high voltage generation experiment

8/17/2019 High Voltage Generation Experiment

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(iv) high transient or impulse voltages of very short duration such as lightning

overvoltages, and

(v)

transient voltages of longer duration such as switching surges.

Normally, in high voltage testing, the current under conditions of failure is limited to a small

value (less than an ampere in the case of d.c. or a.c. voltages and few amperes in the case of

impulse or transient voltages). But in certain cases, like the testing of surge diverters or the

short circuit testing of switchgear, high current testing with several hundreds of ampere is of

importance. Tests on surge diverters require high surge currents of the order of several

kiloamperes. Therefore, test facilities require high voltage and high current generators. High

impulse current generation is also required along with voltage generation for testing purposes.

To generate high voltage, the transformer is used. Transformers are capable of either increasing

or decreasing the voltage and current levels of their supply, without modifying its frequency,

or the amount of Electrical Power being transferred from one winding to another via the

magnetic circuit. A single phase voltage transformer basically consists of two electrical coils

of wire, one called the “Primary Winding” and another called the “Secondary Winding”. When

a transformer is used to “increase” the voltage on its secondary winding with respect to the

primary, it is called a step-up transformer. When it is used to “decrease” the voltage on the

secondary winding with respect to the primary it is called a step-down transformer.

AC voltage dominates transmission and distribution systems. The most common form of

testing HV apparatus is related to high AC voltages, the testing voltages are usually AC single-

phase voltages to ground and sinusoidal. The r.m.s value for a cycle of T:

= 1 ∫

Moreover, a direct test voltage is defined by its arithmetic mean value,

= 1 ∫

Applied voltage to test objects then deviate periodically from the mean value. Therefore, a

ripple is present, show in equation below:


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= 0 . 5 − Impulse voltages are referred to as a T1/T2 impulse. The ‘front time’ T1 is 1.67T when the

impulse is 0.3 and 0.9 of the peak value. ‘Tail-time’ T2 is time to half of peak. T p is time to

peak. From here, we known that 1.2/50 impulse is the accepted standard lightning impulse

voltage (1.2µs = front time, 50 µs = tail time).

Figure 1 General shape and definitions of lightning impulse (LI) voltages.

The design of the above lightning impulse voltage is depending on the following components:

For front time, = 3 ( + ) While for tail or fall time, =0.7 + +

OBJECTIVES

1. To perform simulation of high AC, DC and impulse voltage generation.

2.

To design circuits generating high voltage based on certain requirements.


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RESULT AND DISCUSSION

TEST 1: HIGH AC VOLTAGE GENERATION

Figure 2 Circuit configuration for high a.c. voltage generation.

Graph 1 Graph voltage versus time for high voltage a.c generation with different secondary

inductance value.

-150000

-100000

-50000

0

50000

100000

150000

0 200 400 600 800 1000 1200 1400 1600 1800 2000 V o l t a g e ,

V

Time, t

Voltage versus Time

L2 = 1000H L2 = 10kH L2 = 100kH L2 = 1000kH L2 = 10000kH


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From the output graph, as the value of secondary inductor is increased, the output voltage is

increases. This response can be described from the following transformer identity:

VPV = √ LPL From the above relationship, the increment of secondary inductor will cause the increment of

the output voltage. Due to the square root function, the increment of the inductor gives a rise

of power of two for the output voltage. This therefore cause a small increment of the inductor

magnitude to give a significant increment of output voltage.

Theoretically, the increment of the output of the voltage is due to the increment of the magnetic

flux at the secondary inductor as we increase the value of the inductor, L2. This obey the

Faraday’ law:

V = −N dΦdt The increment of the inductance give a bigger energy storage, therefore, a greater energy can

be used in order to produce an opposing voltage and it is measured as an output voltage.

Questions:

1. Using a suitable diagram, explain the operation of AC voltage generation.

The presence of AC source in the primary side cause the current flow across resistive

load R 1 and producing primary voltage. The inductor that is connected inside the

primary circuit will induce e.m.f. that will influence the secondary inductor by the

principle of mutual inductance as in which the production of the e.m.f. is producedalong with the magnetic flux. As the secondary inductor does influence by the mutual


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inductance, it will store energy due to its inductance value and consequently, the energy

store will induce e.m.f. at the secondary side. This secondary e.m.f. will cause current

flow and the current will flow through R 2 and produce output voltage. The transformer

will step up the voltage according to the value of inductor in the primary winding and

secondary winding. The relationship between the value of inductor and the voltage is

given by:

VPV = √ LPL = The voltage after step up, VS is supplied to the load R 2 thus creating the alternating

output voltage, V0.

