Electrical transients

Electrical transients

This chapter explores the response of capacitors and inductors to sudden changes in DC voltage (called a transient voltage), when wired in series with a resistor. Unlike resistors, which respond instantaneously to applied voltage, capacitors and inductors react over time as they absorb and release energy.

Capacitor transient response

Because capacitors store energy in the form of an electric field, they tend to act like small secondary-cell batteries, being able to store and release electrical energy. A fully discharged capacitor maintains zero volts across its terminals, and a charged capacitor maintains a steady quantity of voltage across its terminals, just like a battery. When capacitors are placed in a circuit with other sources of voltage, they will absorb energy from those sources, just as a secondary-cell battery will become charged as a result of being connected to a generator. A fully discharged capacitor, having a terminal voltage of zero, will initially act as a short-circuit when attached to a source of voltage, drawing maximum current as it begins to build a charge. Over time, the capacitor's terminal voltage rises to meet the applied voltage from the source, and the current through the capacitor decreases correspondingly. Once the capacitor has reached the full voltage of the source, it will stop drawing current from it, and behave essentially as an open-circuit.

When the switch is first closed, the voltage across the capacitor (which we were told was fully discharged) is zero volts; thus, it first behaves as though it were a short-circuit. Over time, the capacitor voltage will rise to equal battery voltage, ending in a condition where the capacitor behaves as an open-circuit. Current through the circuit is determined by the difference in voltage between the battery and the capacitor, divided by the resistance of 10 kΩ. As the capacitor voltage approaches the battery voltage, the current approaches zero. Once the capacitor voltage has reached 15 volts, the current will be exactly zero. Let's see how this works using real values:


The capacitor voltage's approach to 15 volts and the current's approach to zero over time is what a mathematician would call asymptotic: that is, they both approach their final values, getting closer and closer over time, but never exactly reaches their destinations. For all practical purposes, though, we can say that the capacitor voltage will eventually reach 15 volts and that the current will eventually equal zero.
Using the SPICE circuit analysis program, we can chart this asymptotic buildup of capacitor voltage and decay of capacitor current in a more graphical form (capacitor current is plotted in terms of voltage drop across the resistor, using the resistor as a shunt to measure current):

As you can see, I have used the .plot command in the netlist instead of the more familiar .print command. This generates a pseudo-graphic plot of figures on the computer screen using text characters. SPICE plots graphs in such a way that time is on the vertical axis (going down) and amplitude (voltage/current) is plotted on the horizontal (right=more; left=less). Notice how the voltage increases (to the right of the plot) very quickly at first, then tapering off as time goes on. Current also changes very quickly at first then levels off as time goes on, but it is approaching minimum (left of scale) while voltage approaches maximum.

• REVIEW:
• Capacitors act somewhat like secondary-cell batteries when faced with a sudden change in applied voltage: they initially react by producing a high current which tapers off over time.
• A fully discharged capacitor initially acts as a short circuit (current with no voltage drop) when faced with the sudden application of voltage. After charging fully to that level of voltage, it acts as an open circuit (voltage drop with no current).
• In a resistor-capacitor charging circuit, capacitor voltage goes from nothing to full source voltage while current goes from maximum to zero, both variables changing most rapidly at first, approaching their final values slower and slower as time goes on.

Inductor transient response

Inductors have the exact opposite characteristics of capacitors. Whereas capacitors store energy in an electric field (produced by the voltage between two plates), inductors store energy in a magnetic field (produced by the current through wire). Thus, while the stored energy in a capacitor tries to maintain a constant voltage across its terminals, the stored energy in an inductor tries to maintain a constant current through its windings. Because of this, inductors oppose changes in current, and act precisely the opposite of capacitors, which oppose changes in voltage. A fully discharged inductor (no magnetic field), having zero current through it, will initially act as an open-circuit when attached to a source of voltage (as it tries to maintain zero current), dropping maximum voltage across its leads. Over time, the inductor's current rises to the maximum value allowed by the circuit, and the terminal voltage decreases correspondingly. Once the inductor's terminal voltage has decreased to a minimum (zero for a "perfect" inductor), the current will stay at a maximum level, and it will behave essentially as a short-circuit.

When the switch is first closed, the voltage across the inductor will immediately jump to battery voltage (acting as though it were an open-circuit) and decay down to zero over time (eventually acting as though it were a short-circuit). Voltage across the inductor is determined by calculating how much voltage is being dropped across R, given the current through the inductor, and subtracting that voltage value from the battery to see what's left. When the switch is first closed, the current is zero, then it increases over time until it is equal to the battery voltage divided by the series resistance of 1 Ω. This behavior is precisely opposite that of the series resistor-capacitor circuit, where current started at a maximum and capacitor voltage at zero. Let's see how this works using real values:

Just as with the RC circuit, the inductor voltage's approach to 0 volts and the current's approach to 15 amps over time is asymptotic. For all practical purposes, though, we can say that the inductor voltage will eventually reach 0 volts and that the current will eventually equal the maximum of 15 amps.
Again, we can use the SPICE circuit analysis program to chart this asymptotic decay of inductor voltage and buildup of inductor current in a more graphical form (inductor current is plotted in terms of voltage drop across the resistor, using the resistor as a shunt to measure current):

Notice how the voltage decreases (to the left of the plot) very quickly at first, then tapering off as time goes on. Current also changes very quickly at first then levels off as time goes on, but it is approaching maximum (right of scale) while voltage approaches minimum.

