John Santiago

Article / Updated 09-19-2022

Using the Laplace transform as part of your circuit analysis provides you with a prediction of circuit response. Analyze the poles of the Laplace transform to get a general idea of output behavior. Real poles, for instance, indicate exponential output behavior. Follow these basic steps to analyze a circuit using Laplace techniques: Develop the differential equation in the time-domain using Kirchhoff’s laws and element equations. Apply the Laplace transformation of the differential equation to put the equation in the s-domain. Algebraically solve for the solution, or response transform. Apply the inverse Laplace transformation to produce the solution to the original differential equation described in the time-domain. To get comfortable with this process, you simply need to practice applying it to different types of circuits such as an RC (resistor-capacitor) circuit, an RL (resistor-inductor) circuit, and an RLC (resistor-inductor-capacitor) circuit. Here you can see an RLC circuit in which the switch has been open for a long time. The switch is closed at time t = 0. In this circuit, you have the following KVL equation: vR(t) + vL(t) + v(t) = 0 Next, formulate the element equation (or i-v characteristic) for each device. Ohm’s law describes the voltage across the resistor (noting that i(t) = iL(t) because the circuit is connected in series, where I(s) = IL(s) are the Laplace transforms): vR(t) = i(t)R The inductor’s element equation is given by And the capacitor’s element equation is Here, vC(0) = V0 is the initial condition, and it’s equal to 5 volts. Substituting the element equations, vR(t), vC(t), and vL(t), into the KVL equation gives you the following equation (with a fancy name: the integro-differential equation): The next step is to apply the Laplace transform to the preceding equation to find an I(s) that satisfies the integro-differential equation for a given set of initial conditions: The preceding equation uses the linearity property allowing you to take the Laplace transform of each term. For the first term on the left side of the equation, you use the differentiation property to get the following transform: This equation uses IL(s) = ℒ[i(t)], and I0 is the initial current flowing through the inductor. Because the switch is open for a long time, the initial condition I0 is equal to zero. For the second term of the KVL equation dealing with resistor R, the Laplace transform is simply ℒ[i(t)R] = I(s)R For the third term in the KVL expression dealing with capacitor C, you have The Laplace transform of the integro-differential equation becomes Rearrange the equation and solve for I(s): To get the time-domain solution i(t), use the following table, and notice that the preceding equation has the form of a damping sinusoid. Now, you plug in I0 = 0 and some numbers from this figure: Now you’ve got this equation: You wind up with the following solution: i(t) = [-0.01e-400t sin500t]u(t) For this RLC circuit, you have a damping sinusoid. The oscillations will die out after a long period of time. For this example, the time constant is 1/400 and will die out after 5/400 = 1/80 seconds.

Article / Updated 09-19-2022

Using the Laplace transform as part of your circuit analysis provides you with a prediction of circuit response. Analyze the poles of the Laplace transform to get a general idea of output behavior. Real poles, for instance, indicate exponential output behavior. Follow these basic steps to analyze a circuit using Laplace techniques: Develop the differential equation in the time-domain using Kirchhoff’s laws and element equations. Apply the Laplace transformation of the differential equation to put the equation in the s-domain. Algebraically solve for the solution, or response transform. Apply the inverse Laplace transformation to produce the solution to the original differential equation described in the time-domain. To get comfortable with this process, you simply need to practice applying it to different types of circuits such as an RC (resistor-capacitor) circuit, an RL (resistor-inductor) circuit, and an RLC (resistor-inductor-capacitor) circuit. Here is an RL circuit that has a switch that’s been in Position A for a long time. The switch moves to Position B at time t = 0. For this circuit, you have the following KVL equation: vR(t) + vL(t) = 0 Next, formulate the element equation (or i-v characteristic) for each device. Using Ohm’s law to describe the voltage across the resistor, you have the following relationship: vR(t) = iL(t)R The inductor’s element equation is Substituting the element equations, vR(t) and vL(t), into the KVL equation gives you the desired first-order differential equation: On to Step 2: Apply the Laplace transform to the differential equation: The preceding equation uses the linearity property which says you can take the Laplace transform of each term. For the first term on the left side of the equation, you use the differentiation property: This equation uses IL(s) = ℒ[iL(t)], and I0 is the initial current flowing through the inductor. The Laplace transform of the differential equation becomes IL(s)R + L[sIL(s) – I0] = 0 Solve for IL(s): For a given initial condition, this equation provides the solution iL(t) to the original first-order differential equation. You simply perform an inverse Laplace transform of IL(s) — or look for the appropriate transform pair in this table — to get back to the time-domain. The preceding equation has an exponential form for the Laplace transform pair. You wind up with the following solution: The result shows as time t approaches infinity, the initial inductor current eventually dies out to zero after a long period of time — about 5 time constants (L/R).

