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Circuit Surgery Regular clinic by Ian Bell Measuring the frequency response of a circuit using a PC sound card, part 3: using op amps for signal conditioning L ast month, we continued looking at how to measure frequency responses. That followed from the article before, where we concentrated on the principles and theory of frequency responses of linear circuits and how to measure them. In the follow-up article last month, we described how to use the sound card (audio interface) of a PC or laptop together with free software called REW (Room EQ Wizard) for practical frequency response measurement. REW is aimed at acoustic measurements (hence the name) but can be used to analyse purely electronic circuits as well. We already satisfied our objective of measuring the frequency response of an example digital filter from the recently completed DSP series. However, we did not have space to go into details of the circuitry used to interface the DSP filter to the PC’s audio I/O. The processes that manipulate signals so that otherwise incompatible analog circuits/stages can pass signals between them is referred to as ‘signal conditioning’. Signal conditioning commonly requires amplifiers and attenuators, sometimes with variable gains, and these are often implemented using op amps. So this month, we will look at op amp amplifiers, particularly for AC signals, covering a variety of general circuit design and component selection issues, as well as the specific circuits for the frequency response measurements described last month. Before getting into the op amp basics, we will recap the context of the circuitry used to assist the sound-card-based measurement. PC Sound card Line R out L Line L in R In Device under test and signal conditioning Out Fig.1: our configuration for testing an electronic device using a computer sound card. 46 Our measurement system The basic measurement setup is shown in Fig.1. One channel of the sound card’s stereo line (or headphone) outputs is used to produce the input signal to the device under test. The output from the device under test is fed into one channel of the sound card’s line input. REW (or similar software) generates the signal used for testing and processes the resulting response to provide realtime displays of signal levels, waveforms (oscilloscope function) or the signal spectrum. The software can also plot graphs of frequency response and distortion vs frequency and other parameters by running a measurement process. The other stereo channel can be used as a timing reference via a direct outputto-input (loop-back) connection. Computer audio interfaces have a limited range of output voltage, typically close to either the commercial or professional standard audio line levels, depending on the type of sound card; in either case, it is generally in the order of 1V RMS. The audio interface can also only process AC signals centred on 0V, so unlike lab test equipment, DC voltages cannot be output or measured. As a result, the signal amplitudes and DC offsets of the sound card may not match those required by the device under test. We may need amplification/attenuation and DC level shifting at both the input and output of the device under test. Furthermore, even if the voltage levels are compatible, the sound card output may not be able to drive the circuit under test (and vice versa) due to loading effects – the output could be overloaded, or the load may cause instability in the driving circuit. Last month, we discussed in general terms the signal conditioning circuit used for the frequency response measurement From soundcard line output Buffer Out Gain 0 to 3 Device under test of the example digital filter. This structure of the circuit is shown in the block diagram, Fig.2. The circuit could also be used for a range of different devices under test, not just the DSP system, so some circuit parameters, such as gains, may need to be different. We previously discussed the specific signal levels and gains required for interfacing the digital filter to the sound card I used. To recap briefly, the first buffer AC-couples the output from the sound card and has a gain of 2.6 times to obtain 3.2V peak-to-peak (for a sinewave) at the filter’s input from the 1.22V peak-to-peak maximum output from the sound card. The second buffer has a gain of about 0.19 times to obtain 0.61V peak-to-peak from the 3.2V peak-to-peak filter output, which is amplified to 1.22V peak-to-peak by the 2× gain of the audio driver to give an overall unity gain between the sound card input and output. The DSP system under test also has some signal conditioning at its input to shift the signal to a DC level of +1.65V (half the microcontroller’s ADC reference voltage) and provide input over-voltage protection (see the August and September 2025 issues). There is also AC-coupling and a reconstruction filter on the DAC