AN-912 APPLICATION NOTE One Technology Way • P.O. Box 9106 • Norwood, MA 02062-9106, U.S.A. • Tel: 781.329.4700 • Fax: 781.461.3113 • www.analog.com Driving a Center-Tapped Transformer with a Balanced Current-Output DAC by Ken Gentile Note that Figure 1 assumes that the DAC is of the current sourcing variety. In the case of a current sinking DAC, the direction of IA and IB is reversed. Also, a connection to VSUPPLY should replace the ground connections at the transformer center tap and the RO resistors. The use of a center-tapped transformer as the output interface for a balanced current-output DAC offers several benefits. First, transformer coupling offers dc isolation between the DAC output and the final load. It can also aid in the rejection of common-mode signals present at the DAC output. Furthermore, transformer coupling can mitigate the even harmonics that result from an imbalance between the DAC outputs. Finally, all transformers have a limited bandwidth, which can be used to advantage for suppressing the Nyquist images that typically appear in a DAC output spectrum. In this application note, the current flowing through the normal and complementary pins is referred to as IA and IB, respectively. The maximum current that the DAC can deliver is denoted as IMAX and represents the upper limit for both IA and IB. The exact value of IA (or IB) depends on the digital code present at the DAC input. The behavior of IA and IB is such that when the digital code is zero, then IA = 0 and IB = IMAX. Conversely, when the digital code is full scale, then IA = IMAX and IB = 0. For intermediate digital codes, the two output currents are between zero and IMAX, but are balanced such that IA + IB = IMAX at all times. Thus, IA and IB can be expressed as The goal of this application note is two-fold. The first goal is to provide an explanation of the functionality of a balanced output in the context of a current-output DAC. The second goal is to provide formulas that relate the transformer turns ratio (N), the transformer load (RL), the DAC load resistors (RO), and the maximum DAC output current (IMAX). I A = αI MAX BALANCED CURRENT-OUTPUT DAC I B = (1 − α )I MAX Balanced current-output DACs come in two varieties: those with current source outputs and those with current sink outputs. DACs with current source outputs always inject current into the external load, while DACs with current sink outputs always draw current from the external load. In both cases, the DAC output consists of two pins: a normal pin and a complementary pin. The arrows that indicate direction of current flow in the diagrams that follow assume conventional current flow (that is, current flows from a positive potential toward a negative potential). where α is the fractional digital code value, that is, the input digital code value to the DAC divided by the full-scale code value. For example, given a 10-bit DAC with an input code of 200 and an IMAX value of 10 mA, then α = 200/1023 (where 1023 is the full-scale code value given by 210 − 1). This yields IA ≈ 1.955 mA and IB ≈ 8.045 mA. Also, notice that IB can be expressed in terms of IA as IB = IMAX − IA. RO VSUPPLY DIGITAL CODE (1) 1:N IA RL DAC RO 06737-001 IB Figure 1. A Balanced Current-Output DAC with Transformer Coupling Rev. 0 | Page 1 of 12 AN-912 TABLE OF CONTENTS Balanced Current-Output DAC...................................................... 1 Reduction of Even Harmonics.........................................................6 DC Analysis....................................................................................... 3 Conclusion..........................................................................................6 AC Analysis ....................................................................................... 