Amplifier Configurations and Instrumentation Amplifier

Amplifier Configurations and Instrumentation Amplifier:

Properties:

The operational amplifier is an integrated amplifier that is manufactured particularly for the purposes of designing negative feedback amplifiers as shown in figure below. It is a differential amplifier containing two input terminals and hence two separate input voltages can be applied to it. Its major properties are as listed below:

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Figure: An Operational Amplifier

Property:  

Open Loop Gain      Ao        2 x 105 to 106

Input Resistance     Ri         > 2 MΩ to 1014 Ω

Output Resistance   Ro        < 500 Ω

Bandwidth               BW        1 - 100 MHz

Output Current        Iomax     10 mA typical

Note: A very high input resistance signifies that the input current demanded by either input of op-amp, V+ or V-, is negligible and can be taken out as zero for the purpose of analysis. Additionally, when the loop gain, Aoβ >>1, then the error voltage at op-amp input, Ve, can be taken as zero. This signifies that the inverting and non-inverting input terminals can be taken as being at similar potential, that is, V+ = V-. This is as well helpful for the purposes of analysis.
 
Non-Inverting Amplifier:

The schematic diagram of a non-inverting amplifier is as shown in figure below, where the resistors R1 and R2 are employed to give the negative feedback. The non-inverting amplifier is one in which the output signal is in similar sense as the input signal.

2256_2.3.jpg
Figure: Schematic Diagram of a Non-inverting Amplifier

Current flowing into the feedback network is as follows:

Io = Vo/(R1 + R2)

However the feedback voltage is:

Vf = IoR1

By substituting it gives:

Vf = [R1/(R1 + R2)] Vo

But when V+ = V- then,

Vf = Vi

And hence,

Vi = [R1/(R1 + R2)] Vo

Vo = [(R1 + R2)/R1] Vi

The closed loop voltage gain is then as follows:

AV = Vo/ Vi = (R1 + R2)/R1 = 1 + R2/ R1

Unity Gain Buffer:

A special condition of the non-inverting amplifier where R1 = ∞ and R2 → 0 is as shown in figure below. In this condition full feedback is applied to the amplifier with β = 1 and Vf = Vo .

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Figure: Schematic Diagram of a Unity-Gain Buffer

AV = Vo/Vi = [R1 (∞) + R2 (0)]/ R1 (∞) = R1/ R1 → 1

This gives a buffer amplifier, containing a very high input resistance that demands no current and doesn’t cause any attenuation of an input signal applied. The output of op-amp, on other hand, can supply some mA of current to a modest load. The unity-gain buffer can give an interface between a transducer with important internal source resistance and a load of the same resistance to avoid attenuation of signal, even whenever voltage gain is not needed.

The Inverting Amplifier:

The schematic diagram of inverting amplifier is as shown in figure below. This is likely to apply the input signal to inverting input of the op-amp to produce an output signal that is of opposite sense to the input signal. Since the negative feedback must as well be applied to the inverting terminal of the op-amp, the non-inverting input is simply joined to ground. In this condition with Ve → 0V and V+ = V-, the inverting and non-inverting inputs can be considered as being at similar potential and therefore the inverting terminal can be considered to be the ‘virtual earth’ point, that is, V+ = V- → 0V.

2448_2.5.jpg
Figure: Schematic Diagram of Inverting Amplifier

Whenever the input resistance of amplifier is taken as infinite, then this signifies that whatever current flows into the input resistor, R1 , must as well flow via the resistor in the feedback path, R2 and then into output terminal of the amplifier.

Ii = If

But, Ii = (Vi – V-)/R1 and If = (V--Vo)/R2

Then, (Vi – V-)/R1 = (V--Vo)/R2

However if the inverting input terminal can be taken as virtual earth point, V- → 0V and hence:

Vi/R1 = - Vo/R2

And hence,

Av = Vo/Vi = - R2/R1

This provides a gain determined by the values of resistors, R1 and R2 as desired.

Note that when: Ii = Vi/R1

Then, Ri = Vi/Ii = R1

It is the input resistance observe by the source, Vi, that is finite and considerably lower than the input resistance of op-amp itself.