output. Op amp based amplifiers The signal conditioning circuitry was built using op amps. This month, we will discuss aspects of op amp circuit design relevant to the two buffers shown in Fig.2. Figs.3-5 show three widely used op amp circuits. Fig.3 is an inverting amplifier, with a gain of -RF ÷ RI. The inverting amplifier has a phase shift of 180°, which flips the waveform upside down, hence the name. The subscripts are F and I for feedback and input resistor, making it easier to Buffer Audio line driver GainOut 0 to 0.5 Out 2 Gain To soundcard line input Fig.2: the signal conditioning circuitry for testing the DSP filter. Practical Electronics | December | 2025 RF + Vin RI Vin Vin Vout – – V+ + – Vout + RF Vout Vin + RG Fig.3: an inverting op amp amplifier. remember the gain formula than if you designate them R1 and R2, where it is not necessarily obvious which is which. Fig.4 shows a non-inverting amplifier, which has a gain of 1 + RF ÷ RG (G is for grounded resistor) and zero phase shift. That equation is equivalent to (RF + RG ) ÷ RG. Fig.5 shows a unity gain buffer, which is a version of the non-inverting amplifier with RF = 0 and hence a gain of 1. Power supplies and bypassing The circuits in Figs.3-5 do not show the op amp’s power supply connections. This is common practice when discussing circuit configurations, as it keeps the schematics simple, but of course the op amps won’t work without suitable supply rails. Op amp circuits are often run from split supplies, ie, positive and negative voltages of the same value. For example, ±5V, ±12V or ±15V. This generally leads to simpler circuits, particularly when dealing with AC signals centred on 0V, because the circuit can handle positive and negative signal voltages without any DC level shifting. The signal conditioning circuit for the example digital filter uses split supplies. Some op amps have inputs and outputs that can operate almost at their supply rails and such an op amp with a ±5V supply would be able to handle signals of almost 10V peak-to-peak (3.5V RMS for a sinewave) without problems. However, keep in mind that many common op amps can’t do this, and have limitations on the input or output signal swing (often both). An op amp like the NE5532 (which is old, but inexpensive and performs very well) can only handle a signal of around ±3V with ±5V supply rails (2.12V RMS). If this is insufficient for your application, you need higher supply voltages (up to the maximum the op amp can handle) or a ‘rail to rail’ op amp. Some op amps have a rail-to-rail output but not input (eg, the TL971/2/4 series), which is not helpful if you are using them as a buffer but works well in circuits with a gain above unity. Others, like the TSV991/2/4 series, have inputs and outputs that can swing to both rails (rail-to-rail input/output or RRIO), so they can buffer signals that Practical Electronics | December | 2025 + Vout – Fig.4: a noninverting op amp amplifier. Fig.5: an op amp based buffer (gain=1). swing between either supply rail. The trade-off is that they usually cost more, can be noisier and usually don’t have as low signal distortion figures. Power supplies are not perfect, so there will be unwanted voltage variations over a wide range of frequencies, which may include noise from power supply circuitry and perturbations caused by varying voltage drops across the supply resistance and inductance as the circuit’s supply current changes. This is called power supply noise. Circuits such as amplifiers will ideally ignore power supply noise, so it does not affect their outputs, but of course they are not perfect. Op amp data sheets may state a power supply rejection ratio (PSRR) which indicates how good they are at this. They usually have excellent rejection at lower frequencies (most of the audio range) but at higher signal frequencies, some noise and ripple may creep through. The impact of power supply noise can be mitigated by placing capacitors across the power supply – this is called power supply bypassing. The bypass capacitors can be thought of as shorting out high-frequency noise – the effective resistance of an ideal capacitor decreases with increasing frequency, so bypass capacitors short the supplies together at high frequencies and shunt the noise past the circuit. No direct current flows into the capacitors once they have charged to the supply voltage at power-up. Bypass capacitors are also required for stability – op amps without them can act as oscillators, which is usually not what you want! So don’t forget to include them for every op amp package. Typically, the connections from the power supply to the op amp may be quite long (eg, wires inside an enclosure, or from a bench power supply to a prototype, followed by traces on a PCB or breadboard connectors). The wiring is not ideal, not a perfect conductor, so it has unwanted (parasitic) resistance and inductance. When the supply current