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Circuit Surgery Transformers – Part 3 Regular clinic by Ian Bell 𝑣𝑣! = 𝑖𝑖! = 𝑁𝑁! 𝑣𝑣 𝑁𝑁" # 𝑁𝑁# 𝑖𝑖 𝑁𝑁! # Transformers and LTspice – Part 3 T his month, we continue looking at the basics of transformers and some aspects of simulating transformer circuits in LTspice. The first article (June 2021) covered some general transformer concepts, the second (July 2021) looked at power supplies and this Transformers – Part 3 month we will conclude by considering some signal-processing applications. Key aspects of the use of transformers in signal processing are impedance matching and the use of balanced signals. We will discuss some basics related to matching and balanced signals as well as looking at the use of transformers in these contexts. Recap: turns ratio – Part 3 The relationship between the primary and secondary voltages and current for an ideal transform is determined by the turns ratio. For a primary voltage and current (vp, ip) applied to a winding with NP turns, and a secondary with NS turns the secondary voltage and current will be: 𝑣𝑣! = 𝑖𝑖! = 𝑁𝑁! 𝑣𝑣 𝑁𝑁" # 𝑁𝑁# 𝑖𝑖 𝑁𝑁! # An ideal transformer is 100% efficient so the input power will equal the output, although real transformers 𝐿𝐿# are 𝑣𝑣of course 𝑣𝑣 not 100% efficient. ! =% 𝐿𝐿! # comprise a set of coils, Transformers and coils on their own are inductors. Coil inductance (L) is related to the number of turns squared (N2), but the specific 𝑁𝑁#& relationship depends on their type, 𝑅𝑅$% = ( & ) 𝑅𝑅$ structure𝑁𝑁and dimensions. In general, we ! can write L = kN2 so N = (L/k), where k is a constant. For an ideal transformer we can assume k is the same for the 𝑍𝑍$ secondary windings, so: primary and 𝑉𝑉 = 𝑉𝑉 $ (𝑍𝑍' + 𝑍𝑍$ ) ' Simulation files Most, but not every month, LTSpice is used to support descriptions and analysis in Circuit Surgery. The examples and files are available for download from the PE website. 58 𝑣𝑣! = % 𝐿𝐿# 𝑣𝑣 𝐿𝐿! # The turns ratio is equal to the square root of the inductance ratio of the 𝑁𝑁#&(considered as individual windings 𝑅𝑅$% = ( & ) 𝑅𝑅$ 𝑁𝑁! This inductance ratio is inductors). important for setting up transformers in SPICE𝑁𝑁simulations. 𝑣𝑣! = ! 𝑣𝑣# 𝑁𝑁 𝑍𝑍"$ impedance and Reflected 𝑉𝑉$ = 𝑉𝑉 (𝑍𝑍' + 𝑍𝑍$ ) 'model transformer 𝑁𝑁# 𝑖𝑖! = 𝑖𝑖 As mentioned in the first article, 𝑁𝑁! # transformers reflect impedances from one winding to another (Fig.1). For example, if we connect a load resistor (RL) across the secondary then a circuit 𝐿𝐿# driving will see an effective % primary 𝑣𝑣! = the 𝑣𝑣# ! resistance𝐿𝐿(R’ L). The reflected resistance is determined by the square of the turns ratio, so in this case: 𝑁𝑁#& 𝑅𝑅$% = ( & ) 𝑅𝑅$ 𝑁𝑁! L R L Fig.1. Reflected impedance. with a load resistor on the secondary will behave like a resistor, but real transformers have more characteristics than just DC winding resistance. If the secondary is open circuit, then the primary of a real transformer appears (can be modelled) as an inductor (LM) in parallel with the (idealised) primary winding and in series with the primary winding resistance (see Fig.2) – so it does not behave as an open circuit on the primary side. This inductance is called the ‘magnetising inductance’. In non-opencircuit situations its impedance may be sufficiently large to have little effect on the total impedance seen at the primary (if it is in parallel with a much smaller R’ L). Having mentioned magnetising inductance it is worth pointing out that another non-ideal inductance used to model real transformers is the leakage inductance, which is in series with both windings (LLS and LLP) – this accounts for the less-than-perfect flux coupling between the two windings. For an ideal transformer, in this example, the resistance ‘seen’ at the primary is 𝑍𝑍$ just transformers have 𝑉𝑉$ =R’ L , but real 𝑉𝑉' ' + 𝑍𝑍$ ) resistance (also called some (𝑍𝑍 winding ‘DC resistance’ to distinguish it from effects such as leakage inductance). The secondary DC winding resistance (RWS) Matching will appear in series with RL and be The fact that a transformer can change reflected into the primary in the same the effective impedance of a load is a way. The primary DC winding resistance useful property which can be exploited (R WP) will be in series with the total in situations where the values of source and load do not provide workable reflected resistance. If the secondary is or optimal circuit performance, but shorted (so R’L = 0) then the primary will are diffi cult or impossible to change be seen as just the reflected secondary directly. We will look briefly at the plus primary winding resistance. For relationships between source and load example, for a 1:1 transformer with both to help understand the role transformers windings having 100Ω DC resistance, the primary will look like a 200Ω LLP Id ea l tra nsf orm er LLS R W P R W S resistor with the secondary shorted. Despite LM being constructed using coils, if the only imperfection is DC resistance, a t r a n s f o r m e r Fig.2. Simplified electrical model of a transformer. Practical Electronics | August | 2021 – Part 3 𝑣𝑣! = 𝑁𝑁! 