Basics Of Operational Amplifier Operational Amplifier, also called as an Op-Amp, is an integrated circuit, which can be used to perform various linear, non-linear, and mathematical operations. An op-amp is a direct coupled high gain amplifier. You can operate op-amp both with AC and DC signals. This chapter discusses the characteristics and types of op-amps. Construction of Operational Amplifier An op-amp consists of differential amplifier(s), a level translator and an output stage. A differential amplifier is present at the input stage of an op-amp and hence an op-amp consists of two input terminals. One of those terminals is called as the inverting terminal and the other one is called as the non-inverting terminal. The terminals are named based on the phase relationship between their respective inputs and outputs. Characteristics of Operational Amplifier The important characteristics or parameters of an operational amplifier are as follows − Open loop voltage gain Output offset voltage Common Mode Rejection Ratio Slew Rate This section discusses these characteristics in detail as given below − Open loop voltage gain The open loop voltage gain of an op-amp is its differential gain without any feedback path. Mathematically, the open loop voltage gain of an op-amp is represented as − $$A_{v}= frac{v_0}{v_1-v_2}$$ Output offset voltage The voltage present at the output of an op-amp when its differential input voltage is zero is called as output offset voltage. Common Mode Rejection Ratio Common Mode Rejection Ratio (CMRR) of an op-amp is defined as the ratio of the closed loop differential gain, $A_{d}$ and the common mode gain, $A_{c}$. Mathematically, CMRR can be represented as − $$CMRR=frac{A_{d}}{A_{c}}$$ Note that the common mode gain, $A_{c}$ of an op-amp is the ratio of the common mode output voltage and the common mode input voltage. Slew Rate Slew rate of an op-amp is defined as the maximum rate of change of the output voltage due to a step input voltage. Mathematically, slew rate (SR) can be represented as − $$SR=Maximum:of:frac{text{d}V_{0}}{text{d}t}$$ Where, $V_{0}$ is the output voltage. In general, slew rate is measured in either $V/mu:Sec$ or $V/m:Sec$. Types of Operational Amplifiers An op-amp is represented with a triangle symbol having two inputs and one output. Op-amps are of two types: Ideal Op-Amp and Practical Op-Amp. They are discussed in detail as given below − Ideal Op-Amp An ideal op-amp exists only in theory, and does not exist practically. The equivalent circuit of an ideal op-amp is shown in the figure given below − An ideal op-amp exhibits the following characteristics − Input impedance $Z_{i}=inftyOmega$ Output impedance $Z_{0}=0Omega$ Open loop voltage gaine $A_{v}=infty$ If (the differential) input voltage $V_{i}=0V$, then the output voltage will be $V_{0}=0V$ Bandwidth is infinity. It means, an ideal op-amp will amplify the signals of any frequency without any attenuation. Common Mode Rejection Ratio (CMRR) is infinity. Slew Rate (SR) is infinity. It means, the ideal op-amp will produce a change in the output instantly in response to an input step voltage. Practical Op-Amp Practically, op-amps are not ideal and deviate from their ideal characteristics because of some imperfections during manufacturing. The equivalent circuit of a practical op-amp is shown in the following figure − A practical op-amp exhibits the following characteristics − Input impedance, $Z_{i}$ in the order of Mega ohms. Output impedance, $Z_{0}$ in the order of few ohms.. Open loop voltage gain, $A_{v}$ will be high. When you choose a practical op-amp, you should check whether it satisfies the following conditions − Input impedance, $Z_{i}$ should be as high as possible. Output impedance, $Z_{0}$ should be as low as possible. Open loop voltage gain, $A_{v}$ should be as high as possible. Output offset voltage should be as low as possible. The operating Bandwidth should be as high as possible. CMRR should be as high as possible. Slew rate should be as high as possible. Note − IC 741 op-amp is the most popular and practical op-amp. Learning working make money
Category: linear Integrated Circuits Applications
Op-Amp-Applications A circuit is said to be linear, if there exists a linear relationship between its input and the output. Similarly, a circuit is said to be non-linear, if there exists a non-linear relationship between its input and output. Op-amps can be used in both linear and non-linear applications. The following are the basic applications of op-amp − Inverting Amplifier Non-inverting Amplifier Voltage follower This chapter discusses these basic applications in detail. Inverting Amplifier An inverting amplifier takes the input through its inverting terminal through a resistor $R_{1}$, and produces its amplified version as the output. This amplifier not only amplifies the input but also inverts it (changes its sign). The circuit diagram of an inverting amplifier is shown in the following figure − Note that for an op-amp, the voltage at