changes abruptly, the parasitic inductance of the wiring will oppose the change, resulting in a voltage drop proportional to the rate of change. Effectively, the wiring parasitics impede the flow of charge from the + Fig.6: an op amp powered by split supplies, with suitable supply bypass capacitors. V– supply to the op amp. Capacitors across the supply closer to the op amp provide a reservoir of charge, which can respond more quickly than the power supply via the longer wiring. Like the supplies, capacitors are not perfect, so the use of just one type and value capacitor is not fully effective at supply bypassing. A relatively large capacitor (eg, a 10μF to 100μF electrolytic) can provide a good charge reservoir and deal with low frequency noise, but its own parasitics (internal inductance and resistance) make it less effective against high frequency noise. To cover this, smaller capacitors (eg, 10nF to 100nF) are also commonly used. The two types are connected in parallel across the supplies. For split supplies, the capacitors can be connected from each supply to ground, as shown in Fig.6. However, this is not always necessary, so if testing shows good performance is achieved with capacitor(s) between the supply rails, the parts count can be reduced. Commonly for split supplies, a pair of 100nF capacitors would be placed as close as possible to each op amp package, and a single pair of larger capacitors would cover a board or group of chips up to a distance of a few centimetres. Device data sheets may provide specific recommendations on supply bypassing. In a little more detail, the benefits of capacitors between V+ and V- are more charge storage (energy is CV2 and here V is doubled, so energy storage is quadrupled), lower parts count, and better bypassing of the actual supply paths within the op amp (with a split supply, the op amp package has no direct connection to ground anyway). The main advantage of the pair of capacitors, V+ to GND and V- to GND, is that almost all op amps use either V+ or V- as a common terminal for their internal amplifier (usually V-). Therefore, they are more sensitive to noise on that terminal. By bypassing it individually to GND, you effectively make that rail an AC ground, improving the op amp’s PSRR. Which approach is best really depends on the op amp and how it is being used, so if you need the best performance, you may need to experiment. 47 RF RS Vin RF + Vin RI C1 – – A W Vout VS Vin – + + B RI – Vout Vout + A Fig.7: an inverting amplifier with its signal source. RU Inverting or non-inverting? RL If we need amplification (a gain greater than 1) we can use either the non-­inverting or inverting circuit – so how do we choose between them? The circuits have the same number of components, and we can set any reasonable gain greater than one using suitable resistors, so there is no significant difference there. On the other hand, if we specifically need to invert, or avoid inverting, the signal, the choice may be straightforward. In some cases, the non-inverting amplifier will be preferred simply because it does not invert the signal, for example, where phase relationships between multiple channels or signal paths are important. The non-inverting circuit has a minimum gain of one (unity), so it cannot be used to directly implement an attenuator, or variable gain that varies from cut to boost. However, it is straightforward to implement attenuation using a potential divider. The inverting configuration can be used as an amplifier or attenuator. There are significant differences between the input impedances of the two types of amplifier, and somewhat related to this, differences in the consequences of AC coupling their inputs. We will look at this in detail shortly. There are also subtle differences in noise and distortion performance between the two configurations, which we will discuss later. When we go beyond amplifiers to other uses such as summing circuits and filters, different arguments may be relevant to whether the inverting or non-inverting configuration should be employed. In general, it depends on the application, so our discussion here will mainly highlight issues related to AC amplifiers. Virtual earth/ground The input impedance of the inverting amplifier is equal to RI. This is a consequence of the way the feedback operates in the circuit. The negative feedback used in op amp circuits ensures that the voltage difference between the op amp’s inputs is close to zero as long as the op amp is not pushed beyond its limits. The amplifier circuits are effectively control systems that ‘try’ to maintain zero voltage across the inputs. Zero volts between two points is also what you get in a short circuit, so the op amp’s inputs behave almost like they 48 Fig.9: an AC-coupled inverting amplifier. W Fig.8: a basic resistive adjustable attenuator circuit. B are shorted together – referred to as a virtual short circuit. We say ‘almost’ because the op amp is not perfect; ideally, it has infinite gain, but in practice the gain is finite, so the ‘short circuit’ is not perfect. The inverting amplifier has its non-­ inverting input connected to ground, so the inverting input behaves as if it is shorted to ground – referred to as a virtual earth or virtual ground. RI is connected from the input to the virtual earth, so the input impedance is equal to RI. This typically means that a very high input impedance cannot be obtained with the inverting configuration, because using very large resistors to set the gain has some undesirable consequences. These potential problems include: • higher thermal (Johnson) noise from larger resistor values • higher sensitivity to external interference; higher impedance nodes in the circuit are more susceptible to picking up unwanted signals • more likelihood of instability due to the effects of stray capacitance (for a given capacitance, unwanted phase shift in the feedback is larger with larger resistors) • larger DC errors due to bias currents flowing in the resistors (less important with AC-only signals, but may lead to asymmetric clipping) Source loading Fig.7 shows the inverting amplifier connected to a source voltage (vS) with output resistance RS. The source resistance and RI form a potential divider. The voltage at the amplifier’s input (vIN ) is ideally equal to vS, but is reduced to vIN = (RI × vS) ÷ (RI + RS) by the loading imposed by RI (this is the well-known potential divider equation). Alternatively, we can say the effective gain (with respect to vS) is reduced by the same factor. The non-inverting configuration has a very high input impedance as the input is connected directly to the op amp’s input. Op amps typically have an input impedance of megohms, gigaohms or even teraohms. This makes the non-inverting configuration very useful for connecting to high impedance signal sources, where the loading effect would be excessive when using an inverting circuit with typical resistor values, or where the source impedance may be variable or unknown, which would otherwise result in uncertainty or variability in the effective gain. The unity-gain buffer is used where gain is not required but signal loading is a potential problem at a system/stage input, or for driving loads that have too low an impedance for a system/stage output. Potential divider attenuator As mentioned above, the non-inverting amplifier cannot provide a gain below unity (ie, no attenuation). The circuit shown in Fig.8 has a potentiometer between two buffers, which provides a variable attenuation from 0× to 1×. If a fixed attenuation is required, a fixed resistor potential divider can be used, as shown. Resistors can be used in combination with a potentiometer to restrict the attenuation range and give finer control. The op amp circuits do not have to be unity gain buffers – the potentiometer could be driven by the output of any op amp stage, for example, a preceding amplifier or filter. An input buffer is required if the potentiometer is to be connected to a relatively low source resistance, where it will cause loading. The op amp at the output of the potentiometer can be any circuit with a high input impedance (typically one based on the non-inverting configuration). In audio design, it is common practice to AC-couple volume control potentiometers because DC on the potentiometer could (likely will) cause unpleasant crackling noises due to track irregularities as the wiper is moved, resulting in steps in the output voltage. For the frequency response measurement circuit, this is not a concern because there is not the same need to provide a ‘good listening experience’; we are using trimmers to adjust gain to what is likely to remain a fixed value for a given measurement setup. Furthermore, the buffer circuits will be processing purely AC signals due to AC coupling earlier in the signal path, which brings us to the ACcoupling of op amp amplifiers. Practical Electronics | December | 2025 C1 Vin C1 + + Vin Vout – RF RG Vout – RF Fig.10: an AC-coupled noninverting amplifier (incorrect version). AC coupling For the circuits in Figs.3-5, we can assume (in general) that the signals (vIN and vOUT) are a combination of DC and AC. The gain equations apply to DC and AC as long as the op amp’s operating range is not exceeded. If we want to process AC signals only, we typically AC-couple the input using a capacitor. This is shown in Fig.9 for the inverting amplifier. The coupling capacitor attenuates the signal at low frequencies as well as blocking DC. The reactance (effective resistance) of the capacitor (XC1) increases the effective value of the input resistor, reducing the gain magnitude to -RF ÷ √RI2 + XC12. The effective value of the input resistance is the magnitude of the complex impedance of the series combination of RI and XC1. Due to the phase-shifting properties of the