𝑣𝑣 𝑁𝑁" # properties) of the cable into a single Source Loa d / recei ve r component model may not provide 𝑁𝑁# N: 1 𝑖𝑖! = 𝑖𝑖 an accurate picture of how wiring 𝑁𝑁! # ZS ZS behaves when a signal is applied. VS VS ZL VL ZL 2 Zs / N2 N ZL It all depends on the timescales which apply to the situation we are working with. If our signal 𝐿𝐿# cycles times, pulse durations, or 𝑣𝑣! = % 𝑣𝑣# 𝐿𝐿! circuit timescales in general, are Fig.5. Transformer used to match source to load. Fig.3. Source and load connected together. less than, or comparable with the capable of driving, rather than exactly time taken for the signal to travel down the can play in such situations. matching impedance, as is required for wire then the effect of signal propagation Consider a source with impedance ZS 𝑁𝑁#& to a load of impedance ZL, transmission lines. Specific examples of time means we have to consider the connected % 𝑅𝑅$ = ( & ) 𝑅𝑅$ using transformers to modify effective interconnection as a transmission line on as shown Fig.3. The two impedances 𝑁𝑁in ! impedance include microphone output which the signal propagates as a wave. form a potential divider. Thus, the voltage transformers, PA system loudspeaker Transmission lines (Fig.4) have a across the load is given by: transformers and 300Ω-to-75Ω coaxial characteristic impedance (Z0) which is 𝑍𝑍$ matching transformers. The latter requires related to the inductance and capacitance 𝑉𝑉$ = 𝑉𝑉 (𝑍𝑍' + 𝑍𝑍$ ) ' a 2:1 turns ratio to achieve the 300:75 = per unit length of the connection. If a 4:1 impedance ratio. transmission line is connected to or from a source or load whose impedance is If ZS = ZL we say that the source and different from Z0 then wave reflections load are ‘matched’; under this situation Balanced and differential signals maximum power is transferred from source occur. These reflections damage the Returning to Fig.3, it shows a signal to load (in order to prove this, you have to integrity of the signal and must be avoided connection between two circuits, devices, use calculus). This is sometimes, but not – the output, line and input impedances or systems where the signal is carried on a always what is required – often, however, must be matched. As a rule of thumb, we single wire plus a ground connection. This we want to maximise the voltage at the have to take account of transmission lines is a single ended, unbalanced connection. load rather than the power transfer. If we effects when the length of a connection is The term ‘unbalanced’ refers to the fact want VL to be as large as possible then more than about one tenth the wavelength that there are unequal impedances with of the signal. The wavelength is given respect to ground at both the input and ZL must be much larger than ZS (we are by Fvc/f, where f is the frequency, Fv is output – obviously, the ground wire has assuming ZS is fixed). If ZL is very much very low, ideally zero impedance with larger than ZS then the load voltage is the velocity factor of the transmission respect to ground, whereas for the output line and c is the speed of light. For a effectively equal to the source voltage. and input the impedances to ground are connection with velocity factor of 0.5 This situation, with high input / load related to ZS and ZL. this is 1.5km at 10kHz, 15m at 1MHz impedance and low output impedance and just 15mm at 1GHz. is sometimes called ‘voltage matching’. Ground connections are not perfect conductors, so any unwanted voltage difference between the grounds at each Transmission lines Transformer for matching end (noise voltage VN in Fig.6) will affect The circuit in Fig.3 shows two systems In summary, we may need to to: interconnected by two wires – this the voltage input at the receiving system / n Match source impedance to a load represents a generic situation with a signal load. With the single connection in Fig.3 for maximum power transfer wire and return path where typically (modelled as a single copy of Fig.6) it is n Increase the effective impedance the return path is the earth or ground not possible to remove this noise. The of a load or input where the source connection in the system. The situation solution is to carry the signal on two wires impedance is relatively high could apply to a wide variety of situations with both having equal impedances with