the inverting input terminal is equal to the voltage at its non-inverting input terminal. Physically, there is no short between those two terminals but virtually, they are in short with each other. In the circuit shown above, the non-inverting input terminal is connected to ground. That means zero volts is applied at the non-inverting input terminal of the op-amp. According to the virtual short concept, the voltage at the inverting input terminal of an op-amp will be zero volts. The nodal equation at this terminal”s node is as shown below − $$frac{0-V_i}{R_1}+ frac{0-V_0}{R_f}=0$$ $$=>frac{-V_i}{R_1}= frac{V_0}{R_f}$$ $$=>V_{0}=left(frac{-R_f}{R_1}right)V_{t}$$ $$=>frac{V_0}{V_i}= frac{-R_f}{R_1}$$ The ratio of the output voltage $V_{0}$ and the input voltage $V_{i}$ is the voltage-gain or gain of the amplifier. Therefore, the gain of inverting amplifier is equal to $-frac{R_f}{R_1}$. Note that the gain of the inverting amplifier is having a negative sign. It indicates that there exists a 1800 phase difference between the input and the output. Non-Inverting Amplifier A non-inverting amplifier takes the input through its non-inverting terminal, and produces its amplified version as the output. As the name suggests, this amplifier just amplifies the input, without inverting or changing the sign of the output. The circuit diagram of a non-inverting amplifier is shown in the following figure − In the above circuit, the input voltage $V_{i}$ is directly applied to the non-inverting input terminal of op-amp. So, the voltage at the non-inverting input terminal of the op-amp will be $V_{i}$. By using voltage division principle, we can calculate the voltage at the inverting input terminal of the op-amp as shown below − $$=>V_{1} = V_{0}left(frac{R_1}{R_1+R_f}right)$$ According to the virtual short concept, the voltage at the inverting input terminal of an op-amp is same as that of the voltage at its non-inverting input terminal. $$=>V_{1} = V_{i}$$ $$=>V_{0}left(frac{R_1}{R_1+R_f}right)=V_{i}$$ $$=>frac{V_0}{V_i}=frac{R_1+R_f}{R_1}$$ $$=>frac{V_0}{V_i}=1+frac{R_f}{R_1}$$ Now, the ratio of output voltage $V_{0}$ and input voltage $V_{i}$ or the voltage-gain or gain of the non-inverting amplifier is equal to $1+frac{R_f}{R_1}$. Note that the gain of the non-inverting amplifier is having a positive sign. It indicates that there is no phase difference between the input and the output. Voltage follower A voltage follower is an electronic circuit, which produces an output that follows the input voltage. It is a special case of non-inverting amplifier. If we consider the value of feedback resistor, $R_{f}$ as zero ohms and (or) the value of resistor, 1 as infinity ohms, then a non-inverting amplifier becomes a voltage follower. The circuit diagram of a voltage follower is shown in the following figure − In the above circuit, the input voltage $V_{i}$ is directly applied to the non-inverting input terminal of the op-amp. So, the voltage at the non-inverting input terminal of op-amp is equal to $V_{i}$. Here, the output is directly connected to the inverting input terminal of opamp. Hence, the voltage at the inverting input terminal of op-amp is equal to $V_{0}$. According to the virtual short concept, the voltage at the inverting input terminal of the op-amp is same as that of the voltage at its non-inverting input terminal. $$=>V_{0} = V_{i}$$ So, the output voltage $V_{0}$ of a voltage follower is equal to its input voltage $V_{i}$. Thus, the gain of a voltage follower is equal to one since, both output voltage $V_{0}$ and input voltage $V_{i}$ of voltage follower are same. Learning working make money
Arithmetic Circuits In the previous chapter, we discussed about the basic applications of op-amp. Note that they come under the linear operations of an op-amp. In this chapter, let us discuss about arithmetic circuits, which are also linear applications of op-amp. The electronic circuits, which perform arithmetic operations are called as arithmetic circuits. Using op-amps, you can build basic arithmetic circuits such as an adder and a subtractor. In this chapter, you will learn about each of them in detail. Adder An adder is an electronic circuit that produces an output, which is equal to the sum of the applied inputs. This section discusses about the op-amp based adder circuit. An op-amp based adder produces an output equal to the sum of the input voltages applied at its inverting terminal. It is also called as a summing amplifier, since the output is an amplified one. The circuit diagram of an op-amp based adder is shown in the following figure − In the above circuit, the non-inverting input terminal of the op-amp is connected to ground. That means zero volts is applied at its non-inverting input terminal. According to the virtual short concept, the voltage at the inverting input terminal of an op-amp is same as that of the voltage at its non-inverting input terminal. So, the voltage at the inverting input terminal of the op-amp will be zero volts. The nodal