capacitor, we cannot simply add RI and XC1 – the effective resistance is obtained using the root-mean-square value, as shown above. We can apply the commonly used -3dB value to define the low-frequency cutoff (fc) due to inserting the capacitor. -3dB is used because it is the point at which output power falls by half. As voltage is proportional to the square root of power, half-power corresponds to the voltage decreasing by a factor of √2 (0.7071). The gain of the inverting amplifier reduced by √2 can be written -RF ÷ √2RI. Combining this with the equation above, we get √RI2 + XC12 = √2RI, as the RF terms cancel. Squaring both sides gives RI2 + XC12 = 2RI2, so XC12 = RI2 (ie, XC1 = RI ) at the -3dB point. The reactance of a capacitor of value C at frequency f is XC = 1 ÷ (2πfC), so the cut-off frequency at which XC1 = RI is obtained by rearranging RI = 1 ÷ 2πfcC1 to give fc = 1 ÷ 2π RI C1. This is the same as the cut-off frequency of a high-pass RC filter. We could have just assumed this would be the case, but as the circuit topology is not exactly the same, it is worth analysing. Rearranging the cut-off equation to find C1 = 1 ÷ 2πfc RI allows us to choose a suitable AC-coupling capacitor once we have selected the gain-setting resistors. Non-inverting amplifiers Given that AC-coupling generally means inserting a capacitor between Practical Electronics | December | 2025 RB RG Fig.11: an AC-coupled noninverting amplifier (correct version). the signal source and circuit input, it may seem that an AC-coupled version of the inverting amplifier should be as shown in Fig.10. Unfortunately, things are not that simple, and the circuit in Fig.10 will fail to operate correctly. This is because op amps require DC bias currents at their inputs. The rule for op amps is that you must provide a DC path to both inputs, but capacitors block DC, so the circuit in Fig.10 fails to meet this requirement. For the circuit in Fig.10, the op amp will still take its bias current (IB), which for simplicity we can assume to be a constant current. This could be in either direction (in or out of the op amp) depending on the device’s internal circuitry and operating conditions. When a constant current flows into or out of a capacitor, the capacitor will charge or discharge. Thus, in the circuit in Fig.10, the coupling capacitor will charge from the bias current. When a capacitor charges (from zero charge), the DC voltage across it increases, with the polarity of the voltage determined by the current’s direction. Assuming vIN = 0 in Fig.10, the voltage at the op amp’s non-inverting input will be whatever voltage the capacitor has charged to. This will be amplified in the same way as any other input to give a DC output voltage. If we assume the capacitor voltage starts at 0V at power-on, its DC voltage will steadily increase (positively or negatively). The output voltage will follow, proportionately with the gain. Eventually, the output voltage will reach the maximum voltage the op amp can output, which for simplicity we will assume is equal to the supply voltage. At this point, the op amp will saturate and no longer function correctly as an amplifier. If an AC signal is present at the input, this will be superimposed on the DC voltage at the input and amplified. The AC part of the output will start to clip once the total voltage gets sufficiently close to the supply. Providing bias current The solution to the problem is straightforward: we connect a resistor (RB) from ground to the non-inverting input to provide a path for the bias current, as shown in Fig.11. A consequence of this is that C1 and R B form a high-pass RC filter with a -3dB cut-off fc = 1 ÷ 2πRBC1 – a similar scenario to the inverting amplifier. The -3dB point is when the resistor and capacitor impedance magnitudes are equal in both cases. Above the low-frequency cutoff, we would expect the capacitor to have a very low effective resistance, which we can simplify to assuming the capacitor is like a short circuit for AC signals at the frequencies of interest. With this assumption, we see that the input impedance of the circuit in Fig.11 is equal to RB. An advantage with the non-inverting circuit is that RB is independent of the gain setting (via RF and RG) and can be a large value (100s of kilohms to megohms). Typically, RB in the non-inverting circuit can be much larger than RI in the inverting circuit, meaning that the circuit can have a high input impedance. Consequently, much smaller capacitors can be used for the same low-­frequency cut-off. This can reduce the cost and size of the circuit and/or allow better (more linear) capacitors to be used. Alternatively, if we need a specific input resistance, we can just set RB to this value without worrying about interaction with the gain-setting. Op amp bias currents are small (nanoamperes or picoamperes), so large RB values do not prevent the op amp from obtaining sufficient