Basic Analog Circuits

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Document Type: TutorialNI Supported: YesPublish Date: Feb 1, 2006

Overview

This tutorial is part of the National Instruments Measurement Fundamentals series. Each tutorial in this series, will teachyou a specific topic of common measurement applications, by explaining the theory and giving practical examples. Thistutorial introduces and explains the basic fundamentals of analog circuits.

You can also view an interactive presentation that takes you through this tutorial's material at your own pace.

For more information, return to the NI Measurement Fundamentals Main page.

Table of Contents

1. Ohm's Law and Basic Analog Circuit Concepts 2. Capacitance Calculations 3. Inductance Calculations 4. Analog Amplifier Circuits 5. Analog RC Filters 6. Relevant NI Products

Ohm's Law and Basic Analog Circuit Concepts

Resistance: Resistance can be defined as the characteristic of a medium that opposes flow of current through itself.The unit of resistance is Ohms which is represented by the Greek letter (Omega) . The power value associated withresistance is quantified as the amount of power that the resistor can dissipate as heat without overheating itself.

The current I through the resistor R is defined as:I = V/RV = I * R or R = V/I

For a 1 Mega ohm resistance, the current resulting from the application of 10V would be10 micro-amperes.

Figure 1. Simple Representation of Ohm's Law

Ohm’s law is the fundamental equation that describes the above relationship between Voltage potential, the currentflowing in the circuit, and the resistance of a circuit. The power dissipated in a load resistance R is defined as the productof the current and the voltage. Other relationships for power can be easily derived from this by applying Ohm’s law andsubstituting.

The power P dissipated in R is defined as:P = I * V

P = V2 /R or

of 8Basic Analog Circuits- Developer Zone - National Instruments

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P = I2 * R

To calculate the value of resistance that will result in 10 Watts with 10V applied to it, we remember that P = V 2 /R.

Transposing, R = V2 /P. The resistance is 100/10, or 10 ohms. So 10V applied to 10 ohms will yield 10 Watts. Interesting!Any time two of the parameters (V, R, or P) are numerically the same, the third one will be the same. A common way tomeasure resistance is by using a Digital Multimeter (DMM).

Note:The power dissipation P is what sets the limit for how much voltage can be applied to the 50 ohm input of a digitizer.From the equations, we see that 10V into 50 ohms will require the digitizer’s input load to dissipate 2 watts. If you arerunning 2 channels, that’s 4 watts. This amount of power dissipation is definitely something that can’t be ignored. Also

notice that because of the square law effect, if you double that voltage into the digitizer, the power it must dissipate willQUADRUPLE.

Voltage Divider Calculation:

When two resistors are connected in a series configuration, they must share the applied voltage and the same currentflows through both of them.

Figure 2. Voltage Divider Circuit Example

The formula used to calculate the applied voltage is:E1 = I * R1 (E1 = Voltage drop across R1)E2 = I * R2 (E2 = Voltage drop across R2)I = E / Req

where Req = R1 + R2E = E1 + E2E = I * (R1 +R2)

To calculate the voltage across R2:E2 = R2 * IE2 = R2 * (E/Req)E2 = R2 * [E/(R1 + R2)]E2 = E * [R2 / (R1 + R2)]

Note: Voltage divider is described by the equation above.

Current Divider Calculation:

When two resistors are connected in parallel configuration, the same voltage is across each of them. The amount ofcurrent flowing through them depends on the value of the resistances.

Figure 3. Current Divider Circuit Example

The above figure depicts two resistors in a parallel configuration.I = I1 + I2E = I1 * R1E = I2 * R2I = (E/R1) + (E/R2)I = E [(1/R1) + (1/R2)]

Since E = I * ReqReq = (1/R1) + (1/R2)= [(R1 * R2)/(R1 + R2)]

Note: Digital multimeters (DMMs) are the most common measurement devices found in automated test systems. DMMs





are usually simple to use and are often low-cost instruments. Generally, DMMs have built-in conditioning that provides:a) High resolution (commonly measured in digits)b) Multiple measurements (volts, current, resistance, etc)c) Isolation and high voltage capabilities.

View Interactive Presentation

Capacitance Calculations

Capacitors store energy in the form of electrical charge. Amount of charge that the capacitor can hold depends on thearea of the two plates in the figure above and the distance between them. Large plates with a small distance betweenthem have a higher capacity to hold charge. The electric field between the plates of a capacitor resists changes in appliedvoltage. Capacitors decrease their resistance with frequency.

