The voltage measured by the voltmeter is in rms (root-mean-squared) value while the voltages measured by the oscilloscope are in peak-to-peak or just simply peak value. RMS is the DC equivalent of the AC value, equivalent such that it will produce the same heating effect (power output) if it will be in DC operation. The rms-peak value relationship is:
Vrms = Vp χ 21/2
or
Vrms = 0.707 * Vp
Figure 1
Figure 1 shows the secondary and primary voltage of the transformer in Peak and rms values. The voltmeter in the primary winding measured 220V which is in rms, its peak value using the above relationship would be 311.13V which confirms the output of the oscillator which roughly equal to 310V. Same computation will be done for the secondary voltage.
Turns Ratio
Turns ratio of a practical transformer is:
a = Vp/Vs
a = 220 / 9
a = 24.44444
I.
Half-wave Rectifier
An ideal diode is used in the simulation of all the circuits. Supposedly, ideal diode is assumed to be shorted during conduction, but it dont seem to apply in this EWB circuit since there are lot of inconsistencies of output in comparing between the computation and simulated output. EWB does not state how much is the threshold voltage(Vt) of the diode. To find out how much the threshold voltage of diode used in the circuit, the difference between the input and the output waveforms is taken.
Vt = Vi - Vo
Vt = 12.6787 11.9355
Vt = 0.743V
A threshold voltage of 0.7V (threshold voltage of silicon) will be assumed in the computation.
VL = 0.318(Vp 0.7)
Vp = Vrms / 0.707
The value 0.318 and 0.707 is derived from the conduction area of the output waveform with the aid of integral calculus.
Peak voltage computation:
Vp = Vrms / 0.707
Vp = 9 / 0.707
Vp = 12.73V
Voltage output:
VL = 0.318(Vp 0.7)
VL = 0.318(12.73-.7)
VL = 3.83V
Half Wave Rectifier with Filter Capacitor
100΅F Filter Capacitor
Three circuits are shown above with increasing value of capacitance. The load voltage is directly proportional to the value of the filter capacitance. Therefore the smallest capacitance has the least output value. The purpose of the filter capacitor is to flatten the pulsating output of the load resistor which is shown in the oscilloscope with different filter capacitance. The greater the capacitance, the flatter the output will be, eliminating the ripple voltage in the process. This is due to the faster charging and discharging time of the capacitor which is shown in the equation:
t = RC
The DC voltage with filter capacitor cannot be solve by simply using the formula VL = 0.318(Vp 0.7) since the charging and discharging time changes, therefore differing its conduction area from a ordinary half-wave rectifier. The DC voltage increases as the output waveform becomes more flat since the conduction area also increases.
II.
Bridge-type Full-wave Rectifier
Peak voltage computation:
Vp = Vrms / 0.707
Vp = 9 / 0.707
Vp = 12.73V
Voltage output:
VL = 0.318(Vp 0.7)
VL = 0.318(12.73-.7)
VL = 3.83V
Total voltage output:
Vo = VL * 2
Vo = 3.83 * 2
Vo = 7.66V
This circuit is equivalent to a two half-wave rectifiers with diodes in opposite direction so that it would conduct on either pulse. The two half-wave rectifiers will result into a full-wave rectifier with a voltage divider. As shown in the figure the output of each 470 ohm resistor are 3.680V and 3.683V. The voltages are not equal since the peak values of each pulse have some minute difference. The sum of the voltage of the resistors is the output of the full wave rectifier. Similar computations can be done in solving for the DC voltage for each the load resistor with an ordinary half-wave rectifier. Depending on the biasing of the diodes, the full-wave rectifier could either conduct during the negative or positive pulse, depending on the design. Shown below is bridge-type full-wave rectifier which conducts during positive cycle.
Peak voltage computation:
Vp = Vrms / 0.707
Vp = 9 / 0.707
Vp = 12.73V
Voltage output:
VL = 0.637(Vp 2 * 0.7)
VL = 0.637(12.73-2 * 0.7)
VL = 6.99V
Bridge-type Full-wave Rectifier with Capacitor
Three circuits are shown above with increasing value of capacitance. The load voltage is directly proportional to the value of the filter capacitance. Therefore the smallest capacitance has the least output value. The purpose of the filter capacitor is to flatten the pulsating output of the load resistor which is shown in the oscilloscope with different filter capacitance. The greater the capacitance, the flatter the output will be, eliminating the ripple voltage in the process. This is due to the faster charging and discharging time of the capacitor which is shown in the equation:
t = RC
The DC voltage with filter capacitor cannot be solve by simply using the formula VL = 0.637(Vp 2 * 0.7) since the charging and discharging time changes, therefore differing its conduction area from a ordinary half-wave rectifier. The DC voltage increases as the output waveform becomes more flat since the conduction area also increases.
II.
