VLSI Design UPDATED ACTUAL Exam
Questions and CORRECT Answers
Erste Maxwell - CORRECT ANSWER - ε_0 * E_z(z) = ∫ρ dz
Spannung aus E-Feld - CORRECT ANSWER - φ_z(z) = - ∫ E_z(z) dz
Gate-Bulk-Modes - CORRECT ANSWER - 1. Accumulation: V_GB = 0
The majority carriers accumulate under the gate electrode due to certain 2nd order effects.
2. Flat Band Voltage: V_GB = V_FB
The majority carriers are repelled from the gate. They are distributed evenly in the bulk.
3. Depletion: V_FB < V_GB < V_th
The majority carriers are repelled further by the gate. A depletion region is formed under the
gate.
4. Inversion V_GB > V_th
The voltage gets so large, that minority carriers accumulate under the gate and above the
depletion region. They form a conducting channel.
Drain-Source-Modes (V_GS = V_GB) - CORRECT ANSWER - 1. Cut-Off: V_GS = 0,
V_DS = 1V
The Diode between drain and bulk blocks. The electric field enlarges the depletion region
between drain and bulk. No current flows.
2. Triode Region: V_GS = 1V, V_DS = 0.4V
Inversion is established between gate and bulk and by that a channel of minority carriers has
accumulated under the gate and between drain and source. Current flows proportional to the
voltage.
3. Pinch-off-Point: V_GS = 1V, V_DS = 0.8V
With rising more and more minority carriers accumulate at the source and are repelled from
drain. The Pinch-off-Point is the point at which this gets so severe, that the channel is interrupted
at the drain.
4. Saturation: V_GS = 1V, V_DS = 1V
, The part of the bulk without a channel gets larger and larger. The current has reached it's
maximum and stays there. It's now (nearly) independent of
Triode-Current; Long Channel - CORRECT ANSWER - I_D = μ * ε_ox / t_ox * W/L *
[(V_GS-V_th)*V_DS-m/2*V_DS^2]
V_GS > V_th
V_DS < (V_GS-V_th)/m
Saturation-Current; Long Channel - CORRECT ANSWER - I_D = μ * ε_ox / t_ox * W/L
* 1/(2m) * (V_GS-V_th)^2
V_GS > V_th
V_DS > (V_GS-V_th)/m
Threshold Voltage - CORRECT ANSWER - 1. Flat-Band Voltage
- Unwanted Charges trapped in the oxide
- Unwanted states at the SiO2/Si-Barrier
- Implanted ions to modify threshold voltage
V_FB = ΔΦ - Q_ox/C_ox''
C_ox'' = ε_ox / t_ox
2. Potential required to establish inversion
Needed: Density of minority carriers in the channel = density of majority carriers in thermal
equilibrium (deep in the bulk)
V_B = 2φ_F
φ_F = kT/e*ln(N_A/n_i)
Questions and CORRECT Answers
Erste Maxwell - CORRECT ANSWER - ε_0 * E_z(z) = ∫ρ dz
Spannung aus E-Feld - CORRECT ANSWER - φ_z(z) = - ∫ E_z(z) dz
Gate-Bulk-Modes - CORRECT ANSWER - 1. Accumulation: V_GB = 0
The majority carriers accumulate under the gate electrode due to certain 2nd order effects.
2. Flat Band Voltage: V_GB = V_FB
The majority carriers are repelled from the gate. They are distributed evenly in the bulk.
3. Depletion: V_FB < V_GB < V_th
The majority carriers are repelled further by the gate. A depletion region is formed under the
gate.
4. Inversion V_GB > V_th
The voltage gets so large, that minority carriers accumulate under the gate and above the
depletion region. They form a conducting channel.
Drain-Source-Modes (V_GS = V_GB) - CORRECT ANSWER - 1. Cut-Off: V_GS = 0,
V_DS = 1V
The Diode between drain and bulk blocks. The electric field enlarges the depletion region
between drain and bulk. No current flows.
2. Triode Region: V_GS = 1V, V_DS = 0.4V
Inversion is established between gate and bulk and by that a channel of minority carriers has
accumulated under the gate and between drain and source. Current flows proportional to the
voltage.
3. Pinch-off-Point: V_GS = 1V, V_DS = 0.8V
With rising more and more minority carriers accumulate at the source and are repelled from
drain. The Pinch-off-Point is the point at which this gets so severe, that the channel is interrupted
at the drain.
4. Saturation: V_GS = 1V, V_DS = 1V
, The part of the bulk without a channel gets larger and larger. The current has reached it's
maximum and stays there. It's now (nearly) independent of
Triode-Current; Long Channel - CORRECT ANSWER - I_D = μ * ε_ox / t_ox * W/L *
[(V_GS-V_th)*V_DS-m/2*V_DS^2]
V_GS > V_th
V_DS < (V_GS-V_th)/m
Saturation-Current; Long Channel - CORRECT ANSWER - I_D = μ * ε_ox / t_ox * W/L
* 1/(2m) * (V_GS-V_th)^2
V_GS > V_th
V_DS > (V_GS-V_th)/m
Threshold Voltage - CORRECT ANSWER - 1. Flat-Band Voltage
- Unwanted Charges trapped in the oxide
- Unwanted states at the SiO2/Si-Barrier
- Implanted ions to modify threshold voltage
V_FB = ΔΦ - Q_ox/C_ox''
C_ox'' = ε_ox / t_ox
2. Potential required to establish inversion
Needed: Density of minority carriers in the channel = density of majority carriers in thermal
equilibrium (deep in the bulk)
V_B = 2φ_F
φ_F = kT/e*ln(N_A/n_i)
