Sunday, August 28, 2011

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In figs. 1, , and the current rating if the transformer secondary is that flowing through the load since there are no ramifications of the circuits other than the non conducting diode(s) where no current flows, so the current rating can be calculated by simply dividing VL by RL. So the current rating in the secondary is the maximum that can be born by the diodes and that is Vmax/RL.


First, w can see that VDC in circuit 1 is approximately half of that in fig., which is approximately half of that in fig.. First, fig.1 is a half-wave rectifier, that is the signal passed by the diode that reached the load is only the positive part of the 7.5Vrms signal so the mean is reasonably half that of full-wave rectifier of the same input voltage, which will pass the absolute value of the signal giving a mean of (mean of positive part).


Second, the ripple voltage is the same in figs.1 and , 10.5 V, since we are using the same voltage supply which is approximately 7.5 and the voltage drop that is across the diode is the same in both circuits since the current passes by one diode in each case. (In fig. the voltage drop is across the conducting diode). As for fig. the ripple voltage is of 0.7 V, that is approximately the double of that in the previous two circuits. But if we take the ripple facto we can see clearly that the ripple factor is the same in fig. and fig. and is equal to the half of the ripple factor in fig.1, that is also due to the fact that fig.1 is a half-wave rectifier and fig. and are full wave rectifiers.


Third, for average and peak currents, we are the one that fixed the average current to 50 mA in the load but what varies is the peak to peak current in the diode. We can clearly see that the current in fig 1 is far higher than that in figs. and and that is due to the fact since the voltage is half its value the current must be twice its value, and here we have it a little more than twice.


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Fourth, the peak inverse voltage is the voltage across the non-conducting diode or the voltage across the diode when it is non-conducting. In fig.1 since we have only one diode the peak inverse voltage has got to be half that of that in fig. While in fig., we have an input value that is the double of that in fig. but we have in each path one more diode so, at each diode the peak inverse voltage is equal to half its value in the whole circuit that is why it cam out to be equal to the peak inverse voltage measured in fig..


Finally, the percent regulation can be easily compared between the three circuits. As we move from fig.1 then then we see that the percent regulation is shrinking with 4.6 % for fig.1 to .8% for fig. and 1.% for fig.. The regulation is mainly due to the fact that diodes are not ideal and have a certain voltage drop. As for the transformer requirements, we have to be careful to choose a transformer that can take a secondary current rating that matches the peak diode current measured and diodes that can bear that high currents, especially in fig.1.


From table-A we can derive the conditions under which each of these circuits is advantageous. If we want a low DC, average, voltage we have to use fig.1 with a ripple factor of . while if we want a medium average voltage, fig. is the best and finally we can choose fig.. And the regulations are very close to each other even if fig. has the best percent regulation.


For the circuit shown in Fig.4_____________________________________


As it appears in table-1, VC does increase with increasing C, because when C increases Vr decreases and thus VDC=Vmax-Vr/ increases.


Also, Vr decreases when C increases since RC time constant increases thus the capacitor will take more time to discharge and by the time it will begin to charge again it would have lost less charges that with a lower C so the Vr is inversely related to C.


The ratio IF/IDC varies also with C. for values of C not too large we can see from table-1 that as C increases IF/IDC increases but when C reaches a value that is too large it can be charged enough to immediately deliver current that would lead to a decrease in IF.


The approximate relations that are included in the theory section are valid when Vr is small (in the mV range) and that is what appears in table- above. And the ripple voltage is small when the C has large value, in reality when RC has a large value, so we can consider these relations valid for C=1000ìF for R=680 Ù, and for C100 ìF for R=.kÙ.


We can see that with an increasing C, VDC increases so if we take V0 to be the no-load voltage, (VDC-V0)/V0 will decrease in absolute value making the regulation more adequate, and this can be explained by the fact that the ripple becomes smaller and smaller making the signal grow steady.


IF must be compensated for in the capacitor filter, so it must be added to the current rating because we want to get a transformer that can take such a high current as well a s a diode that can take such a high current.


For a certain required Vr and VDC/IDC, the rms voltage rating of the secondary can be calculated from the relation in the theory section _


VDC=Vmax-Vr/ Vmax= VDC+ Vr/ Vrms=( VDC+ Vr/)/?


