Exp2 Magnetic Amplifier

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NAME OF THE EXPERIMENT: MAGNETIC AMPLIFIER OBJECTIVE: DETERMINE THE CONTROL CHARACTERISTIC CURVES OF A MAGNETIC AMPLIFIER AND AN AMPLISTAT. APPARATUS REQUIRED: S.No. Apparatus and their Ratings Quantity 1 Diodes 4 2 Voltmeter( 0 – 300V) 2 3 Ammeter 1) 0 – 5A 2) 0- 500mA 2 1 4 Rheostat( 100ohm, 5A) 2 5 Variable DC Supply (0-40V) 1 6 Variac (for AC Supply)(0-300V) 1 THEORY: Consider a transformer as shown in the Figure1. Figure1: Principle of Magnetic Amplifier It has got a winding supplied by the dc source and is called the ‘Control Winding’. Another winding is supplied by an ac source and is called the ‘Load Winding’. The control winding should have comparatively larger number of turns and the cross- section of the wire can be small, whereas for the load winding, the number of turns is less but the

Transcript of Exp2 Magnetic Amplifier

Page 1: Exp2 Magnetic Amplifier

NAME OF THE EXPERIMENT: MAGNETIC AMPLIFIER

OBJECTIVE: DETERMINE THE CONTROL CHARACTERISTIC CURVES OF A MAGNETIC AMPLIFIER AND AN AMPLISTAT.

APPARATUS REQUIRED:

S.No. Apparatus and their Ratings Quantity1 Diodes 42 Voltmeter( 0 – 300V) 23 Ammeter 1) 0 – 5A

2) 0- 500mA 2

14 Rheostat( 100ohm, 5A) 25 Variable DC Supply (0-40V) 16 Variac (for AC Supply)(0-300V) 1

THEORY: Consider a transformer as shown in the Figure1.

Figure1: Principle of Magnetic Amplifier

It has got a winding supplied by the dc source and is called the ‘Control Winding’. Another winding is supplied by an ac source and is called the ‘Load Winding’. The control winding should have comparatively larger number of turns and the cross-section of the wire can be small, whereas for the load winding, the number of turns is less but the cross-section of the wire is more. A small dc current flows through the control winding, whereas a large ac current flows through the load winding. The output of a magnetic amplifier is taken across a load resistance R2 connected in series with the load winding. A resistance R is connected in series with the control winding. The current flowing in the control winding is varied by varying the value of resistance R.

In order to understand the principle of operation of a magnetic amplifier we consider the B-H curve of a magnetic core shown in Figure2. We know the inductance, L of coil is given by:

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L = d ØdI ......... (1)

Figure2: The B-H Curve

Hence, L is the slope of the B-H curve as B is proportional to the flux(Ø) and H is proportional to the current(I). The slope is high below the point A on the curve, small between the points A and B and almost zero beyond the point B on the curve. The region beyond the point B is called the ‘Saturation’. From Figure2, equation(1) and the above explanation, we can conclude about the inductance of the coils wound on the core of Figure1 that the inductance is high below the point A, it is low between the points A and B, but the inductance is zero beyond the point B. Therefore, the inductance of the coil wound over the core depends on the saturation level of the core. Now we can consider three operating points on the B-H curve for further discussions. These are given below:

i) Below the point A: If the operating point lies below point A on the B-H curve of Figure2, the magnetisation level of the core is low i.e. the value of R is high so that the current flowing through the control winding is small. At this operating point the inductance of the load winding is high. Due to this high value of inductance, most of the voltage supplied by the ac source will appear across the load winding (i.e. L (di/dt)) and a very small voltage drop occurs across the load resistance R2 giving a low power output.

ii) Between the points A and B: If the operating point lies between the points A and B, the magnetization level of the core is moderate and

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the inductance of coil(load winding) is also moderate. This means a part of the voltage is dropped across the load winding and a part of the ac supply voltage will appear across the load resistance R2. This causes the output power to be more than the previous case.

iii) Beyond the point B: If the value of the R i.e. the resistance of the control winding circuit is kept at a very low value, the current through the control winding would be very high, resulting in a very high level of magnetization(‘Saturation’) of the core. Under this situation the inductance of the load winding is very low or even zero. Then, the voltage drop across the load winding would be very small or zero and whole of the ac voltage would appear across the load resistance R2. This would give a very high output power.

