Download Half wave rectifier and Full wave rectifier and more Summaries Circuit Theory in PDF only on Docsity!
Lecture notes: 02
Rectifier – Half wave rectifier and Full wave rectifier
Dr. Mohammad A Rashid December 15, 2020 just.edu.bd/t/rashid
Contents
1 Alternating current (AC) 1
2 Half wave rectifier 4
3 Full wave rectifier 10
4 RC circuit 13
5 Half wave rectifier with capacitor filter 15
6 Full wave rectifier with capacitor filter 15
1 Alternating current (AC)
Alternating current (AC) describes the flow of charge that changes direction periodically. As a result, the voltage level also reverses along with the current. AC is used to deliver power to houses, office buildings, etc. The AC voltage of a periodic waveform may be written as
V (t) = Vm sin(ωt), (1)
where ω = 2π/T is the angular frequency of the waveform or voltage and T is the time period of the voltage.
Figure 1: AC voltage V (t)
Average voltage over the time period
The average value of V (t) over the time period T is defined as
V =
T
∫ T
0
V (t)dt. (2)
Hence
V =
Vm T
∫ T
0
sin(ωt)dt
Vm T
[
cos(ωt) ω
]T
0
Vm ωT
{− cos(ωT ) + cos 0}
=
Vm 2 π
{− cos(2π) + cos 0}
=
Vm 2 π
Therefor, the average value of an AC voltage over the time period of the oscillation is zero.
Hence
V (^) rms^2 =
V (^) m^2 T
∫ T
0
sin^2 (ωt)dt
V (^) m^2 2 T
∫ T
0
2 sin^2 (ωt)dt
V (^) m^2 2 T
∫ T
0
{ 1 − cos(2ωt)} dt
V (^) m^2 2 T
∫ T
0
dt −
V (^) m^2 T
∫ T
0
cos(2ωt)dt
V (^) m^2 2 T
[
T
]T
0
V (^) m^2 2 T
[
sin(2ωt) 2 ω
]T
0
=
V (^) m^2 2
V (^) m^2 4 ωT
sin(2ωT ) − sin(0)
V (^) m^2 2
V (^) m^2 4 ωT
sin(4π) − sin(0)
V (^) m^2 2
V (^) m^2 4 ωT
V (^) m^2 2
Therefore,
Vrms =
Vm √ 2
≈ 0. 707 Vm. (7)
Important relations to remember
A few handy things to keep in mind about RMS values that apply when dealing with a sine wave, are as follows:
- Vrms ≈ 0. 707 × peak AC voltage = 70.7 % of peak voltage
- Peak AC voltage ≈ 1.414 ×Vrms = 141.1 % of Vrms
- Vavg ≈ 0. 637 × peak AC voltage = 63.7 % of peak voltage
Vrms Vavg
π 2
2 Half wave rectifier
A half wave rectifier is a type of rectifier that only allows one half-cycle of an AC voltage waveform to pass, blocking the other half-cycle. Half-wave rectifiers are used to convert
AC voltage to DC voltage, and only require a single diode to construct. A half wave rectifier is the simplest form of rectifier available.
Figure 2 shows the input AC voltage waveform, the circuit diagram and the final output voltage waveform of a half wave rectifier. During the positive half cycle, the diode is forward biased making the current flow through the load resistor. While during the Negative half cycle the diode is reverse biased so it stops the current flow through the load resistor. Since current can not flow through the load during the negative half cycles, the output voltage is equal to zero.
Figure 2: Half wave rectifier circuit diagram and waveform [electrical4u.com].
Therefor, for an AC voltage given by (1) the output voltage of a half wave rectifier will be (for an ideal diode)
Vo(t) =
Vm sin(ωt), 0 ≤ t ≤ T / 2
0 , T / 2 ≤ t ≤ T
Average output voltage of a half wave rectifier
To calculate the average voltage, Vdc, of the pulsating DC output of a half wave rectifier we use the definition (2). Therefore, for the voltage (8) we have
Vdc =
T
∫ T
0
Vo(t)dt
T
∫ T / 2
0
Vm sin(ωt)dt +
T
∫ T
T / 2
0 dt
Vm T
∫ T / 2
0
sin(ωt)dt
Vm T
[
cos(ωt) ω
]T / 2
0
Vm ωT
{− cos(ωT /2) + cos(0)}
=
Vm π
Here we have used the relation ω = 2π/T.