• REVIEW:
• A fully "discharged" inductor (no current through it) initially acts as an open circuit (voltage drop with no current) when faced with the sudden application of voltage. After "charging" fully to the final level of current, it acts as a short circuit (current with no voltage drop).
• In a resistor-inductor "charging" circuit, inductor current goes from nothing to full value while voltage goes from maximum to zero, both variables changing most rapidly at first, approaching their final values slower and slower as time goes on.

Voltage and current calculations

There's a sure way to calculate any of the values in a reactive DC circuit over time. The first step is to identify the starting and final values for whatever quantity the capacitor or inductor opposes change in; that is, whatever quantity the reactive component is trying to hold constant. For capacitors, this quantity is voltage; for inductors, this quantity is current. When the switch in a circuit is closed (or opened), the reactive component will attempt to maintain that quantity at the same level as it was before the switch transition, so that value is to be used for the "starting" value. The final value for this quantity is whatever that quantity will be after an infinite amount of time. This can be determined by analyzing a capacitive circuit as though the capacitor was an open-circuit, and an inductive circuit as though the inductor was a short-circuit, because that is what these components behave as when they've reached "full charge," after an infinite amount of time.
The next step is to calculate the time constant of the circuit: the amount of time it takes for voltage or current values to change approximately 63 percent from their starting values to their final values in a transient situation. In a series RC circuit, the time constant is equal to the total resistance in ohms multiplied by the total capacitance in farads. For a series L/R circuit, it is the total inductance in henrys divided by the total resistance in ohms. In either case, the time constant is expressed in units of seconds and symbolized by the Greek letter "tau" (τ):

The rise and fall of circuit values such as voltage and current in response to a transient is, as was mentioned before, asymptotic. Being so, the values begin to rapidly change soon after the transient and settle down over time. If plotted on a graph, the approach to the final values of voltage and current form exponential curves.
As was stated before, one time constant is the amount of time it takes for any of these values to change about 63 percent from their starting values to their (ultimate) final values. For every time constant, these values move (approximately) 63 percent closer to their eventual goal. The mathematical formula for determining the precise percentage is quite simple:

The letter e stands for Euler's constant, which is approximately 2.7182818. It is derived from calculus techniques, after mathematically analyzing the asymptotic approach of the circuit values. After one time constant's worth of time, the percentage of change from starting value to final value is:

After two time constant's worth of time, the percentage of change from starting value to final value is:

After ten time constant's worth of time, the percentage is:

The more time that passes since the transient application of voltage from the battery, the larger the value of the denominator in the fraction, which makes for a smaller value for the whole fraction, which makes for a grand total (1 minus the fraction) approaching 1, or 100 percent.
We can make a more universal formula out of this one for the determination of voltage and current values in transient circuits, by multiplying this quantity by the difference between the final and starting circuit values:

Let's analyze the voltage rise on the series resistor-capacitor circuit shown at the beginning of the chapter.

Note that we're choosing to analyze voltage because that is the quantity capacitors tend to hold constant. Although the formula works quite well for current, the starting and final values for current are actually derived from the capacitor's voltage, so calculating voltage is a more direct method. The resistance is 10 kΩ, and the capacitance is 100 µF (microfarads). Since the time constant (τ) for an RC circuit is the product of resistance and capacitance, we obtain a value of 1 second:

If the capacitor starts in a totally discharged state (0 volts), then we can use that value of voltage for a "starting" value. The final value, of course, will be the battery voltage (15 volts). Our universal formula for capacitor voltage in this circuit looks like this:

So, after 7.25 seconds of applying voltage through the closed switch, our capacitor voltage will have increased by:

Since we started at a capacitor voltage of 0 volts, this increase of 14.989 volts means that we have 14.989 volts after 7.25 seconds.
The same formula will work for determining current in that circuit, too. Since we know that a discharged capacitor initially acts like a short-circuit, the starting current will be the maximum amount possible: 15 volts (from the battery) divided by 10 kΩ (the only opposition to current in the circuit at the beginning):

We also know that the final current will be zero, since the capacitor will eventually behave as an open-circuit, meaning that eventually no electrons will flow in the circuit. Now that we know both the starting and final current values, we can use our universal formula to determine the current after 7.25 seconds of switch closure in the same RC circuit:

