Circuitry Articles

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 09-27-2021

Before you start working with line voltage in your electronic circuits, you need to understand a few details about how most residential and commercial buildings are wired. The following description applies only to the United States; if you’re in a different country, you’ll need to determine the standards for your country’s wiring. Standard line voltage wiring in the United States is done with plastic-sheathed cables, which usually have three conductors. This type of cable is technically called NMB cable, but most electricians refer to it using its most popular brand name, Romex. Three conductors inside electric cables Two of the conductors in NMB cable are covered with plastic insulation (one white, the other black). The third conductor is bare copper. These conductors are designated as follows: Hot: The black wire is the hot wire, which provides a 120 VAC current source. Neutral: The white wire is called the neutral wire. It provides the return path for the current provided by the hot wire. The neutral wire is connected to an earth ground. Ground: The bare wire is called the ground wire. Like the neutral wire, the ground wire is also connected to an earth ground. However, the neutral and ground wires serve two distinct purposes. The neutral wire forms a part of the live circuit along with the hot wire. In contrast, the ground wire is connected to any metal parts in an appliance, such as a microwave oven or coffee pot. This is a safety feature, in case the hot or neutral wires somehow come in contact with metal parts. Connecting the metal parts to earth ground eliminates the shock hazard in the event of a short circuit. Note that some circuits require a fourth conductor. When a fourth conductor is used, it's covered with red insulation and is also a hot wire. How they're connected to a standard outlet The three wires in a standard NMB cable are connected to the three prongs of a standard electrical outlet (properly called a receptacle). As you can see, the neutral and hot wires are connected to the two vertical prongs at the top of the receptacle (neutral on the left, hot on the right) and the ground wire is connected to the round prong at the bottom of the receptacle. You can plug a two-prong or three-prong plug into a standard three-prong receptacle. Two-prong plugs are designed for appliances that don't require grounding. Most nongrounded appliances are double-insulated, which means that there are two layers of insulation between any live wires and any metal parts within the appliance. The first layer is the insulation on the wire itself; the second is usually in the form of a plastic case that isolates the live wiring from other metal parts. Three-prong plugs Three-prong plugs are for appliances that require the ground connection for safety. Most appliances that use a metal chassis require a separate ground connection. There is only one way to insert a three-prong plug into a three-prong receptacle. But regular two-prong plugs, which lack the ground prong, can be connected with either prong on the hot side. To prevent that from happening, the receptacles are polarized, which means that the neutral prong is wider than the hot prong. Thus, there's only one way to plug a polarized plug into a polarized receptacle. That way, you can always keep track of which wire is hot and which is neutral. You should always place switches or fuses on the hot wire rather than on the neutral wire. That way, if the switch is open or the fuse blows, the current in the hot wire will be prevented from proceeding beyond the switch or fuse into your circuit. This minimizes any risk of shock that might occur if a wire comes loose within your project.

Article / Updated 04-11-2017

The easiest way to provide a voltage source for an electronic circuit is to include a battery. There are plenty of other ways to provide voltage, including AC adapters (which you can plug into the wall) and solar cells (which convert sunlight to voltage). However, batteries remain the most practical source of juice for most electronic circuits. A battery is a device that converts chemical energy into electrical energy in the form of voltage, which in turn can cause current to flow. A battery works by immersing two plates made of different metals into a special chemical solution called an electrolyte. The metals react with the electrolyte to produce a flow of charges that accumulate on the negative plate, called the anode. The positive plate, called the cathode, is sucked dry of charges. As a result, a voltage is formed between the two plates. These plates are connected to external terminals to which you can connect a circuit to cause current to flow. Batteries come in many different shapes and sizes, but for electronics projects, you need concern yourself only with a few standard types of batteries, all of which are available at any grocery, drug, or department store. Cylindrical batteries come in four standard sizes: AAA, AA, C, and D. Regardless of the size, these batteries provide 1.5 V each; the only difference between the smaller and larger sizes is that the larger batteries can provide more current. The cathode, or positive terminal, in a cylindrical battery is the end with the metal bump. The flat metal end is the anode, or negative terminal. The rectangular battery is a 9 V battery. That little rectangular box actually contains six small cells, each about half the size of a AAA cell. The 1.5 volts produced by each of these small cells combine to create a total of 9 volts. Here are a few other things you should know about batteries: Besides AAA, AA, C, D, and 9 V batteries, many other battery sizes are available. Most of those batteries are designed for special applications, such as digital cameras, hearing aids, laptop computers, and so on. All batteries contain chemicals that are toxic to you and to the environment. Treat them with care, and dispose of them properly according to your local laws. Don't just throw them in the trash. You can (and should) use your multimeter to measure the voltage produced by your batteries. Set the multimeter to an appropriate DC voltage range (such as 20 V). Then, touch the red test lead to the positive terminal of the battery and the black test lead to the negative terminal. The multimeter will tell you the voltage difference between the negative and positive terminals. For cylindrical batteries (AAA, AA, C, or D) it should be about 1.5 V. For 9 V batteries, it should be about 9 V. Rechargeable batteries cost more than non-rechargeable batteries but last longer because you can recharge them when they go dead. The easiest way to use batteries in an electronic circuit is to use a battery holder, which is a little plastic gadget designed to hold one or more batteries. Wonder why they sell AAA, AA, C, and D cells but not A or B? Actually, A cell and B cell batteries exist. However, those sizes never really caught on. Check out which tools you'll need for electronics projects.