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Hey friends, welcome to the Kohiki ALL ABOUT ELECTRONICS. So, in this article, we will see, how to design the differential amplifier or subtractor circuit using the op-amp.

Now, we already know that this op-amp is a differential amplifier, which amplifies the difference between the inverting and the non-inverting terminal. But whenever this op-amp is used in the open-loop configuration, then the gain of this op-amp is quite high. And because of that, your output will always get saturated. And to avoid that we always need to use this op-amp in the feedback circuit.

**Op-amp as Differential Amplifier**

So, then we had seen this inverting and the non-inverting op-amp configuration. Where we have applied the input to the either **non-inverting **or the inverting input terminal. So, now let’s see how by combining the inverting and the non-inverting **op-amp** configuration we can design the differential amplifier.

So, by using this configuration, we can use this op-amp as a differential amplifier in a true sense. Because here, we can control the gain of the **op-amp** by controlling the value of this resistor **R2 **and **R1**.

So, let’s see how this configuration will act as a differential amplifier. So, now if you closely look at this configuration, it is the combination of inverting and the non-inverting configuration.

So, if you consider this voltage source **V1 **alone and consider this voltage source **V2 **is zero then it will be an inverting configuration.

Similarly, if you consider only this voltage source **V2**, and if you consider this voltage source **V1 **as zero then it will non-inverting configuration.

So, this configuration is the combination of the inverting and the non-inverting configuration. So, to get the final output, we need to use the superposition theorem.

**Superposition Theorem.**

So, what we will do, we will consider only one voltage source at a time and we will consider the remaining voltage sources as zero. So, first, we will consider this voltage sourceV2 as zero and we will assume that this voltage source V1 is acting alone. So, let’s say in this configuration, the output voltage is Vo1.

- So, this output voltage Vo1 can be given as,(-R2) divide by R1 times this voltage V1, as it inverting configuration right Similarly, let’s say if voltage V1 is zero and let’s say this voltage V2 is acting alone, then, in that case, our output voltage let’s say Vo2 will be equal to 1 plus R2 divided by R1 times this voltage source V2.
- So, now if we consider this voltage sourceV1 and V2 acting simultaneously, then our output voltage Vout will be equal to Vo1 plusVo2.
- That is equal to minus R2 divide by R1 times this voltage V1, plus, one plus R2 divide by R1 times this voltage source V2.
- So, now if you observe here, the gain that is applied to both inputs is different.

suppose if we want this output voltage V out as K times the difference between the two voltage sources, let’s say that is equal to V2 minus v1, then we need to ensure that the gain that is applied to both inputs is equal.

So, suppose in this case, if the gain that is applied to both the input is let’s say that is equal to R2 divide by R1, then ina true sense we can use this configuration as a differential amplifier, where the output will be the difference between the two input voltages, multiplied by some gain factor K. So, now let’s see how we achieve this equal gain.

- So, here, in this second term, instead of this one plus R2 divide by R1 we require the term that is equal to R2 divide by R1.
- So, somehow, we need to arrange this term in such a way that we can get this term R2 divide by R1.
- So, now we can write this term one plus R2divide by R1 as, R1 plus R2 divide by R1.
- And if we want this term R2 divide by R1, then we need to multiply this term by the factor of R2 divided by R1 plus R2.
- So, here, to get this term, what we can do, instead of applying this voltage source V2 directly, we can apply this voltage sourceV2 by using the voltage divider.
- So, let’s see how we can achieve these termR2 divide by R1.

**voltage source**

So, what we will do, we will consider only one voltage source at a time and we will consider the remaining voltage sources as zero. So, first, we will consider this voltage sourceV2 as zero and we will assume that this voltage source V1 is acting alone. So, let’s say in this configuration, the output voltage is Vo1.

So, this output voltage Vo1 can be given as,(-R2) divide by R1 times this voltage V1, as it inverting configuration right Similarly, let’s say if voltage V1 is zero and let’s say this voltage V2 is acting alone, then, in that case, our output voltage let’s say Vo2 will be equal to 1 plus R2 divided by R1 times this voltage source V2.

