RL circuit

A resistor-inductor circuit (RL circuit), or RL filter or RL network, is one of the simplest analogue infinite impulse response electronic filters. It consists of a resistor and an inductor, either in series or in parallel, driven by a voltage source.

Introduction

There are three basic, analogue circuit components: the resistor (R), capacitor (C) and inductor (L). These may be combined in four important combinations: the RC circuit, the RL circuit, the LC circuit and the RLC circuit with the abbreviations indicating which components are used. These circuits, between them, exhibit a large number of important types of behaviour that are fundamental to much of analogue electronics. In particular, they are able to act as passive filters. This article considers the RL circuit, in both series and parallel as shown in the diagrams.

In practice, however, capacitors (and RC circuits) are usually preferred to inductors since they can be more easily manufactured and are generally physically smaller, particularly for higher values of components.

This article relies on knowledge of the complex impedance representation of inductors and on knowledge of the frequency domain representation of signals.

Complex Impedance

The complex impedance Z_L (in ohms) of an inductor with inductance L (in henries) is

Z_L \ = \ Ls

The angular frequency s is, in general, a complex number,

s \ = \ \sigma + j \omega

where

j represents the imaginary unit:

j = \sqrt{-1}

$\sigma \$ is the exponential decay constant (in radians per second), and

$\omega \$ is the sinusoidal angular frequency (also in radians per second).

Eigenfunctions

The complex-valued eigenfunctions of any linear time-invariant (LTI) system are of the form:

V(t) \ = \ \mathbf{A}e^{st} \ = \ \mathbf{A}e^{(\sigma + j \omega) t}  \ = \ Ae^{j \phi}e^{(\sigma + j \omega) t}   \ = \ Ae^{\sigma t}e^{j ( \omega t + \phi )}

From Euler's formula, the real-part of these eigenfunctions are exponentially-decaying sinusoids:

v(t) \ = \ Re \left \{ V(t)  \right \} \ = \ A e^{\sigma t} cos(\omega t + \phi )

Sinusoidal Steady State

Sinusoidal steady state is a special case in which the input voltage consists of a pure sinusoid (with no exponential decay). As a result,

\sigma \ = \ 0

and the evaluation of s becomes

s \ = \  j \omega

Series circuit

By viewing the circuit as a voltage divider, we see that the voltage across the inductor is:

V_L(s) = \frac{Ls}{R+Ls}V_{in}(s)

and the voltage across the resistor is:

V_R(s) = \frac{R}{R+Ls}V_{in}(s)

Transfer functions

The transfer function for the inductor is

H_L(s) = { V_L(s) \over V_{in}(s) }   = { Ls \over R + Ls  }  =  G_L e^{j \phi_L}

Similarly, the transfer function for the resistor is

H_R(s) = { V_R(s) \over V_{in}(s) }   = { R \over R + Ls  }  =  G_R e^{j \phi_R}

Poles and zeros

Both transfer functions have a single pole located at

s = - {R \over L }

In addition, the transfer function for the inductor has a zero located at the origin.

Gain and phase angle

The gains across the two components are found by taking the magnitudes of the above expressions:

G_L =  | H_L(s) | = \left|\frac{V_L(s)}{V_{in}(s)}\right| = \frac{\omega L}{\sqrt{R^2 + \left(\omega L\right)^2}}

and

G_R =  | H_R(s) | = \left|\frac{V_R(s)}{V_{in}(s)}\right| = \frac{R}{\sqrt{R^2 + \left(\omega L\right)^2}}

and the phase angles are:

\phi_L = \angle H_L(s)  =  \tan^{-1}\left(\frac{R}{\omega L}\right)

and

\phi_R =  \angle H_R(s)  = \tan^{-1}\left(-\frac{\omega L}{R}\right)

Phasor notation

These expressions together may be substitued into the usual expression for the phasor representing the output:

V_L = G_{L}V_{in} e^{j \phi_L}

V_R = G_{R}V_{in}e^{j \phi_R}

Current

The current in the circuit is the same everywhere since the circuit is series:

I(s) = \frac{V_{in}(s)}{R + Ls}

Impulse Response

The impulse response for each voltage is the inverse Laplace transform of the corresponding transfer function. It represents the response of the circuit to an input voltage consisting of an impulse or delta function.

The impulse response for the inductor voltage is

h_L(t) = - { R \over L} e^{-tR / L} u(t)   =   - { 1 \over \tau} e^{-t / \tau} u(t)

where u(t) is the Heaviside step function and

\tau = { L \over R}

is the time constant.

Similarly, the impulse response for the resistor voltage is

h_R(t) =  {R \over L} e^{-tR / L} u(t)   =   { 1 \over \tau} e^{-t / \tau} u(t)

Zero input response (ZIR)

The Zero input response, also called the natural response, of an RL circuit describes the behavior of the circuit after it has reached constant voltages and currents and is disconnected from any power source. It is called the zero-input response because it requires no input.

The ZIR of an RL circuit is:

i(t) = i(0)e^{-(R/L) t} = i(0)e^{-t/ \tau} \!\

Frequency domain considerations

These are frequency domain expressions. Analysis of them will show which frequencies the circuits (or filters) pass and reject. This analysis rests on a consideration of what happens to these gains as the frequency becomes very large and very small.

