Hey friends, welcome to the Kohiki.com ALL ABOUT ELECTRONICS. So, in this article, we will understand how to design the Log and Antilog Amplifier using the Opamp.
The applications of log and antilog amplifier
Now, using the logarithmic characteristic of the circuit components it is possible to convert the multiplication into the addition and the division into the subtraction. And similarly, it is possible to convert the powers and the roots into multiplication and division respectively.
For example. let’s say, you have one signal A, and if you pass this signal through the logarithmic amplifier followed by the scaling circuit which scales the input by let’s say some factor b, then the output of this circuit will be equal to times the log of this signal A. And if we pass this signal through the Antilog amplifier then we can raise the original signal to the power of this scaling factor.
Similarly, let’s assume that we are passing the output of the logarithmic amplifier through some division circuit, which divides the output of this (Antilog Amplifiers) logarithmic amplifier by a factor of 2. And then if we pass this signal through the antilog amplifier then we can find the square root of the original signal.
Now, this division circuit can be as simple as the voltage divider circuit. And just by changing the division factor, we can find the cube root or the nth root of the original signal. So, this type of circuit is very useful in finding the RMS value of the signal.
So, in this way, these log and antilog amplifiers are very useful in performing mathematical operations. Or we can say that it is an essential part of the analog computer. So, apart from performing these mathematical operations, these log and antilog amplifiers are very useful for signal compression. So, using this amplifier we can even reduce or enhance the dynamic range of the signal.
Now, these log and antilog amplifiers can be designed by using the diode and transistor. So, first, we will design these log and antilog amplifiers using the diode and later on, we will modify this circuit for further improvements.
Log amplifier circuit using the diode
So, here is the simple circuit of this log amplifier (Antilog Amplifiers) which is designed using the diode and the opamp. So, where the diode is connected in the feedback path between the output and the inverting input terminal. Noe, here let’s assume that IRis the current which is flowing through this resistor R. And Id is the forward current of this diode, and Vd is the forward voltage across this diode.
Now, if we apply the concept of virtual ground, then we can say that this node over here will be also at ground potential. So, now if we apply the KVL in this loop, then we can say that the output voltage Vout will be equal to Vd. And if we apply the KCL at this node, then we can say that this current IR will be equal to the current which is flowing through this diode. And this current IR will be equal to the input voltage divided by this resistor R. Now, here the relationship between the forward current of this diode and the voltage across this diode can be given by this expression.
So, here this Is is the reverse saturation current and Vt is the thermal voltage of this diode. While Vd is the forward voltage across the diode, and η is the ideality factor. Now, generally, the ideality factor used to be in the range of 1 and 2. And it depends upon the forward current of this diode. So, for the ideal diode, this ideality factor should be equal to 1. But in reality, it varies between 1 and 2. Now, here this reverse saturation current and the thermal voltage are the temperaturedependent parameters.
So, with temperature, these two parameters will change. In fact, this reverse saturation current will almost get doubled after every 10degree rise in temperature. While the thermal voltage will change with temperature by this expression. So, here this K is the Boltzmann constant and T is the temperature of the diode. While q is the charge of the electron.
So, if we put all these values and if we evaluate the value of this thermal voltage at T is equal to 300 degrees Kelvin. Or at room temperature, then the approximate value of this thermal voltage will be equal to around 25.8 mV. So, at room temperature, if we put the value of this thermal voltage in this expression then the diode current Id can be given by this expression. Now, here this exponential term will be much larger than 1.
So, approximately we can say that Id is equal to Is times e to the power of 38.6 times this diode voltage Vd, divided by this ideality factor. And in general, at any temperature, we can write this expression Id as, Id is equal to Is times, e to the power Vd divided by times Vt. Now, from this, we can say that Id divided by this Is is equal to e to the power of Vd divided by ita times Vt.
Now, suppose if take the natural log (Antilog Amplifiers) on both sides then we can say that ln (Id/Is)is equal to Vd/(ηVt)That means Vd will be equal to ηVt times ln (Id/Is)Now, earlier we have found that this output voltage Vout is equal to Vd. And this current Ir will be equal to the current that is flowing through the diode. And that will be equal to this input voltage Vin divided by this Resistor R.
So, if we put all these values then we can say that the output voltage Vout will be equal to η*Vt times ln (Vin/Is*R)So, this is the expression of the output voltage in the case of log amplifier. And as you can see over here, the output voltage is proportional to the natural log (Antilog Amplifiers) of the input voltage. Now, we will talk more about this log amplifier in the latter part of the article
Antilog Amplifier circuit using the diode
bit now, first of all, let’s see, similarly how we can design the antilog amplifier. So now here, just by interchanging the position of this diode and the resistor, we can convert the log amplifier into the antilog amplifier.
So, now here let’s assume that the current that is flowing through this diode is equal to Id. And the current that is flowing through this resistor is equal to Ir. And Vd is the voltage across this diode. Now, because of the virtual ground concept, this node will be also at ground potential.
