0
14kviews
Design voltage regulator using IC723 to give Vo= 5V and output current= 2A
1
1.3kviews

It is a Low Voltage High Current circuit.

## Step 1 : Selection of $R_1$ and $R_2$

Let $V_{R2}$ = Regulated DC output = 5V

Current I through $R_1$ & $R_2$ be 1mA $$R_2 = \frac{V_{R2}}{1mA} = \frac{5}{1mA}$$ $$\boxed{R_2 = 5KΩ}$$ $$V_{R1} = V_{REF} - V_{R2}$$ $$V_{R1} = 7.15 - 5$$ $$V_{R1} = 2.15V$$ $$R_1 = \frac{V_{R1}}{1mA} = \frac{2.15}{1mA}$$ $$\boxed{R_1 = 2.15KΩ }$$

## Step 2 : Selection of $R_3$

$$R_3 = R_1 || R_2$$ $$R_3 = 2.1K || 5K$$ $$\boxed{R_3 = 1.5KΩ}$$

## Step 3 : Selection of $R_{SC}$

$$R_{SC} = \frac{V_{SENSE}}{I_{SC}}$$ Assume $V_{SENSE}$ = 0.5V and $I_{SC}$ = 200mA $$R_{SC} = \frac{0.5}{200m}$$ $$\boxed{R_{SC} = 1.5KΩ}$$

## Step 4 : Selection of $Q_3$

$$P_{DT3} = V_{CE3} × I_{C3}$$ $$P_{DT3} = (V_{in} - V_O) × I_{C3}$$ Assume $V_{in}$ = 12V $$P_{DT3} = (12 - 5) × 2$$ $$P_{DT3} = 14W$$ $$β = \frac{I_L}{I_B} = \frac{2A}{150mA}$$ Assume $I_B$ = 150mA

$$β = 13.33$$

under short circuit condition, $V_O$ = 0V $$P_{DT3} = V_{CE3} × I_{SC}$$ $$P_{DT3} = (V_{in} - V_O) × I_{SC}$$ $$P_{DT3} = (12 - 0) × 200mA$$ $$P_{DT3} = 2.4W$$ Select transistor whose $P_{DT3}$ ≥ 2.4W and β ≥ 13.33

## Step 5 : Selection of $R_L$

$$R_L = \frac{V_O}{I_L}$$ $$R_L = \frac{5}{2}$$ $$\boxed{R_L = 2.5Ω}$$

## Step 6 : Designed circuit diagram 