Rotating Amplifiers

(A) Single stage amplification (separately excited d.c. generator):

Voltage amplification = Ut Uf (at no-load) Power amplification = Ut · Il Uf · If

Power amplification may differ in magnitude with the variations in the value of If and Il due to saturation and becomes non-linear. Linearity may be improved with larger air-gap, but it will reduce amplification. Amplification can, however, be increased by increasing the driving speed.

Example: A separately excited d.c. generator has a voltage gain of 2 V/V. The field circuit resistance is 250 Ω and the armature resistance is 10 Ω.

  1. Compute the input voltage and field current for an output voltage of 230 V on open circuit.
  2. With the field excitation constant as in (a), find the output voltage and power amplification for a load current of 2.5 A.
  3. When the drive speed is increased by 150 %, find increase in power amplification for load current of 2.5 A, assuming linear magnetization.


  1. Input voltage = Output voltage / gain = 230/2 = 115 V.
    Field current = 115/250 = 0.46 A
  2. Output volatge = 230 – 2.5 × 10 = 205V
    Power amplification = (205 × 2.5) / (115 × 0.46) = 9.7
  3. Voltage gain increase proportionally with speed and becomes 2 × 150 % = 3
    So input voltage at no-load = 230/3 = 76.6 V
    Corresponding field current = 76.6/250 = 0.306 A
    Power amplification = (205 × 2.5)∕(76.6 × 0.306) = 21.9.

Thus power amplification increases by 21.9/9.7 = 2.26 times due to increase is speed of 50 %.

Example: The figure below shows a simple automatic voltage regulating circuit for the generator of problem above. What amplifier gain will limit the steady state variation of the terminal voltage to 0.5 V only for a load current variation from 0 to 2.5 A? Assume linear magnetization characteristic and constant speed of rotation of the generator.

Solution: Let GA be the necessary amplifier gain and GV be the generator voltage gain.
Error voltage Uε = UrUt          (1)
Open circuit voltage of the generator Ug = Uε·GA·GV          (2)
Terminal voltage Ut = UgIl.Ra          (3)

Now if δUt be the change in the terminal voltage due to change of δIl in load current.
We have δUg = δUε·GA·GV         (4)
But as reference voltage is unchanged i.e. δUr = 0, from (1) δUε = -δUt
But δUt = δUgRa·δIl = - δUt·GA·GVRaIl
So δUt = (Ra / (1 + GA·GV)) . δIl          (5)
Now, δUt = 0.5, GV = 2 (previous example), δIl = 2.5, so from (5) we get GA = 24.5.

Example: In the previous example, if only a fraction of the total voltage, say α.Ut, where α < 1, is obtained for comparison, determine the necessary gain of the electronic amplifier.

Solution: If the measured voltage is reduced, reference voltage will also be reduced proportionally and
Uε = α.(Ur - Ut)
Also δUε = - α.δUt; as δUr = 0
So δUt = (Ra / (1 + GA.GV.α))·δIl

Thus, the gain of amplifier is to be increased by 1/α times. For α = 0.2, GA = 24.5/0.2 = 122.5.

(B) Two stage amplification (two separately excited d.c. generators in cascade):

First stage power amplification = U1 · I1 Uf · If Second stage power amplification = Ut · Il U1 · I1 Overall power amplification = Ut · Il Uf · If

Thus a much larger power amplification (product of those for individual machine) is obtained.

Rotating Amplifiers Based on Heterogeneous Poles


A further principle used in rotating amplifiers is to use different number of poles in the two stages of amplification (instead of cross-field principle). Very often the first stage has two poles and the second stage has four poles. The benefit of such an arrangement is that the fluxes of the different pole systems may be superimposed on each other without inteference.

(a) Two-Stage Rototral

Here, in first stage we have two-poles (P1 and P3) and in second stage we have four-poles (P1, P2, P3, and P4) to avoid interruption. The armature carries a four-pole winding. Schematic of two-stage rototral is shown below—

In the above two-stage rototral, due to field current If flowing in the horizontal poles, a voltage is produced across the vertical brushes. This voltage causes armature current to flow between vertical brushes, which will produce vertical axis armature flux and will cause 'armature reaction'. To counter this armature reaction, compensating windings are used to neutralize vertical axis armature flux.

So, in the first stage, field current If flowing in the horizontal poles, produces voltage across the vertical brushes which circulates a current I2 through the compensating winding; also same current is flowing through the eight field coils of a four-pole excitation system of the second stage. Two field coils, one from each group are placed on each of the four projecting poles of the stator and they are connected in series as shown in the figure above.

Due to this four-pole excitation we get a second amplification output voltage with two vertical brushes working as the common positive '+' and the two horizontal brushes working as the common negative '−'.