Article / Updated 09-19-2022

Using the Laplace transform as part of your circuit analysis provides you with a prediction of circuit response. Analyze the poles of the Laplace transform to get a general idea of output behavior. Real poles, for instance, indicate exponential output behavior. Follow these basic steps to analyze a circuit using Laplace techniques: Develop the differential equation in the time-domain using Kirchhoff’s laws and element equations. Apply the Laplace transformation of the differential equation to put the equation in the s-domain. Algebraically solve for the solution, or response transform. Apply the inverse Laplace transformation to produce the solution to the original differential equation described in the time-domain. To get comfortable with this process, you simply need to practice applying it to different types of circuits such as an RC (resistor-capacitor) circuit, an RL (resistor-inductor) circuit, and an RLC (resistor-inductor-capacitor) circuit. Consider the simple first-order RC series circuit shown here. To set up the differential equation for this series circuit, you can use Kirchhoff’s voltage law (KVL), which says the sum of the voltage rises and drops around a loop is zero. This circuit has the following KVL equation around the loop: -vS(t) + vr(t) + vc(t) = 0 Next, formulate the element equation (or i-v characteristic) for each device. The element equation for the source is vS(t) = VAu(t) Use Ohm’s law to describe the voltage across the resistor: vR(t) = i(t)R The capacitor’s element equation is given as Substituting this expression for i(t) into vR(t) gives you the following expression: Substituting vR(t), vC(t), and vS(t) into the KVL equation leads to Now rearrange the equation to get the desired first-order differential equation: Now you’re ready to apply the Laplace transformation of the differential equation in the s-domain. The result is On the left, the linearity property was used to take the Laplace transform of each term. For the first term on the left side of the equation, you use the differentiation property, which gives you This equation uses VC(s) = ℒ[vC(t)], and V0 is the initial voltage across the capacitor. Using the following table, the Laplace transform of a step function provides you with this: Based on the preceding expressions for the Laplace transforms, the differential equation becomes the following: Next, rearrange the equation: Solve for the output Vc(s) to get the following transform solution: By performing an inverse Laplace transform of VC(s) for a given initial condition, this equation leads to the solution vC(t) of the original first-order differential equation. On to Step 3 of the process. To get the time-domain solution vC(t), you need to do a partial fraction expansion for the first term on the right side of the preceding equation: You need to determine constants A and B. To simplify the preceding equation, multiply both sides by s(s + 1/RC) to get rid of the denominators: Algebraically rearrange the equation by collecting like terms: In order for the left side of the preceding equation to be zero, the coefficients must be zero (A + B = 0 and A – VA = 0). For constants A and B, you wind up with A = VA and B = –VA. Substitute these values into the following equation: The substitution leads you to: Now substitute the preceding expression into the VC(s) equation to get the transform solution: That completes the partial fraction expansion. You can then use the table given earlier to find the inverse Laplace transform for each term on the right side of the preceding equation. The first term has the form of a step function, and the last two terms have the form of an exponential, so the inverse Laplace transform of the preceding equation leads you to the following solution vC(t) in the time-domain: The result shows as time t approaches infinity, the capacitor charges to the value of the input VA. Also, the initial voltage of the capacitor eventually dies out to zero after a long period of time (about 5 time constants, RC).