3 Appendix A ........................................................................................7 Impedance Matching........................................................................ 5 Appendix B.........................................................................................8 Example Calculations....................................................................... 6 Rev. 0 | Page 2 of 12 AN-912 DC ANALYSIS With an understanding of the operation of a balanced currentoutput DAC, a dc analysis of a center-tapped transformer coupled to the DAC output can now be examined. Figure 1 simplifies to the dc equivalent circuit shown in Figure 2, by replacing the DAC with two current sources (one for the normal output and the other for the complementary output). The magnitude of the current delivered by each source is code dependent, as indicated by Equation 1. The current sources have arrows that indicate the direction of current flow. It is assumed that the DAC outputs are of the current source variety. The arrows would be reversed for the current sink variety. In Figure 2, the center-tap connection is redrawn to clearly show that the DAC output circuits are independent current loops (note that the transformer polarity dots have been reoriented to maintain functional equality with Figure 1). IA αIMAX Combining this result with Equation 1, the magnetic flux in the transformer core can be expressed as Φ = kNIMAX(2α − 1). The importance of this result is that for the special case of α = ½ (that is, the middle DAC code), the magnetic flux in the core is 0, whereas any other DAC code results in a build-up of static magnetic flux in the core. The significance of this fact is made apparent in the AC Analysis section that follows. AC ANALYSIS RO IB RO CENTER-TAP 06737-002 (1-α)IMAX dots indicates that the magnetic flux produced by IA is opposed to the magnetic flux produced by IB. The orientation of the dots implies that Φ = kN(IA − IB), instead of kN(IA + IB) as previously stated, for the configuration shown in Figure 1 and Figure 2. Therefore, for the particular center-tapped configuration shown in Figure 1 and Figure 2, the net magnetic flux is proportional to the difference between IA and IB rather than the sum. This is a consequence of the physical connection between the complementary current source and the lower primary winding. Figure 2. DC Equivalent Model Typically, the resistance of the transformer windings is much less than the resistance of the DAC termination resistors (RO). In most applications, this low winding resistance implies that the vast majority of the dc current associated with IA (or IB) flows through the transformer windings instead of through the termination resistors. Thus, the dc power rating for the DAC termination resistors is practically nil. In general, consider a simple magnetic circuit consisting of a single winding with an arbitrary number of turns of wire (N) wrapped on the winding and a static current (I) flowing through the wire. The current flowing in the wire creates a magnetic flux (Φ) concentrated within the winding core that is proportional to the product of the number of wire turns and the current flowing through the wire (that is, Φ = kNI, where k is the constant of proportionality). In the case of the tapped transformer, which has two primary windings, the magnetic flux in the core is the sum of the contributions of each winding (that is, Φ = k(NAIA + NBIB). Given that this analysis is based on a center-tapped transformer, the number of turns in each primary winding is the same (that is, NA = NB), which means that the magnetic flux can be expressed as Φ = kN(IA + IB). Thus, the total magnetic flux in the