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ECE5590 AN
Low Dropout Regulator(LDO)
-Aadit Modi(ID#16037399)
-Altaf Hirani (ID#12197304)
LDO
• Linear Voltage DC regulators.
• Regulation maintained with small differences.
• Output current in range of 50-100mA.
• Pass transistor, error amplifier and voltage
reference.
• Low quiescent current.
•Design a low dropout voltage regulator
to provide an output voltage of 3.3V.
Goals:
For the calculations we assume the following
constants:
• - Pass transistor current = 1mA
• - Vout = 3.3V
• - Dropout voltage
• - VDD=5V
Block Diagram
• Pass transistor & error amplifier.
CALCULATIONS:
Efficiency calculation
Iq (quiescent current) = 112 uA
Io (output current) = 1.39 mA
Vo (output voltage) = 3.37 V
Vi (input voltage) = 5 V
Eff. = Io*Vo/(Io + Iq)*Vi x 100
Using the above equation yields and
efficiency of about 61.1%.
- 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 0.6 100.05 0.6
M2 100 0.6 100.05 0.6
M3 50 0.6 49.95 0.6
M4 50 0.6 49.95 0.6
M5 20 0.6 19.95 0.6
M6 250 0.6 250 0.6
TRANSISTOR SIZE TABLE
Final Schematic
Typical LDO Circuit
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
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
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
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
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
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:
Layout
PRE-LAYOUT DC INPUT TEST
Post-layout Line Regulation (Changing
input voltage)
Post-layout Line Regulation
(Changing input voltage)
THANK YOU.

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Low dropout regulator(ldo)

  • 1. ECE5590 AN Low Dropout Regulator(LDO) -Aadit Modi(ID#16037399) -Altaf Hirani (ID#12197304)
  • 2. LDO • Linear Voltage DC regulators. • Regulation maintained with small differences. • Output current in range of 50-100mA. • Pass transistor, error amplifier and voltage reference. • Low quiescent current.
  • 3. •Design a low dropout voltage regulator to provide an output voltage of 3.3V. Goals:
  • 4. For the calculations we assume the following constants: • - Pass transistor current = 1mA • - Vout = 3.3V • - Dropout voltage • - VDD=5V
  • 5. Block Diagram • Pass transistor & error amplifier.
  • 6. CALCULATIONS: Efficiency calculation Iq (quiescent current) = 112 uA Io (output current) = 1.39 mA Vo (output voltage) = 3.37 V Vi (input voltage) = 5 V Eff. = Io*Vo/(Io + Iq)*Vi x 100 Using the above equation yields and efficiency of about 61.1%.
  • 7. - 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 0.6 100.05 0.6 M2 100 0.6 100.05 0.6 M3 50 0.6 49.95 0.6 M4 50 0.6 49.95 0.6 M5 20 0.6 19.95 0.6 M6 250 0.6 250 0.6 TRANSISTOR SIZE TABLE
  • 10. 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
  • 11. 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
  • 12. 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
  • 13. 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
  • 14. 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
  • 15. 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:
  • 18. Post-layout Line Regulation (Changing input voltage)