rl transients

7/17/2019 RL Transients

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EE 201 RL transient – 1

RL transients

Circuits having inductors:

• At DC – inductor is a short circuit, just another piece of wire.

• Transient – a circuit changes from one DC configuration to anotherDC configuration (a source value changes or a switch flips). Therewill be a transient interval while the voltages and currents in theinductors change.

• AC – currents and voltages are changing continuously, so inductorsare “amping up” and “de-amping” continuously. This requiresspecial techniques and is the next topic for EE 201.




1. Determine the DC currents in the inductors before the changeoccurs. These may be given, or you may have to solve for them fromthe original configuration.

2. Let the change occur instantaneously at time t = 0. The inductorswill maintain their currents into the “instant” just after the change.(Recall: inductor current cannot change instantaneously.)

3. Analyze the circuit. Since inductor voltage depend on diL / dt, the

result will be a differential equation.

4. Solve the differential equation, using the inductor currents frombefore the change as the initial conditions.

5. The resulting equation will describe the “amping” (or “de-amping”)

of the inductor current during the transient and give the final DCvalue once the transient is complete.

Solving a circuit with transient changes

In practice, we will solve only one circuit and try to understand itcompletely. Then, when we encounter other circuit configurations, wewill make those fit the prototype.




Simple RL circuit transient (physics)

In the circuit, I S abruptly changes value from I i to I f at t = 0. Assume

that the source was at I i for a “very long time” before t = 0.

1. For t < 0, iL = I i and vL = 0. Therefore, iR = 0.

2. At t = 0, I S changes. But iL = I i (still), and so any excess current

must flow through the resistor. Consequently, vL jumps abruptly:

3. For t > 0, iL increases. As iL increases, iR and vL will decrease.

4. After a “sufficiently long time”, iL will “amp up” to I f . The current throughthe resistor and voltage across it both drop to zero. The transient is

complete.

iL

I S

R L +

–

vL

I f

I it = 0

() = − ()

=

−




Simple RL circuit transient (math)

In the circuit, I S abruptly changes value from I i to I f at t = 0. Assume

that the source was at I i for a “very long time” before t = 0.

For t < 0, iL = I i and vL = 0.

=

=

−

−

=

/

−

=

/

−

=

/

− ln −

= /

ln

−

− ()

=

/

() = − − exp

−

/

iLI S

R L +

–

v

L

I f

I i

t = 0




• At t = 0, iL = I i, as expected.

• As t ! !, iL ! I f , also as expected.

• In the between the inductor current changes according to a decayingexponential, given above.

• At t = 0, vL jumps to up to a maximum value.

• For t > 0, vL decays away exponentially, as the current approaches itsfinal value.

Finally, note that the equation works just as well for “amping down” as

it does for “amping up”. In the example, we implied I f > I i, but that

was never a requirement in the derivation of the equation.

() = − −

exp

−

/

() = −

() = =

−

exp −

/




Plots of capacitor voltage and current for a simple RL circuit with I f =

15 mA, V i = 5 mA, R = 5 k !, L = 50 mH (L/R = 50 µs).

15 mA

5 mA 5 k ! 50 mHI

S

R L +

–

vL

I f

I i t = 0

0

2

4

6

8

10

12

14

16

-5

0

5

10

15

20

25

i L ( m

A )

t (µs)

0

10

20

30

40

50

60

-5

0

5

10

15

20

25

v L ( V )

t (µs)


http://slidepdf.com/reader/full/rl-transients 7/12EE 201 RL transient – 7

Using a switch

The same transient phenomena occurs when using a switch to changea circuit.

1. For t < 0, the inductor may have a current flowing (depending on thecircuit configuration), iL(t < 0) = I i . Also, vL = 0.

2. At t = 0, the switch closes. The inductor maintains its current, iL(0) = I i.The inductor voltage jumps abruptly to

3. For t > 0, the inductor current rises. The voltage drops as the current rises.

I S

R L

–

vL

+

iL

() = −

−

exp−

/

() = − ()

=

−



• The decay is characterized by a time constant, " = L/R.

(Check the units.)

If R = 1k! and L = 1 H, " = L/R = 1 ms.

If R = 10 k! and L = 10 mH, " = L/R = 1 µs.

RL time constant

• L/R determines the time scale for the transient.

• After about 5 time constants, most of the changed has occurred.

If t = 5(L/R) ! exp(–5) = 0.0067 ! 99.3% of the transition is done.

(Theoretically, iL never really gets to I f . But as engineers, we have to

be practical.)



Example100 mA

680! 75 mH

In the RL circuit above, I S abruptly changes value from I i = 100 mA to I f= 10 mA at t = 0. Assume that the source was at I i for a “very longtime” before t = 0. Find the expression for the inductor current as a

function of time. Find the time at which iL = 50 mA.

Apply the transient function directly, with I i = 100 mA, I F = 10 mA, and L/

R = 75 mH/680 ! = 110 µs.

I SR L

I f

I i

t =0

iL10 mA

() = − −

]exp

−

/

= [] + []exp

−

µ

= [] + []exp

−

µ

= (µ) ln

= µ



Example 2

The voltage source in the above circuit abruptly changes from – 10 V to+10 V at t = 0. Find expressions for the inductor current and inductorvoltage. Find the time at which the inductor current crosses 0 mA.

The circuit is in the wrong form to the equation directly. Instead use asource transformation to change it to the “standard” form.

100!

0.25 H+

–

R

V S

V i

V f

t =0

iLL

–10 V

10 V

I SR LI

f

I i t =0

iL–100 mA

100 mA

100! 0.25 H

= (.) ln

= .

= []− []exp

−

.

() = − −

]exp

−

/

= [.]− [.]exp

−

.

() =

= −

exp

−

/

= []exp−

.



Example 3

In the RL circuit above, V S abruptly changes value from V 1 to V 2 at t =0. Find the expression for the inductor current as a function of time.

Find the Norton equivalent for the circuit attached to the inductor.

27 mH

1.5 k !

2.2 k !

1 k !

0 V

10 V

+

–

R1

R2

R3

V S

t < 0, I i = 0.

t " 0, I f = 3.14 mA.

L/RN = (27 mH)/(1.89 !) = 14.3 µs

+

–

V 1

V 2

t = 0

R1

R2

R3

iL

L

V S

= +

0 VI

N

LI f

I it = 0

iLR

N

3.14 mA

27 mH1.89 k !

I N =V S

R1 +

1 +

R1

R2

R3

= 0.314 V S.

= 1.89 k!.

iL (t) = [3.14mA]1− exp− t

14.3µs



1. Work through the solution to the differential equation (slide 4) andmake sure that you understand it thoroughly.

2. Try to find the solution to Example 3 without using the Theveninequivalent. It can be done, although it might be a bit messy.

3. For the example shown on slide 6, calculate the inductor energybefore and after the transient. Calculate the total energy delivered bythe source during the transient. Show that everything balances.(Don’t forget about the power consumed in the resistor.)

To study:

rl transients

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