2. What is the effect of L2:L1 on the output voltage V0?

Graph 1 shows that when the value of L2:L1 increased, the output voltage, V0 will also

increase. This is due to Faraday’s Law of Induction that states:

= − = − = −− = 1 ℎ > 1, ℎ < 1 By law of conservation of energy, apparent power and reactive power are each

conserved in the input and output.

S = IP

VP

= I

V

2

Combining (1) and (2) yields the ideal transformer identity:

VPV = IIP = NPN = √ LPL = a 3 LLp = [VVP]

This proves that as the ratio LS:LP or L2:L1 increases, the output voltage V0 or VS

increases.


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3. What is the relationship of L1 and L2 with the turn ratio of a transformer?

From equation (3), we can represent the relationship between turn ratios of a

transformer with L1 and L2 as follow:

NN = √ LL From the above relationship, the relation between turn ratios with L1 is proportional in

square root function while for L2, inversely proportional in function of square root. The

overall relationship state that the turn ratios of the transformer is proportional with ratio

between L1 and L2 in function of square root.

4. Using only components available in the market, design a circuit in PSpice, which can

generate an AC voltage, V0 of 100kV peak magnitude using an AC source that can give

an output between 1V to 240V only and a transformer of L2:L1 = 1,000,000. When

designing the circuit, you must ensure that the voltage rating of each component (120

kV) is not exceeded. The cost of the overall circuit must also be as low as possible, i.e.

the circuit cost is higher if more components are used.

Figure 3 Schematic of Design Circuit.


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Graph 2 Result of design circuit simulation

The applied ratio of L2: L1 is specified to be 1000000 and the peak output voltage we

calculated using equation is as below:

|

| = √ |

| = √ 1000000

1 100=100

From the simulation result, we can see that the output voltage obtained is 96.745kV

which is very close to 100kV.


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TEST 2: HIGH DC VOLTAGE GENERATION

Figure 4 Circuit configuration for high d.c. voltage generation.

Graph 3 Graph voltage versus time for high voltage d.c generation with different capacitance

value.

0

20000

40000

60000

80000

100000

120000

0 200 400 600 800 1000 1200 1400 1600 1800 2000

V o l t a g e ,

V

Time, t

Voltage versus Time

C = 1uF C = 10uF C = 100uF C = 1mF


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Graph 2 shows that as the value of the capacitor increases, the ripple magnitude is reduced.

This is due to the stored energy inside the capacitor discharge when the voltage source is drop.

The presence of diode makes the circuit conducts only positive cycle of the A.C. source. The

function of capacitor is to smoothen the fluctuated output voltage to be constant so that it can

represent a constant D.C. output voltage. Whenever the voltage is increases, the charges that

flow across the capacitor will be stored cause the capacitor to charge. Whenever the voltage is

dropped, the capacitor need to compensate with the drop by releasing those charges, therefore,

it is discharged. The higher the capacity of the capacitor, the higher the amount of charges able

to be stored. This therefore cause the discharge process slow and reduce the ripple voltage.

This occurrence can be represented by the following equation:

= This equation is valid for half-wave rectifier since the circuit in this test only use a diode which

only can produce a half-wave rectifier The above relationship explains the result of this test in

which the increment of the capacitance reduce the ripple voltage.

Graph 4 Graph voltage versus time for high voltage d.c generation with different resistance

value.

-20000

0

20000

40000

60000

80000

100000

120000

-200 300 800 1300 1800 2300

V o l t a g e ,

V

Time, t

Voltage versus Time

R = 1k Ohm R = 100 Ohm R = 10 Ohm R = 1 Ohm


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The result above shows that with the decrement of the voltage, the ripple voltage is getting

bigger. This can be represented by the following equation:

= =

The equation states that the relationship between the ripple voltage and the resistance is

inversely proportional. Theoretically, due to the small resistance value, the current flow will

increases as we relate it with Ohm’s law. Due to this high current requirement to flow across

the resistor, the rate of discharge of the capacitor is increases, therefore cause the increment of

ripple voltage. As we refer to the equation above, the increment of the current will increase the

ripple voltage magnitude. This therefore explains the result obtained.

Questions:

1. Using figure below, explain how a DC voltage is generated from an AC voltage source.

The A.C. source will pass across the diode. The diode will only allow one direction of

current flow, therefore, in such configuration in which the positive terminal is

connected to the anode terminal of the diode, therefore, only positive cycle of the

voltage is allowed. Then, the presence of the capacitor will smoothen the ripple voltage

causes a near constant output voltage to be produced due to the charging and

discharging activity of the capacitor. Therefore, a D.C. output voltage is obtained.