From the above discussions two inferences can be drawn. First, the power appearing across the load resistance R2 is not drawn from the dc source connected in the control circuit. It is drawn from the ac source connected in the load circuit. Second, the power across the load resistance R2 depends on the value of the resistance R in the control circuit. If we regard the input to the control circuit as the input to the system, then very large power amplification is evident from the setup as shown in Figure1. Further with a small change of dc power input (control circuit) we can achieve a large change in the power output (load circuit). This is the basic concept used in magnetic amplifier.

Two more aspects need attention. The direction of magnetization of the core by the ac current varies as ac current is positive for one half cycle and negative for the other. So, to minimise the magnetization effect of ac current the number of turns of load winding is kept much less as compared to that of control winding.

The ac voltage applied to the load winding creates ac flux which links with the control winding and induce the ac voltage in control winding. The balanced circuit arrangement as shown in Figure3 is used to avoid this undesirable effect.

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Figure3: Circuit for a Balanced Magnetic Amplifier

In Figure3, the yoke has three limbs. Central limb carries the control winding. Load windings are wound on the outer two limbs and are connected in series opposition. Under the condition of dc current being very small or zero, the inductance of load winding is very high and the load current is small. In the other case when dc current has sufficient magnitude, the core is near saturation condition, the inductance of load winding is low and the load current is high. This would induce ac voltage in the control winding. But, since there are two load windings, the ac fluxes induced in the central limb are in the opposite directions and cancel each other. This results in zero ac emf induced in the control winding. As already mentioned the control winding has many turns. Therefore, a large range of output can be controlled for quite a small change of input. This justifies the name “Magnetic Amplifier”. Figure4 depicts the typical characteristic of a magnetic amplifier which we can observe is symmetrical about the y-axis.

Figure4: Characteristic of a Magnetic Amplifier

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The core material used has typical B-H curve as shown below. B-H curve should saturate rapidly (unlike the transformer core) and thus should be more rectangular in form.

Figure : Hysteresis Curve

Figure: Idealised Hysteresis Curve

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The output of the magnetic amplifier discussed above is ac. If dc output is desired, the ac output can be rectified using the bridge rectifier circuit as shown below.

Figure : Basic DC Amplifier

In an amplistat two diodes are used and so the load current flows through one half of the load winding in one half cycle of ac voltage. In the other half cycle of ac voltage, load current flows through the other half of the load winding. The ac flux’s is not cancelled out (unlike the first connection diagram where ac flux’s effect is cancelled out) and thus it induces an ac emf in control winding. So the control characteristic curve of amplistat which is shown below is not symmetrical about the y-axis.

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AB : Voltage Source region

AC : Transition region

CD : Current Source region

(Circuit can switch from current source to voltage source region just by changing the polarity of the control current)

CONNECTION DIAGRAMS:

Fig : The actual board Diagram present in our lab

(S:-Series MagAmp; P:-Parallel MagAmp; s1, s2, s3, s4, s5, s6, s7 are the switches;

For Parallel MagAmp characteristic:- use 1000T Control Winding SP; keep KNOB at P; keep s1,s3,s5 ON and rest of the switches OFF.

For Amplistat characteristic: - use 100T Control Winding A & BA; keep KNOB at A; keep s1, s3, s6 ON and rest of the switches OFF. )

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Fig: Experimental connections for Parallel Magnetic Amplifier

Fig: Amplistat Circuit Diagram

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EXPERIMENTAL PROCEDURE:

1) Make the connection as shown in the first connection diagram. Give ac supply to the ac input terminals through a variac (autotransformer). Give dc supply to the control signals input terminals making potential divider arrangement to vary control current (I dc) smoothly in steps of 2 mA. Supply constant V ac=230 V (rated value). Rheostat on load winding side (RL) is kept constant (at such a value that rated current of mag-amp flows). Vary rheostat on control winding side i.e. Rdc from zero and record the load current (I L) and dc current in control winding (I dc).

