0
389views
Module 1 - Unit 2
1 Answer
0
2views
  • MOS Inverter Static Characterstics

enter image description here

Scaling

Full Scaling

(Constant Field) (Constant Voltage)
Mag of field is kept constant while dimension are scaled by factor of S. Also, Potentials are propt Scaling of voltage may not be very practical in many lases
Scaled down by same scaling factor dimensions are reduced by factor of S power supply voltage & terminal voltage unchanged
Poission equation be increased by S in order to maintain field conditions Doping conc by $S^2$ to preserve charge field relation
$L^{'} = L/S$ $L^{'} = L/S$
$W^{'} = W/S$ $W^{'} = W/S$
$C_{OX}^{'} = S.C_{OX}$ $C_{OX}^{'} = S.C_{OX}$
$t_{OX}^{'} = t_{OX}/S$ $t_{OX}^{'} = t_{OX}/S$
$X_{j}^{'} = X_{j}/S$ $X_{j}^{'} = X_{j}/S$
$V_{DD}^{'} = V_{DD}/S$ Unchanged
$V_{TO}^{'} = V_{TO}/S$ $V_{TO}^{'} = V_{TO}/S$
$N_{A}^{'} = S.N_A$ $S^2.N_A$
$N_{D}^{'} = S.N_D$ $S^2N_A$
$I_{D}^{'}(lin) = I_D(lin)/S$ SI
$P^{'} = P/S^2$ SP
Power dissipation is low High power dissipation

1.Scaling

- (Constant Field) (Constant Voltage)
- Mag of field is kept constant while dimension are scaled by factor of S. Also, Potentials are propt Scaling of voltage may not be very practical in many lases
- Scaled down by same scaling factor dimensions are reduced by factor of S power supply voltage & terminal voltage unchanged
- Poission equation be increased by S in order to maintain field conditions Doping conc by $S^2$ to preserve charge field relation
$L^{'}$ $L/S$ $L/S$
$W^{'}$ $W/S$ $W/S$
$C_{OX}^{'}$ $S.C_{OX}$ $S.C_{OX}$
$t_{OX}^{'}$ $t_{OX}/S$ $t_{OX}/S$
$X_{j}^{'}$ $X_{j}/S$ $X_{j}/S$
$V_{DD}^{'}$ $V_{DD}/S$ Unchanged
$V_{TO}^{'}$ $V_{TO}/S$ $V_{TO}/S$
$N_{A}^{'}$ $S.N_A$ $S^2.N_A$
$N_{D}^{'}$ $S.N_D$ $S^2N_A$
$I_{D}^{'}(lin)$ $I_D(lin)/S$ SI
$P^{'}$ $P/S^2$ SP

2. Capacitance

enter image description here

  1. Oxide related capacitances:

a) Overlap claps

$C_{GD}(overlap) = C_{OX}w.L_D$

$C_{GS}(overlap) = C_{OX}w.L_D$

b) Gate to channel

  • $C_{gs}, C_{gb}, C_{gh}$

  • cutoff made: $C_{gb} = C_{OX}wL$

$C_{gs} = C_{gd} = 0$

  • Linear mode: $C_{gb} = 0$

$C_{gs} = C_{gd} = 1/2C_{OX}wL$

  • Saturation mode : $C_{gb} = 0$

$C_{gd} = 0$

$C_{gs} = 2/3C_{OX}wL$

Total Cap C L S
$C_{gb}(total)$ $C_{OX}wL$ 0 0
$C_{gd}(total)$ 0 + $C_{OX}wL_D$ 1/2$C_{OX}wL + C_{OX}wL_D$ $C_{OX}wL_D$
$C_{gs}(total)$ $C_{OX}wL_D$ 1/2$C_{OX}wL + C_{OX}wL_D$ $C_{OX}wL_D + 2/3C_{OX}wL$

enter image description here

enter image description here

2. Junction Related Capacitor

$C_{sb}$ and $C_{db}$ -> Voltage dependent

enter image description here

Junction Area Type
1 W.$X_j$ $P/n^{+}$
2 - $n^{+}/P^{+}$
3 - $n^{+}/P^{+}$
4 - $n^{+}/P^{+}$
5 - $n^{+}/P$

$C_{jo} = (\frac{E_{Si}}{2}q(\frac{N_A - N_D}{N_A + N_D})\frac{1}{\phi_o})^{1/2}$

$\phi_o = \frac{KT}{q} ln(\frac{N_A.N_D}{n_i^2})$

3. Channel length Modulation :

enter image description here

$I_D(sat) = \frac{Kn}{2}(V_{GS} - V_{TO})^2 (1 + \lambda V_{DS})$

$L^{'} = L - \Lambda L$ ; $Kn = w/L \mu n C_{OX}$

4. Shrt channel Effect

a) Hot Electron Effect :

  • Hot Elctron is a condition when e gains high kinetic energy and starts penetrating into oxide layer.

  • Normally, In thermal equillibrium, e & holes absorbs and emit photons and hence are stable since avg K.E is 0

  • Now due to impact ionization, e holes paires are generated and process of avalenche breakdown begins.

  • These e tunnel through or are injected into the oxide and cause damage to IC.

These is hot Electron effect

enter image description here

b) Velocity Saturation :

  • One of the short-channel effect.

  • The electric drift velocity '$V_D$' is proportional to electric field.

  • The drift velocity tends to saturate at high channel electric field

$I_D(sat) = \frac{kn}{2}(V_{GS} - V_T)^2$

OR

$I_D(lin) = \frac{kn}{2}[2(V_{GS} - V_T)^2V_{DS} - V_{DS}^2]$

Please log in to add an answer.