Figure 4. Capacitance Circuit Example

Reading Capacitor Values: The unit of capacitance is Farad which is represented by the letter F. The formula to calculate capacitance is:

C = Q/VWhere,C = Capacitance in faradsQ = Accumulated Charge in CoulombsV = Voltage difference between the plates

Series configuration:

Figure 5. Series Capacitor Configuration

The above configuration represents two capacitors in series.

Since the capacitance of a capacitor is inversely proportional to the distance between the plates, the total capacitance CTof any number of capacitances can be calculated by the following:

(1/CT) = (1/C1) + (1/C2) + (1/C3) + ….

For two capacitors connected in series,C1 = Q/V1C2 = Q/V2V1 = Q/C1V2 = Q/C2

V = V1 + V2V = (Q/C1) + (Q/C2) = Q * [(1/C1) + (1/C2)]V = Q/CeqCeq = 1 / [(1/C1) + (1/C2)]Ceq = [(C1 * C2) / (C1+C2)]





Parallel Configuration:

Figure 6. Parallel Capacitor Configuration

Each capacitor charges to the same applied voltage. Total capacitance equals the sum of the individual capacitances ofthe capacitors.The formula used to calculate the capacitance is below:

Q1 = C1 * VQ2 = C2 * VQ = Q1 + Q2Q = V * (C1 + C2)Ceq = C1 + C2


Inductance Calculations

Inductance is defined as the amount of voltage dropped across the inductor for a given rate of change of current flowingthrough it. Inductors increase their resistance with frequency. The unit of inductance is Henry which is represented by theletter H.

Series Configuration:

Figure 7. Series Inductor Configuration

The Figure 7 above is a series configuration of two inductors.When two inductors are connected in series as shown abovethen their total inductance equals the sum of individual inductances.

L1 = E (dI/dt)L2 = E (dI/dt)LT = L1 + L2Where dI/dt is the change of current over time.

However, in the real world if we consider mutual inductance where magnetic field of each inductor affects the other coilthen the total inductance can be calculated using the formula below:

LT = L1 + L2 +/- 2M

Where, M is the mutual inductance between the two coils.

Parallel Configuration:





Figure 8. Parallel Inductor Configuration

The Figure 8 above is a parallel configuration of two inductors. When two inductors are connected in parallel configuration,mutual inductance needs to be considered. Also, mutual inductance will be either added or subtracted from the self-inductance of each coil since current has two paths to flow in.

The total inductance can be calculated using the formula below:

(1/LT) = [1 / (L1 +/- M)] + [1 / (L2 +/- M)]

The NI PXI-4072 FlexDMM and LCR Meter brings digital multimeters one step closer to providing universal measurementcapability by offering engineers the 20 most common automated test measurements, including voltage, current,capacitance, inductance, temperature and resistance.

Impedance: Impedance Z is generally defined as the total opposition a device or circuit offers to the flow of an alternating current AC ata given frequency. Its value is equal to the ratio between the voltage and the current over an element of a circuit.Therefore, the unit of impedance is Ohm .

Impedance is represented as a complex quantity which is graphically shown on a vector plane. An impedance vectorconsists of a real part (resistance, R) and an imaginary part (reactance, X). Impedance can be expressed using therectangular coordinate form R + jX or in the polar form as a magnitude and phase angle:Z .

Admittance: Admittance Y is the reciprocal of impedance. It is also a complex quantity: real part is called conductance ( C ) and theimaginary part is called susceptance ( B ).The unit of admittance is Siemens (S)

Y = G + jBWhere, Y is admittance; G represents conductance and B susceptance.


Analog Amplifier Circuits

The Figure 9 below is a basic op-amp model that consists of three basic stages of an op-amp:

[+] Enlarge Image Figure 9. Basic Operational Amplifier (Op-Amp) Model





1) Differential Amplifier: An amplifier whose output is proportional to the difference between the input signals.

2) Gain/Frequency Response: A filter changes the amplitude or phase characteristics of a signal with respect to frequency.The frequency domain behavior of a filter is mathematically described in terms of a transfer function or a network function.The transfer function H(s) is described as a ratio between output and input signals.