n Match inputs and outputs to including twisted pairs or coaxial cable, respect to ground at all points (modelled transmission lines or to a single trace and ground plane on as two parallel copies of Fig.6 with the n Connect transmission lines of a printed circuit board. As drawn, the same ZS and ZL). Because the impedances different characteristics to together. interconnections imply perfect simple are equal on the two wires the overall conductors, but this is not the case in A range of techniques are applicable, but connection is referred to as ‘balanced’. a real circuit where the wire will have in some situations, transformer-reflected With a balanced connection we some resistance and inductance and impedance can provide the solution. If we effectively have two parallel copies there will be capacitance, and possibly have a transformer with an N:1 ratio, as of the circuit in Fig.6. The ground is insulation ‘leakage’ resistance between shown in Fig.5, then the source sees the common to both, so the same VN will them, particularly if they conductors are load impedance as N2ZL. The reflection affect both connections – we refer to close together. this as ‘common-mode noise’. Looking works both ways – we can also say that At low frequencies we could model such at Fig.6 and taking a reference point the load sees a source of impedance ZS/ imperfections using single components, at the input ground, considering just N2 with the transformer in place. but lumping the capacitance (and other For example, if we had a high impedance source, with Source Input RS = 100kΩ, and needed to drive a load of 1.5kΩ then Syst em or Syst em or ZS ci rcui t 1 ci rcui t 2 an 8:1 transformer would ( so urce) ( l oa d ) VS VL ZL T ra nsm i si on l i ne Z0 2 reflect the load as 1.5 × 8 VN = 96kΩ, providing a much Fig.4. Transmission line model of a connection between better match. This may two systems – if transmission line effects are presented be more about providing Fig.6. Connection with ground noise. a load that the source is then matching must be used. Source Loa d / recei ve r Practical Electronics | August | 2021 59 Fig.7. Differential signal – the voltages on the two individual wires V1 and V2 are shown in the upper plot. The actual signal is the difference between V1 and V2, as shown in the lower plot. VN (assume VS = 0): ZS and ZL form a potential divider with respect to VN, which results in some noise voltage at the input. If the impedances are the same in both copies then the noise voltage at both inputs will be the same. If the received signal or load voltage is taken as the difference between the two signals then the noise voltages at the two inputs will cancel out. Thus, a balanced connection can significantly reduce noise. If the impedances are perfectly balanced, and the difference between the two signals is taken perfectly, then the noise will be removed completely. Of course, this perfection is not possible in a real system, so the degree to which noise is removed is expressed as the common-mode rejection ratio (CMRR). Interconnection options There are two commonly used options for how the two wires in a balanced connection are used. One is to carry the Sh i el d O utput ZS U nb a l a nced si ng l e- end ed O utput ZS/ 2 Sh i el d Input ZL Transformer examples D i f f erenti a l i nput a m pl i f i er V1 V1 – V2 V2 ZS/ 2 B a l a nced si ng l e- end ed Input ZC M b oth i nputs D i f f erenti a l i nput a m pl i f i er Sh i el d V1 ZS/ 2 b oth outputs B a l a nced d i f f erenti a l Input ZC M b oth i nputs Fig.8. Balanced and unbalanced connections. Fig.9. Using a transformer to create a balanced connection. 60 V2 actual signal on just one of the wires, with the other being 0V – equivalent to ground in (ideal) voltage terms but not in terms of impedances. The other approach is to use a differential signal where the two wires carry equal and opposite voltages. Fig.7 shows a differential signal which is a 1kHz sinewave with a peak voltage of 2V (4V peak-to-peak). The two individual voltages (V1 and V2 on the two wires that carry the signal) are shown on the upper plot – these are equal and opposite and have peak voltages of 1V. Because they are opposite, the peak difference between them is 2V, which is the amplitude of the differential signal. The signal itself is shown on the lower plot – this is V1 – V2. It common for the original and final signals to be unbalanced, but balanced connections are used in situations where relatively long wires are required. Typically, these connections are also physically constructed to help reduce noise (eg, use of twisted pairs and shields). Fig.8 shows the three connection scenarios just discussed – unbalanced