equation at the inverting input terminal”s node is $$frac{0-V_1}{R_1}+frac{0-V_2}{R_2}+frac{0-V_0}{R_f}=0$$ $$=>frac{V_1}{R_1}-frac{V_2}{R_2}=frac{V_0}{R_f}$$ $$=>V_{0}=R_{f}left(frac{V_1}{R_1}+frac{V_2}{R_2}right)$$ If $R_{f}=R_{1}=R_{2}=R$, then the output voltage $V_{0}$ will be − $$V_{0}=-R{}left(frac{V_1}{R}+frac{V_2}{R}right)$$ $$=>V_{0}=-(V_{1}+V_{2})$$ Therefore, the op-amp based adder circuit discussed above will produce the sum of the two input voltages $v_{1}$ and $v_{1}$, as the output, when all the resistors present in the circuit are of same value. Note that the output voltage $V_{0}$ of an adder circuit is having a negative sign, which indicates that there exists a 1800 phase difference between the input and the output. Subtractor A subtractor is an electronic circuit that produces an output, which is equal to the difference of the applied inputs. This section discusses about the op-amp based subtractor circuit. An op-amp based subtractor produces an output equal to the difference of the input voltages applied at its inverting and non-inverting terminals. It is also called as a difference amplifier, since the output is an amplified one. The circuit diagram of an op-amp based subtractor is shown in the following figure − Now, let us find the expression for output voltage $V_{0}$ of the above circuit using superposition theorem using the following steps − Step 1 Firstly, let us calculate the output voltage $V_{01}$ by considering only $V_{1}$. For this, eliminate $V_{2}$ by making it short circuit. Then we obtain the modified circuit diagram as shown in the following figure − Now, using the voltage division principle, calculate the voltage at the non-inverting input terminal of the op-amp. $$=>V_{p}=V_{1}left(frac{R_3}{R_2+R_3}right)$$ Now, the above circuit looks like a non-inverting amplifier having input voltage $V_{p}$. Therefore, the output voltage $V_{01}$ of above circuit will be $$V_{01}=V_{p}left(1+frac{R_f}{R_1}right)$$ Substitute, the value of $V_{p}$ in above equation, we obtain the output voltage $V_{01}$ by considering only $V_{1}$, as − $$V_{01}=V_{1}left(frac{R_3}{R_2+R_3}right)left(1+frac{R_f}{R_1}right)$$ Step 2 In this step, let us find the output voltage, $V_{02}$ by considering only $V_{2}$. Similar to that in the above step, eliminate $V_{1}$ by making it short circuit. The modified circuit diagram is shown in the following figure. You can observe that the voltage at the non-inverting input terminal of the op-amp will be zero volts. It means, the above circuit is simply an inverting op-amp. Therefore, the output voltage $V_{02}$ of above circuit will be − $$V_{02}=left(-frac{R_f}{R_1}right)V_{2}$$ Step 3 In this step, we will obtain the output voltage $V_{0}$ of the subtractor circuit by adding the output voltages obtained in Step1 and Step2. Mathematically, it can be written as $$V_{0}=V_{01}+V_{02}$$ Substituting the values of $V_{01}$ and $V_{02}$ in the above equation, we get − $$V_{0}=V_{1}left(frac{R_3}{R_2+R_3}right)left(1+frac{R_f}{R_1}right)+left(-frac{R_f}{R_1}right)V_{2}$$ $$=>V_{0}=V_{1}left(frac{R_3}{R_2+R_3}right)left(1+frac{R_f}{R_1}right)-left(frac{R_f}{R_1}right)V_{2}$$ If $R_{f}=R_{1}=R_{2}=R_{3}=R$, then the output voltage $V_{0}$ will be $$V_{0}=V_{1}left(frac{R}{R+R}right)left(1+frac{R}{R}right)-left(frac{R}{R}right)V_{2}$$ $$=>V_{0}=V_{1}left(frac{R}{2R}right)(2)-(1)V_{2}$$ $$V_{0}=V_{1}-V_{2}$$ Thus, the op-amp based subtractor circuit discussed above will produce an output, which is the difference of two input voltages $V_{1}$ and $V_{2}$, when all the resistors present in the circuit are of same value. Learning working make money
Clampers In the previous chapter, we discussed about clippers. Now, let us discuss about other type of wave shaping circuits, namely clampers. Op-amp based Clampers A clamper is an electronic circuit that produces an output, which is similar to the input but with a shift in the DC level. In other words, the output of a clamper is an exact replica of the input. Hence, the peak to peak amplitude of the output of a clamper will be always equal to that of the input. Clampers are used to introduce or restore the DC level of input signal at the output. There are two types of op-amp based clampers based on the DC shift of the input. Positive Clamper Negative Clamper This section discusses about these two types of clampers in detail. Positive Clamper A positive clamper is a clamper circuit that produces an output in such a way that the input signal gets shifted vertically by a positive DC value. The circuit diagram of a positive clamper is shown in the following figure − In the above circuit, a sinusoidal voltage signal, $V_{i}$ is applied to the inverting terminal of op-amp through a network that consists of a capacitor $C_{1}$ and a resistor $R_{1}$. That means, AC voltage signal is applied to the inverting terminal of the op-amp. The DC reference voltage $V_{ref}$ is applied to the non-inverting terminal of the op-amp. The value of reference voltage $V_{ref}$ can be