Exp1components: resistance(1N4007) pn junction diode ac voltage sources connecting probes ground terminal Theory a rectifier is nothing but a simple diode or group of diodes which converts the ac into dc half wave: a half wave re is a type of re which converts the +ve half cycle of the input signal into pulsating dc output signal. For example if the +ve half cycle is allowed then the -ve half cycle is blocked. Similarly if the -ve half cycle is allowed then +ve half cycle not allow both +ve and -ve half cycle is blocked. Half wave re will not allw both +ve and -ve half cycles at the same time. so the half cycle of the input signal is wasted. parameters :1 supply voltage max value(Vs max) 2 supply max current (Ismax=Vsmax/Rf+Rl) 3 DC output current(Idc=Is max/pie) 4Dc output voltage(Vdc=Idc+Rl) 5 ripple factor y=Iac /Idc CASE 1 Vm=10V F=50 Hz Si diode RL= 100 ohm imax= vs max-0.7/(rf+rl) = 10-0.7/10+100 = 9.3/110 =0.084 A Practical value=83854mA idc= Is max/pie =83.851/pir=0.026A vdc = idc*rl = 0.026*100 =2.6 V Irms = Imax/2=0.0835/2=0.0439A Vrms=irms*Rl= 0.0439*100 = 4.19V ripple factor= root over of {(Imax/2)^2/(Imax/pie)^2)}-1 = 1.2 ripple factor of half wave rectifier rsnge 1.06 to 1.21 Case 2: Vmax =15 V F=60Hz RL=1 kiloohm Ge diode I max= Vmax-0.3/Rf+Rl =15-0.3/1000+12.5 = 0.0145 A Practical value=0.01400 A Idc= Im/ pie = 0.0045 A Vdc= Idc*Rl=4.5 V Irms= Imax/2=0.007 A Vrms = Irms*Rl=7V ripple factor= root over of{(Imax/2)^2/Imax/pie} -1=1.21 Case 3: half wave rectifier using transformer Vm= 100 V W=100pie 2pief=100pie f=100pie/2pie =50Hz I max =100/Rf+Rl= 9.1889 mA= 0.0091 A Idc= 0.0092/pie= 2.9264mA Vdc = Idc*Rl=2.926*1000=2926.4mA Irms=imax/2=4.59445mA Vrms= Irms*IRL = 4.59445*1000=4594.4V Ripple factor=root over of{(Imax/2)/Imax/pie} -1=1.24 Result: The circuit of half wave rectifier are designed and implemented with the components presents n multinism. It can be observed that the frequency of input signals and output signals remain same for half wave rectifier
EXPERIMENT 2 FULL wave rectifier Components: resistance pn junction diode(IN4007), ac voltage source, connecting probes, grounded terminal Theory: A full wave rectifier circuit produces an output voltage or current which is purely DC or has some specified DC component. Full wave rectifiers have some fundamental advantages over their half wave rectifier counterparts. The average (DC) output voltage is higher than for half wave, the output of the full wave rectifier has much less ripple than that of the half wave rectifier producing a smoother output waveform. A circuit that produces the same output waveform as the full wave rectifier circuit is that of the Full Wave Bridge Rectifier. Positive Half-cycle: During the negative half cycle of the supply, diodes D3 and D4 conduct in series, but diodes D1 and D2 switch “OFF” as they are now reverse biased. The current flowing through the load is the same direction as before. Negative Half-cycle: As the current flowing through the load is unidirectional, so the voltage developed across the load is also unidirectional the same as for the previous two diode full-wave rectifier, therefore the average DC voltage across the load is 0.637V Smax . Parameters: Supply voltage max voltage(Vsmax), supply max current , dc output current , dc output voltage , ac or rms current , efficiency(n=Vdc/ Iac), ripple factor Full wave bridge rectifier Vsmax=15V f= 50 Hz, primary to secondary coil ratio 1:1 and Rl=1 kiloohm Vpp= 15+15=30 Vs max=15 V imax= Vsmax-2*0.7/2*(rf+rl)=15-1.4/20+2000=0.0067ohm idc=2Imax/pie=0.00428amp vdc=Idc*Rl=0.0042*1000=4.28V Irms=Imax/root 2 =0.0047 amp Vrms= irms*rl=0.0047*1000=4.7V ripple factor = Irms/idc=1.0981 efficiency= rl/rf+rl=0.401 Result: The circuits of Full Wave Rectifiers are designed and implemented with the components present in MULTISIM. The input and output waveforms are observed properly. It can be observed that the frequency of input signals and output signals remain same for half wave rectifier.
xp 3 half wave full wave bridge rec using rc filter components :Resistance p-n Junction Diodes 4. Capacitor 1 Function Generator 1 Aplab (MSG2M) Connecting Probes specifications:Silicon Diode: Max Forward Current = 1A Max Reverse Current = 5.0µA Max Forward Voltage = 0.8V Max Reverse Voltage = 1000V Max Power Dissipation = 30mW Temperature = -65 to 200° C Theory: A rectifier is a circuit that converts a pure AC signal into a pulsating DC signal or a signal that is a combination of AC and DC components. In DC supplies, a rectifier is often followed by a filter circuit which converts the pulsating DC signal into pure DC signal by removing the AC component. This type of filter is known generally as a “first-order filter” or “one-pole filter”, why first- order or single-pole, because it has only “one” reactive component, the capacitor, in the circuit.