Article / Updated 08-19-2022

There are many applications for an RLC circuit, including band-pass filters, band-reject filters, and low-/high-pass filters. You can use series and parallel RLC circuits to create band-pass and band-reject filters. An RLC circuit has a resistor, inductor, and capacitor connected in series or in parallel. RLC series band-pass filter (BPF) You can get a band-pass filter with a series RLC circuit by measuring the voltage across the resistor VR(s) driven by a source VS(s). Start with the voltage divider equation: With some algebraic manipulation, you obtain the transfer function, T(s) = VR(s)/VS(s), of a band-pass filter: Plug in s = jω to get the frequency response T(jω): The T(jω) reaches a maximum when the denominator is a minimum, which occurs when the real part in the denominator equals 0. In math terms, this means that The frequency ω0 is called the center frequency. The cutoff frequencies are at the –3 dB half-power points. The –3 dB point occurs when the real part in the denominator is equal to Rω/L: You basically have a quadratic equation, which has four roots due to the plus-or-minus sign in the second term. The two appropriate roots of this equation give you cutoff frequencies at ωC1 an ωC2: The bandwidth BW defines the range of frequencies that pass through the filter relatively unaffected. Mathematically, it’s defined as Another measure of how narrow or wide the filter is with respect to the center frequency is the quality factor Q. The quality factor is defined as the ratio of the center frequency to the bandwidth: The RLC series circuit is narrowband when Q >> 1 (high Q) and wideband when Q << 1 (low Q). The separation between the narrowband and wideband responses occurs at Q = 1. Here is a series band-pass circuit and gain equation for an RLC series circuit. The frequency response is shaped by poles and zeros. For this band-pass filter, you have a zero at ω = 0. You start with a gain slope of +20 dB. You hit a cutoff frequency at ωC1, which flattens the frequency response until you hit another cutoff frequency above ωC2, resulting in a slope of –20 dB/decade. RLC series band-reject filter (BRF) You form a band-reject filter by measuring the output across the series connection of the capacitor and inductor. You start with the voltage divider equation for the voltage across the series connection of the inductor and capacitor: You can rearrange the equation with some algebra to form the transfer function of a band-reject filter: When you plug in s = jω, you have poles and zeros shaping the frequency response. For the band-reject filter, you have a double zero at 1 / √LC. Starting at ω = 0, you have a gain of 0 dB. You hit a pole at ωC1, which rolls off at –20 dB/decade until you hit a double zero, resulting in a net slope of +20 dB/decade. The frequency response then flattens out to a gain of 0 dB at the cutoff frequency ωC2. You see how the poles and zeros form a band-reject filter.

Article / Updated 08-19-2022

There are many applications for an RLC circuit, including band-pass filters, band-reject filters, and low-/high-pass filters. You can use series and parallel RLC circuits to create band-pass and band-reject filters. An RLC circuit has a resistor, inductor, and capacitor connected in series or in parallel. You can get a transfer function for a band-pass filter with a parallel RLC circuit, like the one shown here. You can use current division to find the current transfer function of the parallel RLC circuit. By measuring the current through the resistor IR(s), you form a band-pass filter. Start with the current divider equation: A little algebraic manipulation gives you a current transfer function, T(s) = IR(s)/IS(s), for the band-pass filter: Plug in s = jω to get the frequency response T(jω): This equation has the same form as the RLC series equations. For the rest of this problem, you follow the same process as for the RLC series circuit. The transfer function is at a maximum when the denominator is minimized, which occurs when the real part of the denominator is set to 0. The cutoff frequencies are found when their gains |T(jωC)| = 0.707|T(jω)| or the –3 dB point. Therefore, ω0 is The center frequency, the cutoff frequencies, and the bandwidth have equations identical to the ones for the RLC series band-pass filter. Your cutoff frequencies are ωC1 and ωC2: The bandwidth BW and quality factor Q are