transformer core is proportional to the sum of the primary currents. For ac analysis, consider the specific case in which a DAC generates a sinusoidal output signal. In such a case, a time series of digital codes drives the DAC and produces an output current that varies in sinusoidal fashion. The range of the digital input codes is split such that the lower half of the codes (0 to ½ full scale) generates the lower half of the sinusoid and the upper half of the codes (½ full scale to full scale) generates the upper half of the sinusoid. The average value of the DAC-generated sinusoid, therefore, is ½ full scale. The peak amplitude of the sinusoid is also ½ full scale, since this is the amount by which the signal can swing from the midpoint to either zero or full scale. The sinusoidal current waveform at the normal output of the DAC can be expressed as I A = 12 I MAX + 12 I MAX sin(θ) (2) where θ represents the instantaneous phase of the sinusoid. Similarly, because of the relationship between IA and IB, the current waveform at the complementary output of the DAC can be expressed as I B = 12 I MAX − 12 I MAX sin(θ) (3) Inspection of Equation 2 and Equation 3 indicates that IA and IB are both centered at ½IMAX. That is, ½IMAX is the dc term of the sinusoidal waveform. Furthermore, when the magnitude of sine function increases, then IA increases, whereas IB decreases equally. Notice, too, that the sum of the normal and complementary output currents is always IMAX (that is, IA + IB = IMAX, as mentioned previously in the Balanced Current-Output DAC section). Such is the nature of a balanced output signal. Note that Equation 2 and Equation 3 ignore the quantized nature of the sinusoidal DAC output current. It is important to note in Figure 2 that IA flows into the primary winding that is marked with a dot, while IB flows into the primary winding that is opposite the dot. The placement of the Rev. 0 | Page 3 of 12 AN-912 I A = 12 I MAX + 12 I MAX cos(θ) and I B = 12 I MAX − 12 I MAX cos(θ) Given the special case of θ = 0, the sine case yields IA = ½IMAX and IB = ½IMAX, whereas the cosine case yields IA = IMAX and IB = 0. In the DC Analysis section, it was shown that Φ = kN(IA − IB). Thus, if the digital generator stalls at θ = 0, the transformer core carries no magnetic flux for the sine case and kNIMAX for the cosine case. The implication is that if the digital generator is stalled at θ = 0, and is then switched from sine to cosine (or vice versa), the magnetic flux in the core jumps from 0 to kNIMAX (or vice versa). This results in a voltage spike across all of the transformer windings due to the nearly infinite rate of change of flux in the transformer core. The previous paragraphs explore the transient behavior of the transformer when switching between sine and cosine waveforms. To explore the steady state behavior of the transformer in the context of ac analysis, it is necessary to understand how a transformer behaves under the stimulus of a sinusoidal signal. This is covered in the appendices. Appendix A describes the basic ac behavior of an ideal transformer, while Appendix B builds on Appendix A to show the ac operation of a tapped transformer. Figure 3 is the ac equivalent model assuming an ideal, centertapped transformer. Also given in Equation 4 to Equation 7 is a list of the pertinent equations that relate the various circuit parameters. Both the diagram and the equations are a result of the concepts described in Appendix B. 1:N VNORM 1/2 N (IMAX√2)/4 ZCOMP VCOMP ZS VS RL 1/2 The actual voltage that appears across each primary winding is referred to as vA for the upper primary winding and vB for the lower. The value of vA and vB can be derived from vS and the associated turns ratio between the