We have Vr=Vmax/(ðfRLC) C= Vmax/(ðfRLVr) C= (VDC+ Vr /)/(ðfRLVr)


In figs. 1, , and the current rating if the transformer secondary is that flowing through the load since there are no ramifications of the circuits other than the non conducting diode(s) where no current flows, so the current rating can be calculated by simply dividing VL by RL. So the current rating in the secondary is the maximum that can be born by the diodes and that is Vmax/RL.


First, w can see that VDC in circuit 1 is approximately half of that in fig., which is approximately half of that in fig.. First, fig.1 is a half-wave rectifier, that is the signal passed by the diode that reached the load is only the positive part of the 7.5Vrms signal so the mean is reasonably half that of full-wave rectifier of the same input voltage, which will pass the absolute value of the signal giving a mean of (mean of positive part).


Second, the ripple voltage is the same in figs.1 and , 10.5 V, since we are using the same voltage supply which is approximately 7.5 and the voltage drop that is across the diode is the same in both circuits since the current passes by one diode in each case. (In fig. the voltage drop is across the conducting diode). As for fig. the ripple voltage is of 0.7 V, that is approximately the double of that in the previous two circuits. But if we take the ripple facto we can see clearly that the ripple factor is the same in fig. and fig. and is equal to the half of the ripple factor in fig.1, that is also due to the fact that fig.1 is a half-wave rectifier and fig. and are full wave rectifiers.


Third, for average and peak currents, we are the one that fixed the average current to 50 mA in the load but what varies is the peak to peak current in the diode. We can clearly see that the current in fig 1 is far higher than that in figs. and and that is due to the fact since the voltage is half its value the current must be twice its value, and here we have it a little more than twice.


Fourth, the peak inverse voltage is the voltage across the non-conducting diode or the voltage across the diode when it is non-conducting. In fig.1 since we have only one diode the peak inverse voltage has got to be half that of that in fig. While in fig., we have an input value that is the double of that in fig. but we have in each path one more diode so, at each diode the peak inverse voltage is equal to half its value in the whole circuit that is why it cam out to be equal to the peak inverse voltage measured in fig..


Finally, the percent regulation can be easily compared between the three circuits. As we move from fig.1 then then we see that the percent regulation is shrinking with 4.6 % for fig.1 to .8% for fig. and 1.% for fig.. The regulation is mainly due to the fact that diodes are not ideal and have a certain voltage drop. As for the transformer requirements, we have to be careful to choose a transformer that can take a secondary current rating that matches the peak diode current measured and diodes that can bear that high currents, especially in fig.1.


From table-A we can derive the conditions under which each of these circuits is advantageous. If we want a low DC, average, voltage we have to use fig.1 with a ripple factor of . while if we want a medium average voltage, fig. is the best and finally we can choose fig.. And the regulations are very close to each other even if fig. has the best percent regulation.


For the circuit shown in Fig.4_____________________________________


As it appears in table-1, VC does increase with increasing C, because when C increases Vr decreases and thus VDC=Vmax-Vr/ increases.


Also, Vr decreases when C increases since RC time constant increases thus the capacitor will take more time to discharge and by the time it will begin to charge again it would have lost less charges that with a lower C so the Vr is inversely related to C.


The ratio IF/IDC varies also with C. for values of C not too large we can see from table-1 that as C increases IF/IDC increases but when C reaches a value that is too large it can be charged enough to immediately deliver current that would lead to a decrease in IF.


The approximate relations that are included in the theory section are valid when Vr is small (in the mV range) and that is what appears in table- above. And the ripple voltage is small when the C has large value, in reality when RC has a large value, so we can consider these relations valid for C=1000ìF for R=680 Ù, and for C100 ìF for R=.kÙ.


We can see that with an increasing C, VDC increases so if we take V0 to be the no-load voltage, (VDC-V0)/V0 will decrease in absolute value making the regulation more adequate, and this can be explained by the fact that the ripple becomes smaller and smaller making the signal grow steady.


IF must be compensated for in the capacitor filter, so it must be added to the current rating because we want to get a transformer that can take such a high current as well a s a diode that can take such a high current.


For a certain required Vr and VDC/IDC, the rms voltage rating of the secondary can be calculated from the relation in the theory section _


VDC=Vmax-Vr/ Vmax= VDC+ Vr/ Vrms=( VDC+ Vr/)/?