Article / Updated 07-29-2022

To find the total response of an RL parallel circuit such as the one shown here, you need to find the zero-input response and the zero-state response and then add them together. After fiddling with the math, you determine that the zero-input response of the sample circuit is this: Now you are ready to calculate the zero-state response for the circuit. Zero-state response means zero initial conditions. For the zero-state circuit shown earlier, zero initial conditions means looking at the circuit with zero inductor current at t < 0. You need to find the homogeneous and particular solutions to get the zero-state response. Next, you have zero initial conditions and an input current of iN(t) = u(t), where u(t) is a unit step input. When the step input u(t) = 0, the solution to the differential equation is the solution ih(t): The inductor current ih(t) is the solution to the homogeneous first-order differential equation: This solution is the general solution for the zero input. You find the constant c1 after finding the particular solution and applying the initial condition of no inductor current. After time t = 0, a unit step input describes the transient inductor current. The inductor current for this step input is called the step response. You find the particular solution ip(t) by setting the step input u(t) equal to 1. For a unit step input iN(t) = u(t), substitute u(t) = 1 into the differential equation: The particular solution ip(t) is the solution for the differential equation when the input is a unit step u(t) = 1 after t = 0. Because u(t) = 1 (a constant) after time t = 0, assume a particular solution ip(t) is a constant IA. Because the derivative of a constant is 0, the following is true: Substitute ip(t) = IA into the first-order differential equation: The particular solution eventually follows the form of the input because the zero-input (or free response) diminishes to 0 over time. You can generalize the result when the input step has strength IA or IAu(t). You need to add the homogeneous solution ih(t) and the particular solution ip(t) to get the zero-state response: At t = 0, the initial condition is 0 because this is a zero-state calculation. To find c1, apply iZS(0) = 0: Solving for c1 gives you C1 = -IA Substituting c1 into the zero-state response iZS(t), you wind up with

Article / Updated 07-29-2022

A first-order RL parallel circuit has one resistor (or network of resistors) and a single inductor. First-order circuits can be analyzed using first-order differential equations. By analyzing a first-order circuit, you can understand its timing and delays. Analyzing such a parallel RL circuit, like the one shown here, follows the same process as analyzing an RC series circuit. So if you are familiar with that procedure, this should be a breeze. If your RL parallel circuit has an inductor connected with a network of resistors rather than a single resistor, you can use the same approach to analyze the circuit. But you have to find the Norton equivalent first, reducing the resistor network to a single resistor in parallel with a single current source. Start with the simple RL parallel circuit Because the resistor and inductor are connected in parallel in the example, they must have the same voltage v(t). The resistor current iR(t) is based on Ohm’s law: The element constraint for an inductor is given as where i(t) is the inductor current and L is the inductance. You need a changing current to generate voltage across an inductor. If the inductor current doesn’t change, there’s no inductor voltage, which implies a short circuit. Now substitute v(t) = Ldi(t)/dt into Ohm’s law because you have the same voltage across the resistor and inductor: Kirchhoff’s current law (KCL) says the incoming currents are equal to the outgoing currents at a node. Use KCL at Node A of the sample circuit to get iN(t) = iR(t) =i(t). Substitute iR(t) into the KCL equation to give you The RL parallel circuit is a first-order circuit because it’s described by a first-order differential equation, where the unknown variable is the inductor current i(t). A circuit containing a single equivalent inductor and an equivalent resistor is a first-order circuit. Knowing the inductor current gives you the magnetic energy stored in an inductor. In general, the inductor current is referred to as a state variable because the inductor current describes the behavior of the circuit. Calculate the zero-input response for an RL parallel circuit Here is how the RL parallel circuit is split up into two problems: the zero-input response and the zero-state response. Here, you’ll start by analyzing the zero-input response. To simplify matters, you set the input source (or forcing function) equal to 0: iN(t) = 0 amps. This means no input current for all time — a big, fat zero. The first-order differential equation reduces to For an input source of no current, the inductor current iZI is called a zero-input response. No external forces are acting on the circuit except for its initial state (or inductor current, in this case). The output is due to some initial inductor current I0 at time t = 0. You make a reasonable guess at the solution (the natural exponential function!) and substitute your guess into the RL first-order differential equation. Assume the inductor current and solution to be iZI(t) = Bekt This is a reasonable guess because the time derivative of an exponential is also an exponential. Like a good friend, the exponential function won’t let you down when solving these differential equations. You determine the constants B and k next. Substitute your guess iZI(t) = Bekt into the differential equation: Replacing iZI(t) with Bekt and doing some math gives you the following: You have the characteristic equation after factoring out Bekt: The characteristic equation gives you an algebraic problem to solve for the constant k: Use k = –R/L and the initial inductor current I0 at t = 0. This implies that B = I0, so the zero-input response iZI(t) gives you the following: The constant L/R is called the time constant. The time constant provides a measure of how long an inductor current takes to go to 0 or change from one state to another. To analyze the RL parallel circuit further, you must calculate the circuit’s zero-state response, and then add that result to the zero-input response to find the total response for the circuit.