secondary and each primary winding. Thus, vA and vB can be expressed as ⎛ 1 ⎞ ⎛ 2 I MAX v A = v B = v S ⎜ 2 ⎟ = ⎜⎜ 4 ⎝N ⎠ ⎝ RO RL 2RL + 4 RO N 2 Z S = 2 N 2 RO VA ⎛ 2 I MAX v S = ⎜⎜ 2 ⎝ ⎞ ⎞⎛ RO RL ⎟ ⎟⎜ ⎟⎜ R + 2R N 2 ⎟ O ⎠⎝ L ⎠ ⎞⎛ NRO R L ⎟⎜ ⎟⎜ R + 2R N 2 O ⎠⎝ L ⎞ ⎟ ⎟ ⎠ RO RL DAC VB RO Figure 4. Balanced Current-Output DAC with Isolation Switch If the switch is open (see Figure 5), there is effectively no change from an impedance point of view, because the current source internal to the DAC exhibits a very high impedance (ideally infinite). Thus, the complementary output drives the same load regardless of the state of the switch. The voltage at the complementary output (vCOMP) is that given by Equation 6. However, the upper and lower primary windings are mutually coupled with a turns ratio of 1:1. This causes a voltage of the same magnitude to also appear at the upper primary winding (as shown in Figure 5). VCOMP RO RL (4) DAC RO (5) ⎛ 2 I MAX vNORM = vCOMP = ⎜⎜ 8 ⎝ ⎞ ⎟ ⎟ ⎠ Note that vA and vB are twice as large as vNORM and vCOMP. What is the reason for the discrepancy? The answer lies in the fact that the two primary windings interact with each other. Consider Figure 4 where a switch has been added to provide a means to disconnect the normal DAC output pin from the circuit. Figure 3. AC Equivalent Model Using a Center-Tapped Transformer Z NORM = ZCOMP = ⎞⎛ RO R L ⎟⎜ ⎟⎜ R + 2R N 2 O ⎠⎝ L Figure 5. Isolated DAC output (6) (7) VCOMP 06737-005 ZNORM 06737-003 (IMAX√2)/4 Equation 4 through Equation 7 can be used to predict the impedance seen by each DAC output (ZNORM and ZCOMP), the voltage generated by each of the DAC current sources (vNORM and vCOMP), the impedance presented at the transformer secondary (ZS), and the voltage across the secondary (vS). It is important for the reader to understand that vNORM and vCOMP do not represent the voltage that appears across each primary winding, but rather the voltage produced by each current source (IA or IB) as it flows through the reflected impedance of the associated primary winding (ZNORM or ZCOMP). 06737-004 This result has interesting consequences when the DAC is driven by a digital sinusoidal generator like a direct digital synthesizer (DDS), for instance, that can be programmed to deliver either a sine signal or a cosine signal. When a sine signal is generated, Equation 2 and Equation 3 apply directly. When a cosine signal is generated, Equation 2 and Equation 3 become, respectively, The importance of this fact cannot be overstressed. With the normal output of the DAC completely disconnected, there is still a voltage present (vCOMP) across the upper DAC termination resistor (RO). Its presence is due to the mutual coupling of the Rev. 0 | Page 4 of 12 AN-912 Also, Equation 8 can be rewritten for the special case of ZS = RL as two primary windings and the voltage produced by the complementary DAC output driving its associated load (ZCOMP). vA With the switch closed as in Figure 4, the current generated by the normal DAC output produces a voltage across its equivalent load (ZNORM). The magnitude of this voltage is vNORM and is the same as vCOMP. By superposition, this signal sums with the signal produced by the complementary output, that is vA = vNORM + vCOMP. But vNORM = vCOMP, so vA = 2vNORM, which is why vA is twice as large as vNORM. Likewise, vB is twice as large as vCOMP. PL = (8) PL In many applications, it is desirable that ZS be equal to RL. This is especially true when a reconstruction filter is inserted between the secondary and the load, as shown in Figure 6. RO 1:N ZS RL FILTER RL 06737-006 DAC RO Generally, the filter is designed to accommodate equal source and load impedances, which implies that ZS = RL. Equation 32 in Appendix B shows that ZS = 2N2RO. If it is desired that ZS = RL, then RL can be substituted for ZS. Solving for RO yields (9) Z S = RL = ZCOMP = Z S = RL RL 8N 2 ZS = RL v NORM vS Z S = RL = Z S = RL 2 I MAX RL 8N = Z S = RL 2 I MAX RL 32 N 2 2 (15) = Z S = RL RL ⎛ I MAX ⎞ ⎜ ⎟ 2 ⎝ 4N ⎠ 2 (16) 2 (17) In fact, PL is at a minimum when RO = 0 (that is, PL = 0, as expected) and at a maximum when RO = ∞. For the latter, PL MAX (10) (11) = vCOMP v S 2 RL ⎛ I MAX RO N ⎞ ⎜ ⎟ = RL 2 ⎜⎝ RL + 2RO N 2 ⎟⎠ ⎛ ⎞ ⎜ ⎟ RL ⎜ I MAX N ⎟ PL = 2 ⎜ RL + 2N 2 ⎟ ⎜R ⎟ ⎝ O ⎠ With this choice of RO, Equation 4 through Equation 7 can be simplified as follows: Z NORM (14) Equation 16 defines the power delivered to the load for the impedance matched case and Equation 15 for the general case. It is interesting to compare Equation 15 and Equation 16 and to consider the effect on PL in Equation 15 when RO is varied. Recall that there is only one particular value of RO that provides impedance matching; namely, RO = RL/(2N2). If, however, impedance matching is not a requirement, then there is the freedom to choose any arbitrary value for RO. By rewriting Equation 15 in a slightly different form (as shown in Equation 17), the effect on PL due to varying RO becomes apparent. In this form, it is evident that a decrease in RO results in a decrease in the squared term, and vice versa. Figure 6. DAC with Reconstruction Filter R RO = L 2 2N Z S = RL In the case of impedance matching (that is, ZS = RL, which implies RO, as given in Equation 9), the PL equation reduces to IMPEDANCE MATCHING VSUPPLY 2 I MAX RL 16 N 2 = Furthermore, the power delivered to the load is a function of vS, so Equation 7 can be used to express the power delivered to the load as This yields another pair of equations useful for analyzing the center-tapped circuit: v A = 2vNORM and v B = 2vCOMP Z S = RL = vB 2 ⎧ ⎛ ⎞ ⎫ ⎟ ⎪ ⎪ ⎜ 2 I N ⎟ ⎪ RL ⎛ I MAX ⎞ ⎪R = lim ⎨ L ⎜ MAX = ⎜ ⎟ ⎬ RO → ∞ 2 ⎜ RL ⎪ ⎜ + 2N 2 ⎟⎟ ⎪ 2 ⎝ 2N ⎠ R ⎪⎩ ⎝ O ⎠ ⎪⎭ (18) Comparison of Equation 16 and Equation 18 indicates that four times more power (+6 dB) is delivered to the load when RO = ∞ as compared to the impedance matched case. (12) (13) Rev. 0 | Page 5 of 12 AN-912 EXAMPLE CALCULATIONS REDUCTION OF EVEN HARMONICS Here, the previous formulas are used to determine the component values for two different transformer applications. In Example 1, a transformer with a 1:1 turns ratio (N = 1) is employed, while in Example 2, a transformer with a 1:2 turns ratio (N = 2) is employed. Both examples use IMAX = 20 mA, RL = 50 Ω, and assume that impedance matching is employed (that is, ZS = RL). The degree of dc balance between the normal and complementary DAC current sources has a direct impact on the magnitude of even harmonics in the DAC output spectrum. Using a transformer as the output coupling mechanism for the DAC effectively masks any dc imbalance in the DAC outputs. This results in a significant reduction of even harmonics when the spectrum is observed at the output of the transformer. Example 1: IMAX = 20 mA, RL = 50 Ω, and N = 1 Transformer coupling can also mask the effects of a dynamic imbalance between the DAC outputs. However, the ability of the transformer to mask an ac imbalance depends on the inherent longitudinal balance of the transformer. Transformers with a high degree of longitudinal balance require that the manufacturer pay special attention to the physical design of the transformer. The most common factor limiting the longitudinal balance of a transformer is parasitic capacitive coupling within the windings. The transformer must be designed in such a way that the parasitic capacitance is evenly distributed relative to the external contacts of the windings. From Equation 9, RO = 25 Ω (the value of the two DAC termination