Question on Electrical and Electronics Engineering
Design a half-wave rectifier without any smoothing capacitor. The supply voltage to the Germanium diode is 7.07 Volts RMS at a frequency of 50 Hz. The load resistance is 100 z. Calculate the given process. - DC voltage. - Ripple voltage (peak-to-peak). - Ripple percentage. - Current drawn from the transformer. - Line regulation percentage.
Answers
Answered by
0
see the diagram.
R = 100 Ω f= 50Hz and ω = 100π rad/sec
Vi peak = 7.07 V rms * √2 = 10 V
Vi = 10 Sin (100π t) Volts
Assume a voltage drop of 0.7 V across the Germanium diode.
Time period = 1/50 = 20 ms
Ripple voltage = peak-to-peak = 10 - 0.7 = 9.3 Volts
DC voltage = average of the voltage signal over a time period:
=
![Vi\ \textgreater \ =0.7V\ for\ Sin^{-1} 0.07 \le \omega t \le \pi-Sin^{-1} 0.07\\\\i.e.,for\ 0.223ms \le t \le 9.777 ms\\\\V_{avg}=\frac{1}{T}\int \limits_{0}^{T} {V_o} \, dt\\\\=\frac{1}{0.020} \int \limits_{0.000223}^{0.009777} [10 Sin(100\pi t)-0.7]\ dt\\\\=\frac{1}{0.020}[-\frac{10}{100 \pi}\ cos(100\pi t)-0.7\ t ]_{0.000223}^{0.009777}\\\\=\frac{1}{0.020}[\frac{1}{10\pi}(cos(0.0223\pi)-cos(0.9777\pi)) -0.7(0.009777-0.000223)] \\\\=50[0.0568 ]=2.84 Volts](https://tex.z-dn.net/?f=Vi%5C+%5Ctextgreater+%5C+%3D0.7V%5C+for%5C+Sin%5E%7B-1%7D+0.07+%5Cle+%5Comega+t+%5Cle+%5Cpi-Sin%5E%7B-1%7D+0.07%5C%5C%5C%5Ci.e.%2Cfor%5C++0.223ms+%5Cle+t+%5Cle+9.777+ms%5C%5C%5C%5CV_%7Bavg%7D%3D%5Cfrac%7B1%7D%7BT%7D%5Cint+%5Climits_%7B0%7D%5E%7BT%7D+%7BV_o%7D+%5C%2C+dt%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7B0.020%7D+%5Cint+%5Climits_%7B0.000223%7D%5E%7B0.009777%7D+%5B10+Sin%28100%5Cpi+t%29-0.7%5D%5C+dt%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7B0.020%7D%5B-%5Cfrac%7B10%7D%7B100+%5Cpi%7D%5C+cos%28100%5Cpi+t%29-0.7%5C+t+%5D_%7B0.000223%7D%5E%7B0.009777%7D%5C%5C%5C%5C%3D%5Cfrac%7B1%7D%7B0.020%7D%5B%5Cfrac%7B1%7D%7B10%5Cpi%7D%28cos%280.0223%5Cpi%29-cos%280.9777%5Cpi%29%29+-0.7%280.009777-0.000223%29%5D+%5C%5C%5C%5C%3D50%5B0.0568+%5D%3D2.84+Volts "Vi\ \textgreater \ =0.7V\ for\ Sin^{-1} 0.07 \le \omega t \le \pi-Sin^{-1} 0.07\\\\i.e.,for\ 0.223ms \le t \le 9.777 ms\\\\V_{avg}=\frac{1}{T}\int \limits_{0}^{T} {V_o} \, dt\\\\=\frac{1}{0.020} \int \limits_{0.000223}^{0.009777} [10 Sin(100\pi t)-0.7]\ dt\\\\=\frac{1}{0.020}[-\frac{10}{100 \pi}\ cos(100\pi t)-0.7\ t ]_{0.000223}^{0.009777}\\\\=\frac{1}{0.020}[\frac{1}{10\pi}(cos(0.0223\pi)-cos(0.9777\pi)) -0.7(0.009777-0.000223)] \\\\=50[0.0568 ]=2.84 Volts")
If we neglect the voltage drop across the diode, then the average voltage
= 10 V / π = 3.18 V
The current from the transformer
I(t) = [10 Sin (100 π t) - 0.7 ] /100 during 0.223 ms <= t <= 9.777 ms
= 0 during the rest of the time period from 0 to 20 ms
Average current drawn from the transformer = 2.84/100 = 28.4 mA
Average current if we neglect the voltage drop across the diode: 31.8 mA
Line regulation% = (change in Vo / Change in Vi) * 100
= (9.3 - 0.7)*100 /(10 - (-10)) = 860/20 = 43%
If we neglect the drop across the diode then, line regulation = 10*100/20 = 50%
R = 100 Ω f= 50Hz and ω = 100π rad/sec
Vi peak = 7.07 V rms * √2 = 10 V
Vi = 10 Sin (100π t) Volts
Assume a voltage drop of 0.7 V across the Germanium diode.
Time period = 1/50 = 20 ms
Ripple voltage = peak-to-peak = 10 - 0.7 = 9.3 Volts
DC voltage = average of the voltage signal over a time period:
=
If we neglect the voltage drop across the diode, then the average voltage
= 10 V / π = 3.18 V
The current from the transformer
I(t) = [10 Sin (100 π t) - 0.7 ] /100 during 0.223 ms <= t <= 9.777 ms
= 0 during the rest of the time period from 0 to 20 ms
Average current drawn from the transformer = 2.84/100 = 28.4 mA
Average current if we neglect the voltage drop across the diode: 31.8 mA
Line regulation% = (change in Vo / Change in Vi) * 100
= (9.3 - 0.7)*100 /(10 - (-10)) = 860/20 = 43%
If we neglect the drop across the diode then, line regulation = 10*100/20 = 50%
Attachments:
Similar questions