Cheat Sheet / Updated 01-26-2022

When doing circuit analysis, you need to know some essential laws, electrical quantities, relationships, and theorems. Ohm’s law is a key device equation that relates current, voltage, and resistance. Using Kirchhoff’s laws, you can simplify a network of resistors using a single equivalent resistor. You can also do the same type of calculation to obtain the equivalent capacitance and inductance for a network of capacitors or inductors. For more complicated circuits, the node-voltage analysis and mesh current techniques come in handy. And when you want to try different loads for a particular source circuit, you can use the Thévenin or Norton equivalent.

Article / Updated 02-09-2017

In circuits, inductors resist instantaneous changes in current and store magnetic energy. Inductors are electromagnetic devices that find heavy use in radiofrequency (RF) circuits. They serve as RF “chokes,” blocking high-frequency signals. This application of inductor circuits is called filtering. Electronic filters select or block whichever frequencies the user chooses. Describe an inductor Unlike capacitors, which are electrostatic devices, inductors are electromagnetic devices. Whereas capacitors avoid an instantaneous change in voltage, inductors prevent an abrupt change in current. Inductors are wires wound into several loops to form coils. In fact, the inductor’s symbol looks like a coil of wire, as shown here. Current flowing through a wire creates a magnetic field, and the magnetic field lines encircle the wire along its axis. The concentration, or density, of the magnetic field lines is called magnetic flux. The coiled shape of inductors increases the magnetic flux that naturally occurs when current flows through a straight wire. The greater the flux, the greater the inductance. If you needed a circuit that stored more magnetic energy, you could get even larger inductance values by inserting iron into the wire coil. Here’s the defining equation for the inductor: where the inductance L is a constant measured in henries (H). Here is the same equation in graphical form. The figure shows the i-v characteristic of an inductor, where the slope of the line is the value of the inductance. The preceding equation says that the voltage across the inductor depends on the time rate of change of the current. In other words, no change in inductor current means no voltage across the inductor. To create voltage across the inductor, current must change smoothly. Otherwise, an instantaneous change in current would create one humongous voltage across the inductor. Think of inductance L as a proportionality constant, like a resistor acts as a constant in Ohm’s law. This notion of Ohm’s law for inductors (and capacitors) becomes useful when you start working with phasors. To express the current through the inductor in terms of the voltage, you integrate the preceding equation as follows: The second term in this equation is the initial current through the inductor at time t = 0. Find the energy storage of an attractive inductor To find the energy stored in the inductor, you need the following power definition, which applies to any device: The subscript L denotes an inductor device. Substituting the voltage for an inductor into the power equation gives you the following: The energy wL(t) stored per unit time is the power. Integrating the preceding equation gives you the energy stored in an inductor: The energy equation implies that the energy in the inductor is always positive. The inductor absorbs power from a circuit when storing energy, and the inductor releases the stored energy when delivering energy to the circuit. To visualize the current and energy relationship shown here, which shows the current as a function of time and the energy stored in an inductor. This also shows how you can get the current from the inductor relationship between current and voltage. Calculate total inductance for series and parallel inductors Inductors connected in series or connected in parallel can be reduced to one single inductor. Take a look at the circuit with three series inductors shown in the top diagram. Because the inductors are connected in series, they have the same currents: i1(t) = i2(t) =i3(t) = i(t) Add up the voltages from the series inductors to get the net voltage v(t), as follows: For a series inductors, you have an equivalent inductance of LEQ = L1 + L2 + L3 For a parallel connection of inductors, apply Kirchhoff’s current law (KCL) in the bottom diagram of the figure. KCL says the sum of the incoming currents and outgoing current at a node is equal to 0, giving you Because you have the same voltage v(t) across each of the parallel inductors, you can rewrite the equation as This equation shows how you can reduce the parallel inductors to one single inductor:

