1. Low Voltage Dropout Regulator

2. Goal:
Design a voltage regulator to
provide an output voltage of 3.3V

3. For the calculations we assume the
following constants:
- Pass transistor current = 1ma
- Vout = 3.3V
- Dropout voltage =
- VDD=5V
-

4. Calculations:
- Calculation of a range of Vbias1
1. To find Ibias1:
From the desired a photodiode range, the minimum
value of Ibias1:
VGS3
=Vphmin
Ibias1 = ½ K1(W/L)3
(VGS3
-VTHN
)2
= ½ * 50 * 10-6 A/V2
*
3µm/0.6µm * (0.8V – 0.617)2
= 4.186µA =4µA
The maximum value of Ibias1:
Ibias1 = ½ K1(W/L)3
(VGS3
-VTHN
)2
= ½ * 50 * 10-6 A/V2
*
3µm/0.6µm * (3.0V – 0.617)2
= 0.7mA

5. Calculations:
- Calculation of a range of Vbias1
2. To find Vbias1:
Next we find the value of Vbias1 given by
Vbias1
= VDD
– VGS0
= VDD
- √[(2Ibias1)/(K2
(W/L)0
] –
VTHp
Vbias1
= VDD
– VGS0
= VDD
- √[
(2Ibias1)/(K2
(W/L)0
] – VTHp
p
The maximum value of Vbias1:
Vbias1(max) = 5V - √[(2*4µA)/(19.1µA/V2*
20µm/0.6µm)] – 0.915V =1.026 = 4V
The minimum value of Vbias1:
Vbias1(min) = VDD
– VGS0
= 5V- √[(2*0.7*10-3
)/25*
10-6
/V2 * 20µm/0.6µm) – 0.915V = 2.8V

6. Calculations:
- Calculation of sizes of the transistors M5, M4
1. To determine W5
From requirement to keep M5 in saturation
region:
VTH
≤VGS5
= Vbias1(min) + VTHp
– Vph
(max) =
2.8V +0.9V – 3.0V = 0.7V
W5 = (2InL5
)/(K1
(VGS5
-VTHN
)2
) = (2 * 1.2µA *
0.6µm)/(50µA/V2
* (0.7V – 0.617V)2
) = 4µm

7. Calculations:
- Calculation of sizes of the transistors M5, M4
2. To determine W4
VDS4
≥VGS4
– VTHN
VDS4
= Vph
(min) = 0.8V
Assumed VGS4
= 0.75V
W4 = (2InL4
)/(K1
(VGS4
-VTHN
)2
) = (2 * 1.2µA *
0.6µm)/(50µA/V2
* (0.75V – 0.617V)2
) = 1.60µm

8. Calculations:
- Calculation of the gain for the current mirror transistors
M1, M2, M7
1. To find VGS
for M1, M2, M7
VGS1
= VDS1
= VGS2
= VGS1
= √[(2Iout)/(K2
(W/L)2,7
] + VTHp
= √(2 *
1.2µA)/(25µA/V2
* (20/2.4)) + 0.915V = 0.107V + 0.915V = 1V

9. Calculations:
- Calculation of the gain for the current mirror transistors
M1, M2, M7
2. To find VDS
for current mirror:
Next we find VDS2
and VDS7
(which are the same in value)
VDS2,7
= VDD
– VDS6
= VDD
- √[(2Iout)/(K1
(W/L)6
] - VTHN
=
5V - √(2 * 1.2µA)/(50µA/V2
* (1.5/8.55)) - 0.617V = 3.85V

10. Calculations:
- Calculation of the gain for the current mirror transistors
M1, M2, M7
3. To determine W1:
Finally, we calculate the size of transistor M1. It's required that Iin =
Iout. Consequently, the current conveyor ought to have I1 = I2,7.
Assuming L1= L2,7:
W1/L1* (1 + ƛpDS2,7) = W2,7/L2,7(1 + ƛpDS2,7)
W1 = 2(1 + ƛpDS2,7)/(1 + ƛpDS1)
W1 = (20µm*(1+0.2*3.85V)/(1+0.2*1V) = 29.5µm

11. Summary of Transistor Sizes:
- Summary of calculated transistor sizes vs the
transistor simulation sizes
TransistTor Calculated Size Actual Size Used
Width(µm) Length(µm) Width(µm) Length(µm)
M1 100 0.6 19.55 0.6
M2 100 0.6 21.3 2.4
M3 20 0.6 19.55 0.6
M4 20 0.6 3 0.6
M5 300 0.6 3 1.5

12. Final Schematic

13. - Test Schematic
- Test Schematic
Test Schematic

14. - Pre-Layout Simulation
- Pre-Layout Simulation
PRE-LAYOUT DC INPUT TEST

15. - Pre-Layout Simulation- Pre-Layout Simulation
PRE-LAYOUT PHASE AND GAIN

16. LDO LAYOUT

17. - Post-Layout Simulation
POST LAYOUT DC FIXED INPUT

18. - Post-Layout Simulation
POST LAYOUT GAIN AND PHASE