# ECET210 Wk 7 Homework

This is a Regulated DC Power Supply design problem. The students must choose the values of the following parameters for the design:

VOut = 6.0 V or 10.0 V or 12.0 V

RL = ?? Ω and R1 = ?? Ω

Wattage ratings of R1 and RL to be determined

IL = 20 mA; IZD = 20 mA; IR1 = ?? mA

Voltage VA_DC = VB + 3 V

Turns ratio, n = ??

Pick an output voltage number from (a)

Choose the appropriate Zener Diode from the data sheet given below.

Determine the Ω values and the wattage ratings of R1 and RL.

Determine the peak value of the transformer secondary voltage VS.

Determine the transformer turns ratio n.

**Assume the following:**

Rectifier diode drop = 0.7 V for each. The DC Voltage (after smoothing by the capacitor) is equal to the peak value of the peak value of the full wave rectified pulses.

The Zener voltages are the Nominal Values

Answers

a) we choose $Vout = 12 V$

The nominal regulated voltage of the zener diode is 12 V. From the attached table the diode is 1N5242B.

b) from $Il =20 mA$ it results $RL = Vout/IL =12/0.02 =600 ohm$

The internal resistance of the voltage source (need to be at least 10 times smaller than the resistance of the load.

$R1 =RL/10 =600/10 =60 ohm$

d) $IR1 = IL + Izd = 20+20 =40 mA$

c) the power on a resistor is $P = I^2*R$

$P(RL) = 0.02^2*600 = 0.24 W$

$P(R1) =0.04^4*60 = 0.096 W$

e) $V(A_DC) = Vout +I1*R1 = 12 +0.04*60 = 14.4 V$

$V(A_DC) = V(S) + 3V$

3 V is the voltage loss on the bridge rectifiers $0.7*4 =2.8 V$

$V(S) = 14.4 – 3 = 11.4 V$

Since V(A_DC) is equal to the peak voltage of the bridge rectifier it implies that $V(S) = 11.4 V$ is the peak value.

f) V(S) is the peak value.

The peak value for V(primary) is

$V(primary) = V(RMS)*sqrt{(2)} =120*sqrt{(2)} = 169.7 V$

The transformer turns ratio is

$n = V(primary)/V(S) =169.7/11.4 = 14.72$