6 of 20 The input voltage applied to a voltage tripler is 12
6 of 20
The input voltage applied to a voltage tripler is 120 Vrms. What is the ideal dc output voltage?
509 V
360 V
170 V
56.56 V
The dc current flowing through the diode in a half-wave rectifier equals
twice the dc load current
half the dc load current
the dc load current
one-fourth the dc load current
The dc current through each diode in a bridge rectifier equals
twice the dc load current
half the dc load current
the load current
one-forth the dc load current
A half-wave rectifier has a dc load current of 50 mA and a filter capacitance of 470 F. The peak-to-peak ripple voltage equals
17.7 Vp-p
3.6 Vp-p
1.77 Vp-p
0.886 Vp-p
For forward-biased diodes, the voltage where the current increases rapidly is called the
knee voltage
breakdown voltage
zener voltage
turn-off voltage
6 of 20
The input voltage applied to a voltage tripler is 120 Vrms. What is the ideal dc output voltage?
509 V
360 V
170 V
56.56 V
The dc current flowing through the diode in a half-wave rectifier equals
twice the dc load current
half the dc load current
the dc load current
one-fourth the dc load current
The dc current through each diode in a bridge rectifier equals
twice the dc load current
half the dc load current
the load current
one-forth the dc load current
A half-wave rectifier has a dc load current of 50 mA and a filter capacitance of 470 F. The peak-to-peak ripple voltage equals
17.7 Vp-p
3.6 Vp-p
1.77 Vp-p
0.886 Vp-p
For forward-biased diodes, the voltage where the current increases rapidly is called the
knee voltage
breakdown voltage
zener voltage
turn-off voltage
| 509 V | |
| 360 V | |
| 170 V | |
| 56.56 V |
Solution
+1) Vdc=3*(Vpeak)=3*120*sqrt(2)=509v
2) option c:Id=I load for half wave rectifier
3) option 2:Id=half the dc load current because each diode carries the load current for half cycle, gives the average current value half of the load current.
4)peak to peak ripple voltage=(Vp*T)/(R*C)=I/(f*C)
=>(50*10^-3)/(60*470*10^-6)=1.77 volts
5) knee voltage

