# Counter EMF (also known as Back EMF)

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Counter EMF (also known as Back EMF)
When the current through an inductor changes, the magnetic field also changes. This changing magnetic field causes there to be an induced voltage across the terminals of the inductor. This induced voltage’s polarity is in a direction opposite to that of the original voltage applied. This is basically Lenz’s law in a nutshell.

Current through inductors in a DC circuit when switch closes

What do you think voltage across an inductor looks like in a DC circuit when the switch is closed?

Voltage through inductors in a DC circuit when switch closes
Inductance has it greatest effect only when there is a change in current.

What would happen in this circuit?

Inductance Summary Inductors react against changes in current by consuming or providing voltage in a polarity necessary to oppose the change When an inductor is faced with an increasing current, it acts as a load, by consuming voltage. (series opposing) When an inductor is faced with a decreasing current, it acts as a source, by sourcing voltage. (series aiding) The ability of an inductor to store energy in the form of a magnetic field (and consequently opposing changes in current) is called inductance. The unit of inductance is the Henry – H (Denoted as the letter L on a schematic)

Random Video of the Day Another RVOTD

Inductor Schematic Symbol
Typically, the range of inductors are from micro-Henry’s to milli-Henry’s (H to mH)

In the picture of the inductor below, since current takes the path of least resistance, why wouldn’t the current just bypass all the turning and jump straight across the tops of all the coils? Because there is a clear enamel coating on the wires that acts like an insulator

Factors Affecting Coil Inductance

Factors Affecting Coil Inductance
N – Number of turns The design of the inductor is most important in determining the value of inductance. 1. The amount of turns in creating the coil affect how much voltage can be linked between the turns. In a later equation, this figure is denoted by (N).

Coil Inductance Factors (cont.)
A – Area of coil 2. The diameter of the coil affects inductance. A larger diameter core results in more magnetic lines of force than compared to a smaller diameter coil. Denoted by (A) for area=πr2. r

Coil Inductance Factors (cont.)
l – length of coil 3. As the length (l) of the inductor grows, the distance between turns increases, which causes the magnetic field to get weaker. Denoted by l, for length in meters, not I for current! l

Coil Inductance Factors (cont.)
µ - permeability of core 4. And finally, inductance is affected by the core material in which the coil is wrapped around. Higher permeability enables more flux to form. More flux means more inductance. This is denoted by (µ) pronounced “mju”, not known as micro in this case. Table 16.2 from book Material Permeability, µ (H/m) Relative Permeability, µr (No Units) Air or Vacuum 1.26 x 10-6 1 Nickel 6.28 x 10-5 50 Cobalt 7.56 x 10-5 60 Cast Iron 1.1 x 10-4 90 Machine Steel 5.65 x 10-4 450 Transformer iron core 6.9 x 10-3 5500 Silicon Iron 8.8 x 10-3 7000 Permalloy 0.126 100,000 Supermalloy 1.26 1,000,000

Inductance equation 18.3.1 L= µN2A l L= Inductance in Henrys, H
Table 16.2 from book Material Permeability, µ (H/m) Relative Permeability, µr (No Units) Air or Vacuum 1.26 x 10-6 1 Nickel 6.28 x 10-5 50 Cobalt 7.56 x 10-5 60 Cast Iron 1.1 x 10-4 90 Machine Steel 5.65 x 10-4 450 Transformer iron core 6.9 x 10-3 5500 Silicon Iron 8.8 x 10-3 7000 Permalloy 0.126 100,000 Supermalloy 1.26 1,000,000 L= Inductance in Henrys, H µ=permeability of core, µ0µr µ0= 1.26 x (The absolute permeability of air) µr= relative permeability of core material, refer to Table 16.2  N= number of coil turns A= area of cross-section of coil wire, πr2. l= length of core material (m)

Inductance problem worked out
Given: l=10cm r=1.5cm N=200 turns l= .1m r=.015m N=200 turns It’s an air core that we’re dealing with. So relative permeability is µr =1. L= (µ) (200)2 [π(.0152)] .1 L= µN2A l L= [(1.26 x 10-6)] (40000) ( ) .1 L= H or 356.3 µH L= .1

Another Inductance problem!
Given: l=15cm r=.02m N=500 turns Machine Steel Core Find Inductance Value

Another Inductance problem!
Given: l = 15cm r = .02m L = 3mH Nickel Core Find the number of turns

Another Inductance problem!
What is the inductance if you take a wire and wrap it around a pencil 15 times?