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2. What is the effect of C value on the ripple voltage? Explain why.

As the value of capacitance increases, there is a higher capacity to store the charges.

This therefore, during the charging, the voltage magnitude is reduced in slow rate due

to the discharging of the same rate of charges over a bigger charges stored during

charging process. This caused the ripple voltage to reduce. This relationship can also

be represented by the following equation:

= The equation states that the ripple voltage magnitude is inversely proportional to the

capacitance value.

3. What is the effect of R L value on the ripple voltage? Explain why.

As the value of resistor decreases, the value of current that pass through the resistor is increases.

This is due to the Ohm’s law relationship. In order to compensate with this high curr ent flow,

the capacitance will discharge more in which the rate of the discharge will be increases. This

will cause the ripple voltage to increase significantly. The relationship also can be represented

by the following equation:

= From the above equation, the ripple voltage is inversely proportional to the resistance.

4.

The diode D has a maximum voltage rating of 140kV. In Figure 2, do you think that

having only 1 diode D is sufficient in this circuit? Explain your answer.

Diode must withstand a peak reverse voltage, 2Vmax. So if the diode D has a maximum

voltage rating of 140kv, it is insufficient in this circuit. Due to the source is V AC=100

sin (2π*50t), it will exceed the limits of the diode, 2Vmax.


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5. Using only components available in the market (refer to “Other Information” in page

1), design a circuit in PSpice, which can generate a DC voltage, V0 of +200kV peak

magnitude and allowable 2kV ripple using an AC source (sinusoidal) that can give an

output between 1V to 140kV only. When designing the circuit, you must ensure that

the voltage rating of each component (120kV) is not exceeded. The power dissipated

by each resistor must not exceed 1kW. The cost of the overall circuit must also be as

low as possible, i.e. the circuit cost is higher if more components are used.

Figure 5 Schematic of DC voltage generation.

Graph 5 Output Voltage of DC voltage generation.


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A DC voltage can be generated by cascading the half-wave rectifier circuit. It can

produce an output DC voltage of 200kV with 140kV of input voltage. From figure 16,

during negative half cycle, the diode D2 is forward bias and the capacitor C1 is charged

up to peak value of input voltage 100kV. During positive half cycle, diode D1 is forward

bias to charge up the capacitor C2 while diode D2 blocks the capacitor C1 from

discharging. Now the capacitor C1 has same peak voltage as input voltage. Therefore,

capacitor C2 is charged up to twice the peak voltage value and produce 200kV.


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TEST 3: HIGH IMPULSE VOLTAGE GENERATION

Figure 6 Circuit configuration for high impulse voltage generation.

Graph 6 Graph voltage versus time for high impulse voltage generation with different load

capacitance value.

The result above shows that the magnitude of the voltage is decreases as the value of the

capacitance, C2 is increases. Capacitor C2 behaves as a load capacitance. The discharge of the

load capacitor, C2 is dependent to the discharge resistance, R 2. Therefore, as what we can see,

the rate of discharge is the same even though we change the capacitance of C2. Since the

0

10000

20000

30000

40000

50000

60000

70000

80000

90000

100000

0 20 40 60 80 100 120

V o l t a g e ,

V

Time, t

Voltage versus Time

C2 = 1pF C2 = 10pF C2 = 100pF C2 = 1nF


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discharge resistance is fixed, therefore, the quantity of charges is fixed. The increment of the

load capacitance will result in a lower output voltage. This relationship can be described from

the equation of capacitance in between two plates:

= Where q is charges ad V equal to voltage. As the capacitance increases, the voltage is reduced

due to its inversely proportional relationship.

Graph 7 Graph voltage versus time for high impulse voltage generation with different

discharge resistance value.

R 2 is a discharge resistance in which its discharges the capacitors and control the tail wave. As

R 2 reduces, the time to start discharge is quicker. A quicker time to start discharge causes the

magnitude of the output voltage to decrease. This is due to the Ohm’s law that explains as

resistance is getting smaller, current will increase. Therefore, the following equation may

applied in order to determine the time to discharge.

= −

0

10000

20000

30000

40000

50000

60000

70000

80000

90000

100000

0 20 40 60 80 100 120

V o l t a g e ,

v

Time, t

Voltage versus Time

R2 = 10k Ohm R2 = 1k Ohm R2 = 100 Ohm R2 = 10 Ohm


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From the equation, a bigger current will cause a shorter time to discharge.

