Hello, and welcome to the lecture for the TI Precision Lab discussing Instrumentation Amplifier Topologies, specifically two-amp IAs. In a previous lecture, we discussed and derived the one-amp IA, commonly known as a difference amplifier. While this IA topology satisfies the two main characteristics of an idealized IA, it does have its own drawbacks. This IA topology heavily relies on matched resistors, and it has low input impedance. We resolved both of these issues by introducing the three-amp IA.

In this three-amp IA circuit, we've added two resistors and two amplifiers to form a balanced input, balanced output stage. Resistors R1 through R4 are ratiometrically matched in production to achieve the desired gain, and resistors Rf1 and Rf2 are absolutely matched since their interaction with Rg determines the gain of the entire circuit. This circuit resolves the first issue of a one-amp IA, which relies on precision matched resistors, and also resolves the second issue regarding input impedance since these two input amplifiers, A1 and A2, have very high input impedance.

But even this topology has its drawbacks. While the drawbacks are typically unrelated to precision, this topology requires a complex design, three amplifiers and six passive components. This complexity may result in larger die size, higher current consumption, and higher manufacturing cost. If you have an application where any of these drawbacks cause a concern, you may consider a two-amp IA topology.

In this configuration, we have an IA comprised of two amps and four resistors. This is considerably smaller than the three-amp design, which in turn reduces IC size and manufacturing cost. This configuration satisfies the two main characteristics of an idealized IA and yields high input impedance, but it does have its own set of drawbacks. Let us analyze a circuit to prove that it satisfies the idealized model of an IA and expose its drawbacks.

Let us first derive the output of amplifier A2 using superposition. If we ground Ref, this now looks like the familiar non-inverting configuration with VO2 equal to 1 plus R4 over R3 multiplied by V1. Let us call this equation V1 star.

Now, let us grab V1. A2 now looks like an inverting amplifier with VO2 equal to negative R4 over R3 multiplied by Ref. We will call this equation Ref star. Let us now combine equations V1 star and Ref star to yield VO2. VO2 simplifies to the following equation. We will use this equation later.

Let's work our way down the circuit and derive the output of A1 using superposition. First, we will ground VO2 and find the output due to V2. Amplifier A1 looks like a non-inverting amplifier with the input equal to V2. So Vout is equal to 1 plus R2 over R1 multiplied by V2. We will call this equation V2 star.

Now, let's ground V2 and find the output due to VO2. Amplifier A1 looks like an inverting amplifier with the input equal to VO2. So Vout is equal to negative R2 over R1 multiplied by VO2. We will call this equation VO2 star.

Let's combine V2 star and VO2 star to yield the output of amplifier A1. Vout becomes the following. We previously calculated VO2 as the following equation. Substituting this into the first equation, we get Vout is equal to the following equation.

Assuming R4 matches R1 and R3 matches R2, this equation reduces to the following. If we simplify this even further, we get Vout is equal to 1 plus R2 over R1 multiplied by V2 minus V1 plus Ref, where 1 plus R2 over R1 is the differential gain Ad, and V2 minus V1 is the differential voltage Vd.

This topology yields high input impedance, but precision resistor matching still remains a concern. As with the one-amp IA topology, if we wanted to change the gain of the entire circuit, we would need to change four resistors with focus on resistor matching. In order to rectify this, we would like to control the gain of the entire circuit with just one added resistor, as with the three-amp IA topology.

Without going into too much derivation, adding resistor Rg allows us to change the gain of the entire circuit with one resistor. In many integrated designs, R1 through R4 are integrated, and Rg is left as an external addition by the user. As we covered earlier, R4 matches R1, and R3 matches R2. R1 through R4 are also absolutely matched in production since their absolute value interacts with Rg to determine the gain of the entire circuit.

These pins should look familiar to the one-amp and three-amp IA topologies. As with the other topologies, it's important to drive the reference pin carefully. When the Ref pin needs to be lifted above ground, a low impedance source, such as a buffer or voltage reference, is usually needed to drive the reference pin. Otherwise, the CMRR, the instrumentation amplifier circuit, will degrade.

Let us now analyze the circuit's ability to reject the common mode signal. By definition, a common mode voltage is the same on both inputs, and if the two input terminals are tied together to a common mode input voltage, then there is no input voltage difference, so Vd is 0 volts, and the expected output is 0 volts. But in actuality, there will be some small voltage at the output.