So, now if we consider this voltage sourceV1 and V2 acting simultaneously, then our output voltage Vout will be equal to Vo1 plusVo2. That is equal to minus R2 divide by R1 times this voltage V1, plus, one plus R2 divide by R1 times this voltage source V2. So, now if you observe here, the gain that is applied to both inputs is different.

suppose if we want this output voltage V out as K times the difference between the two voltage sources, let’s say that is equal to V2 minus v1, then we need to ensure that the gain that is applied to both inputs is equal.

So, suppose in this case, if the gain that is applied to both the input is let’s say that is equal to R2 divide by R1, then ina true sense we can use this configuration as a differential amplifier, where the output will be the difference between the two input voltages, multiplied by some gain factor K.

- So, now let’s see how we achieve this equal gain.
- So, here, in this second term, instead of this one plus R2 divide by R1 we require the term that is equal to R2 divide by R1.
- So, somehow, we need to arrange this term in such a way that we can get this term R2 divide by R1.
- So, now we can write this term one plus R2divide by R1 as, R1 plus R2 divide by R1.
- And if we want this term R2 divide by R1, then we need to multiply this term by the factor of R2 divided by R1 plus R2.
- So, here, to get this term, what we can do, instead of applying this voltage source V2 directly, we can apply this voltage sourceV2 by using the voltage divider.
- So, let’s see how we can achieve these termR2 divide by R1.
- So, here what we have done now, we have applied this voltage source V2 using this voltage divider circuit.
- So, now if you see at this point, then at this point let’s say the voltage is V plus.
- So this V plus will be equal to R2 divided by R1 plus R2 times this voltage V2 right !!So, our voltage V2 will be equal to this term.
- Now, we know that at the output voltage Vout, because of this voltage source V2 will be equal to 1 plus R2 divide by R1 times this voltage Vplus.
- So, now if we put this value of Vplus, then we will get one plus R2 divide by R1 times, R2 divided by (R1 plus R2) times this voltageV2.
- So, if we simplify it then we will get term, R2 divide by R1 times this voltage V2.

So, now if you see, if we consider this voltage source V1 and V2 simultaneously, then our output Vout will be equal to -R2 divide byR1 times this voltage V1, plus R2 divide by R1 times this voltage source V2. Or we can say that our output voltage will be equal to R2 divide by R1 times the difference between the voltage source V2 and V1. So, now if you see this configuration, here the output voltage is the difference between the two input voltages, multiplied by the factor of R2 divided by R1. So, in this way, using this configuration, in a true sense we can use this op-amp as a differential amplifier.

**Application of Differential Amplifier**

So, now let’s see where this differential amplifier can be used. So, apart from performing mathematical operations, this differential amplifier can also be used along with the sensors. So, let’s say here we have RTD and it is connected to a balanced bridge circuit. And the output of the two arms of the balanced bridge is connected as the input to the two ends of the differential amplifier.

- So, here initially, this bridge is balanced but as the temperature changes, the value of this resistance R will change.
- And because of that, the voltage that appears at the two inputs will also change.
- So, when the bridge is balanced, then the input at both ends of this differential amplifier will be equal.
- that is V1 is equal to V2.
- But as the temperature changes, the value of this resistance R will change. And because of that, the voltage that appears at this end will also change.
- And we can amplify the difference between the V1 and V2 and we can use this output to control another circuitry.
- So, let’s say, if you want to control the temperature of some room, then we can use this sensor along with this differential amplifier.
- So, by triggering another circuity based on this output voltage, we can control the temperature of the room.
- So, this is how this differential amplifier can be used along with the sensors.

Now, here the problem with this differential amplifier is that the input impedance of this differential amplifier is very low, and it depends upon the value of the R1 and R2. So, if you see in this configuration, the input impedance that is seen through this end will be equal to R1. While, at this end, if you see, the input impedance is equal to R1 plus R2.

**General Term**

the general, term, let’s say if you are applying the input voltage between the two ends, then, in this case, the input impedance Zin will be equal to 2 times R1.

So, as you can see here, the input impedance of this differential amplifier is very low. And because of that, it is quite possible that your source might get loaded.