2) Reverse the dc supply terminals and repeat step1.

3) Now, keeping both I dc and RL constant (i.e. both the rheostat are kept at fixed position), vary the autotransformer i.e. V ac is varied. Record V L and I L.

4) Now, keeping constant I dc and constant ac supply voltage (V ac) equal to 230V, vary the rheostat on the load winding side. Record V L and I L. Find RLby using RL=V L/I L. Plot RL vs. I L.

5) Make the connection for Amplistat as in second connection diagram. Keep the load resistance at such a value that the rated current of the mag-amp flows through it when the whole of the supply voltage is across it. Supply constant V ac=230 V (rated value). Vary rheostat on control winding side i.e. Rdc from zero and record the load current (I L) and dc current in control winding (I dc).

6) Reverse the dc supply terminals and repeat step5.

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TABLES AND GRAPHS:

1) (RLand V ac are constant)

S.No I dc(mA) I L(A)

On reversing the dc supply terminals:

(RLand V ac are constant)

S.No I dc(mA) I L(A)

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2) (RLand I dc are constant)

S.No I L(A) V L(V)

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3) (I dc and V ac are constant)

S.No I L(A) V L(V) RL=V L/I L (Ω)

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4) For Amplistat (RLand V ac are constant)

S.No I dc(mA) I L(A)

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On reversing the dc supply terminals:

(RLand V ac are constant)

S.No I dc(mA) I L(A)

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INFERENCE: Magnetic Amplifier is an electromagnetic device for amplifying electrical signals. The amount of control current fed into the control winding sets the point in the AC winding waveform at which either core will saturate. In saturation, the AC winding on the saturated core will go from a high impedance state ("off") into a very low impedance state ("on") - that is, the control current controls at which voltage the mag amp switches "on". The control characteristics of the magnetic amplifier and amplistat have been plotted.

PRECAUTIONS AND NOTES:

1) Choose the range of the instruments and the value of the load resistance very carefully knowing the ratings of the magnetic amplifier.

2) Rated ac supply voltage is equal to the absorption voltage of each saturable reactor.

3) Note the polarity of connections of winding. This is shown by dots (marked on the sides of the windings) in the Figures.

4) The load resistance must always be of high value to limit the load current to safe rated value before putting on the ac supply.

SAMPLE READINGS AND CORRESPONDING GRAPHS:

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1) (RLand V ac are constant)

S.No I dc(mA) I L(A)123456789

10111213

050

100125150175200225250275300325350

0.251.401.852.202.502.803.103.303.603.804.004.254.50

On reversing the dc supply terminals:

(RLand V ac are constant)

S.No I dc(mA) I L(A)123456789

101112

50100125150175200225250275300325350

1.001.451.752.102.452.803.053.353.603.854.054.25

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-400 -300 -200 -100 0 100 200 300 4000

0.5

1

1.5

2

2.5

3

3.5

4

4.5

Idc(mA)

Iload(A)

2) (RLand I dc are constant)

S.No I L(A) V L(V)12345678

0.000.501.201.602.052.553.003.80

020406080

100120128

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0 0.5 1 1.5 2 2.5 3 3.5 40

20

40

60

80

100

120

140

Iload(A)

Vload(V)

3) (I dcand V ac are constant)

S.No I L(A) V L(V) RL=V L/I L (Ω)12345678

3.803.904.004.104.204.304.404.45

1241081009284726454

32.6327.6925.0022.4420.0016.7414.5512.13

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10 15 20 25 30 350

0.5

1

1.5

2

2.5

3

3.5

4

4.5Iload(A)

Rload(ohm)

4)For Amplistat (RLand V ac are constant)

S.No I dc(mA) I L(A)12345

50100150200225

1.001.452.453.954.70

On reversing the dc supply terminals:

(RLand V ac are constant)

S.No I dc(mA) I L(A)123

0100150

0.250.801.20

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456789

10

200250300350400450475

1.601.852.102.452.853.153.35

-500 -400 -300 -200 -100 0 100 200 3000

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

Idc(mA)

Iload(A)