Center-tapped Full-wave Rectifier
Since the center the transformer is grounded, the input peak voltage of the center-tapped transformer is half the ordinary practical transformer used for the other rectifiers. Similar computations with the bridge-type rectifier for deriving its output, only, the peak voltage must be first divided into two.
Peak voltage computation:
Vp = Vrms / 0.707
Vp = 9 / 0.707
Vp = 12.73V
Peak voltage of the center-tapped transformer:
Vp = Vp / 2
Vp = 12.73 / 2
Vp = 6.37V
Voltage output:
VL = 0.637(Vp 0.7)
VL = 0.637(12.73 - 0.7)
VL = 3.61V
Center-tapped Full-wave Rectifier with Filter Capacitor
The purpose of the filter capacitor is to flatten the pulsating output of the load resistor which is shown in the oscilloscope with different filter capacitance from above examples. The greater the capacitance, the flatter the output will be, eliminating the ripple voltage in the process. This is due to the faster charging and discharging time of the capacitor which is shown in the equation:
t = RC
The DC voltage with filter capacitor cannot be solve by simply using the formula VL = 0.637(Vp 0.7) since the charging and discharging time changes, therefore differing its conduction area from a ordinary half-wave rectifier. The DC voltage increases as the output waveform becomes more flat since the conduction area also increases.
III.
Clippers
Series Clipper Forward-Negative
The DC supply 3V causes the diode to be in reversed biased, therefore clipping the output voltage by 3V and the 0.7 threshold voltage, as shown by the following equation:
Peak voltage computation:
Vp = Vrms / 0.707
Vp = 9 / 0.707
Vp = 12.73V
Peak voltage of the load resistor:
VL(peakvalue) = Vp 3 0.7
VL(peakvalue) = 12.37 3 0.7
VL(peakvalue) = 9.03V
Load Voltage in DC:
Vdc = VL(peakvalue) * 0.318
Vdc = 9.03 * 0.318
Vdc = 2.87V
Series Clipper Forward-Positive
The DC supply 3V causes the diode to conduct positively, and adding the amplitude of the peak value of the positive pulse of the input. It is careful to take note that there is always a voltage drop across the diode.
Peak voltage computation:
Vp = Vrms / 0.707
Vp = 9 / 0.707
Vp = 12.73V
Peak voltage of the load resistor:
VL(peakvalue) = Vp + 3 0.7
VL(peakvalue) = 12.37 + 3 0.7
VL(peakvalue) = 15.03V
Load Voltage in DC:
Vdc = VL(peakvalue) * 0.318
Vdc = 15.03V * 0.318
Vdc = 4.78V
Series Clipper Reverse-Negative
The DC supply 3V causes the diode to conduct negatively, and adding the amplitude of the peak value of the negative pulse of the input. The diode does not conduct during the positive pulse since it will be reverse biased.
Peak voltage computation:
Vp = Vrms / 0.707
Vp = 9 / 0.707
Vp = 12.73V
Peak voltage of the load resistor:
VL(peakvalue) = Vp - 3 + 0.7
VL(peakvalue) = -12.37 - 3 + 0.7
VL(peakvalue) = -14.67V
Load Voltage in DC:
Vdc = VL(peakvalue) * 0.318
Vdc = -15.03V * 0.318
Vdc = -4.67V
Series Clipper Reverse-Positive
The diode will conduct only during the negative pulse of the supply. The DC supply 3V causes the diode to be in reversed biased, therefore clipping the output voltage by 3V and the 0.7 threshold voltage, as shown by the following equation:
Peak voltage computation:
Vp = Vrms / 0.707
Vp = 9 / 0.707
Vp = 12.73V
Peak voltage of the load resistor:
VL(peakvalue) = Vp + 3 + 0.7
VL(peakvalue) = -12.37 + 3 + 0.7
VL(peakvalue) = -8.67V
Load Voltage in DC:
Vdc = VL(peakvalue) * 0.318
Vdc = -8.67 * 0.318
Vdc = -2.75V
Parallel Clipper Forward-Negative
The diode will conduct during the positive pulse, but the it will be limited only to 3 volts( 3.7 including the threshold voltage of diode) since the output is parallel to the diode and DC supply. Also, the output of the circuit is parallel to its input, so if the diode does not conduct, which is during the negative cycle, the output voltage will be in phase its input voltage.
The DC output value cannot be solved by directly substituting to a formula since the time of its conduction is not 180 degrees, therefore varying its conduction area. Its area is an indefinite shape which is a function of the time of its conduction. The DC output voltage can be computed with the aid of integral calculus.
Parallel Clipper Forward-Positive
The diode will conduct during the positive cycle of the input. During its conduction, the output would be equal to -3volts (threshold voltage is assumed to be zero) since the output is parallel to the DC supply. If the diode is on off state which is during negative cycle, the output will then be in phase with input voltage but starting on the reference 3 volts, since then again, the output will be paralleled to its input supply.