resistors) From Equation 10, ZNORM = ZCOMP = 6.25 Ω (the load driven by each DAC output pin) From Equation 14, vA = vB = 88.39 mV rms (the voltage across each primary) From Equation 13, CONCLUSION vS = 176.8 mV rms (the voltage across the secondary) A center-tapped transformer can be used to advantage as the coupling element for a balanced current-output DAC. Formulas have been presented to determine the load (ZNORM and ZCOMP) and voltage (vA and vB) at each DAC output pin, the voltage (vS) across the load (RL), and the power (PL) delivered to the load (RL). Furthermore, the relationship between the DAC termination resistors (RO), the load resistance (RL), and the transformer turns ratio (N) was defined. From Equation 16, PL = 0.625 mW (the power in the load) Example 2: IMAX = 20 mA, RL = 50 Ω, and N = 2 From Equation 9, RO = 6.25 Ω (the value of the two DAC termination resistors) From Equation 10, ZNORM = ZCOMP = 1.5625 Ω (the load driven by each DAC output pin) From Equation 14, vA = vB = 22.10 mV rms (the voltage across each primary) From Equation 13, vS = 88.39 mV rms (the voltage across the secondary) From Equation 16, PL = 0.156 mW (the power in the load) Rev. 0 | Page 6 of 12 AN-912 SINUSOIDAL VOLTAGE SOURCE Transformer Basics VSRC 06737-007 Figure 7. Basic Transformer In Figure 8, a transformer is shown with its primary winding driven by a voltage source of VSRC (volts rms) that has a series resistance of RSRC (ohms). The secondary is terminated with an arbitrary resistance of RTERM. When a transformer is driven by an ac signal, the ratio of the voltage across the secondary winding to the voltage across the primary winding is the same as the turns ratio; that is, vS/vP = N. This gives rise to the concept of voltage transformation. That is, the primary voltage is transformed to a secondary voltage (or vice versa) based on the turns ratio. SECONDARY 1:N RTERM VSRC PRIMARY SECONDARY IMPEDANCE TRANSFORMED TO THE PRIMARY IMPEDANCE TRANSFORMED TO THE SECONDARY 1:N 1:N ZS = (N2)RSRC ZS ZP VSRC RTERM 06737-009 VSRC RTERM Note that when selecting a transformer, the reader should be aware that some manufacturers specify the impedance transformation ratio rather than the turns ratio. The turns ratio (N) is found by taking the square root of the impedance transformation ratio. RSRC RSRC VS Furthermore, conservation of energy requires that the power exhibited in the primary winding must equal the power appearing in the load of the secondary winding (RTERM). Alternatively, the power exhibited in the secondary winding must equal the power appearing in the load of the primary winding (RSRC). This knowledge makes it possible to treat RTERM as though it appears in the primary circuit as ZP (that is, the secondary impedance is transformed to an equivalent primary impedance). On the other hand, RSRC can be treated as though it appears in the secondary circuit as ZS, (that is, the primary impedance is transformed to an equivalent secondary impedance). This property of impedance transformation is related to the turns ratio and is expressed as: ZP = (1/N2)RTERM and ZS = (N2)RSRC. The concept of impedance transformation is demonstrated by the equivalent circuits shown in Figure 9. NS SECONDARY ZS ZP Figure 8. Transformer Driven by an AC Source A:B PRIMARY VP PRIMARY N = NS/NP = B/A NP 1:N RSRC The basic behavior of a transformer is governed by its turns ratio (or winding ratio). The turns ratio, N, is the ratio of the number of turns of wire in the secondary windings (NS) to the number of turns of wire in the primary windings (NP); that is, N = NS/NP. The turns ratio is often denoted on schematics by two colon-separated numbers (for example, 3:5). An example appears in Figure 7 in which an arbitrary turns ratio