To calculate Vr(rms), the RMS value of the AC component present in the output of the half wave rectifier we write the output voltage as
Vo(t) = Vac + Vdc, (13)
where Vac is the AC component remaining when converting the AC voltage waveform into a DC waveform. The RMS value of the AC component present in the output of the half wave rectifier is given by
Vr(rms) =
[
T
∫ T
0
V (^) ac^2 dt
] 1 / 2
Therefore,
V (^) r(rms)^2 =
T
∫ T
0
(Vo − Vdc)^2 dt
T
∫ T
0
(V (^) o^2 − 2 VoVdc + V (^) dc^2 ) dt
T
∫ T
0
V (^) o^2 dt −
2 Vdc T
∫ T
0
Vo dt + V (^) dc^2
= V (^) rms^2 − 2 V (^) dc^2 + V (^) dc^2 = V (^) rms^2 − V (^) dc^2. (15)
Hence the formula to calculate the ripple factor can be written as
γ =
Vr(rms) Vdc
Vrms Vdc
Using the values of Vdc and Vrms given in (9) and (11) respectively for the half wave rectifier we find the the ripple factor as
γ =
Vm 2
×
π Vm
π 2
Note that to construct a good rectifier, one should keep the ripple factor as low as possible. This is why capacitors and inductors as filters are used to reduce the ripples in the circuit.
Efficiency of half wave rectifier
The ratio of the DC power available at the load to the applied input AC power is known as the efficiency, η. Mathematically it can be given as:
η =
DC power output AC power input
Pdc Pac
Let rf and RL be the forward resistance and load resistance of the diode. The voltage appearing across the secondary of the power transformer is given by (1). The waveform diagram at the right side of the Figure 2 shows only a positive waveform at the output and a suppressed negative waveform. During the conduction period the instantaneous value of the current is given by the equation:
I(t) =
V (t) RL + rf
Vm RL + rf
sin(ωt) = Im sin(ωt), (19)
with Im = Vm/(rf + RL) being the maximum current.
Now, the AC power input to the load is given as,
Pac = Irms^2 (RL + rf ) =
V (^) rms^2 RL + rf
Since the output is obtained across RL, the DC power output is given by
Pdc = Idc^2 RL =
V (^) dc^2 RL
The half wave rectifier efficiency is then
η =
Pdc Pac
=
V (^) dc^2 RL
×
RL + rf V (^) rms^2
=
V (^) dc^2 V (^) rms^2
×
RL + rf RL
=
Vdc Vrms
×
rf RL
Vm/π Vm/ 2
×
rf RL
rf RL
In reality rf is much smaller then RL. If we neglect rf compare to RL then the efficiency of the rectifier is maximum. Therefore,
ηmax ≈ 0 .4053 = 40.53%. (23)
This indicates that the half wave rectifier can convert maximum 40.53% of AC power into DC power, and the remaining power of 59.47% is lost in the rectifier circuit. In fact, 50% power in the negative half cycle is not converted and the remaining 9.47% is lost in the circuit.
Advantages of half wave rectifier
The main advantage of half-wave rectifiers is in their simplicity. As they do not require as many components, they are simpler and cheaper to setup and construct. As such, the main advantages of half-wave rectifiers are:
- Simple (lower number of components)
- Cheaper up front cost (as their is less equipment. Although there is a higher cost over time due to increased power losses)
Disadvantages of half wave rectifier
The disadvantages of half-wave rectifiers are:
- They only allow a half-cycle through per sinewave, and the other half-cycle is wasted. This leads to power loss.