So, now if we apply the KVL in this particular (Antilog Amplifiers) loop then we can say that this input voltage Vin will be equal to the voltage across the diode. And if we apply the KCL at this node, then we can say that this diode current Id will be equal to the current that is flowing through this resistor R.
That is equal to Vout/R. Or simply we can say that the output voltage V out will be equal to R times the current that is flowing through this diode. Now, we know that the diode current Id can be approximately given as Is times e to the power of Vd divided by η*Vt. And earlier as we have seen, the output voltage V out will be equal to R times this diode current.
So, we can say that the output voltage Vout will be equal to R times Id. Or we can say that that is equal to (R) times Is into e to the power of Vd divided by η*Vt. Now, as we have seen, this input voltage Vin will be equal to the voltage across the diode. So, we can say that this output voltage Vout will be equal to, R times Is into e to the power Vin divided by η*Vt.
So, this is the expression of the output voltage in the case of an antilog amplifier. So, here are the two expressions for the log and the antilog amplifiers. And as you can see over here, the output voltage V out also depends on the other factors, like the ideality factor η, the thermal voltage Vth, and the reverse saturation current Is. Now, like said before, this reverse saturation current and the thermal voltage are the temperaturedependent parameters.
So, with temperature, they will change. Now, we will talk more about these factors at the later part of the article, but before that first of all let’s talk about this ideality factor. Now, as I said before, this ideality factor used to be in the range of 1 and 2. And because of this ideality factor, the dynamic range of the circuit used gets reduced. So, that is the disadvantage of this diodebased log and antilog amplifier. And that problem can be solved by using the transistorbased (Antilog Amplifiers) log and antilog amplifier.
Log Amplifier circuit using a transistor
So, here is the circuit of the (Antilog Amplifiers) log amplifier which is designed using the transistor and the opamp. Now, here, in fact, this transistor is used as a diode. And this circuit is also known as the transdiode configuration. So, if you observe this circuit, where this base is at ground potential. And because of the virtual ground, this collector of this transistor will also at ground potential.
So, basically, this collectorbase junction is at ground potential. So, only this baseemitter junction will be active in this transistor. So, effectively this transistor will act as a diode.
So, now let’s write down the different currents which are flowing through this circuit. So, here this IR is the current which is flowing through this resistor R. And Ic & IE is the collector and the emitter currents of this transistor. Now, approximately we can say that this collector current will be equal to the emitter current. And if we apply the KCL at this node, then we can say that this resistor current IR will be equal to this collector current IC.
And that will be equal to this input voltage Vin divided by the resistor R. Now, here let’s assume that the voltage across this baseemitter junction is equal to VBE. And if we apply the KVL around this loop, then we can say that this output voltage Vout will be equal to VBE. So, using this now let’s find out the output voltage for this log amplifier.
Now, here in the given circuit as this baseemitter junction is acting as a diode, so we can write the relationship between this VBE and in a similar fashion that we have written for the diode. And here, as this collector current is approximately equal to this emitter current.
so we can write the relationship as, Ic is equal to Is times, e to the power of VBE divided by this thermal voltage VT, minus1. Approximately we can write this expression as Is times, e to the power of VBE divided by VT. Because this exponential term is much larger than 1. Now, from this expression, we can easily find the relationship between the output voltage and the input voltage.
So, if we rearrange this expression then can write this expression as, VBE that is equal to VT times, ln (Ic/Is). Now, earlier we have written that this output voltage Vout will be equal to VBE. And this collector current Ic will be equal to the current that is flowing through this resistor R.
That is equal to Vin divided by this resistor R.So from this, we can say that the output voltage Vout will be equal to VT * ln (Vin /R*Is). So, this is the expression of the output voltage for the (Antilog Amplifiers) log amplifier, which is designed using the transistor. And as you can see over here, for the transistor the ideality factor is equal to 1. So, there is no dependency on the ideality factor.
Antilog amplifier circuit using a transistor
So, using this transistorbased log (Antilog Amplifiers) amplifier, we can increase the dynamic range of the circuit. So, now similarly, let’s find out how we can design the antilog amplifier using this transistor. So, here, we have just interchanged the position of the resistor and transistor.
Now, using the concept of virtual ground we can say that this collector is also at ground potential. But here to make this baseemitter function positive, this input voltage Vin should be less than zero. Or we can say that the input voltage should be negative. Because here we are using the NPN transistor.
So, if we want to apply the positive input voltage, then instead of this NPN transistor, we should use a PNP transistor. So, now in this PNP transistor, if we apply the positive voltage at this node, then this emitterbase junction will become forward biased. And we can use this circuit as an antilog amplifier. So, now let’s write down the different currents for the given circuit.
So, here once again let’s assume that IE and the IC are the emitter and the collector current for the given transistor. And here we are assuming that IR is the current which is flowing through this resistor R. Now, considering the virtual short, this node will be at ground potential.
So, now let’s apply the KVL around this loop. So, we can say that this input voltage Vin will be equal to the voltage across this emitter and base junction. (Antilog Amplifiers) Now, considering we are using the PNP transistor, instead of writing VBE, we are writing VEB. And now, if we apply the KCL at this node, then we can say that this current IC will be equal to the current IR.