of A:B is shown. This leads to the relationship N = NS/NP = B/A. 06737-008 APPENDIX A ZP = (1/N2)RTERM Figure 9. Transformed Impedance Rev. 0 | Page 7 of 12 AN-912 Since the DAC is assumed to be of the balanced, current-output variety, Figure 10 can be redrawn as shown in Figure 11. The DAC is replaced by current sources IA and IB. These represent sinusoidal current sources with a peak-to-peak amplitude of IMAX (the maximum output current of the DAC). Also, the center-tap connection is drawn differently than in Figure 10 to clearly show that the signal sources exist as separate current loops. APPENDIX B A Balanced Current-Output DAC Driving a Tapped Transformer Figure 10 shows the general case for a DAC coupled to a tapped transformer. For completeness, the two primary windings are not assumed to be symmetrical (that is, the primary tap does not split the primary winding into two equal halves) and the two DAC termination resistors are not assumed to be equal (RA and RB). ZA consists of two parallel impedances. The first is the transformed impedance of the secondary resistor (RL), which is referred to as Z1. The second is the transformed impedance of RB, which is referred to as Z2. Note that the DAC output impedance can be ignored under the assumption that the internal current sources exhibit an infinite impedance (ideally), which means that the internal impedance of the DAC output does not impact the parallel combination of Z1 and Z2. Therefore, ZA can be expressed as (see Appendix A regarding impedance transformation): The primary winding is split into two separate circuits as a result of the ground connection at the primary tap. The upper winding is referred to as Primary A and the lower winding as Primary B. The windings are labeled A, B, and C to indicate the number of turns associated with each winding (Primary A, Primary B, and secondary, respectively). The overall turns ratio of the transformer is 1:N, where N = C/(A + B). The tapped transformer exhibits the following three independent accoupled networks (the associated turns ratios appear in parentheses): {( ) R }|| {( ) R } A 2 C A 2 B L 1:N VA A ZA TAP DAC VB IB ZB ZS C VS RL B RB 06737-010 RA N = C/(A+B) Figure 10. A Balanced Current-Output DAC Coupled to a Tapped Transformer 1:N A IA RA VA ZA C ZS VS RL B IB RB VB (19) Note that the symbol || in this and all subsequent equations can be read as "in parallel with.” Also shown in Figure 10 are the transformed impedances at Primary A (ZA), Primary B (ZB), and the secondary (ZS) along with the voltages that appear across each winding (vA, vB, and vS). IA B ⎧ R A2 ⎫ ⎧ R A 2 ⎫ RL RB = ⎨ L 2 ⎬ || ⎨ B 2 ⎬ = 2 B 2 C B ( ) R + RB ( CA ) ⎩ ⎭ ⎩ ⎭ L A Primary A and the secondary (A:C) Primary B and the secondary (B:C) Primary A and Primary B (A:B) ZB TAP Figure 11. DAC Shown as a Dual Current Source Rev. 0 | Page 8 of 12 06737-011 • • • Z A = Z1 || Z2 = AN-912 Furthermore, in the context of ac analysis, the sine function can be replaced by its rms equivalent, √2/2, which yields Likewise, the value of ZB consists of two parallel impedances. The first is the transformed impedance of the secondary resistor (RL), which is referred to as Z3. The second is the transformed impedance of RA, which is referred to as Z4. Therefore, ZB can be expressed as: B 2 A RA ⎧⎪ R B 2 ⎫⎪ ⎧⎪ R B 2 ⎫⎪ RL RA = ⎨ L 2 ⎬ || ⎨ A 2 ⎬ = 2 2 ⎪⎩ C ⎪⎭ ⎪⎩ A ⎪⎭ RL ( AB ) + RA ( CB ) {( ) R }|| {( ) R } A C 2 B B ⎧ R C2 ⎫ ⎧ R C2 ⎫ R A RB = ⎨ A 2 ⎬ || ⎨ B 2 ⎬ = 2 B 2 A B R ( ) + RB ( CA ) ⎭ ⎩ ⎭ ⎩ A C (21) Referring to Figure 11, the sinusoidal current delivered by the normal and complementary DAC outputs is given by IA = ½IMAX + ½IMAXsin(θ) and IB = ½IMAX − ½IMAXsin(θ), respectively. However, for ac analysis, the dc term in both equations