single-ended, balanced single-ended, and balanced differential. Fig.8 shows balanced signals being handled by differential amplifiers (differential input or output as required). This approach is commonly used, but transformers can also be used to convert between balanced and unbalanced signals, with a basic example shown in Fig.9. Transformers used for this purpose are often referred to as ‘baluns’. V1 – V2 Fig.10 shows an LTspice circuit to illustrate how a transformer converting a balanced (bal1, bal2) to unbalanced (unbal) signal removes common-mode noise. The balanced 250mV 1kHz signal is generated by V1, which has low output resistance of 1Ω each side. The ideal transformer is inherently symmetrical so there is no imbalance here. The common-mode noise takes the form of a 20mV 50kHz signal generated by V2. Using a centre-tapped transformer allows this to be added equally to the two signals, as would happen with ground noise in a real system – the noise can be seen in the upper plot plane in the simulation results in Fig.11. The lower plot plane shows the output signal on the secondary of the transformer – the 1kHz signal is present without any noise. The two aspects of the use of signal transformers discussed so far can be combined – that is a transformer can be used to simultaneously provide balanced/unbalanced signal conversion and impedance conversion/matching. However, the transformer may not always Practical Electronics | August | 2021 be used to (or be able to) provide full matching, but it must be taken into account. For example, if a 50Ω line (eg, coax cable) is connected to a transformer input the reflected impedance of what is connected to the secondary must be taken into account when terminating the line. Fig.12 shows a 50Ω line driving 500Ω via a 1:1 transformer. The reflected 50Ω in parallel with the 55.6Ω resistor provides a combined input impedance of 50Ω (1/55.5 + 1/500 = 1/50) to correctly terminate (match input impedance to) the line. The discussion so far can apply to a variety of circuits. Transformers are used in both audio and radio frequency circuits. Balanced/unbalanced conversion is commonly used on the inputs of high-frequency analogueto-digital converters (ADCs) with differential inputs and the outputs of digital-to-analogue converters (DACs) and direct digital synthesis (DDS) ICs. As is usually the case, design of high-performance circuit is limited by the non-ideal characteristics of components. The previous discussion on the transformer model in Fig.2 gave some hint of this. This model can be redrawn in symmetrical form (sharing the series resistance and inductance between two connections of the primary and/or secondary) to accommodate use with balanced circuits. However, Fig.2 is far from the whole story. For example, there is capacitance across the windings and between the secondary and primary. This, together with the inductance and resistance leads to complex frequencydependent behaviour which limits the bandwidth over which a transformer can be used and may require additional components to be connected to achieve the required characteristics from the circuit as a whole (eg, specific resistors across the transformer). Phantom power Fig.10. LTspice schematic for balanced-to-unbalanced signal conversion. Fig.11. Simulation results for the circuit in Fig.10. appears as a common-mode signal on the balanced line and can be obtained from the centre-tapped secondary of the transformer. Given that the balanced output here is a differential signal, the signal voltage at the centre tap is zero and so the preamplifier power supply, taken from the centre tap only sees the DC power – there is no signal fed to the amplifier supply. The shield is used as the return path for the DC power to the preamplifier. Finally, a specific and well-known use of transformers in audio is as part of the phantom power circuit used with condenser microphones with builtin preamplifiers. The circuit enables power to be delivered to the preamplifier over the same wires used to convey the signal. T 1 M i croph one A simplified phantom power circuit is shown in Fig.13. The preamplifier unbalanced output is converted to a balanced signal by the transformer and conveyed to P rea m pl i f i er the audio system in this form via the microphone cable, which has two signal conductors plus a shield. The power (typically 48V) is connected to both signal wires via a couple of resistors – thus it Fig.13. Phantom power circuit. Practical Electronics | August | 2021 0Ω ine . Ω 00Ω Fig.12. Example line termination taking reflected impedance into account. + 4 8 V Sh i el d P h a ntom pow er sw i tch R 1 . kΩ T o b a l a nced i nput R 2 . kΩ 61