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
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%.
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
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: