Cockroft Walton Voltage Multiplier Circuit
Q. Why is Cockcroft-Walton circuit preferred for voltage multiplier circuits? Explain its working with a schematic diagram.
Q. Derive an expression for ripple voltage of a Multistage Cock Croft-Walton circuit.
- It is possible to generate very high DC voltages from single supply transformers by extending the voltage doubler circuit. This is simple and compact when load current requirement is less than 1 mA
- Voltage multiplier circuit using the Cockroft Watlon principle is shown in Figure (A).
- The first stage i.e. D1,
D1’, C1 and C1’ and transformer T are identical as in voltage doubler circuit For higher output voltage of 4, 6, … 2n of the input voltage V, the circuit is repeated with cascade or series connection. Thus capacitor C4
is charged to 4Vmax and C2n to 2nVmax above earth potential. But voltage max across any individual capacitor or rectifier is only 2Vmax.
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| Figure A |
Working of Cockroft Walton Voltage Multiplier Circuit
- The portion ABM’MA is exactly identical and voltage across C becomes 2Vmax when M attains a voltage 2Vmax. (Figure (A)).
- During the next half cycle when B becomes positive with respect to A, potential of M falls and therefore potential of N also falls becoming less than potential at M’. Hence C2 is charged through D2. (Figure(B)).
- Next half cycle A becomes more positive and potential of M and N rise bris thus charging C2 ‘ through D2‘
- Finally all capacitors C1’, C2’, C3’ , C1, C2 and C3 are charged.
- The voltage across the column of capacitors consisting of C1, C2 and C3 keeps on oscillating as the supply voltage alternates therefore, this column is known as oscillating column.
- However voltage across the capacitances C1’, C2’, C3’ remains constant and is known as smoothening column.
- The voltages at M’, N’, and O’ are 2Vmax, 4Vmax, and 6Vmax. Therefore voltage across all the capacitors is 2Vmax except for C1 where it is Vmax only.
- The total output voltage is 2nVmax where n is the number of stages. The equal stress of the elements (both capacitors and diodes) used is very helpful and promotes a modular design of such generators.
- Generator loaded condition : When the generator is loaded, the output voltage will never reach the value 2nVmax. Also the output wave will consists of ripples on the voltage.
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| Figure B |
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| Figure C |
- Say potential of point O’ is now 6Vmax. This discharges through the load resistance and say the charge lost is q=IT over the cycle.
- This must be regained during the charging cycle (Figure (A)) for stable operation of graph.
- C3, is therefore supplied a charge ‘q’ from C2.
- For this C2, must acquire a charge of 2q so that it can supply q charge to the load and q to C3, in the next half cycle, termed as transformer cycle (Figure (b)).
Similarly C’1 must acquire for stability reasons a charge 3q so that it can supply a charge q to the load and 2q to capacitance C2, in next half cycle (transformer half cycle).
2𝛿V = I/f (1/C’n + 2/C’n-1
+ 3/C’n-2 + … + n/C’1)
or 𝛿V
= I/f (1/C’n + 2/C’n-1 + 3/C’n-2 + … + n/C’1) ….(1)
For 3 stage
𝛿V = I/f (1/C’3 + 2/C’2+
3/C’1)m
From Equation (1) it is clear that in a multistage circuit the lowest capacitors are responsible for most ripple and it is therefore desirable to increase the capacitance in lower stages. However it is objectionable from the view point of high voltage circuit where if the load is large and the load voltage goes down , the smaller capacitors (within the column) would be overstressed.
Therefore capacitors of equal value are used in practical circuit i.e. C’n = C’n-1 = C’1 = C and ripple is given by.
𝛿V = I/2fC n(n+1)/2 = In(n+1)/4fC ….(2)
The second quantity to be evaluated is voltage drop △V which is difference between theoretical NL voltage 2n Vmax. and the on load voltage.
Here C’1 is not charged upto full voltage 2 Vmax but only to 2Vmax
– 3q/C because of the charge given up through C1, in one cycle which gives a voltage drop of 3q/C = 3I/fC.
The voltage drop in transformer is assumed to be negligible. Thus C2 is charged to the voltage.2Vmax – 3q/C = 3I/fC
Since the
drop in voltage across C’3 is again 3I/fC
∴ C’2
atteain the voltage = 2Vmax– (3I + 3I + 2I / fC)
Hence, in a 3
stage generator
△ V1
= 3I/fC
△V2
= [2 x3 + (3 – 1)] I/fC
And △V3
= (2 x 3 + 2 x 2 + 1) I/fC
In general
for n stage generator
△Vn
= nI/fC
△Vn-1
= I/fC (2n + (n+1))
△V1
= I/fc [2n + 2(n-2)+ …2 x 3 +2 x 2 +1]
△V = △Vn
+ △Vn-1+…+△V1After omitting I/fC, the
series can be rewritten as
Tn =
n
Tn-1 =
2n + (n-1)
T1 = 2n + 2(n-1)+ 2(n-2)+…+2 x 3+2 x 2+1
∴ T =
Tn+Tn-1+Tn-2+…+T1
Taking once
again all terms and simplifying it, we get
|| T = 2n3/3
+ n2/2 – n/6 || ….(3)
Thus, here again the lowest capacitors contribute most to the voltage drop △V and so it is advantageous to increase their capacitance in suitable steps.
However, only by doubling value of C1 decreases the △V1 by an amount nI/fC which further reduces △V of every stage by the same amount i.e. by
n . nI/2fC
Hence, △V = I/fC (2/3n3-n/6 ….(4)
If n ≥ 4 the linear terms can be neglected and therefore the voltage drop can be approximated to
△V ≈ I / fc 2/3 n3 ….(5)
Thus, maximum output voltage (under loaded condition) is given by
V0 max = 2n Vmax –I/fC 2/3 n3 ….(6)
It is to be noted that in general it is more economical to use high frequency and smaller value of capacitance to reduce the ripples or the voltage drop rather than low frequency and high capacitance.
For a given load, V0 = Vmax however,may rise initially with the number max of stages η, and reaches a maximum value but decays beyond an optimum number of stages. The optimum number of stages assuming a constant Vmax I, f and C can be obtained for maximum value of V by differentiating equation (6) w.r.t.n equating it to zero.
d / Vmax = 2 Vmax 2/3 I / fc = 3η2 = 0
= Vmax I / fcη2 = 0
∴ ηopt = √Vmax fc/I
Substituting ηopt in Eqn (6) we have
(V0 max)max
= √Vmax fc/ I (2 Vmax 2I/3Fc Vmax fC/ I)
= √Vmax fc/ I (2 Vmax 2/3 Vmax) ….(8)
= √Vmax fc/ I . 4/3 Vmax