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Figure 8.7. Charging the capacitor C2
The voltage across the capacitor C2 is twice the input voltage provided that the current through the load RL is very small. If the stream is too large, ie the small resistance RL, the capacitor C2 will be discharged rapidly through, and the voltage across it remains enough time to double the expected level. For this reason we have the desired effect with a normal load, the capacitors C1 and C2 are chosen large. Typical values of these capacitors is approximately 100 mg or even higher.
8.6.1 Voltage Multiplier
The Sch.8.8 shows a multiplier circuit voltage for the case of 6-plasiasmou voltage. The circuit can be extended to give any multiple of the input voltage. The comparison of this circuit to circuit voltage dipalsiasmou shows that comprises a sequence of doubling circuits in series.
Where the ac input voltage gradually charged capacitors C2, C4, C6 until they each create a potential difference 2Vp so, because we associate the corresponding double-sentation in series, the final output will be 6Vp. Finally, the capacitors C1, C3, C5 ac capacitors act as couplers, carrying the ac input to the rectifiers D2, D4, D6, so that the capacitors C2, C4 and C6 to find ¬ Dai in contact with the tape.
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The doubling and multiplication circuits susceptible to voltage load current relatively low price and are mainly used in devices such as cathode rays (CRT), fotopollapla siastes-ionization tubes Geiger-Muller.
In large volumes, the ripple and regulation regarding the burden is too poor for two reasons. First, the storage capacitors consist of a number of capacitors in series, so the effective value of the closing. Second, it takes several cycles to ac recharge all capacitors are in a multiple voltage, unlike the full rectifier, where the storage capacitors are recharged twice in each ac cycle.
Input circuit
(Rms) voltage ac
Output
chorisforto Output Voltage with Reverse Burdens
Trend
rectifier current rectifier
Imianorthotis
Full Rectifier
Bridge rectifier with
2-Voltage plasiastis
P.8.1 List. Comparative characteristics of rectifier
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8.7 Smoothing filters
The process by which removes partially or totally on the production ¬ ac component of the rectified waveform and thus normalize the smooth the waveform to be close as much as possible to a constant dc output, called normalization. In Sch.8.9 depicted a fully rectified voltage waveform with a corresponding partially streamlined (thick line) and a completely streamlined dc voltage (dotted line).
Figure 8.9. Typical rectified waveform (a), some streamlined (b), and fully streamlined (c)
In the general case, the normalization will not be complete, so the final voltage n0 displays periodic variation, as shown. The range Vnc = AV this variation called ripple and is an important feature of a feed device. Often the ripple is expressed in percentage form, ie by the factor:
Vdc
8.7.1 Smoothing capacitor
For normalizing erectile waveforms and the meta ¬ trepsoume a constant dc output voltage, we use a large-capacity capacitor C, connected in parallel to the load after the rectifier. The value of this capacitor on the grid frequency (50 Hz), ranges from 100 to 30000 and MF depends on the load current and the degree of normalization, which we desire.
The Sch.8.10 shows the circuits of single and double
ripple r = AV I
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equipped with the rectifier smoothing capacitor, and the Sch.8.11 depicts the waveforms of the output circuits. The smoothed waveforms are those corresponding to the charge and discharge voltages of the figure. The output voltage is a smooth continuous line, while the rectified output is the dashed line the average dc value assigned by the dotted (crumbled) line.
(A) (b)
Figure 8.10. Simple circuits (a) and dual rectifier (b)
smoothing filter capacitor
Vp
1 Charging Discharging ripple r = AV
i mean t.
h The Season
NP
(A) (b)
Figure 8.11. Waveforms output rectifier with filter capacitor.
Imianorthosis cases (a) full recovery (b)
The smoothing operation is carried out as follows: During the latter part of the first half of each pulse rectified output voltage charging the capacitor and the voltage is increased to the peak voltage Vp. During the remainder of the rectified output of the cycle, the capacitor is discharged through the load RL. Therefore, the longer each cycle only provides peak load.