chosen by varying the resistor $R_{2}$. In this case, we will get a reference voltage $V_{ref}$ of a positive value. The above circuit produces an output, which is the combination (resultant sum) of the sinusoidal voltage signal $V_{i}$ and the reference voltage $V_{ref}$. That means, the clamper circuit produces an output in such a way that the sinusoidal voltage signal $V_{i}$ gets shifted vertically upwards by the value of reference voltage $V_{ref}$. The input wave form and the corresponding output wave form of positive clamper are shown in above figure − From the figure above, you can observe that the positive clamper shifts the applied input waveform vertically upward at the output. The amount of shift will depend on the value of the DC reference voltage. Negative Clamper A negative clamper is a clamper circuit that produces an output in such a way that the input signal gets shifted vertically by a negative DC value. The circuit diagram of negative clamper is shown in the following figure − In the above circuit, a sinusoidal voltage signal $V_{i}$ is applied to the inverting terminal of the op-amp through a network that consists of a capacitor C1 and resistor $R_{1}$. That means, AC voltage signal is applied to the inverting terminal of the op-amp. The DC reference voltage $V_{ref}$ is applied to the non-inverting terminal of the op-amp.The value of reference voltage $V_{ref}$ can be chosen by varying the resistor $R_{2}$. In this case,we will get reference voltage $V_{ref}$ of a negative value. The above circuit produces an output, which is the combination (resultant sum) of sinusoidal voltage signal $V_{i}$ and reference voltage $V_{ref}$. That means, the clamper circuit produces an output in such a way that the sinusoidal voltage signal $V_{i}$ gets shifted vertically downwards by the value of reference voltage $V_{ref}$. The input wave form and the corresponding output wave form of a negative clamper are shown in the following figure − We can observe from the output that the negative clamper shifts the applied input waveform vertically downward at the output. The amount of shifting will depend on the value of DC reference voltage. Learning working make money
Log And Anti Log Amplifiers The electronic circuits which perform the mathematical operations such as logarithm and anti-logarithm (exponential) with an amplification are called as Logarithmic amplifier and Anti-Logarithmic amplifier respectively. This chapter discusses about the Logarithmic amplifier and Anti-Logarithmic amplifier in detail. Please note that these amplifiers fall under non-linear applications. Logarithmic Amplifier A logarithmic amplifier, or a log amplifier, is an electronic circuit that produces an output that is proportional to the logarithm of the applied input. This section discusses about the op-amp based logarithmic amplifier in detail. An op-amp based logarithmic amplifier produces a voltage at the output, which is proportional to the logarithm of the voltage applied to the resistor connected to its inverting terminal. The circuit diagram of an op-amp based logarithmic amplifier is shown in the following figure − In the above circuit, the non-inverting input terminal of the op-amp is connected to ground. That means zero volts is applied at the non-inverting input terminal of the op-amp. According to the virtual short concept, the voltage at the inverting input terminal of an op-amp will be equal to the voltage at its non-inverting input terminal. So, the voltage at the inverting input terminal will be zero volts. The nodal equation at the inverting input terminal’s node is − $$frac{0-V_i}{R_1}+I_{f}=0$$ $$=>I_{f}=frac{V_i}{R_1}……Equation 1$$ The following is the equation for current flowing through a diode, when it is in forward bias − $$I_{f}=I_{s} e^{(frac{V_f}{nV_T})} ……Equation 2$$ where, $I_{s}$ is the saturation current of the diode, $V_{f}$ is the voltage drop across diode, when it is in forward bias, $V_{T}$ is the diode’s thermal equivalent voltage. The KVL equation around the feedback loop of the op amp will be − $$0-V_{f}-V_{0}=0$$ $$=>V_{f}=-V_{0}$$ Substituting the value of $V_{f}$ in Equation 2, we get − $$I_{f}=I_{s} e^{left(frac{-V_0}{nV_T}right)} ……Equation 3$$ Observe that the left hand side terms of both equation 1 and equation 3 are same. Hence, equate the right hand side term of those two equations as shown below − $$frac{V_i}{R_1}=I_{s}e^{left(frac{-V_0}{nV_T}right)}$$ $$frac{V_i}{R_1I_s}= e^{left(frac{-V_0}{nV_T}right)}$$ Applying natural logarithm on both sides, we get − $$Inleft(frac{V_i}{R_1I_s}right)= frac{-V_0}{nV_T}$$ $$V_{0}=-{nV_T}Inleft(frac{V_i}{R_1I_s}right)$$ Note that in the above equation, the parameters n, ${V_T}$ and $I_{s}$ are constants. So, the output voltage $V_{0}$ will be proportional to the natural logarithm of the input voltage $V_{i}$ for a fixed value of resistance $R_{1}$. Therefore, the op-amp based logarithmic amplifier circuit discussed above will produce an output, which is proportional to the natural