This is called the common mode output voltage or VOCM. VOCM is usually much less than the applied input voltage, so the ratio of the common mode output voltage to the common mode input voltage is much less than 1. This ratio is called the common mode gain or attenuation of the instrumentation amplifier or ACM. For an ideal instrumentation amplifier, ACM is equal to zero. The goal with any IA is to make this common mode output voltage error term as small as possible, ideally 0 volts.

Let us analyze this topology's ability to suppress the output error term when we apply a common mode voltage at its inputs. Let us apply one volt at V1 and V2. Let us also ground the reference voltage and assume all resistors are 1 kiloohm.

If V1 is equal to 1 volt, then V1 prime is also 1 volt, and the current flowing through R3 is equal to 1 volt divided by 1 kiloohm, which is equal to 1 milliamp. If V2 is equal to 1 volt, then V2 prime is also equal to 1 volt. Since V1 prime is equal to V2 prime, which is 1 volt, there is no current flowing through Rg, so iRg is equal to 0 amps.

The current flowing through R4 is equal to the sum of currents through R3 and Rg. In this example, that's 1 milliamp. If 1 milliamp is flowing through R4, which is 1 kiloohm, then the voltage drop across R4 is 1 volt, so VO2 is equal to 2 volts.

Since VO2 is equal to 2 volts and V2 prime is equal to 1 volt, then the current through R1 is 1 volt divided by 1 kiloohm or 1 milliamp. The current flowing through R2 is the sum of currents through Rg and R1, which is equal to 1 milliamp. The voltage drop across R2 is 1 volt, which means Vout is equal to 0 volts. The two-amp IA was able to reject the common mode voltage.

To recap, the two-amp IA topology yields the following output equation, where the differential voltage Vd is amplified by the differential gain Ad. You may notice that the gain of this circuit cannot be unity or less than 1 due to the addition of 1 in the gain term. That is one disadvantage of this design.

Another disadvantage of the two-amp IA design is the common mode input voltage range must be traded off against gain, assuming that the reference voltage is grounded. This means amplifier A2 must amplify the signal at V2 by 1 plus R4 over R3. If R4 is much greater than R3, A2 will saturate if the V2 common mode signal is too high, leaving no A2 headroom to amplify the wanted differential signal. For high gains, where R4 is much less than R3, there is correspondingly more headroom at node VO2, allowing larger common mode input voltages.

Let's see an example of what this actually looks like. We have two circuits here. Both are configured as a two-amp IA powered by plus minus 15 volt supplies but with different differential gains.

The top circuit is a low gain example where R4 is much greater than R3. The gain of this circuit is 1.1 volts per volt. We are trying to process a 1-volt input differential voltage in the presence of a 5-volt common mode voltage, and our reference voltage is grounded. If we follow the output equation, we expect to see 1.1 volts on the output of this circuit.

The bottom circuit is a high gain example where R4 is much less than R3. The gain of this circuit is 11 volts per volt. Following the same input differential voltage, VCM, and reference, we expect to see 11 volts on the output.

If we calculate the internal nodes, specifically the output of amplifier A2, we see that in the top circuit, VO2 is 49.5 volts. On a 15-volt amplifier, we would not be able to achieve this output and therefore would be railing the amplifier's output, which is a nonlinear condition. Having an inaccurate voltage level at VO2 will diminish the precision at the output of A1. As we covered in previous TIPL videos, we must aim for our designs to conform to the input and output limitations of every amplifier in the circuit, not just the input and output of the whole circuit. Now, if we look at the bottom circuit, configured in high gain, you can see all nodes are within linear range, and we can process differential voltages in the presence of larger common mode voltages.

This input range versus gain trade-off is especially important for applications where the reference voltage is grounded. If there is a reference voltage applied, then the differential voltage seen at the input of amplifier A2 will not be as pulled apart, and therefore, a lower gain may be achieved linearly. Here's an example of a low gain circuit we analyzed previously. Keeping every circuit condition the same, we will now only change the reference voltage to 4 volts.

Now, since there isn't a large differential voltage between the non-inverting input of amplifier A2 and Ref, the output of A2 is no longer saturating, and we're able to process linear voltages from the input to the output of the entire circuit. Let's now compare this low gain example to a three-amp IA with the same conditions, when Ref is grounded and not grounded.

