{DS} is less than VGSVTV_{GS} - V_T in transistor Q2Q_2?
  4. How would increasing RR affect the required VDDV_{DD}?
  5. How can you ensure that both transistors remain in saturation under different load conditions?

Tip:

When designing MOSFET circuits, always verify the operating region by comparing VDSV_{DS} with VGSVTV_{GS} - V_T.

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Math Problem Analysis

Mathematical Concepts

MOSFET circuit design
Saturation region of MOSFET
Ohm's law
Voltage-current relationships in transistors

Formulas

I_D = (k'n (W/L)/2) (V_GS - V_T)^2
V_R = I_O * R
V_DD = V_R + V_GS
V_O ≥ V_GS2 - V_T

Theorems

MOSFET saturation condition (V_DS ≥ V_GS - V_T)

Suitable Grade Level

Undergraduate Electrical Engineering