The *voltage measured
by the voltmeter is in rms*

V_{rms} = V_{p} χ 2^{1/2}

^{
or}

V_{rms }= 0.707 *_{ }V_{p}

^{
}

^{ }

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 = V_{p}/V_{s}

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 = V_{i }- V_{o}

Vt = 12.6787_{ } 11.9355

Vt = 0.743V

A threshold voltage of **0.7V
**(threshold voltage of silicon) will be assumed in the computation.

V_{L }= 0.318(Vp_{ } 0.7)

V_{p} = V_{rms }/_{ }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:*

V_{p} = V_{rms }/_{ }0.707
_{ }

V_{p} = 9 / 0.707

V_{p} = 12.73V

*
Voltage output:*

V_{L }= 0.318(Vp_{ } 0.7)

V_{L} = 0.318(12.73-.7)

V_{L} =
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
V_{L }= 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

** **

**
**

_{
}

*
Peak voltage computation:*

V_{p} = V_{rms }/_{ }0.707
_{ }

V_{p} = 9 / 0.707

V_{p} = 12.73V

*
Voltage
output:*

V_{L }= 0.318(Vp_{ } 0.7)

V_{L} = 0.318(12.73-.7)

V_{L} = 3.83V

*
Total voltage
output:*

*
*V_{o} = V_{L} * 2

V_{o} = 3.83 * 2

V_{o }= 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:*

V_{p} = V_{rms }/_{ }0.707
_{ }

V_{p} = 9 / 0.707

V_{p}
= 12.73V

*
Voltage output:*

V_{L }= 0.637(Vp_{
} 2 * 0.7)

V_{L} = 0.637(12.73-2 * 0.7)

V_{L} = 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
V_{L }= 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

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:*

V_{p} = V_{rms }/_{ }
0.707 _{ }

V_{p} = 9 / 0.707

V_{p} = 12.73V

*Peak voltage of the
center-tapped transformer:*

Vp = V_{p} / 2

Vp = 12.73 / 2

Vp = 6.37V

*
Voltage output:*

V_{L }= 0.637(Vp_{ } 0.7)

V_{L} = 0.637(12.73 - 0.7)

V_{L}
= 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
V_{L }= 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:*

V_{p} = V_{rms }/_{ }0.707
_{ }

V_{p} = 9 / 0.707

V_{p} = 12.73V

*Peak
voltage of the load resistor:*

V_{L}(*peakvalue*) = V_{p}
3 0.7

V_{L}(*peakvalue*) = 12.37 3
0.7

V_{L}(*peakvalue*) = 9.03V

*Load
Voltage in DC:*

V_{dc }= V_{L}(*peakvalue*) *
0.318

V_{dc }= 9.03 * 0.318

V_{dc }= 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:*

V_{p} = V_{rms }/_{ }0.707
_{ }

V_{p} = 9 / 0.707

V_{p}
= 12.73V

*Peak
voltage of the load resistor:*

V_{L}(*peakvalue*) = V_{p} +
3 0.7

V_{L}(*peakvalue*) = 12.37 + 3
0.7

V_{L}(*peakvalue*) = 15.03V

*Load
Voltage in DC:*

V_{dc }= V_{L}(*peakvalue*) *
0.318

V_{dc }= 15.03V * 0.318

V_{dc }= 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:*

V_{p} = V_{rms }/_{ }0.707
_{ }

V_{p} = 9 / 0.707

V_{p} = 12.73V

*Peak
voltage of the load resistor:*

V_{L}(*peakvalue*) = V_{p} -
3 + 0.7

V_{L}(*peakvalue*) = -12.37 - 3 +
0.7

V_{L}(*peakvalue*) = -14.67V

*Load
Voltage in DC:*

V_{dc }= V_{L}(*peakvalue*) *
0.318

V_{dc }= -15.03V * 0.318

V_{dc }= -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:*

V_{p} = V_{rms }/_{ }0.707
_{ }

V_{p} = 9 / 0.707

V_{p}
= 12.73V

*Peak
voltage of the load resistor:*

V_{L}(*peakvalue*) = V_{p} +
3 + 0.7

V_{L}(*peakvalue*) = -12.37 + 3 +
0.7

V_{L}(*peakvalue*) = -8.67V

*Load
Voltage in DC:*

V_{dc }= V_{L}(*peakvalue*) *
0.318

V_{dc }= -8.67 * 0.318

V_{dc }= -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.

**
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 V_{L }= 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 - V_{L }= 0

V_{L} = -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 V_{L }= 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 - V_{L }= 0

V_{L} = 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 V_{L }= 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 - V_{L }= 0

V_{L} = 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 V_{L }= -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 - V_{L }= 0

V_{L} = 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 V_{L }= 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 - V_{L }= 0

V_{L} = 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 V_{L }= -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 - V_{L }= 0

V_{L} = 21.74V

** **