Note that the figure obtained for change is negative, not positive! This tells us that current has decreased rather than increased with the passage of time. Since we started at a current of 1.5 mA, this decrease (-1.4989 mA) means that we have 0.001065 mA (1.065 µA) after 7.25 seconds.
We could have also determined the circuit current at time=7.25 seconds by subtracting the capacitor's voltage (14.989 volts) from the battery's voltage (15 volts) to obtain the voltage drop across the 10 kΩ resistor, then figuring current through the resistor (and the whole series circuit) with Ohm's Law (I=E/R). Either way, we should obtain the same answer:

The universal time constant formula also works well for analyzing inductive circuits. Let's apply it to our example L/R circuit in the beginning of the chapter:

With an inductance of 1 henry and a series resistance of 1 Ω, our time constant is equal to 1 second:

Because this is an inductive circuit, and we know that inductors oppose change in current, we'll set up our time constant formula for starting and final values of current. If we start with the switch in the open position, the current will be equal to zero, so zero is our starting current value. After the switch has been left closed for a long time, the current will settle out to its final value, equal to the source voltage divided by the total circuit resistance (I=E/R), or 15 amps in the case of this circuit.
If we desired to determine the value of current at 3.5 seconds, we would apply the universal time constant formula as such:

Given the fact that our starting current was zero, this leaves us at a circuit current of 14.547 amps at 3.5 seconds' time.
Determining voltage in an inductive circuit is best accomplished by first figuring circuit current and then calculating voltage drops across resistances to find what's left to drop across the inductor. With only one resistor in our example circuit (having a value of 1 Ω), this is rather easy:

Subtracted from our battery voltage of 15 volts, this leaves 0.453 volts across the inductor at time=3.5 seconds.

• REVIEW:
• Universal Time Constant Formula:
• 
• To analyze an RC or L/R circuit, follow these steps:
• (1): Determine the time constant for the circuit (RC or L/R).
• (2): Identify the quantity to be calculated (whatever quantity whose change is directly opposed by the reactive component. For capacitors this is voltage; for inductors this is current).
• (3): Determine the starting and final values for that quantity.
• (4): Plug all these values (Final, Start, time, time constant) into the universal time constant formula and solve for change in quantity.
• (5): If the starting value was zero, then the actual value at the specified time is equal to the calculated change given by the universal formula. If not, add the change to the starting value to find out where you're at.

Why L/R and not LR?

It is often perplexing to new students of electronics why the time-constant calculation for an inductive circuit is different from that of a capacitive circuit. For a resistor-capacitor circuit, the time constant (in seconds) is calculated from the product (multiplication) of resistance in ohms and capacitance in farads: τ=RC. However, for a resistor-inductor circuit, the time constant is calculated from the quotient (division) of inductance in henrys over the resistance in ohms: τ=L/R.
This difference in calculation has a profound impact on the qualitative analysis of transient circuit response. Resistor-capacitor circuits respond quicker with low resistance and slower with high resistance; resistor-inductor circuits are just the opposite, responding quicker with high resistance and slower with low resistance. While capacitive circuits seem to present no intuitive trouble for the new student, inductive circuits tend to make less sense.
Key to the understanding of transient circuits is a firm grasp on the concept of energy transfer and the electrical nature of it. Both capacitors and inductors have the ability to store quantities of energy, the capacitor storing energy in the medium of an electric field and the inductor storing energy in the medium of a magnetic field. A capacitor's electrostatic energy storage manifests itself in the tendency to maintain a constant voltage across the terminals. An inductor's electromagnetic energy storage manifests itself in the tendency to maintain a constant current through it.
Let's consider what happens to each of these reactive components in a condition of discharge: that is, when energy is being released from the capacitor or inductor to be dissipated in the form of heat by a resistor:

In either case, heat dissipated by the resistor constitutes energy leaving the circuit, and as a consequence the reactive component loses its store of energy over time, resulting in a measurable decrease of either voltage (capacitor) or current (inductor) expressed on the graph. The more power dissipated by the resistor, the faster this discharging action will occur, because power is by definition the rate of energy transfer over time.
Therefore, a transient circuit's time constant will be dependent upon the resistance of the circuit. Of course, it is also dependent upon the size (storage capacity) of the reactive component, but since the relationship of resistance to time constant is the issue of this section, we'll focus on the effects of resistance alone. A circuit's time constant will be less (faster discharging rate) if the resistance value is such that it maximizes power dissipation (rate of energy transfer into heat). For a capacitive circuit where stored energy manifests itself in the form of a voltage, this means the resistor must have a low resistance value so as to maximize current for any given amount of voltage (given voltage times high current equals high power). For an inductive circuit where stored energy manifests itself in the form of a current, this means the resistor must have a high resistance value so as to maximize voltage drop for any given amount of current (given current times high voltage equals high power).
This may be analogously understood by considering capacitive and inductive energy storage in mechanical terms. Capacitors, storing energy electrostatically, are reservoirs of potential energy. Inductors, storing energy electromagnetically (electrodynamically), are reservoirs of kinetic energy. In mechanical terms, potential energy can be illustrated by a suspended mass, while kinetic energy can be illustrated by a moving mass. Consider the following illustration as an analogy of a capacitor:

The cart, sitting at the top of a slope, possesses potential energy due to the influence of gravity and its elevated position on the hill. If we consider the cart's braking system to be analogous to the resistance of the system and the cart itself to be the capacitor, what resistance value would facilitate rapid release of that potential energy? Minimum resistance (no brakes) would diminish the cart's altitude quickest, of course! Without any braking action, the cart will freely roll downhill, thus expending that potential energy as it loses height. With maximum braking action (brakes firmly set), the cart will refuse to roll (or it will roll very slowly) and it will hold its potential energy for a long period of time. Likewise, a capacitive circuit will discharge rapidly if its resistance is low and discharge slowly if its resistance is high.
Now let's consider a mechanical analogy for an inductor, showing its stored energy in kinetic form:

This time the cart is on level ground, already moving. Its energy is kinetic (motion), not potential (height). Once again if we consider the cart's braking system to be analogous to circuit resistance and the cart itself to be the inductor, what resistance value would facilitate rapid release of that kinetic energy? Maximum resistance (maximum braking action) would slow it down quickest, of course! With maximum braking action, the cart will quickly grind to a halt, thus expending its kinetic energy as it slows down. Without any braking action, the cart will be free to roll on indefinitely (barring any other sources of friction like aerodynamic drag and rolling resistance), and it will hold its kinetic energy for a long period of time. Likewise, an inductive circuit will discharge rapidly if its resistance is high and discharge slowly if its resistance is low.
Hopefully this explanation sheds more light on the subject of time constants and resistance, and why the relationship between the two is opposite for capacitive and inductive circuits.

Complex voltage and current calculations

There are circumstances when you may need to analyze a DC reactive circuit when the starting values of voltage and current are not respective of a fully "discharged" state. In other words, the capacitor might start at a partially-charged condition instead of starting at zero volts, and an inductor might start with some amount of current already through it, instead of zero as we have been assuming so far. Take this circuit as an example, starting with the switch open and finishing with the switch in the closed position:

Since this is an inductive circuit, we'll start our analysis by determining the start and end values for current. This step is vitally important when analyzing inductive circuits, as the starting and ending voltage can only be known after the current has been determined! With the switch open (starting condition), there is a total (series) resistance of 3 Ω, which limits the final current in the circuit to 5 amps:

So, before the switch is even closed, we have a current through the inductor of 5 amps, rather than starting from 0 amps as in the previous inductor example. With the switch closed (the final condition), the 1 Ω resistor is shorted across (bypassed), which changes the circuit's total resistance to 2 Ω. With the switch closed, the final value for current through the inductor would then be:

So, the inductor in this circuit has a starting current of 5 amps and an ending current of 7.5 amps. Since the "timing" will take place during the time that the switch is closed and R₂ is shorted past, we need to calculate our time constant from L₁ and R₁: 1 Henry divided by 2 Ω, or τ = 1/2 second. With these values, we can calculate what will happen to the current over time. The voltage across the inductor will be calculated by multiplying the current by 2 (to arrive at the voltage across the 2 Ω resistor), then subtracting that from 15 volts to see what's left. If you realize that the voltage across the inductor starts at 5 volts (when the switch is first closed) and decays to 0 volts over time, you can also use these figures for starting/ending values in the general formula and derive the same results:

Solving for unknown time

Sometimes it is necessary to determine the length of time that a reactive circuit will take to reach a predetermined value. This is especially true in cases where we're designing an RC or L/R circuit to perform a precise timing function. To calculate this, we need to modify our "Universal time constant formula." The original formula looks like this:

However, we want to solve for time, not the amount of change. To do this, we algebraically manipulate the formula so that time is all by itself on one side of the equal sign, with all the rest on the other side:


The ln designation just to the right of the time constant term is the natural logarithm function: the exact reverse of taking the power of e. In fact, the two functions (powers of e and natural logarithms) can be related as such:


If e^x = a, then ln a = x.


If e^x = a, then the natural logarithm of a will give you x: the power that e must be was raised to in order to produce a.
Let's see how this all works on a real example circuit. Taking the same resistor-capacitor circuit from the beginning of the chapter, we can work "backwards" from previously determined values of voltage to find how long it took to get there.

The time constant is still the same amount: 1 second (10 kΩ times 100 µF), and the starting/final values remain unchanged as well (E_C = 0 volts starting and 15 volts final). According to our chart at the beginning of the chapter, the capacitor would be charged to 12.970 volts at the end of 2 seconds. Let's plug 12.970 volts in as the "Change" for our new formula and see if we arrive at an answer of 2 seconds:

Indeed, we end up with a value of 2 seconds for the time it takes to go from 0 to 12.970 volts across the capacitor. This variation of the universal time constant formula will work for all capacitive and inductive circuits, both "charging" and "discharging," provided the proper values of time constant, Start, Final, and Change are properly determined beforehand. Remember, the most important step in solving these problems is the initial set-up. After that, its just a lot of button-pushing on your calculator!