The Inverting Summing Amplifier:

This is an amplifier structure that is basically a multiple version of Inverting Amplifier, containing several inputs as shown in figure below. Each input is joined through a corresponding resistor to the inverting input terminal of op-amp. As this input is considered to contain a very high input resistance, it can be supposed that no current flows into inverting input. This signifies that, applying Kirchhoff’s Current Law, all the input currents from the sources, I1, I2 and I3 should join at the inverting input terminal to flow out of this node via the feedback resistor, Rf. In this condition the inverting input terminal of op-amp is termed to as a current-summing node. This can still as well be considered as a virtual earth, as Ve → 0V and the non-inverting input terminal is joined to ground.
Then,

If = I1 + I2 + I3

78_2.6.jpg
Figure: Schematic diagram of Inverting Summing Amplifier

And hence,

(V- - Vo)/Rf = (V1 – V-)/R1 + (V2– V-)/R2 + (V3– V-)/R3

However as V- → 0V then,

- Vo/Rf = V1/R1 + V2/R2 + V3/R3

And hence finally:

Vo = - [(Rf/R1) V1+ (Rf/R2) V2 + (Rf/R3) V3]

This exhibits that this circuit structure essentially adds or sums the input source voltages V1, V2 and V3. The total output is inverted however this can simply be corrected by following this phase with a single inverting amplifier stage containing unity gain. Each and every input source voltage is scaled by a coefficient that is determined by the related input resistor and hence individual input voltages can be scaled by various factors.

Difference Amplifier:

The Difference Amplifier can be observe to be a combination of an Inverting Amplifier and a Non-inverting Amplifier, with the addition of resistor network comprising of R3 and R4 on the non-inverting input as shown in figure below.

1176_2.7.jpg
Figure: Schematic diagram of a Difference Amplifier

If the two input sources are independent and don’t influence with each other, then the effects of each and every can be assessed independently and the Principle of Superposition can be applied to determine the overall output. In this case, the contribution of V1 source to the output can be found while making V2 = 0V and then the contribution of V2 source to the output can be recognized with V1 = 0V.

Consider at first the effect of V1 source with V2 = 0V. In this case, the V2 input is joined to ground, in which case the resistors R3 and R4 can be ignored as they contain no voltage across them and no current flows via them. In this case, the configuration is similar to an inverting amplifier with V1 source as input and hence:

Vo1/V1 = - R2/R1 and hence, Vo1 = - (R2/R1) V1

Secondly, consider the contribution of V2 source with V1 = 0V. In this case, the V1 input terminal is joined to ground, that signifies that the input end of the resistor R2 is joined to ground. This provides a circuit structure identical to the non-inverting amplifier, however with additional resistor network comprising of R3 and R4 at non-inverting input of op amp, as shown in figure below.

The best approach is to consider the amplifier to be efficient; a non-inverting amplifier as far as the input voltage to its non-inverting terminal, V+ is regarded. The voltage at non-inverting terminal, V+, can then be determined in terms of input voltage V2.

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Figure: Equivalent diagram of a Difference Amplifier with V1 = 0V

Then for the voltage at non-inverting terminal of op-amp:

Vo2/V+ = 1 + R2/R1 = (R1 + R2)/R1

And hence, Vo2 = [(R1 + R2)/R1] V+

Resistors R3 and R4 simply act as a potential divider so that:

V+ = [R4/(R3 + R4)] V2

Then by substituting it gives,

Vo2 = [(R1 + R2)/R1] [R4/(R3 + R4)] V2

Combining the contributions to output from each of the input sources provides:

Vo = Vo1 + Vo2 = [{(R1 + R2)/R1} {R4/(R3 + R4)} V2] – (R2/R1) V1

If a selection is made and hence R3 = R1 and R4 = R2 then:

Vo = [{(R1 + R2)/R1} {R2/(R1 + R2)} V2] - (R2/R1) V1

And hence finally a difference expression is received as:

Vo = (R2/R1) (V2 – V1)

Instrumentation Amplifier:

The modification of difference amplifier can be made that significantly enhances on the performance of the latter and overcomes few of its limitations. This is termed as the standard 3 op-amp instrumentation amplifier and is shown in figure below. The structure around op-amp A3 is that of difference amplifier discussed above. The modification is that the two input-stage op-amps have been added, viz. A1 and A2 that act as buffer amplifiers containing high input impedance and are configured as non-inverting gain phases however with cross coupling between them through the resistor, R1.