subhamkar1203 says:
Theory a rectifier is nothing but a simple diode or group of diodes which converts the ac into dc
half wave: a half wave re is a type of re which converts the +ve half cycle of the input signal into pulsating dc output signal. For example if the +ve half cycle is allowed then the -ve half cycle is blocked. Similarly if the -ve half cycle is allowed then +ve half cycle not allow both +ve and -ve half cycle is blocked. Half wave re will not allw both +ve and -ve half cycles at the same time. so the half cycle of the input signal is wasted.
parameters :1 supply voltage max value(Vs max) 2 supply max current (Ismax=Vsmax/Rf+Rl) 3 DC output current(Idc=Is max/pie) 4Dc output voltage(Vdc=Idc+Rl) 5 ripple factor y=Iac /Idc
CASE 1
Vm=10V F=50 Hz Si diode RL= 100 ohm
imax= vs max-0.7/(rf+rl) = 10-0.7/10+100 = 9.3/110 =0.084 A Practical value=83854mA
idc= Is max/pie =83.851/pir=0.026A
vdc = idc*rl = 0.026*100 =2.6 V
Irms = Imax/2=0.0835/2=0.0439A
Vrms=irms*Rl= 0.0439*100 = 4.19V
ripple factor= root over of {(Imax/2)^2/(Imax/pie)^2)}-1 = 1.2 ripple factor of half wave rectifier rsnge 1.06 to 1.21
Case 2: Vmax =15 V F=60Hz RL=1 kiloohm Ge diode
I max= Vmax-0.3/Rf+Rl =15-0.3/1000+12.5 = 0.0145 A Practical value=0.01400 A
Idc= Im/ pie = 0.0045 A
Vdc= Idc*Rl=4.5 V
Irms= Imax/2=0.007 A
Vrms = Irms*Rl=7V
ripple factor= root over of{(Imax/2)^2/Imax/pie} -1=1.21
Case 3: half wave rectifier using transformer
Vm= 100 V W=100pie 2pief=100pie
f=100pie/2pie =50Hz
I max =100/Rf+Rl= 9.1889 mA= 0.0091 A
Idc= 0.0092/pie= 2.9264mA
Vdc = Idc*Rl=2.926*1000=2926.4mA
Irms=imax/2=4.59445mA
Vrms= Irms*IRL = 4.59445*1000=4594.4V
Ripple factor=root over of{(Imax/2)/Imax/pie} -1=1.24
Result: The circuit of half wave rectifier are designed and implemented with the components presents n multinism. It can be observed that the frequency of input signals and output signals remain same for half wave rectifier
subhamkar1203 says:
Components: resistance pn junction diode(IN4007), ac voltage source, connecting probes, grounded terminal
Theory: A full wave rectifier circuit produces an output voltage or current which is purely DC or has
some specified DC component. Full wave rectifiers have some fundamental advantages over
their half wave rectifier counterparts. The average (DC) output voltage is higher than for half
wave, the output of the full wave rectifier has much less ripple than that of the half wave
rectifier producing a smoother output waveform. A circuit that produces the same output
waveform as the full wave rectifier circuit is that of the Full Wave Bridge Rectifier.
Positive Half-cycle:
During the negative half cycle of the supply, diodes D3 and D4 conduct in series, but diodes
D1 and D2 switch “OFF” as they are now reverse biased. The current flowing through the
load is the same direction as before.
Negative Half-cycle:
As the current flowing through the load is unidirectional, so the voltage developed across the
load is also unidirectional the same as for the previous two diode full-wave rectifier, therefore
the average DC voltage across the load is 0.637V Smax .
Parameters: Supply voltage max voltage(Vsmax), supply max current , dc output current , dc output voltage , ac or rms current , efficiency(n=Vdc/ Iac), ripple factor
Full wave bridge rectifier
Vsmax=15V f= 50 Hz, primary to secondary coil ratio 1:1 and Rl=1 kiloohm
Vpp= 15+15=30
Vs max=15 V
imax= Vsmax-2*0.7/2*(rf+rl)=15-1.4/20+2000=0.0067ohm
idc=2Imax/pie=0.00428amp
vdc=Idc*Rl=0.0042*1000=4.28V
Irms=Imax/root 2 =0.0047 amp
Vrms= irms*rl=0.0047*1000=4.7V
ripple factor = Irms/idc=1.0981
efficiency= rl/rf+rl=0.401
Result: The circuits of Full Wave Rectifiers are designed and implemented with the components
present in MULTISIM. The input and output waveforms are observed properly. It can be
observed that the frequency of input signals and output signals remain same for half wave
rectifier.
subhamkar1203 says:
components :Resistance p-n Junction Diodes 4. Capacitor 1
Function Generator 1 Aplab (MSG2M)
Connecting Probes
specifications:Silicon Diode:
Max Forward Current = 1A
Max Reverse Current = 5.0µA
Max Forward Voltage = 0.8V
Max Reverse Voltage = 1000V
Max Power Dissipation = 30mW
Temperature = -65 to 200° C
Theory:
A rectifier is a circuit that converts a pure AC signal into a pulsating DC signal or a signal
that is a combination of AC and DC components. In DC supplies, a rectifier is often followed
by a filter circuit which converts the pulsating DC signal into pure DC signal by removing the
AC component. This type of filter is known generally as a “first-order filter” or “one-pole filter”, why first-
order or single-pole, because it has only “one” reactive component, the capacitor, in the circuit.