• REVIEW:
• To determine the time it takes for an RC or L/R circuit to reach a certain value of voltage or current, you'll have to modify the universal time constant formula to solve for time instead of change.
• 

Battery Runtime

Calculating the Battery Runtime

If the battery were a perfect power source and behaved linearly, the discharge time could be calculated according to the in-and-out flowing currents. “What is put in should be available as an output in the same amount” goes the argument, and “a one-hour charge at 5A should deliver a one-hour discharge at 5A, or a 5-hour discharge at 1A." This is not possible because of intrinsic losses. The output is always less than what has been put in, and the losses escalate with increasing load. High discharge currents make the battery less efficient. To learn about the coulomb counter, see Inner Workings of a Smart Battery.

The efficiency factor of a discharging battery is expressed in the Peukert Law. W. Peukert, a German scientist (1897), was aware of this loss and devised a formula that expresses the loss at a given discharge rate in numbers. Because of sluggish behavior of lead acid, the Peukert numbers apply mostly to this battery chemistry and help in calculating the capacity when loaded at various discharge rates.

The Peukert Law takes into account the internal resistance and recovery rate of a battery. A value close to one (1) indicates a well-performing battery with good efficiency and minimal loss; a higher number reflects a less efficient battery. The Peukert Law of a battery is exponentialand the readings for lead acid are between 1.3 and 1.4. Nickel-based batteries have low numbers and lithium-ion is even better. Figure 1 illustrates the available capacity as a function of ampere drawn with different Peukert ratings.

Figure 1: Available capacity of a lead acid battery at Peukert numbers of 1.08–1.50 A value close to 1 has the smallest losses; higher numbers deliver lower capacities. Source: von Wentzel (2008)

The lead acid battery prefers intermittent loads to a continuous heavy discharge. The rest periods allow the battery to recompose the chemical reaction and prevent exhaustion. This is why lead acid performs well in a starter application with brief 300A cranking loads and plenty of time to recharge in between. All batteries require recovery, and with nickel- and lithium-based system, the electrochemical reaction is much faster than with lead acid. Read more about the Basics About Charging.

The runtime of batteries in portable devices relates to the specific energy marked in Ah (mAh in personal devices). Ah as a performance indicator works best at low discharge currents. At higher loads, the internal resistance begins to play a larger role in the ability to deliver power. Resistance acts as the “gatekeeper.” Energy in Ah presents the available storage capacity of a battery and is responsible for the runtime; power governs the load current. These two attributes are critical in digital devices that require long runtimes and must deliver high-current pulses.

Ah alone is not a reliable runtime indicator and the relationship between capacity and the ability to deliver current can best be illustrated with the Ragone Chart. Named after David V. Ragone, the Ragone chart evaluates batteries not on energy alone but also represents power. 

Figure 2 illustrates the Ragone chart on a digital camera that is powered by an Alkaline, Lithium (Li-FeS2) or NiMH battery drawing 1.3W. (1.3W at 3V draws 433mA.) The horizontal axis displays energy in Watt/hours and the vertical axis displays power in Watts. The scale is logarithmic to allow a wide selection of battery sizes.

Figure 2: Ragone chart illustrates battery performance with various load conditions. Digital camera loads NiMH, Li-FeS2 and Alkaline with 1.3W pulses according to ANSI C18.1 (dotted line). The results are: - Li- FeS2 690 pluses - NiMH 520 pulses - Alkaline 85 pulses Energy = Capacity x V Power = Current x V Courtesy of Exponent

The dotted line represents the power demand of the digital camera. All three batteries have similar Ah rating: NiMH delivers the highest power but has the lowest specific energy. This battery works well at high loads such as power tools. The Lithium Li-FeS2 offers the highest specific energy but has moderate loading conditions. Digital cameras and personal medical instruments suit the system well. Alkaline offers an economic solution for lower current drains such as flashlights, remote controls and wall clocks, but a digital camera is stretching the capability of Alkaline. Read more about the Choices of Primary Batteries.

Series motor

Series motor

A series motor is identical in construction to a shunt motor except for the field. The field is connected in series with the armature and must, therefore, carry the full armature current (Fig. 5.10a). This series field is composed of a few turns of wire having a cross section sufficiently large to carry the current.

Although the construction is similar, the properties of a series motor are completely different from those of a shunt motor. In a shunt motor, the flux F per pole is constant at all loads because the shunt field is connected to the line. But in a series motor the flux per pole depends upon the armature current and, hence, upon the load. When the current is large, the flux is large and vice versa. Despite these differences, the same basic principles and equations apply to both machines.