In the top left-hand circuit, we have a two-amp IA where the VCM and Ref are 5 volts and 0 volts, respectively. The difference between Ref and the voltage seen at the non-inverting input is 4.5 volts, which is gained up by the first stage by 11 volts per volt, yielding a saturated and nonlinear output. In the top right-hand circuit, we have a two-amp IA where the VCM and Ref are 5 volts and 4 volts, respectively. The difference between Ref and the voltage seen at the non-inverting input is 0.5 volts, which is gained up by the first stage by 11 volts per volt, yielding a linear output within the supply rails.

In the bottom two circuits, we have a three-amp IA where the first stage provides all the differential gain, irrespective of the reference voltage. So the output of these two input amplifiers are within linear range of the power supplies. In summary, for a two-amp IA, the further VCM is from Ref, the more gain you need for linear operation. If you pull Ref closer to VCM, then you can achieve lower gains. When considering the topology for your application, you must take into account all system requirements.

Another limitation of a two-amp IA circuit is the difficulty of achieving high AC common mode rejection. This limitation results in poor CMRR because the signal path from V2 to Vout has the additional phase shift of amplifier A2. Assume a sinusoidal common mode voltage, VCM, at a frequency, FCM, as applied to both inputs V1 and V2. Ideally, the common mode error should be 0 volts, independent of the frequency.

If the AC common mode error is 0, amplifier A1 must see 0 instantaneous difference between the common mode voltage applied directly to V1 and the version of the common mode voltage that is amplified by amplifier A2. However, any phase shift introduced by A2 will cause the phase of VO2 to lag behind the phase of the directly applied common mode voltage at input V1. This difference in phase will result in an instantaneous difference in VO2 and V1, which will cause a frequency-dependent common mode voltage error at the circuit's output, Vout.

Let's take everything we've learned and apply it to an example with the INA126, one of our high-precision two-amp instrumentation amplifiers. Assume we're powering this device by plus minus 10 volts, we've grounded the reference voltage, our input differential voltage is 10 millivolts in the presence of a 2-volt common mode signal, and we want to amplify this differential voltage to 3 volts on the output. Before we make any connections, let's analyze this problem by following these four design steps.

First, we'll determine the gain needed given the input and output requirements. Second step is to analyze the INA126 boundary plot to make sure we can satisfy the circuit requirements. Third, we'll use the INA126 data sheet to calculate the gain resistor required. And lastly, we can build and measure our circuit with confidence.

Step one is to determine the gain required for the circuit. Gain is defined as the change in output voltage divided by the change in input voltage. In this case, that's 3 volts divided by 10 millivolts, or 300 volts per volt.

Step two is to check the boundary plot for this IA to make sure we are operating linearly. Remember, the boundary plot for an IA analyzes every internal node to represent the input and output ranges by plotting the combination of every headroom limit, including the common mode input voltage and the output voltage. This plot is available with all of our IA releases through a calculator tool called analog engineer's calculator under INA VCM vs Vout. You can select the IA part number, in our example, INA126, enter your specifications, including supply, plus minus 10 volts, gain, 300 volts per volt, reference voltage, 0 volts, common mode voltage, 2 volts, and the calculate will generate the boundary plot to guide you through linear design and operation.

In this plot, we can see that the allowable input differential voltage range is approximately negative 30 millivolts to positive 30 millivolts, and our allowable output swing is negative 9.2 volts to positive 9.2 volts. Our input differential voltage is 10 millivolts. That's within range. Our expected output voltage is 3 volts. That's also within range. The INA126 should work for our conditions.

Step three is to determine the resistor value for our circuit. Using the gain equation for the INA126 as specified in the data sheet, we need an Rg resistor value of 271 ohms. We can now build this circuit with confidence.

As with other IA topologies, the two-amp IA has its benefits and drawbacks. The two-amp IA requires less amplifiers and resistors, which in turn can reduce AC size, current consumption, and manufacturing cost. It also satisfies the need for high input impedance.

In terms of drawbacks, to summarize, the two-amp IA can only be configured in gains greater than 1 volt per volt. High gain configurations allow for higher common mode voltages, and CMRR is frequency-dependent and will degrade with increasing frequency. As with the three-amp IA, a two-amp IA must also have a common mode input voltage between the power supply rails of the amplifier. If you have an application which can accept these drawbacks in return for possibly smaller solution size, lower power consumption, and lower manufacturing cost, you can consider the two-amp IA topology.

That concludes this video. Please try the quiz to test your understanding of this video's content.