Graph 8 Graph voltage versus time for high impulse voltage generation with different

discharge capacitance value.

The increment of the discharge capacitance, C1 causes the discharge slope is reduced. This is

due to bigger capacitance will store more charges and during the discharge process, the

capacitance will take longer time to discharge due to more charges stored. The voltage is

constant because it is parallel to the voltage source which is constant when the circuit is not yet

open.

0

10000

20000

30000

40000

50000

60000

70000

80000

90000

100000

0 20 40 60 80 100 120

V o l t a g e ,

V

Time, t

Voltage versus Time

C1 = 1nF C1 = 10nF C1 = 100nF C1 = 1uF


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Questions:

1. Using figure below, explain how impulse voltage is generated from a DC voltage

source.

From figure above, the capacitor C1 is slowly charge from a D.C. source until the spark

gap G break down. This spark gap acts as a voltage-limiting and voltage-sensitive

switch, whose ignition time (time to voltage break down) is very short in comparison

to TS. As such single-stage generator may be used for charging voltages from some kV

up to about 1MV, the sphere gap will offer proper operating conditions. The resistor R 1

and R 2 and the capacitance C2 from wave shaping network R 1 will primary damp the

circuit and control the front time TS. R 2 will discharge the capacitor and therefore

essentially control the wave tail. The capacitance C2 represents the full load, example

the object test as well as all other capacitive elements, which are in parallel to the test

object.

2. What is the effect of C2 value on the rise and fall time of V0? Explain why.

From equation of front time,

= 3 (

+ ) 1

As the value of C2 is reduced, the front time is also reduced.

While for tail or fall time, =0.7 + + 2 The increment of C2 will cause the tail time to increase.


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3. What is the effect of R 2 value on the rise and fall time of V0? Explain why.

As we refer to equation front time, t1, the value of R 2 does not affect the front time. For

the tail or fall time as mentioned in equation 2, the increment of the R 2 will cause the

increment of the fall time.

4. What is the effect of C1 value on the rise and fall time of V0? Explain why.

From equation 1, as we increase the value of C1, the front time will decrease. While

from equation 2, the increment of C1 will cause the increment in fall time.

5.

Using only components available in the market, design a circuit in PSpice or OrCAD,

which can generate an impulse voltage, V0 of +80 kV peak magnitude, 8us rise time

and 20us fall time, using a DC source that can give an output between 1V to 140kV

only. When designing the circuit, you must ensure that the voltage rating of each

component (120kV) is not exceeded. The power dissipated by each resistor must not

exceed 1kW. The cost of the overall circuit must also be as low as possible, i.e. the

circuit cost is higher if more components are used.

Figure 7 Design for Impulse Voltage.


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Graph 9 Result of Impulse Voltage design.

In this open-ended task, a DC source with 100kV is used for two-stage impulse circuit

in order to generate an impulse voltage, V0 of 200 kV peak magnitude, 8µs rise time

and 20µs fall time. From Figure 21, we noticed that an impulse voltage, V0 of 200 kV

peak magnitude is produced. However, I failed to get the result of 8µs rise time and

20µs fall time


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CONCLUSION

In conclusion, there are three generation of high voltage been conducted in this experiment

which are A.C. generation, D.C. generation and impulse generation. In A.C. generation, themagnitude of the output voltage is varied depend on the transformer ratio between primary and

secondary side. In D.C. generation, the smoothness of the D.C. output is dependent to the size

of the capacitor and the resistor. In impulse generation, it involves three components to vary

the output which are discharge capacitor, discharge resistor and load capacitor. All of these

type of high voltage generation is successfully being simulated by using Pspice Schematic.

REFERENCES

1.

Fitzpatrick, R. (2 Feb. 2006). Mutual inductance. Retrieved on 10 Dec. 2015 from

http://farside.ph.utexas.edu/teaching/em/lectures/node83.html

2. All About Circuit (n.d.). Mutual Inductance and Basic Operation: Chapter 9 –

Transformers. Retrieved on 10 Dec. 2015 from

http://www.allaboutcircuits.com/textbook/alternating-current/chpt-9/mutual-

inductance-and-basic-operation/

3. Gibbs, K. (2013). The charge and discharge of a capacitor. Retrieved on 10 Dec. 2015

from http://www.schoolphysics.co.uk/age16-

19/Electricity%20and%20magnetism/Electrostatics/text/Capacitor_charge_an

d_discharge/index.html

4. Naidu, M. S., & Kamaraju, V. (2009). High Voltage Engineering: Tata McGraw Hill

Education Private Limited.

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