logarithm of the input voltage ${V_T}$, when ${R_1I_s}=1V$. Observe that the output voltage $V_{0}$ has a negative sign, which indicates that there exists a 1800 phase difference between the input and the output. Anti-Logarithmic Amplifier An anti-logarithmic amplifier, or an anti-log amplifier, is an electronic circuit that produces an output that is proportional to the anti-logarithm of the applied input. This section discusses about the op-amp based anti-logarithmic amplifier in detail. An op-amp based anti-logarithmic amplifier produces a voltage at the output, which is proportional to the anti-logarithm of the voltage that is applied to the diode connected to its inverting terminal. The circuit diagram of an op-amp based anti-logarithmic amplifier is shown in the following figure − In the circuit shown above, the non-inverting input terminal of the op-amp is connected to ground. It means zero volts is applied to its non-inverting input terminal. According to the virtual short concept, the voltage at the inverting input terminal of op-amp will be equal to the voltage present at its non-inverting input terminal. So, the voltage at its inverting input terminal will be zero volts. The nodal equation at the inverting input terminal’s node is − $$-I_{f}+frac{0-V_0}{R_f}=0$$ $$=>-frac{V_0}{R_f}=I_{f}$$ $$=>V_{0}=-R_{f}I_{f}………Equation 4$$ We know that the equation for the current flowing through a diode, when it is in forward bias, is as given below − $$I_{f}=I_{s} e^{left(frac{V_f}{nV_T}right)}$$ Substituting the value of $I_{f}$ in Equation 4, we get $$V_{0}=-R_{f}left {{I_{s} e^{left(frac{V_f}{nV_T}right)}}right }$$ $$V_{0}=-R_{f}{I_{s} e^{left(frac{V_f}{nV_T}right)}}……Equation 5$$ The KVL equation at the input side of the inverting terminal of the op amp will be $$V_{i}-V_{f}=0$$ $$V_{f}=V_{i}$$ Substituting, the value of 𝑉𝑓 in the Equation 5, we get − $$V_{0}=-R_{f}{I_{s} e^{left(frac{V_i}{nV_T}right)}}$$ Note that, in the above equation the parameters n, ${V_T}$ and $I_{s}$ are constants. So, the output voltage ${V_0}$ will be proportional to the anti-natural logarithm (exponential) of the input voltage ${V_i}$, for a fixed value of feedback resistance ${R_f}$. Therefore, the op-amp based anti-logarithmic amplifier circuit discussed above will produce an output, which is proportional to the anti-natural logarithm (exponential) of the input voltage ${V_i}$ when, ${R_fI_s}= 1V$. Observe that the output voltage ${V_0}$ is having a negative sign, which indicates that there exists a 1800 phase difference between the input and the output. Learning working make money
Basics Of Linear Integrated Circuits Applications An electronic circuit is a group of electronic components connected for a specific purpose. A simple electronic circuit can be designed easily because it requires few discrete electronic components and connections. However, designing a complex electronic circuit is difficult, as it requires more number of discrete electronic components and their connections. It is also time taking to build such complex circuits and their reliability is also less. These difficulties can be overcome with Integrated Circuits. Integrated Circuit (IC) If multiple electronic components are interconnected on a single chip of semiconductor material, then that chip is called as an Integrated Circuit (IC). It consists of both active and passive components. This chapter discusses the advantages and types of ICs. Advantages of Integrated Circuits Integrated circuits offer many advantages. They are discussed below − Compact size − For a given functionality, you can obtain a circuit of smaller size using ICs, compared to that built using a discrete circuit. Lesser weight − A circuit built with ICs weighs lesser when compared to the weight of a discrete circuit that is used for implementing the same function of IC. using ICs, compared to that built using a discrete circuit. Low power consumption − ICs consume lower power than a traditional circuit,because of their smaller size and construction. Reduced cost − ICs are available at much reduced cost than discrete circuits because of their fabrication technologies and usage of lesser material than discrete circuits. Increased reliability − Since they employ lesser connections, ICs offer increased reliability compared to digital circuits. Improved operating speeds − ICs operate at improved speeds because of their switching speeds and lesser power consumption. Types of Integrated Circuits Integrated circuits are of two types − Analog Integrated Circuits and Digital Integrated Circuits. Analog Integrated Circuits Integrated circuits that operate over an entire range of continuous values of the signal amplitude are called as Analog Integrated Circuits. These are further classified into the two types as discussed here − Linear Integrated Circuits − An analog IC is said to be Linear, if there exists a linear relation between its voltage and current. IC 741, an 8-pin Dual In-line Package (DIP)op-amp, is an example of Linear IC. Radio Frequency Integrated Circuits − An analog IC is said to be Non-Linear, if there exists a non-linear relation between its voltage and current. A Non-Linear IC is also called as Radio Frequency IC. Digital Integrated Circuits If the integrated circuits operate only at a few pre-defined levels instead of operating for an entire range of continuous values of the signal amplitude, then those are called as Digital Integrated Circuits. In the coming chapters, we will discuss about various Linear Integrated Circuits and their applications. Learning working make money