bias or cause disruptive voltage drops (100pA through 100kΩ drops 10μV). However, any such voltage drop will be amplified and appear at the output. It is worth noting that the same thing happens with the gain-setting resistors, so it is the balance between the amplified bias-current drops on both inputs that affects the output. In some circuits (particularly for DC), a resistor may be added to specifically balance the drops caused by bias currents. The bias resistor will add some noise to the circuit – all resistors generate thermal or Johnson noise, which increases with both temperature and resistance. The larger the resistor, the more noise, so as always there are design trade-offs, in this case with noise and offsets vs input impedance and capacitor size. Taking some time to fail Returning to what happens if you fail to use a grounded resistor to supply bias, we can calculate the time taken for an amplifier to saturate due to the coupling capacitor charging. A fundamental equation for capacitors is Q = CV; the stored charge (Q) is equal to the capacitance times the voltage across the capacitor. Current is the flow of change, so a constant current (I) flowing for time Δt onto a 49 Fig.12: an LTspice schematic to illustrate the use of a bias resistor. capacitor will add charge ΔQ = IΔt to that stored on the capacitor (Δ [delta] means ‘change in’). This will cause a voltage change of ΔV = (I/C)Δt, or we can write the time taken for the voltage to change by ΔV as Δt = (C/I )ΔV. Consider a non-inverting amplifier (Fig.10) with a gain of 10, a bypass capacitor of 100nF, and an op amp with a 2nA bias current. If this was operating on a ±5V supply, it would need 0.5V on the capacitor to saturate (5V ÷ 10), which would occur about 25s after power-on, calculated as (100 x 10-9F ÷ 2 × 10-9A) × 0.5V. Given that in other circuits, the capacitance could be much larger and the bias currents much lower than these figures, the time for the circuit to fail could run to minutes or even hours. This could be a major problem for the unwary, as a test of the circuit for a much shorter period may not reveal the problem. Modern high-precision and bipolar low-bias op amps often have internal biasing circuits, which might seem to imply that the problem with bias currents charging the coupling capacitor would be avoided. However, external currents are not completely eliminated in these devices, so the same problem may arise, but take longer to manifest. The bias current voltage-drop balancing resistors mentioned above are often not recommended for these op amps. Simulating the problem Fig.12 shows an LTspice simulation to demonstrate the effect of not using a bias resistor. The values are as in the example calculation above (gain=10, 2nA bias current). There are two copies of a noninverting amplifier, with and without the bias resistor. The input is a 100Hz, 0.1V peak sinewave, which should produce a 1V peak output centred on 0V for the duration of the 30s simulation. The op amp is modelled using LTspice’s UniversalOpAmp5 model, which allows Fig.13: the simulation results for Fig.12. 50 you to set the input bias current for both inputs (right-click symbol to access the parameters). Fig.13 shows the results of the full 30 second simulation. The circuit with the bias resistor maintains its output correctly centred on 0V throughout the simulation, whereas the one without the resistor drifts towards the negative supply, taking about 25s as calculated above. Figs.14 & 15 show a few cycles of the signals near the start and end of the simulation. After 50ms (Fig.14), the signals are both centred on zero, although there is a slight shift down for the circuit without the resistor. At the end of the simulation (Fig.15), the output from the circuit with the bias resistor is unchanged, but the one without the resistor has drifted to a significant negative offset and the waveform is badly distorted. The output does not completely saturate, as predicted by simply assuming the bias current is an unchanging ideal constant current source. The behaviour is, as might be expected, more complex, but the time estimate is sufficiently close given that this is a failing circuit. Choice of op amp There are a huge number of op amps available, and it can be daunting trying to choose one. Device manufacturers and component suppliers provide online interactive selection tables that can be very useful if you know what ranges of parameter values are appropriate. Manufacturers often classify op amps according to application type or performance (eg, high precision, high speed, audio), which can help narrow the search. Since the signal conditioning circuits we are considering here are operating on AC signals in the audio range and need to provide good signal integrity so as not to influence the measurements too much, devices listed by manufacturers for audio applications should be a good fit. This classification