A half-wave rectifier circuit with a 1-kΩ load operates from a 120-V(rms) 60-Hz household supply

1. The peak voltage of the rectified output: The transformer steps down the voltage by a factor

A half-wave rectifier circuit with a 1-kΩ load operates from...

A half-wave rectifier circuit with a $1-\mathrm{k} \Omega$ load operates from a $120-\mathrm{V}(\mathrm{rms}) 60-\mathrm{Hz}$ household supply through a $10-$ to-1 step-down transformer. It uses a silicon diode that can be modeled to have a $0.7-\mathrm{V}$ drop for any current. What is the peak voltage of the rectified output? For what fraction of the cycle does the diode conduct? What is the average output voltage? What is the average current in the load?

Transcript

00:01 Mario student so because of some technical issues i am recording this lecture again so we have been given a half -wave rectifier so this is the circuit of the highway rectifier this is the source voltage here at the output we have a diode and a resistor this is primary coil this is secondary coil so basically this is a transformer the transformer is used for the half -wave rectifier so in a transformer the ratio of the voltage is equal to the ratio of the turns v2 is the output voltage the rms value of the output voltage divided by the rms value of the input voltage or the source voltage is equal to the ratio of the number of tons at the secondary to the number of turns at the primary so he from here the source or the output voltage the rms value of the output voltage is equal to the turns ratio that is 1 by 10 times the input voltage that is 120 volts all right so from here my dear student we get the value of the rms value rms value of the voltage at the output that is 12 volts all right now the requirement of the question is we have to find the peak value peak value is equal to the square root of 2 times the rmsm s value is 12 volts when it is multiplied with square root of 2 it gives us 18 .3 at volts it means this is the output voltage of this half -wave rectifier and this is the rectified voltage now if we wish to find that total voltage that is coming out it means the voltage which we extract so the output voltage is taken across the resistor let's call this v0 so this is equal to the total voltage which is going to the output circuit that is to the second recoil minus the voltage that is dropped across the diode so the voltage becomes so the equation becomes the output voltage should be equal to the total voltage at the output side minus the voltage drop at the diode so the output voltage is this one 18 .38 volts the peak voltage minus the voltage drop across the diode is 0 .7 watts because this is a silicon diode so it means that the output voltage across the resistor comes out to be 17 .68 volts so this is the peak value now is this is a half -wave rectifier so this diode will conducts only for the positive half cycle because for positive values of the voltage this is forward biased and when this is forward bias we will get the output voltage but for the negative values of the voltages this is reversed biased when this is reverse bias we will not get any value of the current or any value of the voltage at output side so it means that this diode conducts for only 50 % all idea students now in part c we have been asked to find the average output voltage that equals the output voltage divided by pi so this is the average output voltage of the half -wave rectifier so the output voltage is basically equal to 17 .68 volts divided by pi so this equals the average output voltage equals 5 .63 volts all right so in part d we have been asked to find the average current so average current should be basically equal to the average voltage divided by the resistance.

03:41 The average voltage has already been calculated that is 5 .36 volts divided by the value of the resistance.

03:48 The value of the resistance is basically one kilo -ooms...

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