Question

MODULE I: PRACTICAL No. 3 Based on Chapter 5 - Study Unit 1 1) Purpose Design and build a full-wave rectifier with a smoothing capacitor that can deliver 31 V to a 10 k load with a ripple voltage of 0,31 Vpp}The secondary voltage of the transformer can be taken as V2 22 V. 2) Components Apparatus 4 x IN4007 diodes 1x 10 k2 resistor 1x smoothing capacitor Oscilloscope Variable ac source Multimeter Breadboard 3) Circuit Diagram 4) Method Build the circuit and calculate: VRL(p); Vpc; Ii, and a value for the capacitor 5) Results Measure the following parameters with the multimeter/oscilloscope and compare it with the calculated values. Calculated value Measured Parameter VRI (P) VDC IRI

Answer 1

From the given data

From the given data Transformer secondary V_rms = 22V Then, V_max = squareroot = 2 V_rms = squareroot 2 * 22 = 31.11 V V_dc = 2V_m/pi = 19.79 V V_dc = V_RL = 19.79 V I_RL = V_RL/R_L = 19.79/10 k = 1.979 Ma Now capacitor value have to be find The most important formula for calculating the smoothing capacitor is: C = O delta t/ delta U The smoothing capacitor formula, alternatively: I = C * delta U/ delta t Clarification: C = capacity of the capacitor in mu F I = Charge current in mA delta t = half-period in ms = 1/2*F = 0.5F = 0.5*1/50 = 10 ms delta U = = ripple voltage in V = Vm - 0 = 31.11-0 = 31.11V C = I delta t/delta U = 63.6 mu F

Transformer secondary V_rms = 22V

Then, V_max = $\sqrt{2}$ Vrms =$\sqrt{2}$ * 22= 31.11 V

V_dc = $\frac{2V_{m}}{\prod }$ =19.79V

Vdc= V_RL=19.79_V

I_RL =$\frac{V_{RL}}{R_{L} }$ = $\frac{19.79}{10K }$ =1.979 mA

Now capacitor value have to be find

The most important formula for calculating the smoothing capacitor is:

C=I _{$\tfrac{\Delta t}{\Delta U}$}

The smoothing capacitor formula, alternatively:

I=C*ΔU/Δt

Clarification:
C = capacity of the capacitor in μF
I = Charge current in mA
Δt = half-period in ms =1/2*F=0.5F=0.5*1/50= 10 ms
ΔU = ripple voltage in V =Vm - 0 = 31.11-0=31.11V

C = I _{$\tfrac{\Delta t}{\Delta U}$} = 63.6 μF