Chapter 2, Problem 2.3P

A half−wave rectifier such as shown in Figure 2.2(a) has a $2\text{k}\Omega$ load. The input is a $\text{12}0\text{ V}\left(\text{rms}\right)$ , 60 Hz signal and the transformer is a 10:1 step−down transformer. The diode has a cut−in voltage of $V_{\gamma }=0.7\text{V}\left(r_f=0\right)$ . (a) What is the peak output voltage? (b) Determine the peak diode current. (c) What is the fraction (percent) of a cycle that ${\upsilon }_O>0$ . (d) Determine the average output voltage. (e) Find the average current in the load.

To determine

The peak output voltage of a half-wave rectifier.

Answer to Problem 2.3P

Peak output voltage is 16.27 V

Explanation of Solution

Given information:

Load resistance $R=2\text{ k}\Omega$

Input RMS voltage, $V_I=120\text{ V}$

Input frequency, $f=60\text{ Hz}$

Transformer turn ratio $(N_1:N_2)=10:1$

Diode cut-in voltage $V_{\gamma }=0.7\text{ V }(r_f=0)$

Calculation:

Since the turn ratio of the transformer is 10:1 the secondary voltage is

  $v_{s(rms)}=12\text{ V}$

Then the peak secondary voltage,

  $v_{s(\max )}=12\times \sqrt{2}=16.97\text{ V}$

The peak output voltage,

  $\begin{array}{c}{v}_{o(\max )}=v_{s(\max )}-V_{\gamma }\\ =16.97-0.7\text{V}\\ \text{=16}\text{.27}\text{V}\end{array}$

Conclusion:

The peak output voltage is $16.27\text{ V}$ .

To determine

The peak diode current of a half-wave rectifier

Answer to Problem 2.3P

Peak diode current is 8.14 mA.

Explanation of Solution

Given information:

Load resistance $R=2\text{ k}\Omega$

Input RMS voltage, $V_I=120\text{ V}$

Input frequency, $f=60\text{ Hz}$

Transformer turn ratio $(N_1:N_2)=10:1$

Diode cut-in voltage $V_{\gamma }=0.7\text{ V }(r_f=0)$

Calculation:

Since the turn ratio of the transformer is 10:1 the secondary voltage is

  $v_{s(rms)}=12\text{ V}$

Then the peak secondary voltage,

  $v_{s(\max )}=12\times \sqrt{2}=16.97\text{ V}$

The peak output voltage,

  $\begin{array}{c}{v}_{o(\max )}=v_{s(\max )}-V_{\gamma }\\ =16.97-0.7\\ =16.27\text{ V}\end{array}$

The diode current is maximum when the output voltage is maximum, hence the peak diode current is,

  $\begin{array}{c}{i}_{d(\max )}=\frac{v_{o(\max )}}{R}\\ =\frac{16.27}{2\times {10}^3}\\ =8.14\text{mA}\end{array}$

Conclusion:

The diode maximum current is 8.14 mA.

To determine

The fraction of a cycle where the output voltage always greater than zero.

Answer to Problem 2.3P

The fraction of a cycle (where output voltage is positive) is 48.7%

Explanation of Solution

Given information:

Load resistance $R=2\text{ k}\Omega$

Input RMS voltage, $V_I=120\text{ V}$

Input frequency, $f=60\text{ Hz}$

Transformer turn ratio $(N_1:N_2)=10:1$

Diode cut-in voltage $V_{\gamma }=0.7\text{ V }(r_f=0)$

Calculation:

When diode is forward biased and started to conduct then there will be an output voltage which is greater than zero. This concept is used to find where diode starts conducting. Similarly, the time where the diode is going to reverse bias can be calculated. Then the diode conduction time can be found as a fraction from the output cycle.

Since the turn ratio of the transformer is 10:1 the secondary voltage is

  $v_{s(rms)}=12\text{ V}$

Then the peak secondary voltage,

  $v_{s(\max )}=12\times \sqrt{2}=16.97\text{ V}$

The output voltage can be written as,

  $v_o=v_{s(\max )}\sin (\omega t)-V_{\gamma }\left(1\right)$

When diode is forward biased and started to conduct, $v_o>0$ . At the time diode start conducting, at $\left(t_1\right)$ , $v_o=0$ and from the equation (1),

  $\omega t_1={\sin }^{-1}\left(\frac{V_{\gamma }}{v_{s(\max )}}\right)$

Substituting the values $V_{\gamma }=0.7\text{ V}$ and $v_{s(\max )}=16.97\text{ V}$ ,

  $\begin{array}{c}\omega t_1=2.364°\\ =\text{0}\text{.01313}\pi \end{array}$

By symmetry it can be found where $v_o$ goes zero again $\left(t_2\right)$ ,

  $\begin{array}{c}\omega t_2=180°-2.364°\\ =177.64°\\ =0.9869\pi \end{array}$

Hence the diode conduction period (where $v_o>0$ ) is

  $\begin{array}{c}\omega t_2-\omega t_1=177.64°-2.364°\\ =175.276°\end{array}$

So, the fraction of a cycle such that $v_o>0$ ,

Fraction of a cycle = $\frac{175.272°}{{360}^0}\times 100\%=48.7\%$

Conclusion:

The fraction of cycle the diode is conducting is 48.7%.

To determine

The average output voltage for the half-wave rectifier

Answer to Problem 2.3P

Average output voltage is5.06 V.