If two groups of coils have same number of turns, then load current IL does not produce any resultant excitation because one coil of each pole carries IL∕2 in each direction and the other coil on the same pole carries IL∕2 in the reverse direction. In this case, the machine behaves like an cross-field machine with full compensation of armature reaction due to load current i.e. an amplidyne .

If partial compensation is desired, the coils of one group may have greater of lesser number of turns than the coils of other group.

(b) Three-Stage Rototral

Here main field winding is same as in two-stage rototral, but vertical axis flux produced is not compensated. Schematic of three-stage rototral is given below—

In the first-stage, the current If flowing through field coils on horizontal poles produces voltage across vertical brushes and causes current I2 to flow in armature and produces vertical flux. This vertical flux produced is not compensated in three-stage rototral, and produces voltage across horizontal brushes as second-stage amplification.

This voltage produced across horizontal brushes is used to circulate current I3 through eight filed coils of four-pole excitation system of third-stage. Due to this four-pole excitation we get amplified output voltage with two vertical brushes working as the common positive '+' and the two horizontal brushes working as the common negative '−', across which load is connected.

(c) Magnavolt

This machine differs from all the previous constructions in that it has two distinct armature windings on the same rotor and in the same slots; one being a two-pole lap winding and the other a four-pole wave winding. Two commutators are used for these windings—one on each side of the armature. The first-stage winding is a two-pole winding rotating in a two-pole magnetic field and the second stage or the output stage is a four-pole winding rotating in a four pole magnetic field. It is generally used in two-stage.

Note: In other constructions, if armature is wound for four-pole system, then the two-pole winding will be seriously short-pitched, so voltage gain will be less.

The four-pole field has no effect on the two-pole armature winding and the two-pole field has no effect on four-pole armature winding, as exlained in the figure below—

  1. Full-pitched armature winding can be used for each stage, resulting is greater utilization of armature winding and more output i.e. higher voltage and power gain in each stage.
  2. Commutation is very good because the armature reaction of the load current does not affect the two-pole field or the commutation at the two-pole brushes and armature reaction is also avoided.
  3. A variety of feedbacks is possible.

Cross Field Generators

Only one machine with two sets of brushes in quadrature can serve the purpose of two individual machines as shown below:

Various parameters of machine are:
Ra — armature winding resistance;
Rf — field winding resistance;
Rq = Ra + R'q — total resistance along q-axis;
Kqf — quadrature axis voltage coefficient (Voltage/Field current);
Kdq — direct axis voltage coefficient (Voltage/Armature current);
Kqd — quadrature axis voltage coefficient due to armature current (Voltage/Armature current);

The average voltage produced across the quadrature-axis brushes of the armature due to ϕf at no-load condition is:

Uq = ϕf n P Za = 2 ϕf n P Z2a = 2 ϕf n P N = 2 ϕf n P N If If = Kqf If

Here, Z is the number of active conductors and a is number of parallel paths. Because it takes two coil sides to make a turn, number of series turns N=Z2a. P is the number of poles and n is revolution per second.

This voltage circulates a current Iq=UqRq; which produces a flux ϕq in the q-axis. Owing to this flux ϕq, average voltage produced across direct-axis brushes, Udo = Kdq·Iq.

Under load conditions, current Id produces a flux ϕd, which opposes the main flux ϕf. This flux ϕd induces a voltage Kqd·Id across q-axis brushes, which again circulates a current Kqd·IdRq, producing additional q-axis flux. Due to this additional q-axis flux, an opposing voltage is set-up across the direct axis brushes whose magnitude is given by KdqKqd·IdRq, and is called armature reaction drop. Thus the net voltage across the load terminals under loaded condition is given by—

Ud = Udo - armature reaction drop - resistance drop = Kqd Iq - Kqd Kdq Rq Id - Id Ra = Kqf Kdq Rq If - Kqd Kdq Rq Id - Id Ra = Kqf Kdq Rq If - K1 + Ra Id

Where— K1 = (KqdKdq)/Rq.

Voltage amplification (at no-load) GV,0 = Udo Uf = Kqf Kdq Rq If If Rf = Kqf Kdq Rf Rq

Power amplification will be—

GP = Ud Id Uf If

Compensated type cross-field machines

To neutralize the armature reaction, a compensating winding may be used on the poles to produce a flux in direct-axis opposing ϕd (helping ϕf) as shown in figure below:

Now, if we define C=mmf of compensating winding per polemmf of armature winding per pole, then the ratio uncompensated armature mmf per poletotal armature mmf per pole becomes (1-C).