Figure 5.10a. Series motor connection diagram b. Schematic diagram of a series motor

When a series motor operates at full-load, the flux per pole is the same as that of a shunt motor of identical power and speed. However, when the senes motor starts up, the armature current is higher than normal, with the result that the flux per pole is also greater than normal It follows that the starting torque of a senes motor is considerably greater than that of a shunt motor.This can be seen by comparing the T versus / curves of Figs 5. 8 and 5.11.

On the other hand, if the motor operates at less than full-load, the armature current and the flux per pole are smaller than normal. The weaker field causes the speed to rise in the same way as it would for a shunt motor with a weak shunt field. For example, if the load current of a senes motor drops to half its normal value, the flux diminishes by half and so the speed doubles. Obviously, if the load is small, the speed may nse to dangerously high values. For this reason we never permit a senes motor to operate at no-load It tends to run away, and the resulting centnfugal forces could tear the windings out of the armature and destroy the machine

Series motor speed control

When a senes motor carries a load, its speed may have to be adjusted slightly. Thus, the speed can be increased by placing a low resistance in parallel with the series field. The field current is then smaller than before, which produces a drop in flux and an increase in speed.

Figure 5.11 Typical speed-torque and current-torque characteristic of a series motor.

Conversely, the speed may be lowered by connecting an external resistor in senes with the armature and the field The total IR drop across the resistor and field reduces the armature supply voltage, and so the speed must fall.

Typical torque-speed and torque-current characteristics are shown in Fig. 5.11. They are quite different from the shunt motor characteristics given in Fig. 5.8b.

Example 5-5
A 15 hp, 240 V, 1780 r/min dc senes motor has a full-load rated current of 54 A. Its operating characteristics are given by the per-unit curves of Fig. 5.11.

Calculate

a. The current and speed when the load torque is
24 N×m
b. The efficiency under these conditions


Solution
a. We first establish the base power, base speed, and base current of the motor. They correspond to the full-load ratings as follows:

P_B = 15 hp = 15 X 746 = 11 190 W

n_B = 1780 r/min

/_B= 54 A

The base torque is, therefore,

A load torque of 24 N-m corresponds to a per-unit torque of

T(pu) = 24/60 = 0.4

Referring to Fig. 5.11, a torque of 0.4 pu is attained at a speed of 1.4 pu. Thus, the speed is

n = n(pu) X n_B = I.4 X 1780
= 2492 r/min

From the T vs / curve, a torque of 0.4 pu requires a current of 0.6 pu. Consequently, the load current is

/ = /(pu) X /_B = 0.6 X 54 = 32.4 A

b. To calculate the efficiency, we have to know P_o and P_i.

P_i = EI = 240 X 32.4 = 7776 W

P_o = nT/9.55 = 2492 X 24/9.55

= 6263 W

h = P_o/P_i = 6263/7776 = 0.805 or 80.5%

  < /div >

Applications of the series motor

Series motors are used on equipment requiring a high starting torque. They are also used to drive devices which must run at high speed at light loads. The series motor is particularly well adapted for traction purposes, such as in electric trains. Acceleration is rapid because the torque is high at low speeds. Furthermore, the series motor automatically slows down as the train goes up a grade yet turns at top speed on flat ground. The power of a series motor tends to be constant, because high torque is accompanied by low speed and vice versa. Series motors are also used in electric cranes and hoists: light loads are lifted quickly and heavy loads more slowly.

 

Magnetic core

A magnetic core is a piece of magnetic material with a high permeability used to confine and guide magnetic fields
in electrical,electromechanical and magnetic devices such as electromagnets, transformers, electric motors,
inductors and magnetic assemblies. It is made of ferromagnetic metal such as iron, or ferrimagnetic compounds
such as ferrites. The high permeability, relative to the surrounding air, causes the magnetic field lines to be
concentrated in the core material. The magnetic field is often created by a coil of wire around the core that
carries a current. The presence of the core can increase the magnetic field of a coil by a factor of several thousand
over what it would be without the core.

The use of a magnetic core can enormously concentrate the strength and increase the effect of magnetic fields produced
by electric currents and permanent magnets. The properties of a device will depend crucially on the following factors:

• the geometry of the magnetic core.
• the amount of air gap in the magnetic circuit.
• the properties of the core material (especially permeability and hysteresis).
• the operating temperature of the core.
• whether the core is laminated to reduce eddy currents.               

Air core

A coil not containing a magnetic core is called an air core coil. This includes coils wound on a plastic or ceramic 

form in addition to those made of stiff wire that are self-supporting and have air inside them. Air core coils generally 

have a much lower inductance than similarly sized ferromagnetic core coils, but are used in radio frequency circuits

 to prevent energy losses called core losses that occur in magnetic cores. The absence of normal core losses permits

 a higher Q factor, so air core coils are used in high frequency resonant circuits, such as up to a few megahertz. 

