Question

Question 12: [5] The system shown in figure 13 is a basic switching regulator. If the frequency at which the transistor is “switching", is 100Hz with an OFF time of 6 milli seconds, what output voltage is delivered? Also calculate the duty cycle of the transistor. +12 Vo 000 OVOUT } R 4.7k Oscillator fo= 100 Hz (R2 10 k12 For 6 ms R; 15 ΚΩ D2 2.7 V FIGURE 13

Answer 1

Q +12 VO 000 VOUT } R 4.7k Oscillator fo= 100 Hz for 6 ms R2 10 kn 벌 ų R3 15k D 2.7 V =

From the circuit,

For an Opamp;

V1 = V2 = 2.7V

Given:

Source Voltage,

OFF-time,

toft 6ms

Frequency, $f_0=100Hz$

Time, $t=\frac{1}{f_0}=0.01=10ms$

ON-time, $t_{on}=t-t_{off}=10-6=4ms$

Duty cycle, $D=\frac{t_{on}}{t_{on}+t_{off}}=\frac{4}{10}=0.4$

$D=\frac{V_0-V_2}{V_{in}}$

$\rightarrow 0.4=\frac{V_0-2.7}{12}$

$\rightarrow V_0-2.7=4.8$

$\Rightarrow V_0=7.5V$

The Output Voltage is 7.5V