should imply low noise and distortion characteristics, with sufficient bandwidth and slew rates for audio signals. We do not need very high precision DC operation or very high bandwidth into the GHz range. Device packaging should be considered. I wanted to be able to use solderless breadboard to quickly construct and modify circuits as I was experimenting with REW. Therefore, I was looking for devices in DIP packages. This mainly means older devices as more recent ones are often only available in SMD packages. This does not prevent their use with breadboards but means that you have to solder them to adaptors. Power supply requirements are another consideration. I had a board providing ±5V supply available and preferred not to Practical Electronics | December | 2025 gain circuits. Not all high-bandwidth op amps have this capability, though; check the data sheets. Noise, distortion and CMRR Fig.14: the initial simulated waveforms for Fig.12. have the supply not go too far beyond the 3.3V used by the microcontroller implementing the filter. Different op amps have different minimum and maximum supply voltages, so this needs to be checked on the data sheet (and recall the earlier discussion about rail-to-rail types). Op amps that you regularly see used in relevant applications may help guide your choice. For example, the NE5532 and NE5534 are widely used audio op amps that are available in DIP packages. However, here these devices would be operating at their minimum supply of ±5V, which may not give optimal performance. [Editor’s note – our experience is that they still perform very well with a ±5V supply as long as the signals remain below about 2.2V RMS.] The Texas Instruments page for the NE5534 suggested the OPA134 “high-performance audio op amp” as an alternative; this device can operate as low as ±2.5V and has other specifications that are suitable. Several other devices were also considered before selecting the OPA134. When using unity gain buffers or other circuits with low gain, it is important to check the minimum circuit gain the op amp can be used with. Negative feedback provides significant advantages in compensating op amp imperfections and allowing us to accurately set gain just via resistor selection (we do not need to know the gain of the op amp itself). The more negative feedback is applied, the easier it is for an unwanted phase shift (eg, due to stray capacitance on the input) to change the feedback to positive, causing instability (unwanted oscillation). Low-gain op amps have the most feedback (percentage of signal fed back) so are more likely to be unstable. The gain of op amps is deliberately reduced at high frequencies to help overcome this, but that reduces the available bandwidth in higher-gain circuits. Therefore, some op amps are not designed to be stable in low-gain circuits, as this improves performance in higher-gain applications. The NE5532 and OPA134 are unitygain stable, but the NE5534 is stable for gains of three or more. A capacitor can be connected to the compensation pins of the NE5534 to facilitate use in lower Fig.15: the simulated waveforms for Fig.12 after 30 seconds. Practical Electronics | December | 2025 The noise gain is higher than the circuit gain for inverting amplifiers, but the gains are the same for the non-inverting configuration. This means that, generally, the non-inverting configuration has lower noise. However, for AC-coupled circuits, the bias resistor is an additional source of noise in the non-inverting configuration that would not be present in a DC circuit. Op amps are differential amplifiers, so ideally they ignore signals that are the same on both inputs (common-mode signals). Of course, they are not perfect, and the measure of their ability to do this is called the common-mode rejection ratio (CMRR). Op amps also have a maximum common-mode input voltage at which they will work correctly (specified as common-mode input range). Although the inverting and non-­inverting amplifiers built from the op amps are not differential, the common-mode characteristics of the op amp do impact circuit performance, particularly for the non-­ inverting configuration. The virtual short circuit between the op amp’s inputs in the non-inverting configuration means that a common-mode signal equal to the amplifier’s input signal is applied to the op amp. This will have some influence on the output, so the effective gain will be different from that set by the designed resistor values (by a factor of 1/CMRR) and the signal distortion will also be slightly higher. As most op amps have a high CMRR (eg, 70-100dB), the gain error is small, typically smaller than resistor tolerances (the error is about 0.03% for a 70dB CMRR). For the signal-conditioning circuit, the gain error does not matter, as the gain is manually adjusted to suit the signals. The data sheet for the OPx134 (OP134, OP2134 & OP4134) discusses the fact that