- They produces a low output voltage.
- The output current we obtain is not purely DC, and it still contains a lot of ripple (i.e. it has a high ripple factor)
3 Full wave rectifier
A full wave rectifier converts both halves of each cycle of an alternating wave (AC signal) into pulsating DC signal. Figure 3 shows the input AC voltage waveform, the circuit diagram and the final output voltage waveform of a center tapped full wave rectifier.
Figure 3: Center tapped full wave rectifier circuit diagram and waveform [electrical4u.com].
For an AC voltage given by (1) the waveform of the output voltage of a full wave rectifier can be written as (for an ideal diode)
Vo(t) =
Vm sin(ωt), 0 ≤ t ≤ T / 2
Vm sin(ωt − π), T / 2 ≤ t ≤ T
Average output voltage of a full wave rectifier
Vdc =
T
∫ T
0
Vo(t)dt
T / 2
∫ T / 2
0
Vm sin(ωt)dt
2 Vm T
∫ T / 2
0
sin(ωt)dt
2 Vm π
RMS value of the output voltage of a full wave rectifier
Vrms =
[
T
∫ T
0
V (^) o^2 (t)dt
] 1 / 2
[
V (^) m^2 T / 2
∫ T / 2
0
sin^2 (ωt)dt
] 1 / 2
[
V (^) m^2 T
∫ T / 2
0
2 sin^2 (ωt)dt
] 1 / 2
Vm √ 2
Ripple factor of full wave rectifier
γ =
Vrms Vdc
π 2
- It can be used to supply polarized voltage in welding.
- The Bridge Rectifier circuits are widely used in power supply for various appliances, as they are capable of converting the High AC voltage into Low DC voltage.
Advantages of full wave rectifier
- Full wave rectifiers have higher rectifying efficiency than half-wave rectifiers. This means that they convert AC to DC more efficiently.
- They have low power loss because no voltage signal is wasted in the rectification process.
- The output voltage of center tapped full wave rectifier has lower ripples than a half wave rectifiers.
Disadvantages of full wave rectifier
- The center tapped rectifier is more expensive than half-wave rectifier and tends to occupy a lot of space.
A comparison of different parameters related to the half and full wave recti- fiers are given below:
Parameters Half wave rectifier Full wave rectifier
Number of diodes 1 2 or 4
Maximum efficiency 40.53% 81.06 % Peak inverse voltage Vm Vm or 2Vm Average voltage no load Vm/π 2 Vm/π
Vrms no load Vm/ 2 Vm/
Ripple factor 1.21 0.
Form factor 1.57 1. Output frequency f 2 f
4 RC circuit
An RC circuit (also known as an RC filter or RC network) stands for a resistor-capacitor circuit. An RC circuit is defined as an electrical circuit composed of the passive circuit components of a resistor (R) and capacitor (C), driven by a voltage source or current
source. In an RC series circuit, a pure resistor having resistance R in ohms and a pure capacitor of capacitance C in Farads are connected in series.
Figure 4: RC circuit diagram [electrical4u.com].
RC time constant
The RC time constant indicates the rate of charge or discharge. RC specifies the time it takes C to charge to 63% of the charging voltage. Similarly, RC specifies the time it takes C to discharge 63% of the way down to the value equal to 37% of the initial voltage across C at the start of discharge.
Figure 5: C charges through R to 63% of VT in one RC time constant.
Figure 6: C to discharge through R, VC drops to 37% of its initial voltage in one time constant.
the maximum value of the input voltage. At this point, the supply voltage is equal to capacitor voltage.
Figure 9: Waveform of full wave rectifier RC filter [electrical4u.com].
When the applied AC voltage starts decreasing and less than the capacitor, the capacitor starts discharging slowly but this is slower when compared to the charging of the capacitor and it does not get enough time to discharge entirely and the charging starts again. So around half of the charge present in the capacitor gets discharged. During the negative cycle, the diode D 2 starts conducting, and the above process happens again. This will cause the current to flow through the same direction across the load.