Thus, equation for output terminal voltage for a cross-field machine using compensating winding becomes—

Ud = Udo - K1 1-C Id - Ra + Rc Id

Where, Rc is the resistance of compensating winding.

With 100 % compensation, C = 1, and Ud = Udo - Ra + Rc Id . The machine is then termed as amplidyne and the terminal voltage remains practically constant.

With C = 0, the load current remains practically constant and the machine is then called a metadyne. The terminal voltage of a metadyne is Ud = Udo - K1 + Ra Id .

Voltage-current curve of cross-field machine are below—

The power amplification of an amplidyne is very high and can be of the order of 2 × 104 for a 2 kW to 5 kW machine. On the other hand, an uncompensated metadyne supplies practically a constant output current to load for a given field current or excitation and the machine behaves somewhat like a Constant-current generator; Rosenberg Generator is an example of this mode of operation (used for train lighting and battery-charging purposes).

Example: A 5 kW, 250 V, 2000 rpm amplidyne has the following constants—
Rf = 50 Ω, Ra = 5 Ω, Rc = 1 Ω, Kqf = 300 V∕A, Kdq = 100 V∕A, Kqd = 70 V∕A
Calculate the field current and power gain at rated output. Also obtain these values when the compensation is zero.


Voltage amplification GV = Kqf Kdq Rq Rf = 300 ×100 5×50 = 120

For an amplidyne, Ud = Udo - Ra + Rc Id
250 = Udo − (5000∕250) × (5 + 1)
Udo = 250 + 120 = 370 V
GvUf = 370 V
so, Uf ≈ 3.1 V
and If = UfRf = 3.1∕50 ≈ 0.062 A

Power amplification = Ud Id Uf If = 5000 3.1×0.062 =26 000

With zero compensation, Ud = Udo - K1 + Ra Id
K1 = (KqdKdq)∕Rq = (100 × 70)∕5 = 1400
so, 250 = GvUf − (5000∕250) × (1400 + 5)
so, Uf = 236.25 V
and If = UfRf = 236.25∕50 = 4.725 A

Power amplification = Ud Id Uf If = 5000 236.25×4.725 =4.48

Transfer Function of Amplidyne

Transfer function is defined as the ratio of Laplace transform of the output quantity to the Laplace transform of the input quantity with initial values set at zero.

The approximate transfer function for an amplidyne can be obtained assuming that—

  1. the machine is equivalent to two separately excited d.c. generators in cascade;
  2. the second generator does not load the first generator appreciably.

Voltage Transfer Function

uf = Rf + s Lf if , where Lf is inductance of field circuit.

Now, uq = Kqf if , thus first stage transfer function is—

uq uf = Kqf Rf + s Lf = Kqf Rf 1 1 + s Lf Rf = GV1 1 1 + sτf

Where, GV1 is first stage voltage gain and τf is time constant of field circuit.

Similiary, second stage voltage transfer function will be—

ud uq = GV2 1 1 + sτa

Where, GV2 is second stage voltage gain and τa is time constant of armature circuit and so overall transfer is—

ud uf = GV 1 1 + sτf 1 + sτa

Where, GV is overall voltage gain.

Power Transfer Function

First stage power transfer function will be—

uq iq uf if = 2 ϕf nPN iq if2 Rf +s Lf = 2nP Nf ϕf if 1 Rf +s Lf N iq Nf if = 2nP Lf Rf +s Lf Fm,q Fm,f = 2nP τf 1+s τf Fm,q Fm,f

Similiarly second stage power transfer function—

ud id uq iq = 2nP τq 1+s τq Fm,d Fm,q

So overall power transfer function becomes—

4n2P2 τf τq 1+s τf 1+s τq Fm,d Fm,f

Where τf and τq are time constants of field circuit and armature circuit in q-axis respectively; and Fm,f, Fm,q, and Fm,d are field mmf, armature mmf in q-axis, and armature mmf in d-axis (on per pole basis) respectively.

Thus the power amplification of a cross-field generators may be increased -

  1. By increasing the speed. However sparking at the brushes and mechanical stresses in the rotor bring a limitation to the speed.
  2. Decreasing magnetomotive force of field circuit by reducing the air-gap. However to avoid saturation, lower values of flux density is used in the d-axis.
  3. Time constants τf and τq may be increased for increasing power amplification, but at the same time quicker transient response necessitates lower values of time constants. It has been found that optimum result is obtained when τf = τq.

Other Operational Features of Amplidyne

As a machine used in control system, the transient performance of an amplidyne is of considerable importance. Time delays are introduced due to the inductances of the windings resulting in hunting of the output circuit. The time constant of the field system is reduced by supplying it from an electronic power amplifier (for decreasing inductance of field circuit).