For the complete recovery period of the triangular output ripple is T = 1/2f. Therefore, the ripple at full lift is given by:
(8.7.6)
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221
capacitor in each time you close the charge-discharge cycle. If the current through the load was zero, the capacitor would remain stable charged to peak voltage V input signal.
The existence of the load current results in even slightly discharged the capacitor at r = AV for each cycle of the input signal. This gives rise to an abnormal periodic dc output voltage waveform shown as approximately triangular waveform.
It turns out that for the case of simply righting, Sch.8.10 the ripple r is given by the equations:
(8.7.2)
where, Vnc Vdc and ac and dc the corresponding component of the output voltage, while I represents the average value of the dc load current equal to:
Vdc
(8.7.3)
T is the period and the frequency f of the ac input voltage. The average value of the dc output voltage is:
IVdc V-(8.7.4)
For the simple rectifier ripple is given by the relations
Vac 1 Vdc 2 ^ 3RLCf (8.7.5)
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and the output voltage from the relationship
(8.7.7)
As shown in Sch.8.11, the ripple in the double lift is twice the frequency in simple recovery. This is another reason I prefer the double lift.
Example 8-1
A power supply uses a full recovery given by dc current of 150 mA and 20 V dc voltage ripple with no more than 1%. Calculate the load resistor and capacitor smoothing.
Solution
In Eq. (8.7.3), the load resistor will be:
20 V
R, == 133 Z
L 0.15A
Finally, from Eq. (8.7.5), we find:
8.7.2 Smoothing with LC Filter
As we saw in the previous paragraph, the rectifier output filter capacitor has a certain ripple which, as shown by Eq. (8.7.4) or (8.7.6) is proportional to the load current I. In other words, if you grow the current drawn by the load, and the ripple grows. This dependence is illustrated in Sch.8.12. The discharge of the capacitor is the main cause of this problem.
To counteract the above disadvantages make use of the smoothing filter Sch.8.13, which uses two capacities C1 and C2 and an inductance L as shown. To inductor L (Chuck) opposes any change in power and because the load current flowing through the coil tends to keep it stable.
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O first capacitor, C,, acts as a filter capacitor in the manner we saw in the previous paragraph. With the addition of the second capacitor C2, the output of the filter further filtered and becomes almost completely free from ripple. Typical values of the circuit are: C, = C2 = 200 MP and L = 1 έως 30 H.
(A) low current (b) average current (c) large currents
Figure 8.12. Change the ripple depending on the load current
From the rectifier
Figure 8.13. LC smoothing filter to improve the ripple
The usefulness of the coil can be seen if we calculate the inductive resistance, XL, the ripple frequency. For complete recovery, in which the ripple frequency is f = 2 x 50 = 100 Hz and coil 30 H, we have:
XL = 2 pi f L = 2 pi x 100 x 30 19 KW
In comparison, this value ohmic (dc) resistance of the coil, which is about some 10-don Oh, it's very small. So, for the comparable currents dc dc voltage drop in the coil will be much smaller than the ac voltage drop, so the remaining ac output voltage Vnc is
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much smaller than the corresponding Vdc. Thus, according to Eq. (8.7.1), the ripple is very low r.
Finally, the output LC filter is called a resistance drain resistance, Rn, and serves to quickly capacitors discharged through it, rather than the operator, when "cut" the tendency of the network. The current drain is typically 10% of the total load current.
It turns out that the ripple consideration for the smoothing filter, which is of type P is given by:
(8.7.8)
where the capacities C1 and C2 are expressed in me, the inductance in H and resistance in ohms
Example 8-2
Uses a power rectifier which gives full current of 100 mA and dc voltage of 12 V. The ripple of output we want to be 0.1%. Calculate the appropriate smoothing filter press P.
Solution
The burden is
Assuming that, for symmetry, C1 = C2 = C, after solving the Eq. (8.7.8) we have:
(8.7.9)
Assuming L for a normal market value L = 20 H, compute the
C = 49 wIth
V 0.001 x 20x 120
Tantalum electrolytic capacitors use a 100 with.