However, losses such as proximity effect and dielectric losses are still present.

• 
  
  

  Pot core

  Usually ferrite or similar. This is used for inductors and transformers. The shape of a pot core is round with an 
  internal hollow that almost completely encloses the coil. Usually a pot core is made in two halves which fit together 
  around a coil former (bobbin). This design of core has a shielding effect, preventing radiation and reducing electromagnetic interference.
  
  
  *From Wikipedia, the free encyclopedia

Problems

Transformer Core Losses

The abi l i ty  of   i ron  or s teel   to carry  m agnet i c  f l ux   i s   m uch greater  than  i t   i s   i n  ai r,  and  thi s  abi l i ty   to al l ow m agnet i c  f l ux   to  f l ow  i s

cal l ed  permeabi l ity .   M os t   trans form er  cores   are  cons tructed  from   l ow  carbon  s teel s   whi ch  can  hav e  perm eabi l i t i es   i n  the

order  of   1500  com pared  wi th  j us t   1.0  for  ai r.   Thi s   m eans   that   a  s teel   l am i nated  core  can  carry   a  m agnet i c  f l ux   1500  t im es

bet ter  than  that   of   ai r.   Howev er,   when  a  m agnet i c  f l ux   f l ows   i n  a  trans form ers   s teel   core,   two  ty pes   of   l os s es   occur  i n  the

s teel .  One  term ed "eddy  current   l os s es " and  the other  term ed "hy s teres i s   l os s es ".

Hys te re s i s   Los s e s

Trans form er Hy s teres i s  Los s es   are caus ed becaus e of   the  fri ct i on of   the m ol ecul es   agai ns t   the  f l ow of   the  m agnet i c  l i nes  of

force  requi red  to  m agnet i s e  the  core,  whi ch  are cons tant l y   changi ng  i n  v al ue  and di rect i on  f i rs t   i n  one di rect i on  and  then  the

other  due  to  the  i nf l uence  of   the  s i nus oi dal   s uppl y   v ol tage.   Thi s   m ol ecul ar  fri ct i on  caus es   heat   to  be  dev el oped  whi ch

repres ents   an  energy   l os s   to  the  trans form er.   Ex ces s i v e  heat   l os s   can  ov ert im e  s horten  the  l i fe  of   the  i ns ul at i ng  m ateri al s

us ed  i n  the m anufacture of   the wi ndi ngs  and s tructures .  Therefore,  cool i ng of  a  trans form er  i s   im portant .

A l s o,   trans form ers  are des i gned  to operate at  a part i cul ar s uppl y   frequency .  Loweri ng  the  frequency  of   the s uppl y  wi l l  res ul t   i n

i ncreas ed hy s teres i s  and  hi gher  tem perature  i n  the  i ron  core.  So reduci ng  the s uppl y   frequency   from  60 Hertz   to  50 Hertz  wi l l

rai s e  the am ount  of  hy s teres i s  pres ent ,  decreas ed  the V A  capaci ty  of   the  trans form er.

Eddy Curre nt  Los s e s

Trans form er  Eddy   Current   Los s es   on  the  other  hand  are  caus ed  by   the  f l ow  of   ci rcul at i ng  currents   i nduced  i nto  the  s teel

caus ed  by   the  f l ow of   the  m agnet i c  f l ux   around  the core.   Thes e  ci rcul at i ng  currents   are  generated becaus e  to  the  m agnet i c

caus ed  by   the  f l ow of   the  m agnet i c  f l ux   around  the core.   Thes e  ci rcul at i ng  currents   are  generated becaus e  to  the  m agnet i c

f l ux   the  core  i s   act i ng  l i ke a  s i ngl e  l oop  of  wi re.  Si nce  the  i ron  core  i s  a  good conductor,   the eddy  currents   i nduced by   a s ol i d

i ron  core  wi l l   be  l arge.   Eddy   currents   do  not   contri bute any thi ng  towards   the  us eful nes s   of   the  trans form er  but   i ns tead  they

oppos e  the  f l ow  of   the  i nduced  current  by   act i ng  l i ke  a  negat i v e  force  generat i ng res i s t i v e  heat i ng  and power  l os s  wi thi n  the

core.

La mi na ti ng  the   I ron Core

Eddy   current   l os s es   wi thi n  a  trans form er  core  can  not   be  el im i nated  com pl etel y ,   but   they   can  be  great l y   reduced  and

control l ed by  reduci ng  the  thi cknes s  of   the s teel   core.   Ins tead of  hav i ng one bi g  s ol i d  i ron core as   the  m agnet i c core m ateri al

of   the  trans form er or coi l ,   the m agnet i c path  i s  s pl i t  up  i nto m any   thi n pres s ed s teel  s hapes  cal l ed "l am i nat i ons ".

