Tutorial 15 - Inductive Transients

Learning Objective

To understand the significance of inductive transients.

To learn how they can be damaging and how to minimise the risk.

To study inductive rise quantitatively and graphically.

Key Question

What are transients in inductors?

What happens when we turn an inductor on?

What is inductive rise?

Introduction to inductive transients

If we connect a transformer, or other inductor, to a DC supply, then turn off the supply with a switch, we can get a reverse voltage spike that has a high value.  If we happened to be touching the wires to the inductor, we could feel it as a short sharp shock.  The spike can ruin electronic components, so we need to protect them against this using a reverse-biased diode. The reverse biased diode D1 is there to protect the transistor from high voltage spikes that can occur when the relay turns off.

There are two ways we can explain this observation:

1.      The rapid collapse of the magnetic field. The field drops from a high value to 0 in a short time.  So the dt term (time interval) is small, making the reverse voltage very big.

2.      The rapid fall in current. Again, if the dt term is very small, the reverse voltage is big.

There is a lot of energy contained in an inductor, given by the equation: These inductive transients can be a damned nuisance, especially if you are dealing with big DC motors.  They have big coils which mean big magnetic fields.  In turn this means big voltage spikes, resulting in flash-overs and burned out wiring.  Large motors are started with a low voltage to start with, and the voltage is slowly increased over a period of several seconds.  This is called ramping up.  A similar process of ramping down is used when the motor is turned off.  The idea of this is to increase the dt term, so reducing the reverse voltage.  In old installations, this was done manually, with a man winding a control handle.  Now there is electronic control gear that does this automatically.

The train here is taking a current of 2000 A DC at a voltage of about 660 V.  When the contact between the conductor skate and the conductor rail is broken, the reverse voltage spike can strike a large arc like this.  Sparks of red hot metal can be thrown from the rail. Picture by kind permission of  SPSmiler (Wikimedia Commons) These transients happen with DC, so reactance of an inductor to an alternating current does not come into it.

 Calculate the magnetic energy contained a DC motor of inductance 150 mH that takes a current of 50 A. Question 2 The motor in question 1 is turned off improperly by opening the switch.  What is the value of the reverse voltage spike, if the time taken for the current to fall to zero is 20 ms?

### What happens when we switch an inductor on?

Consider this circuit that consists of a DC power supply of p.d. V, a switch, S, an inductor, L, and a resistor, R.  The circuit looks like this: When the switch is first closed, the inductor has a current I passing through it.  The magnetic field is being built up according to the relationship: This is the combination of Faraday’s and Lenz’s Laws.  Lenz’s Law gives us the minus sign that tells us that the induced EMF opposes the voltage from the battery.  Immediately the switch is turned on, the induced voltage has the same value as the battery voltage, but the opposite sign. We also know that the EMF is related to the current by: It is a series circuit, so the current is the same all the way round the circuit.  By Kirchhoff II we can write: The battery voltage V remains the same.  The voltage across the inductor has the same value as the induced voltage, E.  We can write: And we can rewrite this as: The voltage drop across the resistor is IR, so we can write the Kirchhoff II statement as: Immediately the switch is turned on, the voltage across the resistor is 0 as all the voltage is across the inductor.

A short time later the balance is altered, as the induced emf is reduced, the voltage across the resistor is increased.  The sum of the two voltages is V.

After a couple of seconds or so, the reverse voltage made by the inductor is zero.  Therefore the voltage across the resistor is the battery voltage, V.

Measuring the transient is not at all easy unless you have a data-logger.  The graph for the induced voltage looks like this: For the voltage across the resistor, it looks like this: The graph of current against time follows the same pattern as the voltage across the resistor to give: Since the equation has a differential term (dI/dt), the mathematical trick of calculus tells us that the form of the graphs will be exponential

Consider the reverse voltage decay.  As with all exponential functions, there is a constant fraction of the voltage remaining for each time period.  We can say that there is a half-life, the time taken for the voltage to drop from say 12 V to 6 V.  In the next half-life, the voltage falls from 6 V to 3 V and so on.