Rectifiers AC and DC are two frequent terms that you encounter while studying the flow of electrical charge. Alternating Current (AC) has the property to change its state continuously. For example, if we consider a sine wave, the current flows in one direction for positive half cycle and in the opposite direction for negative half cycle. On the other hand, Direct Current (DC) flows only in one direction. An electronic circuit, which produces either DC signal or a pulsated DC signal, when an AC signal is applied to it is called as a rectifier. This chapter discusses about op-amp based rectifiers in detail. Types of Rectifiers Rectifiers are classified into two types: Half wave rectifier and Full wave rectifier. This section discusses about these two types in detail. Half wave Rectifier A half wave rectifier is a rectifier that produces positive half cycles at the output for one half cycle of the input and zero output for the other half cycle of the input. The circuit diagram of a half wave rectifier is shown in the following figure. Observe that the circuit diagram of a half wave rectifier shown above looks like an inverting amplifier, with two diodes D1 and D2 in addition. The working of the half wave rectifier circuit shown above is explained below For the positive half cycle of the sinusoidal input, the output of the op-amp will be negative. Hence, diode D1 will be forward biased. When diode D1 is in forward bias, output voltage of the op-amp will be -0.7 V. So, diode D2 will be reverse biased. Hence, the output voltage of the above circuit is zero volts. Therefore, there is no (zero) output of half wave rectifier for the positive half cycle of a sinusoidal input. For the negative half cycle of sinusoidal input, the output of the op-amp will be positive. Hence, the diodes D1 and D2 will be reverse biased and forward biased respectively. So, the output voltage of above circuit will be − $$V_0=-left(frac{R_f}{R_1}right)V_1$$ Therefore, the output of a half wave rectifier will be a positive half cycle for a negative half cycle of the sinusoidal input. Wave forms The input and output waveforms of a half wave rectifier are shown in the following figure As you can see from the above graph, the half wave rectifier circuit diagram that we discussed will produce positive half cycles for negative half cycles of sinusoidal input and zero output for positive half cycles of sinusoidal input Full wave Rectifier A full wave rectifier produces positive half cycles at the output for both half cycles of the input. The circuit diagram of a full wave rectifier is shown in the following figure − The above circuit diagram consists of two op-amps, two diodes, D1 & D2 and five resistors, R1 to R5. The working of the full wave rectifier circuit shown above is explained below − For the positive half cycle of a sinusoidal input, the output of the first op-amp will be negative. Hence, diodes D1 and D2 will be forward biased and reverse biased respectively. Then, the output voltage of the first op-amp will be − $$V_{01}=-left(frac{R_2}{R_1}right)V_i$$ Observe that the output of the first op-amp is connected to a resistor R4, which is connected to the inverting terminal of the second op-amp. The voltage present at the non-inverting terminal of second op-amp is 0 V. So, the second op-amp with resistors, R4 and R4 acts as an inverting amplifier. The output voltage of the second op-amp will be $$V_0=-left(frac{R_5}{R_4}right)V_{01}$$ Substituting the value of $V_{01}$ in the above equation, we get − $$=>V_{0}=-left(frac{R_5}{R_4}right)left { -left(frac{R_2}{R_1}right)V_{i} right }$$ $$=>V_{0}=left(frac{R_2R_5}{R_1R_4}right)V_{i}$$ Therefore, the output of a full wave rectifier will be a positive half cycle for the positive half cycle of a sinusoidal input. In this case, the gain of the output is $frac{R_2R_5}{R_1R_4}$. If we consider $R_{1}=R_{2}=R_{4}=R_{5}=R$, then the gain of the output will be one. For the negative half cycle of a sinusoidal input, the output of the first op-amp will be positive. Hence, diodes D1 and D2 will be reverse biased and forward biased respectively. The output voltage of the first op-amp will be − $$V_{01}=-left(frac{R_3}{R_1}right)V_{i}$$ The output of the first op-amp is directly connected to the non-inverting terminal of the second op-amp. Now, the second op-amp with resistors, R4 and R5 acts as a non-inverting amplifier. The output voltage of the second op-amp will be − $$V_{0}=left(1+frac{R_5}{R_4}right)V_{01}$$ Substituting the value of $V_{01}$ in the above equation, we get $$=>V_{0}=left(1+frac{R_5}{R_4}right) left{-left(frac{R_3}{R_1}right)V_{i}right } $$ $$=>V_{0}=-left(frac{R_3}{R_1}right)left(1+frac{R_5}{R_4}right)V_{i}$$ Therefore, the output of a full wave rectifier will be a positive half cycle for the negative half cycle of sinusoidal input also. In this case, the magnitude of the gain of the output is $left(frac{R_3}{R_1}right)left(1+frac{R_5}{R_4}right)$. If we consider $R_{1}=2R_{3}=R_{4}=R_{5}=R$ then the gain of the output will be one. The input and output waveforms of a full wave rectifier are shown in the following figure As you can see in the above figure, the full wave rectifier circuit diagram that we considered will produce only positive half cycles for both positive and negative half cycles of a sinusoidal input. Learning working make money
Phase Locked Loop IC Phase Locked Loop (PLL) is one of the vital blocks in linear systems. It is useful in communication systems such as radars, satellites, FMs, etc. This chapter discusses about the block diagram of PLL and IC 565 in detail. Block Diagram of PLL A Phase Locked Loop (PLL) mainly consists of the following three blocks − Phase Detector Active Low Pass Filter Voltage Controlled Oscillator (VCO) The block diagram of PLL is shown in the following figure − The output of a phase detector is applied as an input of active low pass filter. Similarly, the output of active low pass filter is applied as an input of VCO. The working of a PLL is as follows − Phase detector produces a DC voltage, which is proportional to the phase difference between the input signal having frequency of $f_{in}$ and feedback (output) signal having frequency of $f_{out}$. A Phase detector is a multiplier and it produces two frequency components at its output − sum of the frequencies $f_{in}$ and $f_{out}$ and difference of frequencies $f_{in}$ & $f_{out}$. An active low pass filter produces a DC voltage at its output, after eliminating high frequency component present in the output of the phase detector. It also amplifies the signal. A VCO produces a signal having a certain frequency, when there is no input applied to it. This frequency can be shifted to either side by applying a DC voltage to it. Therefore, the frequency deviation is directly proportional to the DC voltage present at the output of a low pass filter. The above operations take place until the VCO frequency equals to the input signal frequency. Based on the type of application, we can use either the output of active low pass filter or output of a VCO. PLLs are used in many applications such as FM demodulator, clock generator etc. PLL operates in one of the following three modes − Free running mode Capture mode Lock mode Initially, PLL operates in free running mode when no input is applied to it. When an input signal having some frequency is applied to PLL, then the output signal frequency of VCO will start change. At this stage, the PLL is said to be operating in the capture mode. The output signal frequency of VCO will change continuously until it is equal to the input signal frequency. Now, it is said to be PLL is operating in the lock mode. IC 565 IC 565 is the most commonly used phase locked loop IC. It is a 14 pin Dual-Inline Package (DIP). The pin diagram of IC 565 is shown in the following figure − The purpose of each pin is self-explanatory from the above diagram. Out of 14 pins, only 10 pins (pin number 1 to 10) are utilized for the operation of PLL. So, the remaining 4 pins (pin number 11 to 14) are labelled with NC (No Connection). The VCO produces an output at pin number 4 of IC 565, when the pin numbers 2 and 3 are grounded. Mathematically, we can write the output frequency, $f_{out}$ of the VCO as. $$f_{out}=frac{0.25}{R_VC_V}$$ where, $R_{V}$ is the external resistor that is connected to the pin number 8 $C_{V}$ is the external capacitor that is connected to the pin number 9 By choosing proper values of $R_{V}$ and $C_{V}$, we can fix (determine) the output frequency, $f_{out}$ of VCO. Pin numbers 4 and 5are to be shorted with an external wire so that the output of VCO can be applied as one of the inputs of phase detector. IC 565 has an internal resistance of $3.6KOmega$. A capacitor, C has to be connected between pin numbers 7 and 10 in order to make a low pass filter with that internal resistance. Note that as per the requirement, we have to properly configure the pins of IC 565. Learning working make money