The  l am i nat i ons   us ed  i n  a  trans form er  cons truct i on  are  v ery   thi n  s tri ps

of   i ns ul ated  m etal   j oi ned  together  to  produce a  s ol i d but   l am i nated core

as  we s aw abov e.  Thes e  l am i nat i ons  are  i ns ul ated  from  each other by  a

coat   of   v arni s h  or  paper  to  i ncreas e  the  ef fect i v e  res i s t i v i ty   of   the  core

thereby   i ncreas i ng  the  ov eral l   res i s tance  to  l im i t   the  f l ow  of   the  eddy

currents .   The  res ul t   of   al l   thi s   i ns ul at i on  i s   that   the  unwanted  i nduced

eddy   current   power-l os s   i n  the  core  i s   great l y   reduced,   and  i t   i s   for  thi s

reas on  why   the  m agnet i c  i ron  ci rcui t   of   ev ery   trans form er  and  other

el ectro-m agnet i c  m achi nes   are  al l   l am i nated.   Us i ng  l am i nat i ons   i n  a

trans fom er cons truct i on reduces  eddy  current   l os s es .

The  l os s es   of   energy ,   whi ch  appears   as   heat   due  both  to  hy s teres i s   and  to  eddy   currents   i n  the

m agnet i c  path,   i s   known  com m onl y   as   "trans form er  core  l os s es ".   Si nce  thes e  l os s es   occur  i n  al l

m agnet i c  m ateri al s   as   a  res ul t   of   al ternat i ng m agnet i c  f i el ds .   Trans form er  core  l os s es   are  al way s

pres ent   i n  a  trans form er  whenev er  the  prim ary   i s   energi z ed,   ev en  i f   no  l oad  i s   connected  to  the

s econdary   wi ndi ng.   A l s o  thes e  hy s teres i s   and  the  eddy   current   l os s es   are  s om et im es   referred  to

as  "trans form er  i ron  l os s es ",  as   the m agnet i c  f l ux  caus i ng  thes e  l os s es   i s  cons tant  at  al l   l oads .

Eddy Curre nt  Los s e s

But   there  i s   al s o  another  ty pe  of  energy   l os s   as s oci ated  wi th  trans form ers   cal l ed  "copper  l os s es ".

Trans form er  Copper  Losses  are  m ai nl y   due  to  the  el ectri cal   res i s tance  of   the  prim ary   and

s econdary   wi ndi ngs .   M os t   trans form er  coi l s   are  m ade  from   copper  wi re  whi ch  has   res i s tance  i n

Ohm s ,  ( Ω  ).  Thi s  res i s tance oppos es   the m agnet i s i ng currents   f l owi ng  through  them .

When  a  l oad  i s   connected  to  the  trans form ers   s econdary   wi ndi ng,   l arge  el ectri cal   currents   f l ow  i n

both  the  prim ary   and  the  s econdary   wi ndi ngs ,   el ectri cal   energy   and  power  (  or  the  I   R  )  l os s es

occur as   heat .  General l y  copper  l os s es   v ary  wi th  the  l oad  current ,  bei ng alm os t  z ero  at  no-l oad,  and

at   a  m ax im um   at   ful l -l oad  when  current   f l ow  i s   at   m ax im um .   Trans form ers   wi th  hi gh  v ol tage  and  current   rat i ngs   requi re

conductors  of   l arge cros s -s ect i on  to hel p m i nim i s e  thei r copper  l os s es .

Then we can def i ne an  i deal   trans form er as  hav i ng:

1.  No Hy s teres i s   l oops  or Hy s teres i s   l os s es    →  0

2.   Inf i ni te Res i s t i v i ty  of  core m ateri al  gi v i ng z ero Eddy  current   l os s es    →  0

3.  Zero wi ndi ng res i s tance gi v i ng z ero  I R copper  l os s es    →  0

In  the  nex t   tutori al   about   T ransf ormers  we  wi l l   l ook  at   Tr a n s f o r m e r   L o a d i n g   of   the  s econdary   wi ndi ng  wi th  res pect   to  an

el ectri cal   l oad and s ee  the ef fect  a "NO-l oad" and a "ON-l oad" connected  trans form er has  on  the prim ary  wi ndi ng current 

Destination

I.E.S. Paper 1

                                                     

                                           I.E.S. (Infinite Energy Source)

As the name implies the project itself is an attempt to achieve self sustainability for its operation i.e. infinite energy. I have been trying to make a SIMO (Single Input Multiple Output) system. Input given to the system is in form of electrical energy and output energy obtained is actually a proper combination of mechanical and electrical energy. Block diagram shown below would make it clear to all.

It is a project which needs deep study of the concepts of Electromagnetism (Induction, Flux lines, Faraday Law and etc.) and sort of expertise to select the appropriate material. I could successfully achieved what I wanted to.  Soon I’ll update the blog with the original images and videos of my project to let you know about it more.

Continue……….

Written By-

Navin Chand Paneru

Project Assistant 

E-mail-navinpan1@gmail.com