As with capacitors, there is a time constant.  With capacitors we saw that the product of ohms and farads was seconds.  With inductance, we see that the henries divided by ohms makes seconds.  Let’s give the time constant the physics code t (“tau”, a Greek lower case letter ‘t’). So we can write an equation for the decay of the reverse voltage.  For a capacitor, we saw that the equation was of the form: So we can write a similar equation for the inductor. This tidies up to give us: There is no reason at all that we can’t write the equation as: The term t is the time constant, which we work out by dividing the inductance by the resistance.  In this form, the equation is identical to the equation for the capacitor.

As with capacitance, the time constant can be defined as the time taken for the voltage to fall to about 37 % of its original value.

It takes about 5 time constants for the reverse voltage to be close to zero, i.e., the full supply voltage to be applied.  This is often called the transient time.  A real inductor (a coil of wire) will not be perfect; it will have a definite resistance, so there will be a time constant.  If we have an external resistor, we would need to take the resistance of the inductor into account by adding it to the resistance of the inductor.

 Worked Example A universal motor has a resistance of 60 ohms and an inductance of 100 mH. It is connected to a 230 V DC supply. Work out: (a)    The time constant of the drop in reverse voltage; (b)   The time needed for the forward voltage to reach 100 V; (c)    The time needed for the forward voltage to be about 230 V. Answer           1          t = 100 × 10-3 ÷ 60 = 1.67 × 10-3 s             2          Reverse voltage = 130 V 130 = 230 × e-t/1.67× 10^-3 Þ ln 130 = ln 230 + -t/1.67 × 10-3 -t = (ln 130 – ln 230) × 1.67 × 10-3  t =  (0.571) × 1.67 × 10-3 =  9.53 × 10-4 s = 0.95 ms             3          Time = 5 t = 8.35 × 10-3 s.

You can see from this that the inductive fall in the reverse voltage is a very short-term phenomenon. When doing a calculation like this, make sure that you know whether you are dealing with a reverse voltage (back emf) or the forward voltage.  It was easy to put the 100 V into the equation in the example above.

In reality, it is impossible to measure the reverse voltage directly.  This is because we are assuming that the inductor is perfect.  In reality it is not.  It will always have a resistance associated with it.  If we measure the voltage, we would see an exponential rise in the forward voltage.  So we have to infer the exponential fall in the reverse voltage.

Another problem about inductive rise is that it takes place over a very short time period.

 An inductor of 0.50 mH has a resistance of 10 W.  What is the time constant? ### Inductive rise

Kirchhoff II tells us that the reverse voltage and the voltage across the resistance add up to the supply voltage, V0. So we can rearrange the equation to write: If we were to measure the voltage across the real inductor, we would observe the inductive rise which is given by the formula: The graph for the voltage looks like this: For an exponential rise, the voltage at the time constant is 63 % of the final voltage.

The current graph looks like this: Both look the same, although the numbers on the vertical axis are different.  This stands to reason, since I = V/R.  Since R is constant, we can rewrite the formula: The data that were used to generate these graphs were:

·         Supply voltage = 12 V;

·         Inductance = 0.15 H;

·         Resistance = 9.0 W.

These are typical values you may see in an ordinary physics lab.  A huge inductor as found at CERN has a value of only 14 H.  Suppose the resistance was 1 ohm, the time constant would be 14 s.  The point is that it is difficult to capture transients unless we use data capture equipment with appropriate software.

With a capacitor, you can increase the time constant simply by increasing the resistance.

 You can carry out the experiment to measure the exponential rise in voltage across a capacitor easily in the lab with basic physics equipment.  Explain why you cannot investigate inductive transients in this way.  Consider the time constant in your argument. Question 5 What would happen to the time constant if you increased the resistance in an inductor? Question 6 An inductor of inductance 0.47 mH has a resistance of 4.0 W and is connected to a DC supply of 20 V. a.       What is the steady state current? b.      What is the time constant? c.       What is transient time? d.      What is the value of the instantaneous reverse voltage after 10 ms? e.      What is the value of the instantaneous current after 10 ms? 