- So, one way we can avoid this by using the buffer before the input voltages. So, as you can see here, before applying this voltage V1, suppose if we use buffers then, the input impedance of this buffer is very very high.
- in this way, we can avoid the loading of these source voltages.
- But in most of the practical cases, where you are dealing with the very small input voltage which is coming from the sensors, in such cases instead of this differential amplifier, the instrumentation amplifier is used.
- And we will talk more about this instrumentation amplifier in a separate article.

So, this instrumentation amplifier is a very high gain differential amplifier, and the main advantage of this instrumentation amplifier is that it is having very high input impedance. So, that is particularly useful, when we are dealing with the sensors. But in general applications, we can use this differential amplifier along with these buffers.

**Example 1**

So, that is being said, now let’s see some examples based on this differential amplifier. So, here we have one differential amplifier and we need to find the output voltage of this differential amplifier. And here assume that the biasing voltages, which are being applied to this op-amp are +15 and -15 V.

- So, here to find the output voltage what we will do, we will consider one voltage source at a time.
- So, first, let’s assume that this 2V voltage source is acting alone.
- So, if we consider this 2Vvoltage source acting alone, in that case, let’s say the output voltage is Vo1.
- And that will be equal to minus 5 divided by 1 time this 2V.
- And that will come out as minus -10V.
- Similarly, if we assume that this 3V voltage source is acting alone and this 2V voltage source is equal to zero, then in the case of our output voltage let’s say Vo2 will be equal to one plus 5 divided by 1 times this voltage Vplus.
- And here this voltage V plus is equal to 8divided by 8 plus one, times this input voltage that is equal to 3V.
- So, our output voltage Vo2 will be equal to16V. So, now if we combine these two input voltages and assume that these two input voltages are acting simultaneously then, in that case, our output voltage Vout will be equal to Vo1 plus Vo2.
- So our output voltage will be equal to 16-10, which is equal to 6V.

So, this is a very simple and basic example based on this differential amplifier. Now, here, you also need to consider the biasing voltage that is being applied to the differential amplifier. So, let’s take one more example and we will see how this biasing voltage can affect your output.

**Example 2 (Effect of Biasing Voltages on output)**

So, if you see this example, this is very similar to the previous example but here, in this case, let’s assume that our bias voltage is +12 V and -12 V, and we want to find the output voltage. so, here once again, what we will do, we will assume only one voltage source is acting at a time and we will find the output voltage. And then we will combine the individual responses.

- So, let’s say output Vo1 is the output voltage when this 2V voltage source is acting alone.
- So, our output voltage will be equal to (-5)divided by 1 time this 2V voltage source.
- So, that will come out again as -10V.
- And now let’s assume this 5V voltage source is acting alone.
- So, in that case, our output voltage Vo2 will be equal to one plus this 5 divided by one time this voltage V plus.
- That will be equal to 6 times V plus.
- Now, this voltage V plus will be equal to the voltage divider that is applied at this point.
- so, it will be equal to 4 divided by 4 plus one’s times this 5V input voltage.
- And it will come out as 24V.
- So, our output voltage Vo2 will be equal to24V.

So, now in this case, if you see, our total output voltage V out will be equal to Vo1 +Vo2. And if you see, it will come out as 24 -10, which is equal to 14V. But in this example, we have applied plus-minus12V. So, our output voltage cannot go beyond this plus-minus 12V right!!! So, our output voltage will get saturated at 12V. So, apart from the input voltages, you also need to consider the biasing voltages which are being applied to the op-amp.

**Example 3 (For practice )**

So, that is being said, now let’s see the third example and this example is for your exercise. So, here instead of a single input voltage, we have applied the multiple input voltages at the inverting and the non-inverting terminal. And we need to find the output voltage of this configuration. And here once again assume that the biasing voltages for this circuit are +15V and -15V. So, try to solve this example by yourself, and do let me know your answer in the comment section below.

So, I hope in this article you understood, how to use the op-amp as a differential amplifier or as a subtractor.

So, if you have any questions or suggestions, do let me know in the comment section below.

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