That is equal to Vout divided by this resistor. Or simply we can say that the output voltage V out will be equal to – R times collector current Ic. Now, earlier as we have seen, this collector current can be given by the expression, Is times e to the power VEB divided by VT. And as we have just seen the output voltage V out will be equal to R times this collector current Ic.
So, we can say that output voltage Vout will be equal to R times this Is * e^(Veb/Vt). Now, just we have also seen that this input voltage Vin is equal to VEB. So, from this we can say that the output voltage V out will be equal to R* Is* e^(Vin/Vt). So, this is the expression of the output voltage in the case of a log amplifier which is designed using the transistor. (Antilog Amplifiers) So, here are the expressions for this log and antilog amplifiers which are designed using the transistor.
So, using this transistor the dependency on the ideality factor has been removed. But still, if you observe these expressions, the output voltage still depends upon the thermal voltage VT and the reverse saturation current Is. So, with temperature, the output voltage Voutwill change. So, now let’s find out how we can modify this circuit so that the dependency on the temperature can be removed.
Practical Log amplifier circuit (Temperature compensated Log amplifier Circuit)
So, here is the circuit of the temperature compensated (Antilog Amplifiers) log amplifier. So, now let’s find out how this circuit works and using this circuit how we can compensate for the effect of temperature on the output voltage. So, now in this circuit, instead of a single log amplifier, the two log amplifiers have been used. So, in the first log amplifier, the input voltage is given, while in the second log amplifier some reference voltage has been given. And in these two log amplifiers, the matched pair of transistors are used.
It means that both transistors have the same thermal voltage and the same reverse saturation current. So, now let’s say that the output of this first log amplifier (Antilog Amplifiers) is equal to Va. So, we can say that the output voltage Va will be equal to Vt*ln (Vin/R*Is) And similarly now let’s say the output of this second log amplifier is equal to Vb. So, this Vb will be equal to Vt *ln (Vr/R*Is)Now, these input voltages Va and Vb are given to the differential amplifier.
Now, by selecting the value of this R1, R2, R3, and R4, we can ensure that the output voltage of this differential amplifier is equal to VbVa. Or we can say that output voltage Vout will be equal to Vt*ln (Vin/R*Is) – Vt* ln(Vr/R*Is)And if we simplify it then we can say that Vb Va will be equal to Vt* ln ((Vin/R*Is)*(R*Is/Vr))Or we can say that VbVa will be equal to Vt* ln (Vin/Vr)So if we observe this expression of this Vb – Va, it is independent of the reverse saturation current. So, by using this reference voltage Vr, and this differential amplifier circuit block, we can remove the effect of this reverse saturation current.
But still, this output voltage depends upon this thermal voltage. So, using this third block we can eliminate the effect of this thermal voltage. So, now let’s write down the expression for the output voltage. So, the output voltage Vout can be given as(1+ R6/R5)*(VbVa) And that is equal to Vt* ln (Vin/Vr)Now, here this third block is nothing but the noninverting opamp. But in this third block, this resistor R5is the temperaturedependent resistor.
So, with temperature, the value of this resistorR5 will change. So, now if we select the temperature coefficient of resistance for this R5 in such a way that it can nullify the effect of this Vt. And in a way, we can nullify the effect of temperature from this log amplifier (Antilog Amplifiers) circuit. So, this is one of the ways, by which we can design this temperature compensated log amplifier. So, in this way, we can also design the temperaturecompensated antilog amplifier. And if you are interested in knowing that circuit, then do let me know here in this poll. You can also comment on me in the comment section below.
FAQ

What is log and antilog amplifier?
Advertisements. The electronic circuits that perform the mathematical operations like power ANd antilogarithm (exponential) with an amplification square measure referred to as as power electronic equipment and AntiLogarithmic electronic equipment severally.

What are the applications of log amplifier?
In general, the principal application of log amps is to live signal strength, as against police investigation signal content. The log amp’s sign, which may represent a manydecade dynamic vary of highfrequency signaling amplitudes by a comparatively slender vary, is usually accustomed regulate gain.

What is the output voltage of the antilog amplifier?
Antilog amplifier using diode and opamp
V_{o} is the output voltage. The noninverting terminal of the opamp is connected to the ground. This means that the voltage of the noninverting terminal is zero volts.

What is the role of a diode in logarithmic amplifier?
Opampdiode log electronic equipment.
The voltage across the diode are going to be continually proportional to the log of the present through it associated once a diode is placed within the feedback path of an opamp in inverting mode, the output voltage are going to be proportional to the negative log of the input current. 
What is comparator and its application?
A comparator is associate degree electronic element that compares 2 input voltages. Comparators square measure closely associated with operational amplifiers, however a comparator is intended to control with feedback and with its output saturated at one power rail or the opposite.
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This is all about the log and antilog amplifiers which are designed using the opamp. So, if you have any questions or suggestions, do let me know in the comment section below.
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