The DC output value cannot be solved by directly substituting to a formula since the time of its conduction is not 180 degrees, therefore varying its conduction area. Its area is an indefinite shape which is a function of the time of its conduction. The DC output voltage can be computed with the aid of integral calculus.
Parallel Clipper Reverse-Negative
At any given time, the diode will conduct since it is forward biased by the 3V DC supply. Since again the output is parallel to the diode and DC supply, the output will be equal to 2.3 volts ( 3 - 0.7) since the diode and the supply is on opposite polarities. The circuit is therefore limited to the least value of 2.3 volts peak value. During nonconduction, the output will be in phase to the input voltage since both will then be paralleled to each other.
The DC output value cannot be solved by directly substituting to a formula since the time of its conduction is not 180 degrees, therefore varying its conduction area. Its area is an indefinite shape which is a function of the time of its conduction. The DC output voltage can be computed with the aid of integral calculus.
Parallel Clipper Reverse-Positive
The diode will conduct during the negative cycle with an output voltage of -2.3V(-3 + 0.7), since the diode and the DC supply is parallel to the output. The least value of voltage will then be limited to -2.3V. During nonconduction, the output will be paralled to the input thus both voltages will be in phase to each other.
The DC output value cannot be solved by directly substituting to a formula since the time of its conduction is not 180 degrees, therefore varying its conduction area. Its area is an indefinite shape which is a function of the time of its conduction. The DC output voltage can be computed with the aid of integral calculus.
IV. Clampers
Clamper Forward
During conduction of diode, the output voltage would be equal to 0V( for this case, -0.7V). From that given time wherein VL = 0V, the capacitor will then start to charge up until it is equal to -12.37V(-9 * 0.707) wherein on that given time, Vin = 0V. The capacitor will then discharge during the negative cycle, in which then, the diode is in off state.
Note: the threshold voltage is assumed to be zero for convenience.
Peak voltage of RL after discharge:
-12.37 -12.37 - VL = 0
VL = -24.7V
Clamper Reverse
If the diode is on on state which is during the negative cycle, the output voltage would be equal to 0V( for this case, -0.7V). From that given time wherein VL = 0V, the capacitor will then start to charge up until it is equal to 12.37V(9 * 0.707) wherein on that given time, Vin = 0V. The capacitor will then discharge during the positive cycle, in which then, the diode is in off state.
Note: the threshold voltage is assumed to be zero for convenience.
Peak voltage of RL after discharge:
12.37 +12.37 - VL = 0
VL = 24.7V
Clamper Forward Negative
During conduction of diode, the output voltage would be equal to 3V since the DC supply is parallel to the load. From that given time wherein VL = 3V, the capacitor will then start to charge up until it is equal to 9.37V(9 * 0.707 - 3) wherein on that given time, Vin = 0V. The capacitor will then discharge during the negative cycle, in which then, the diode is in off state.
Note: the threshold voltage is assumed to be zero for convenience.
Charge of the capacitor:
12.37 Vc 3 = 0
Vc = 9.37V
Peak voltage of RL after discharge:
+12.37 +9.37 - VL = 0
VL = 21.7V
Clamper Forward Positive
The diode will conduct during the positive cycle, the output voltage would be equal -3V(-2.3V if the threshold voltage is considered) since the DC supply is parallel to the load. From that given time wherein VL = -3V, the capacitor will then start to charge up until it is equal to 15.37Vwherein on that given time, Vin = 0V. The capacitor will then discharge during the negative cycle, in which then, the diode is in off state.
Note: the threshold voltage is assumed to be zero for convenience.
Charge of the capacitor:
12.37 Vc + 3 = 0
Vc = 15.37V
Peak voltage of RL after discharge:
12.37 +15.37 - VL = 0
VL = 27.74V
Clamper Reverse Negative
The diode is on state during the negative cycle, the output voltage would be equal to 3V since the DC supply is parallel to the load. From that given time wherein VL = 3V, the capacitor will then start to charge up until it is equal to 15.37V wherein on that given time, Vin = 0V. The capacitor will then discharge during the positive, in which then, the diode is in off state.
Note: the threshold voltage is assumed to be zero for convenience.
Charge of the capacitor:
-12.37 + Vc - 3 = 0
Vc = 15.37V
Peak voltage of RL after discharge:
+12.37 + 15.37 - VL = 0
VL = 27.74V
Clamper Reverse Positive
The diode is on state during the negative cycle, the output voltage would be equal to -3V since the DC supply is parallel to the load. From that given time wherein VL = -3V, the capacitor will then start to charge up until it is equal to -9.37V(9 * 0.707 - 3) wherein on that given time, Vin = 0V. The capacitor will then discharge during the positive, in which then, the diode is in off state.
Note: the threshold voltage is assumed to be zero for convenience.
Max Charge of the capacitor:
-12.37 + Vc + 3 = 0
Vc = 9.37V
Peak voltage of RL after discharge:
+12.37 + 9.37 - VL = 0
VL = 21.74V