The armature time-constant can be reduced either by decreasing number of armature turns (so that inductance is reduced) or by connectiong an external resistance between the quadrature-axis brushes (thus increasing the effective resistance of the armature circuit). Both these measures will however decrease the overall gain of the amplidyne.

The reduction of time constant of the field and armature circuit can also be achieved by increasing the air-gap length of the machine (thus reducing the inductances of field and armature circuits. This would simultaneously help in extending the range of linear operation. The decrease in gain due to larger air-gap can be compensated by increasing the speed of operation which will not affect the time constant.

Due to the two major time-constants, the machine has an inherent tendency to unstable operation. Such tendency can be minimized by connecting one of the control fields to the output terminals of amplidyne through a stabilizing device (which may be in the form of an anti-hunt transformer or a simple R-C circuit). The anti-hunt device is ineffective if the output is purely d.c. voltage. Whenever there is an oscillation superimposed on the output d.c. voltage, the alternating component is fed back to a control field through the stabilizing device resulting in an induced voltage in the armature of opposite sense to finally decrease the oscillations. These two schemes are shown in figure below—

Application of Amplidyne

The amplidyne is widely used in closed loop control system. For such applications, generally, the amplidyne posseses more than one field winding, one of which is used as the reference field and the other for the feedback circuit. The excitation of the fields are such that their fluxes oppose each other.

1. Voltage Control Schemes

(a) Voltage Control of d.c. generator

The scheme of voltage control with both one-field and two-field amplidyne is as below—

Problem: For one-field amplidyne used for voltage control, various constants are—amplidyne voltage amplification factor = 100; d.c. generator output voltage = 200 V∕field current in A; field winding resistance of d.c. generator = 125 Ω. The feedback potentiometer is tapped to give Ub = 0.1 Ut and reference potentiometer is set to give Ur = 50 V. Determine terminal voltage of the generator.

Uf = UrUb = (50 − 0.1 Ut)
amplidyne output voltage = 100 (50 − 0.1 Ut)
generator field current = 100 (50 − 0.1 Ut) ∕ 125
so generator terminal voltage Ut = 100 50 - 0.1 Ut 125 × 200
from this we get Ut = 470.6 V

Problem: In the previous problem, if a terminal voltage of 400 V is desired, what should be the reference voltage?

Ub = 0.1 × 400 = 40 V
Uf = (Ur − 40)
amplidyne output voltage = 100 (Ur − 40)
generator field current = 100 (Ur − 40) ∕ 125
generator terminal voltage Ut = 100 Ur - 40 125 × 200 = 400V
from this we get Ur = 42.5 V

(b) Voltage regulation of a.c. generator

2. Speed Control of d.c. Motor

3. Constant Current Schemes

Motors used with dredgers, ships windlasses etc. are likely to be stalled and in view of this it is advantageous to operate such motors from constant current systems (avoiding any chances of overloading). One of the methods of obtaining constant current system with the help of amplidyne is given below—

Constructional Aspects

The pole structure of a cross-field generator is different from that of a d.c. machine, the pole configuration (for a 2-pole machine) being shown in the figure given. If ordinary pole structure is used (as in a d.c. machine) then the coil undergoing commutation under the d-axis brushes would be cutting the maximum d-axis flux ϕf and would have therefore maximum rotational voltage generated in it, giving large circulating current and excessive sparking at the brushes. This may be reduced by making a cut in the middle of the pole faces as shown in figure (a). Interploes are normally required along the output brushes (under d-axis) and with this simple arrangement there is no space to fit interpoles. To facilitate the use of interpoles, each interpole is splitted into two separate poles, giving the appearance of 4-pole for a 2-pole machine as shown in figure (b). The paths of direct and quadrature axes fluxes have been shown in both the cases. A number of different arrangement of the poles may be adopted to fulfil the above basic needs, two more such examples are shown in figures (c) and (d).

The compensating winding may be in the form of a single coil on the poles (like the main field winding) or may be distributed in slots over the pole faces. The former is simple and the latter provides full compensation. The distributed compensating winding is the one generally used.

The control field winding is generally made up of several coils for convenience in applying signals and for stabilization purposes.

In addition to the control field windings and the compensating windings, other windings may also be provided on the poles. One such is the quadrature series winding. This is connected to assist the production of quadrature-axis flux. The quadrature axis armature current magnitude then reduces, reducing the heating of armature and reducing the sparking at the brushes in q-axis. Sometimes a variable resistor R'q is connected in series with the q-axis series winding to alter the time-constant and the amplification of the machine.

A quadrature-axis shunt winding is also sometimes provided (connected across d-d brushes) to increase the q-axis flux. The poles are laminated to reduce eddy-current losses.


Post a comment