A) Explain the roll of freewheeling diode (FD) in the input...

a) Explain the roll of freewheeling diode (FD) in the input and output electrical characteristics of a half wave diode rectifier for R-L loads. Draw relevant waveforms to justify your explanation. b) A battery of constant voltage 'E' is charged from a single phase have wave diode rectifier through a limiting resistor R in series. For the source voltage 220V, 50 Hz, and for R = 10?, ? = 120V, Find out, (i) Average charging current (ii) Diode conduction angle (iii) Average output voltage (9 + 9 = 18 Marks)

Transcript

00:01 So here we are considering about a half wave rectifier that has a voltage is given that is virms which is equals to 240 volt and frequency is given that is equals to 60 hertz and the rl load is given whose resistance is equals to 15 ohm and its inductance is given that is equals to 80 milli henry.

00:23 So here we have in the first part we have to determine the expression for the load current.

00:28 So from here we are considering the value of the omega which is greater than and equals to 0 and less than equals to beta.

00:34 So vi from here become equals to so vi from here is equals to under root 2 multiply by the vi multiply by the sine of omega t that is equals to l of di which is divided by t plus r of i naught.

00:47 Let's say this is our equation number 1.

00:49 So i of omega t is equals to 0 which is equals to i of omega t that is equals to b which is further equals to 0.

00:57 So here we can say that the value of i naught become equals to i naught e raised to the power minus omega t which is divided by the tangent of phi plus under root 2 of vi divided by z multiplied by the sine of omega t minus phi.

01:11 Where we are considering that tangent of phi from here is equals to omega of l which is divided by r.

01:17 So z from here is equals to under root r square plus omega of l to its whole square.

01:22 So i naught become equals to under root 2 divided by z vi multiplied by the sine of phi.

01:30 So this is the value here.

01:33 Now the average load current that is i of dc is equals to v of dc which is divided by the r.

01:39 So the value of v of dc is equals to vm which is divided by the 2 pi multiplied by the 1 minus cos of pi plus beta.

01:48 We are given the value of r that is equals to 50 ohm.

01:51 L is given that is equals to 80 millihenry.

01:54 And we are also given the value of inductance is given that is equals to 80.

02:02 Frequency is given that is equals to 50 hertz.

02:05 And vi for rms is given that is equals to 240 volt.

02:13 So in the first part we have to find out the expression.

02:16 So vdc become equals to 240 multiplied by the under root 2 that is divided by the 2 pi multiplied by the 1 minus cos of pi plus beta.

02:26 Beta is from here is equals to tangent inverse of omega of l which is divided by r.

02:33 So that from here is equals to tangent inverse of 2 pi f of l which is divided by r.

02:39 So beta become equals to tangent inverse of 2 pi multiplied by the 60 which is multiplied by the 80 multiplied by the 10 raised to the power minus 3 which is divided by 15.

02:50 So solving the term for beta that become equals to 63 .55.

02:56 So the value of beta is here.

02:59 So we can simplify the term from here that become equals to v of dc become equals to 240 multiplied by the under root 2 that is divided by 2 pi multiplied by the 1 minus cos of 180 plus 63 .55...

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