DAC Example Problem In the previous chapter, we discussed the two types of DACs. This chapter discusses an example problem based on R-2R ladder DAC. Example Let us find the value of analog output voltage of R-2R Ladder DAC for a binary input, $b_{2}b_{1}b_{0}$ = 100. Circuit Diagram and its Simplification The circuit diagram of a 3-bit R-2R Ladder DAC when binary input, $b_{2}b_{1}b_{0}$ = 100 applied to it is shown in the following figure − In the above circuit, there exists series and parallel combinations of resistors to the left of point A with respect to ground. So, we can replace that entire resistor network with a single resistor having resistance of $2ROmega$. The simplified circuit diagram is shown in the following figure − We can replace the part of the network that is connected to the left of point B with respect to ground by using a Thevenin’s equivalent circuit. The modified circuit diagram is shown in the following figure − In the above circuit, there exist a series combination of two resistors. Replace this combination with a single resistor. The final circuit diagram after simplification is shown in the following figure − Now, the above circuit diagram looks like an inverting amplifier. It is having an input voltage of $-frac{V_{R}}{2}$ volts, input resistance of $2ROmega$ and feedback resistance of $2ROmega$. The output voltage of the circuit shown above will be − $$V_{0}=-frac{2R}{2R}left(-frac{V_{R}}{2}right)$$ $$V_{0}=frac{V_{R}}{2}$$ Therefore, the output voltage of 3-bit R-2R Ladder DAC is $frac{V_{R}}{2}$ volts for a binary input, $b_{2}b_{1}b_{0}$ = 100. Learning working make money
Converters Of Electrical Quantities Voltage and current are the basic electrical quantities. They can be converted into one another depending on the requirement. Voltage to Current Converter and Current to Voltage Converter are the two circuits that help in such conversion. These are also linear applications of op-amps. This chapter discusses them in detail. Voltage to Current Converter A voltage to current converter or V to I converter, is an electronic circuit that takes current as the input and produces voltage as the output. This section discusses about the op-amp based voltage to current converter. An op-amp based voltage to current converter produces an output current when a voltage is applied to its non-inverting terminal. The circuit diagram of an op-amp based voltage to current converter is shown in the following figure. In the circuit shown above, an input voltage $V_{i}$ is applied at the non-inverting input terminal of the op-amp. According to the virtual short concept, the voltage at the inverting input terminal of an op-amp will be equal to the voltage at its non-inverting input terminal . So, the voltage at the inverting input terminal of the op-amp will be $V_{i}$. The nodal equation at the inverting input terminal”s node is − $$frac{V_i}{R_1}-I_{0}=0$$ $$=>I_{0}=frac{V_t}{R_1}$$ Thus, the output current $I_{0}$ of a voltage to current converter is the ratio of its input voltage $V_{i}$ and resistance $R_{1}$. We can re-write the above equation as − $$frac{I_0}{V_i}=frac{1}{R_1}$$ The above equation represents the ratio of the output current $I_{0}$ and the input voltage $V_{i}$ & it is equal to the reciprocal of resistance $R_{1}$ The ratio of the output current $I_{0}$ and the input voltage $V_{i}$ is called as Transconductance. We know that the ratio of the output and the input of a circuit is called as gain. So, the gain of an voltage to current converter is the Transconductance and it is equal to the reciprocal of resistance $R_{1}$. Current to Voltage Converter A current to voltage converter or I to V converter is an electronic circuit that takes current as the input and produces voltage as the output. This section discusses about the op-amp based current to voltage converter. An op-amp based current to voltage converter produces an output voltage when current is applied to its inverting terminal. The circuit diagram of an op-amp based current to voltage converter is shown in the following figure. In the circuit shown above, the non-inverting input terminal of the op-amp is connected to ground. That means zero volts is applied at its non-inverting input terminal. According to the virtual short concept, the voltage at the inverting input terminal of an op-amp will be equal to the voltage at its non-inverting input terminal. So, the voltage at the inverting input terminal of the op-amp will be zero volts. The nodal equation at the inverting terminal”s node is − $$-I_{i}+frac{0-V_0}{R_f}=0$$ $$-I_{i}=frac{V_0}{R_f}$$ $$V_{0}=-R_{t}I_{i}$$ Thus, the output voltage, $V_{0}$ of current to voltage converter is the (negative) product of the feedback resistance, $R_{f}$ and the input current, $I_{t}$. Observe that the output voltage, $V_{0}$ is having a negative sign, which indicates that there exists a 1800 phase difference between the input current and output voltage. We can re-write the above equation as − $$frac{V_0}{I_i}=-R_{f}$$ The above equation represents the ratio of the output voltage $V_{0}$ and the input current $I_{i}$, and it is equal to the negative of feedback resistance, $R_{f}$. The ratio of output voltage $V_{0}$ and input current $I_{i}$ is called as Transresistance. We know that the ratio of output and input of a circuit is called as gain. So, the gain of a current to voltage converter is its trans resistance and it is equal to the (negative) feedback resistance $R_{f}$ . Learning working make money