the input capacitance of the FET input stages used varies with the common-mode voltage, which increases distortion if the parallel combination of gain setting resistors is greater than 2kΩ, so adding a balancing resistance is recommended. If the input signal amplitude of a noninverting amplifier exceeds the maximum common-mode input voltage, problems that may occur including distortion, phase reversal (the output flips polarity), saturation, a significant increase in input current and even damage to the device. The common-mode input range for the OPx134 is to within 2.5V of the supply rails, so the signal amplitude should be less than 2.5V peak on ±5V supplies. For the digital filter setup, the largest 51 Fig.16: an LTspice schematic of the buffer 1 circuit from Fig. 2 (with no supply bypass capacitors). Fig.17: the buffer 2 circuit from Fig.2, including the supplies and signal source. signal is 1.65V peak, so well within the required range. To handle larger signals, a higher supply voltage can be used, or a rail-to-rail op amp. Like most modern op amps, the OPx134 has protection against phase reversal, even for several volts beyond the specified common mode range. Buffer circuits and simulation LTspice simulation schematics for the buffer 1 and buffer 2 circuits from Fig.2 are shown in Figs.16 & 17. This is one simulation file in which the output of buffer 1 is connected to buffer 2 (there is no model of the device under test, as this is just a simulation of the buffers). Fig.17 also includes the power supplies, signal source and simulation commands. The supply bypass capacitors (Fig.6) included in the practical circuit (100μF electrolytic for both supplies on the breadboard and pairs of 100nF ceramics for each chip) are not included because the simulation supplies are ideal voltage sources. The model of the OPx134 op amp was downloaded from Texas Instruments and added to LTspice. Buffer 1 includes a non-inverting amplifier (like Fig.11) with a gain of three times (+9.5 dB), a bias resistor of 100kΩ and a 2.2μF coupling capacitor, giving a low-frequency cutoff of 0.72Hz, well below the 20Hz minimum frequency we were measuring for the digital filter. As an aside, electrolytic capacitors work well for AC-coupling as long as the -3dB point is kept sufficiently low (usually no more than a few Hz). They are somewhat nonlinear, but this arrangement keeps the distortion due to non-linearities outside of the audible range. The amplifier is followed by a variable attenuator (like Fig.8). The practical circuit uses a trimmer potentiometer, but the simulation has fixed resistors for simplicity, with values selected for an overall gain of 2.6× (+8.3 dB), as discussed last month and recapped above. A 100kΩ trimmer could have been used in place of R1 to save an op amp, but for experimenting with these circuits, it was helpful to have one operation per op amp, and I didn’t have a suitable trimpot to hand! A capacitor added across the feedback resistor reduces the gain to unity (1) at high frequencies. The -3dB point (buffer 1 gain 5.3dB) is at around 255kHz. Feedback capacitors are commonly used to improve amplifier stability. The gain reduction will help if there are unwanted high frequencies in the input. More usually, an RC low-pass filter and/ or ferrite bead would be included in the input circuit to address this if needed. Buffer 2 is simply another attenuator, again with fixed resistors to model the trimmer and set the attenuation as required (to 0.195 or -14.2dB). This gives an overall gain for the two buffers of 2.6 × 0.195 = 0.507 (8.3 – 14.3 = -5.9dB). With the gain of the line driver (not included here) being about 1.97× (+5.9dB), the overall gain for the circuit in Fig.2 is unity (0dB), as required. 10kΩ resistor R4 provides finer control of the gain in the 0-0.5× range than just using a trimmer. The input is from buffer 1, so an input op amp (as shown in Fig.8) is not needed. The output op amp is needed, as the line driver has a low input impedance. The simulation results (AC Analysis) in Fig.18 confirm that the circuits have the required gain values and are flat over the 20Hz to 20kHz range required to measure the digital filter frequency response. When the key interest in a simulation is AC analysis to confirm gain and frequency response, it is worth running a transient simulation to check that there are not unexpected problems with the waveforms. It is not impossible to make mistakes with the schematic drawing or design that will not be obvious in an AC analysis. The schematic in Fig.17 can be configured for a sinewave simulation. Next month Fig.18: the results of simulating the circuits in Figs.16 & 17. Next month, we will continue to look at aspects of the circuit design theory and practice for the signal conditioning circuitry used with the frequency response measurement of the example PE digital filter. 52 Practical Electronics | December | 2025