Article / Updated 03-26-2016

Here’s your chance to convert light into electricity using simple operational circuits. You can apply a similar approach to develop instruments that measure other physical variables in the environment, such as temperature and pressure. You use an input transducer to turn a physical variable into an electrical variable. A photoresistor is an input transducer that converts light energy into a change in resistance, resulting in a change in the current flowing in the circuit. The light is, in fact, an electrical signal. Suppose you’re dealing with a photoresistor that has a resistance value between 20 MW in total darkness and 20 kW in bright light. If the photoresistor is a linear device, then doubling the amount of light doubles the amount of voltage. You can therefore model a photoresistor as a variable resistor that changes resistance according to the amount of light. The following figure shows a photoresistor and a complete design of an operational-amplifier (op-amp) circuit to produce an output voltage vO. Use circuit analysis to show that the op-amp output voltage vO is 0 volts in total darkness and 5 volts in bright light. In other words, show that the voltage range of the output varies from 0 to 5 volts. Here’s how: Determine the output voltage v2 from the transducer. To determine the range of the output v2 from the transducer (that is, the voltage across Terminals A and B), you can use the voltage divider equation. This equation sets the output voltage equal to the input voltage multiplied by the ratio of the resistance of the output device (R2) to the total series resistance (R1 + R2): Determine the lower bound of v2. The lower bound of the output voltage v2 occurs in bright light, when the photoresistor’s resistance is at a minimum. When R2 = 20 kW, the lower voltage v2L is Determine the upper bound of v2. The upper bound of the output voltage v2 occurs when the photoresistor’s resistance is highest. In total darkness, R2 = 20 MW = 20,000 kW , so the upper voltage v2U is The voltage of v2 ranges from 5 to 10 volts. Simplify the transducer (the source circuit) using the Thévenin technique. The Thévenin technique reduces a source circuit to one single resistor RT and one single voltage source vT. By using the Thévenin equivalent to simplify the transducer, you get a Thévenin voltage vT that varies from 5 to 10 volts and a Thévenin resistance RT that varies from 10 kW to 20 kW. Analyze the op-amp circuit (inverting summer). The op-amp circuit in the following figure is a typical configuration of an inverting op-amp summer circuit. In this circuit, you have two inputs: one coming from the transducer and another coming from a voltage source of –10 volts. The equation for the inverting operational amplifier is Because the voltage range of v2 varies between 5 and 10 volts, the output voltage range of the inverting summer goes from 0 to 5 volts.