As $\omega \to \infty$ :

G_L \to 1

G_R \to 0

As $\omega \to 0$ :

G_L \to 0

G_R \to 1

This shows that, if the output is taken across the inductor, high frequencies are passed and low frequencies are attenuated (rejected). Thus, the circuit behaves as a high-pass filter. If, though, the output is taken across the resistor, high frequencies are rejected and low frequencies are passed. In this configuration, the circuit behaves as a low-pass filter. Compare this with the behaviour of the resistor output in an RC circuit, where the reverse is the case.

The range of frequencies that the filter passes is called its bandwidth. The point at which the filter attenuates the signal to half its unfiltered power is termed its cutoff frequency. This requires that the gain of the circuit be reduced to

G_L = G_R = \frac{1}{\sqrt{2}}

Solving the above equation yields

\omega_{c} = \frac{R}{L}

rad/s

f_c = \frac{R}{2\pi L}

which is the frequency that the filter will attenuate to half its original power.

Clearly, the phases also depend on frequency, although this effect is less interesting generally than the gain variations.

As $\omega \to 0$ :

\phi_L \to 90^{\circ} = \pi/2^{c}

\phi_R \to 0

As $\omega \to \infty$ :

\phi_L \to 0

\phi_R \to -90^{\circ} = -\pi/2^{c}

So at DC (0 Hz), the resistor voltage is in phase with the signal voltage while the inductor voltage leads it by 90°. As frequency increases, the resistor voltage comes to have a 90° lag relative to the signal and the inductor voltage comes to be in-phase with the signal.

Time domain considerations

This section relies on knowledge of e, the natural logarithmic constant.

The most straightforward way to derive the time domain behaviour is to use the Laplace transforms of the expressions for $V_L$ and $V_R$ given above. This effectively transforms $j\omega \to s$ . Assuming a step input (i.e. $V_{in} = 0$ before $t = 0$ and then $V_{in} = V$ afterwards):

V_{in}(s) = V\frac{1}{s}

V_L(s) = V\frac{sL}{R + sL}\frac{1}{s}

and

V_R(s) = V\frac{R}{R + sL}\frac{1}{s}

Partial fractions expansions and the inverse Laplace transform yield:

\,\!V_L(t) = Ve^{-tR/L}

\,\!V_R(t) = V\left(1 - e^{-tR/L}\right)

Thus, the voltage across the inductor tends towards 0 as time passes, while the voltage across the resistor tends towards V, as shown in the figures. This is in keeping with the intuitive point that the inductor will only have a voltage across as long as the current in the circuit is changing — as the circuit reaches its steady-state, there is no further current change and ultimately no inductor voltage.

These equations show that a series RL circuit has a time constant, usually denoted $\tau = L/R$ being the time it takes the voltage across the component to either fall (across L) or rise (across R) to within $1/e$ of its final value. That is, $\tau$ is the time it takes $V_L$ to reach $V(1/e)$ and $V_R$ to reach $V(1- 1/e)$ .

The rate of change is a fractional $\left(1 - \frac{1}{e}\right)$ per $\tau$ . Thus, in going from $t=N\tau$ to $t = (N+1)\tau$ , the votage will have moved about 63% of the way from its level at $t=N\tau$ toward its final value. So the voltage across L will have dropped to about 37% after $\tau$ , and essentially to zero (0.7%) after about $5\tau$ . Kirchoff's voltage law implies that the voltage across the resistor will rise at the same rate. When the voltage source is then replaced with a short-circuit, the voltage across R drops exponentially with t from $V$ towards 0. R will be discharged to about 37% after $\tau$ , and essentially fully discharged (0.7%) after about $5\tau$ . Note that the current, $I$ , in the circuit behaves as the voltage across R does, via Ohm's Law.

The delay in the rise/fall time of the circuit is in this case caused by the back-EMF from the inductor which, as the current flowing through it tries to change, prevents the current (and hence the voltage across the resistor) from rising or falling much faster than the time-constant of the circuit. Since all wires have some self-inductance and resistance, all circuits have a time constant. As a result, when the power supply is switched on, the current does not instantaneously reach its steady-state value, $V/R$ . The rise instead takes several time-constants to complete. If this were not the case, and the current were to reach steady-state immediately, extremely strong inductive electric fields would be generated by the sharp change in the magnetic field — this would lead to breakdown of the air in the circuit and electric arcing, probably damaging components (and users).

These results may also be derived by solving the differential equation describing the circuit:

V_{in} = IR + L\frac{dI}{dt}

and

\,\!V_R = V_{in} - V_L

The first equation is solved by using an integrating factor and yields the current which must be differentiated to give $V_L$ ; the second equation is straightforward. The solutions are exactly the same as those obtained via Laplace transforms.

Parallel circuit

The parallel RL circuit is generally of less interest than the series circuit unless fed by a current source. This is largely because the output voltage $V_{out}$ is equal to the input voltage $V_{in}$ — as a result, this circuit does not act as a filter for a voltage input signal.

With complex impedances:

I_R = \frac{V_{in}}{R}

and

\,\!I_L = \frac{V_{in}}{j\omega L} = -\frac{jV_{in}}{\omega L}

This shows that the inductor lags the resistor (and source) current by 90°.

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