We have Vr=Vmax/(ðfRLC) C= Vmax/(ðfRLVr) C= (VDC+ Vr /)/(ðfRLVr)


In figs. 1, , and the current rating if the transformer secondary is that flowing through the load since there are no ramifications of the circuits other than the non conducting diode(s) where no current flows, so the current rating can be calculated by simply dividing VL by RL. So the current rating in the secondary is the maximum that can be born by the diodes and that is Vmax/RL.


First, w can see that VDC in circuit 1 is approximately half of that in fig., which is approximately half of that in fig.. First, fig.1 is a half-wave rectifier, that is the signal passed by the diode that reached the load is only the positive part of the 7.5Vrms signal so the mean is reasonably half that of full-wave rectifier of the same input voltage, which will pass the absolute value of the signal giving a mean of (mean of positive part).


Second, the ripple voltage is the same in figs.1 and , 10.5 V, since we are using the same voltage supply which is approximately 7.5 and the voltage drop that is across the diode is the same in both circuits since the current passes by one diode in each case. (In fig. the voltage drop is across the conducting diode). As for fig. the ripple voltage is of 0.7 V, that is approximately the double of that in the previous two circuits. But if we take the ripple facto we can see clearly that the ripple factor is the same in fig. and fig. and is equal to the half of the ripple factor in fig.1, that is also due to the fact that fig.1 is a half-wave rectifier and fig. and are full wave rectifiers.


Third, for average and peak currents, we are the one that fixed the average current to 50 mA in the load but what varies is the peak to peak current in the diode. We can clearly see that the current in fig 1 is far higher than that in figs. and and that is due to the fact since the voltage is half its value the current must be twice its value, and here we have it a little more than twice.


Fourth, the peak inverse voltage is the voltage across the non-conducting diode or the voltage across the diode when it is non-conducting. In fig.1 since we have only one diode the peak inverse voltage has got to be half that of that in fig. While in fig., we have an input value that is the double of that in fig. but we have in each path one more diode so, at each diode the peak inverse voltage is equal to half its value in the whole circuit that is why it cam out to be equal to the peak inverse voltage measured in fig..


Finally, the percent regulation can be easily compared between the three circuits. As we move from fig.1 then then we see that the percent regulation is shrinking with 4.6 % for fig.1 to .8% for fig. and 1.% for fig.. The regulation is mainly due to the fact that diodes are not ideal and have a certain voltage drop. As for the transformer requirements, we have to be careful to choose a transformer that can take a secondary current rating that matches the peak diode current measured and diodes that can bear that high currents, especially in fig.1.


From table-A we can derive the conditions under which each of these circuits is advantageous. If we want a low DC, average, voltage we have to use fig.1 with a ripple factor of . while if we want a medium average voltage, fig. is the best and finally we can choose fig.. And the regulations are very close to each other even if fig. has the best percent regulation.


For the circuit shown in Fig.4_____________________________________


As it appears in table-1, VC does increase with increasing C, because when C increases Vr decreases and thus VDC=Vmax-Vr/ increases.


Also, Vr decreases when C increases since RC time constant increases thus the capacitor will take more time to discharge and by the time it will begin to charge again it would have lost less charges that with a lower C so the Vr is inversely related to C.


The ratio IF/IDC varies also with C. for values of C not too large we can see from table-1 that as C increases IF/IDC increases but when C reaches a value that is too large it can be charged enough to immediately deliver current that would lead to a decrease in IF.


The approximate relations that are included in the theory section are valid when Vr is small (in the mV range) and that is what appears in table- above. And the ripple voltage is small when the C has large value, in reality when RC has a large value, so we can consider these relations valid for C=1000ìF for R=680 Ù, and for C100 ìF for R=.kÙ.


We can see that with an increasing C, VDC increases so if we take V0 to be the no-load voltage, (VDC-V0)/V0 will decrease in absolute value making the regulation more adequate, and this can be explained by the fact that the ripple becomes smaller and smaller making the signal grow steady.


IF must be compensated for in the capacitor filter, so it must be added to the current rating because we want to get a transformer that can take such a high current as well a s a diode that can take such a high current.


For a certain required Vr and VDC/IDC, the rms voltage rating of the secondary can be calculated from the relation in the theory section _


VDC=Vmax-Vr/ Vmax= VDC+ Vr/ Vrms=( VDC+ Vr/)/?


We have Vr=Vmax/(ðfRLC) C= Vmax/(ðfRLVr) C= (VDC+ Vr /)/(ðfRLVr)





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