This configuration can be analyzed in a similar way to the other simpler circuits, employing the principle that the error voltage approaches zero for each and every individual op-amps and that no current flows into any of the input terminal. This signifies that the potential at inverting and non-inverting terminals of the op-amps can be supposed to be similar even when they are not ground. As a result, the potential at both input terminals of op-amp A1 in figure below can be taken as being at the potential of input V1, whereas both input terminals of op-amp A2 can be taken as at the potential of input V2. Additionally, the current flowing via resistors, R2A, R1 and R2B can be taken as similar current, i. This can be employed in analysing the input phase of the overall amplifier structure.

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Figure: Schematic diagram of the Standard 3 Op-Amp Instrumentation Amplifier

By using this principle:

i = (Vo1 – V1)/R2A = (V1 – V2)/R1 = (V2 – Vo2)/R2B

Taking:

(Vo1 – V1)/R2A = (V1 – V2)/R1

Vo1/R2A - V1/R2A = V1/R1 – V2/R1

Vo1/R2A = (1/R2A + 1/R1) V1 – (1/R1) V2

Vo1/R2A = [(R1 + R2A)/( R2A R1)] V1 – [(1/R1) V2]

Vo1 = R2A [(R1 + R2A)/( R2A R1)] V1 – [(R2A/R1) V2]

Vo1 = [1 + (R2A/R1)] V1 - [(R2A/R1) V2]..................... (a)

Likewise with:

[(V1 –V2)/R1] = [(V2 – Vo2)/R2B]

(Vo2 - V2)/ R2B = (V2 - V1)/R1

Vo2/ R2B = [{(1/ R2B) + (1/R1)} V2] – [1/R1]V1

And hence, Vo2 = [1 + (R2B/R1)] V2 – (R2B/R1) V1............ (b)

Examining Eq. (a) and Eq. (b), if R2A = R2B = R2 then:

Vo1 = = [1 + (R2/R1)] V1 – (R2/R1) V2........................... (c)

Vo2 = = [1 + (R2/R1)] V2 – (R2/R1) V1............................ (d)
                                  
By using the expressions obtained for the difference amplifier considered above the total output voltage from the third phase op-amp A3 in figure above is given as:

Vo = [(R3B + R4B)/R3B] [(R4A)/(R3A + R4A)] Vo1 – (R4B/R3B)Vo2

When R3A = R3B = R3 and R4A = R4B = R4 then,

Vo = [(R3 + R4)/R3] [(R4)/(R3 + R4)] Vo1 - (R4/R3)Vo2

(R4/R3) (Vo1 - Vo2)

By combining this with Eq. (c) and Eq. (d) gives finally:

Vo = (R4/R3) [1 + (R2/R1)] (V1 – V2)......................... (e)

This exhibits that the 3 op-amp instrumentation amplifier just amplifies the difference among the two inputs V1 and V2. This makes it ideal for utilization with transducers that have a Wheatstone bridge arrangement in their construction. For such kinds of transducer, similar to the SX50 pressure transducer examined former, the output voltage is similar at both terminals whenever no input pressure is applied. This common voltage can be quite big at both terminals, amounting to half of the bridge supply voltage. Though, when the potential is similar at both terminals, the output voltage from the instrumentation amplifier is zero as needed. Whenever input pressure is applied, the potential at one terminal will rise by a small amount, whereas that at the other terminal will reduce by a small amount. The instrumentation amplifier will sense this small difference in the potentials among the two terminals and amplify this by the gain recognized by the combined 3 op-amp phases given by the expression in Eq. (e) above.

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