LDO project
18
Low Voltage Dropout Regulator
Transcript of LDO project
Low Voltage Dropout Regulator
Goal:Design a voltage regulator to provide an output voltage of 3.3V
For the calculations we assume the following constants:
- Pass transistor current = 1ma
- Vout = 3.3V
- Dropout voltage =
- VDD=5V
-
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(V
GS3-V
THN)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(V
GS3-V
THN)2 = ½ * 50 * 10-6 A/V2 *
3µm/0.6µm * (3.0V – 0.617)2 = 0.7mA
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)/(K
2(W/L)
0] – V
THpp
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
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)/(K
1(V
GS5-V
THN)2) = (2 * 1.2µA *
0.6µm)/(50µA/V2* (0.7V – 0.617V)2) = 4µm
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)/(K
1(V
GS4-V
THN)2) = (2 * 1.2µA *
0.6µm)/(50µA/V2* (0.75V – 0.617V)2) = 1.60µm
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] + V
THp = √(2 *
1.2µA)/(25µA/V2* (20/2.4)) + 0.915V = 0.107V + 0.915V = 1V
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] - V
THN =
5V - √(2 * 1.2µA)/(50µA/V2* (1.5/8.55)) - 0.617V = 3.85V
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
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
Final Schematic
- Test Schematic- Test SchematicTest Schematic
- Pre-Layout Simulation
- Pre-Layout Simulation
PRE-LAYOUT DC INPUT TEST
- Pre-Layout Simulation- Pre-Layout Simulation
PRE-LAYOUT PHASE AND GAIN
LDO LAYOUT
- Post-Layout Simulation
POST LAYOUT DC FIXED INPUT
- Post-Layout Simulation
POST LAYOUT GAIN AND PHASE