Types of Inductors Types of Inductors Air Core Low value L
High frequency applications Iron-Core Large value L Used in transformers Powdered iron-core Mid value L Used to reduce losses such as eddy currents. Used for higher DC current applications Ferrite Core Low to High value L Great Magnetic Conductor, Low Electrical Conductor. Suitable for high and low freq applications Toroidal core Mid to High value L Very low flux leakage loss because of donut shape. Variable Core Variable L Used for tuning circuits like radios Printed Circuit board core Very low L Coil printed Directly on circuit board. For Freq above 500Mhz (Very high

PCB Core Inductors

Measuring the induced voltage across an inductor
Equation 18.1: This equation says that the higher the change in current, and the faster that change occurs, the higher the voltage across the inductor will be.

How much self-induced voltage occurs across a 4H inductor when the current going through it changes by 10Amps in 1 second?

How much self-induced voltage occurs across a 4H inductor when the current going through it changes by 10Amps in 1ms? 1ms) = 400V

Calculating total inductance LT
What is the total inductance of this circuit? 62H Calculating the total inductance in a circuit is done the same way as calculating the total resistance

Calculating total inductance LT
What is the total inductance of this circuit? 6H Calculating the total inductance in a circuit is done the same way as calculating the total resistance

Calculating total inductance LT
What is the total inductance of this circuit? LT = 15H Calculating the total inductance in a circuit is done the same way as calculating the total resistance

THIS IS CALLED MUTUAL INDUCTANCE
However! There is a subtle difference in calculating LT that you don’t have to consider when calculating RT What if the magnetic field from one inductor interferes with the magnetic field from the inductor next to it? In other words, if the magnetic field from one inductor cuts the across the coils of the inductor located physically next to it, then extra inductance will be introduced, (or some inductance will be cancelled out.) THIS IS CALLED MUTUAL INDUCTANCE

MUTUAL INDUCTANCE

Mutual Inductance

Calculating total inductance if there is mutual inductance involved
In the circuit above, if the inductor’s magnetic fields didn’t interfere with each other, aka have no mutual inductance, then the total inductance is just 13H. However, suppose the inductors magnetic fields were completely overlapping each other, then we would have to consider this.

k is the coefficient of coupling and it is a number ranging from 0% to 100% If the coils of 2 inductors were wrapped around each other, then the magnetic fields of each have no choice but to couple themselves to the other inductor. In this case the coefficient of coupling, k, is 100%. If the inductors’ magnetic fields are not coupled, then k = 0.

MUTUAL INDUCTANCE k = .9 k = .3 k = 0

Calculating total inductance if there is mutual inductance involved
Suppose the inductors were connected in a series aiding arrangement with a coefficient of coupling of 90%.

Calculating total inductance if there is mutual inductance involved
Suppose the inductors were connected in a series opposing arrangement with a coefficient of coupling of 90%.

2 coils have inductances of 8mH and 4. 7mH
2 coils have inductances of 8mH and 4.7mH. If the coefficient of coupling between them is .82, what is the mutual inductance? Now find the total inductance if they are series aiding and if they are series opposing LT = 8mH mH + 2·5.03mH = 22.76mH (If in series aiding arrangement) LT = 8mH mH - 2·5.03mH = 2.64mH (If in series opposing arrangement)

Stray Inductance Talk about stray inductance

Go over Cap/Inductor Handout

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