H(s) = Vout(s) / Vin(s)

Where, Vout(s) and Vin(s) are the output and input voltage signals and s is the complex frequency variable.The magnitude of transfer function is called amplitude response or frequency response especially in radio applications.

3) Output Buffer

Inverting Amplifier: An Inverting is as simple as it sounds, it simply reverses the polarity of the input signal. For example,if the voltage going into the amplifer is positive, it is negative when it comes out.

Figure 10. Basic Inverting Amplifier

Calculating the gain of an inverting amplifier:(Vs – V1)/R1 = (V – Vo)/R2Since V1 = V = 0 (Virtual ground)Vs/R1 = -Vo/R2Gain = Vo/Vs= -R2/R1

Non-Inverting Amplifier: The gain of the amplifer is determined by the ratio of R1 and R2.

Figure 11. Basic Non-Inverting Amplifier

Calculating the gain of a non-inverting amplifier:VoR1 = (VsR1 +VsR2)(Vo – Vs) R1 = VsR2[(Vo/Vs) – 1] = (R2/R1)Gain – 1 = (R2/R1)Gain = 1 + (R2/R1)Where Gain = Vo/Vs

Note: All the Data Acquisition (DAQ) products offered by National Instruments have built-in amplifiers.


Analog RC Filters

RC Low Pass Filter: A common circuit to attenuate high-frequency components in an analog signal is the RC Low PassFilter. Examine the diagram below, where Vin is the applied voltage and the voltage Vout across C1 is the output.





Figure 12. Simple RC Low Pass Filter

The RC Low Pass Filter passes low frequency and DC signals to the output, but blocks out high frequency signals. Thiscould be either desirable or undesirable.

You get the equivalent of this circuit even when you do not expect it. For example, C could be the input capacitance of aDigital Multimeter(DMM), Digitizer, etc. R in this example might be the source resistance of the device under test (DUT).The DUT has to charge and discharge C as the signal varies. As the variation increases in frequency, the impedance of Ceventually becomes lower than R and starts to attenuate the signal. The frequency where the value of Vout is at 0.707 ofVin, is defined as the –3dB frequency or the half-power point, because the output is down –3dB of the input signal at thatpoint.

Single-Pole RC Low Pass

t = R * Cf3dB

= 1/(2pt)

RC High Pass Filter: A circuit that attenuates low-frequency components in an analog signal is called a RC High PassFilter. Notice that the circuit is similar to the one above, but Vout is now measured across R1.

Figure 13. Simple RC High Pass Filter

A very important typical application of this circuit is in the input coupling circuits of a Digitizer or Digital Multimeter (DMM) .With the capacitor in, it is “AC Coupled.” With the capacitor shorted, it would be “DC Coupled.”

Note: Low pass and High pass filters are also used in Dynamic Signal Acquisition (DSA) devices


Relevant NI Products

Customers interested in this topic were also interested in the following NI products:

LabVIEW

NI-4072 FlexDMM and LCR Meter

Data Acquisition (DAQ)

Signal Conditioning

For more tutorials, return to the NI Measurement Fundamentals Main page.

Reader Comments | Submit a comment »

Your formula for calculate the total inductance in parallel is wrong. You assume that the voltage acros the inductor is v =L1 di1/dt + Mdi1/dt and v = L2 di2/dt + Mdi2/dt. But the voltage generate by the mutual depend on the current variation





on the other inductor. v = L1di1/dt + Mdi2/dt and v = L2di2/dt + Mdi1/dt. The right formula is: LT = (L1L2 - M(square))/(L1+ L2 - 2M).- Jan 26, 2007

this document is a good basic introduction for beginners, I just regret that the units do not seem to be mentioned theright way (Ohm and Siemens are first names, the units are ohm and siemens, with the symbols as indicated)- [email protected] - Nov 2, 2006

Equation typo The 'Req' equation under 'Current Divider' topic, should show '1/Req = .....'- Nov 22, 2005

all information are clear and easy tanks- irving said hernandez, SIDI. [email protected] - Oct 6, 2005

presentation My comment is specifically on the audio visual presentation- denden tagle . [email protected] - Jun 29, 2005

Low Pass Filter and Inductancethe demonstration seems to me as an attenuator, the "2M" in inductance topic is not discussed.- Denden tagle, Teradyne Phils. Inc.. [email protected] - Jun 29, 2005

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