Explanation of Solution

Given information:

Load resistance $R=2\text{ k}\Omega$

Input RMS voltage, $V_I=120\text{ V}$

Input frequency, $f=60\text{ Hz}$

Transformer turn ratio $(N_1:N_2)=10:1$

Diode cut-in voltage $V_{\gamma }=0.7\text{ V }(r_f=0)$

Calculation:

Since the turn ratio of the transformer is 10:1 the secondary voltage is

  $v_{s(rms)}=12\text{ V}$

Then the peak secondary voltage,

  $v_{s(\max )}=12\times \sqrt{2}=16.97\text{ V}$

The output voltage can be written as,

  $v_o=v_{s(\max )}\sin (\omega t)-V_{\gamma }\left(1\right)$

When diode is forward biased and started to conduct, $v_o>0$ . At the time diode starts conducting, at $\left(t_1\right)$ , $v_o=0$ and from the equation (1),

  $\omega t_1={\sin }^{-1}\left(\frac{V_{\gamma }}{v_{s(\max )}}\right)$

Substituting the values $V_{\gamma }=0.7\text{ V}$ and $v_{s(\max )}=16.97\text{ V}$ ,

  $\begin{array}{c}\omega t_1=2.364°\\ =\text{0}\text{.01313}\pi \end{array}$

By symmetry it will be found where $v_o$ goes zero again $\left(t_2\right)$ ,

  $\begin{array}{c}\omega t_2=180°-2.364°\\ =177.64°\\ =0.9869\pi \end{array}$

Hence the diode conduction period (where $v_o>0$ ) is,

  $\omega t_2-\omega t_1=175.272°$ .

The average output voltage can be determined by integrating the equation (1) over a one period,

  $v_{o(avg)}=\frac{1}{2\pi }{\displaystyle \underset{0}{\overset{\pi }{\int }}(v_{s(\max )}\sin (\omega t)-V_{\gamma })d}\omega t$

During the period $\omega t_2-\omega t_1$ , the output voltage is $v_o=16.97\sin (\omega t)-0.7$ otherwise zero.

Hence

  $\begin{array}{c}{v}_{o(avg)}=\frac{1}{2\pi }{\displaystyle \underset{\text{0}\text{.01313}\pi }{\overset{0.9869\pi }{\int }}(16.97\sin (\omega t)-0.7)d}\omega t\\ =\frac{1}{2\pi }\left[\left(-16.97\right)\cos (\omega t)\underset{0.01313\pi }{\overset{0.9869\pi }{|}}-0.7(\omega t)\underset{0.01313\pi }{\overset{0.9869\pi }{|}}\right]\\ =\frac{1}{2\pi }\left[\left(-16.97\right)(-0.99915-0.99915)-0.7(0.9738\pi )\right]\\ =5.06\text{ V}\end{array}$

Conclusion:

Average voltage is $v_{o(avg)}=5.06\text{ V}$

To determine

The average output current the half-wave rectifier.

Answer to Problem 2.3P

Average output current is2.53 mA

Explanation of Solution

Given information:

Load resistance $R=2\text{ k}\Omega$

Input RMS voltage, $V_I=120\text{ V}$

Input frequency, $f=60\text{ Hz}$

Transformer turn ratio $(N_1:N_2)=10:1$

Diode cut-in voltage $V_{\gamma }=0.7\text{ V }(r_f=0)$

Calculation:

Since the turn ratio of the transformer is 10:1 the secondary voltage is

  $v_{s(rms)}=12\text{ V}$

Then the peak secondary voltage,

  $v_{s(\max )}=12\times \sqrt{2}=16.97\text{ V}$

Output voltage can be written as,

  $v_o=v_{s(\max )}\sin (\omega t)-V_{\gamma }\to \left(1\right)$

When the diode is forward biased and started to conduct, $v_o>0$ . At the time diode starts conducting, at $\left(t_1\right)$ , $v_o=0$ and from the equation (1),

  $\omega t_1={\sin }^{-1}\left(\frac{V_{\gamma }}{v_{s(\max )}}\right)$

Substituting the values $V_{\gamma }=0.7\text{ V}$ and $v_{s(\max )}=16.97\text{ V}$ ,

  $\begin{array}{c}\omega t_1=2.364°\\ =\text{0}\text{.01313}\pi \end{array}$

By symmetry,

  $\begin{array}{c}\omega t_2=180°-2.364°\\ =177.64°\\ =0.9869\pi \end{array}$

Hence the diode conduction period (where $v_o>0$ ) is $\omega t_2-\omega t_1=175.272°$ .

The average output voltage can be determined by integrating the equation (1) over a one period,

  $v_{o(avg)}=\frac{1}{2\pi }{\displaystyle \underset{0}{\overset{\pi }{\int }}(v_{s(\max )}\sin (\omega t)-V_{\gamma })d}\omega t$

During the period $\omega t_2-\omega t_1$ , the output voltage is $v_o=16.97\sin (\omega t)-0.7$ otherwise zero.

Hence

  $\begin{array}{l}{v}_{o(avg)}=\frac{1}{2\pi }{\displaystyle \underset{\text{0}\text{.01313}\pi }{\overset{0.9869\pi }{\int }}(16.97\sin (\omega t)-0.7)d}\omega t\\ v_{o(avg)}=\frac{1}{2\pi }\left[\left(-16.97\right)\cos (\omega t)\underset{0.01313\pi }{\overset{0.9869\pi }{|}}-0.7(\omega t)\underset{0.01313\pi }{\overset{0.9869\pi }{|}}\right]\\ v_{o(avg)}=\frac{1}{2\pi }\left[\left(-16.97\right)(-0.99915-0.99915)-0.7(0.9738\pi )\right]\\ v_{o(avg)}=5.06\text{ V}\end{array}$

Then the average diode current is,

  $I_{d(avg)}=\frac{v_{o(avg)}}{R}=\frac{5.06}{2\times {10}^3}$

  $I_{d(avg)}=$ $2.53\text{ mA}$

Conclusion:

Then the average diode current is $I_{d(avg)}=$ $2.53\text{ mA}$ .