can be eliminated, yielding IA = ½IMAX sin(θ) and IB = −½IMAX sin(θ). IA = IB = 2 4 I MAX A RA VA ZA C ZS VS RL B –(IMAX√2)/4 RB VB ZB TAP Figure 12. Modified AC Equivalent Model 1:N A (IMAX√2)/4 RA VA ZA C ZS VS RL B (IMAX√2)/4 RB VB (22) (23) 1:N (IMAX√2)/4 I MAX With the modification in Figure 13 it is no longer necessary to treat IA and IB separately because it is now apparent that 06737-012 C 2 A 2 4 The upper current source drives the side of Primary A marked with a dot, while the lower current source drives the side of Primary B that is not marked with a dot. However, the dot associated with Primary B can be moved to the other side of the Primary B winding without impacting the functionality as long as the connection to the signal source is reversed. Reversal of the signal source is equivalent to simply changing its sign. This is shown in Figure 13, where the sign of the lower current source is changed and the dot is moved to the other side of the Primary B winding. (20) Similarly, the value of ZS consists of two parallel impedances. The first is the transformed impedance of RA, which is referred to as Z5. The second is the transformed impedance of RB, which is referred to as Z6. Therefore, ZS can be expressed as: ZS = Z5 || Z6 = I MAX and I B = − Note that IB = −IA. From these results, Figure 11 can be redrawn by replacing IA with its rms equivalent and by replacing IB with –IA (see Figure 12). {( ) }|| {( ) } B 2 C RL 2 4 ZB TAP Figure 13. AC Equivalent Model with the Primary B Circuit Modified Rev. 0 | Page 9 of 12 06737-013 Z B = Z3 || Z 4 = IA = AN-912 Notice that the load as seen by the current source driving Primary A is the parallel combination of RA and ZA. Likewise, the load as seen by the current source driving Primary B is the parallel combination of RB and ZB. These loads are referred to as ZNORM and ZCOMP, since they are the loads as seen by the normal and complementary outputs of the DAC, respectively, and are given as ⎫⎪ ⎧⎪ RL RB Z NORM = RA || Z A = {RA } || ⎨ C 2⎬ B 2 ⎪⎩ RL ( A ) + RB ( A ) ⎪⎭ = R A R B RL 2 2 RA RL ( AB ) + RA RB ( CA ) + RB RL ⎛ 2 I MAX A v A = v S ⎛⎜ ⎞⎟ = ⎜⎜ 4 ⎝C ⎠ ⎝ RA RB RL = 2 A 2 RB RL ( B ) + RA RB ( CB ) + RA RL B ⎛ 2 I MAX v B = v S ⎛⎜ ⎞⎟ = ⎜⎜ 4 ⎝C⎠ ⎝ This result makes it possible to express the voltage generated by each DAC output, which is the product of the DAC output current and the load as seen by the DAC output. The normal and complementary DAC output voltages are expressed as vNORM = I A ZNORM The equations derived thus far will work for any generalized case of a tapped transformer. However, in practice, there are two simplifications that significantly reduce the complexity of these equations. The first is to use a center-tapped transformer (that is, A = B). The second is to use equal DAC termination resistors (that is, RA = RB) of value RO. Additionally, recall that N = C/(B + A). With the stipulation that A = B, it is possible to show that C/B = C/A = 2N. Applying these concepts to the previous equations yields the following simplified equations: (26) ⎞ ⎞⎛ RA RB RL ⎟ ⎟⎜ ⎟⎜ R R ( A )2 + R R ( C )2 + R R ⎟ ⎠⎝ B L B A B B A L ⎠ (27) where IA and IB have been replaced based on Equation 23. The secondary voltage (vS) is made up of the contribution of each of the primary voltages multiplied by the associated turns ratio. Specifically, ⎛ 2 I MAX C C vS = v NORM ⎛⎜ ⎞⎟ + vCOMP ⎛⎜ ⎞⎟ = ⎜⎜ A B 4 ⎝ ⎠ ⎝ ⎝ ⎠ Z NORM = ZCOMP = RO RL 2RL + 4 RO N 2 Z S = 2N 2 RO vCOMP = I B ZCOMP ⎛ 2 I MAX = ⎜⎜ 4 ⎝ ⎞ ⎟× ⎟ ⎠ ⎞ ⎛ RA RB RL ( AB ) RA RB RL ⎟ ⎜ 2 ⎜ R R ( B ) + R R ( C )2 + R R + R R ( A )2 + R R ( C )2 + R R ⎟ B L B L B A B A A B B A L ⎠ ⎝ A L A (30) (25) ⎞ ⎞⎛ RA RB RL ⎟ ⎟⎜ ⎟⎜ R R ( B )2 + R R ( C )2 + R R ⎟ ⎠⎝ A L A B L ⎠ A B A ⎞ ⎟× ⎟ ⎠ ⎞ ⎛ RA RB RL ( AB ) RA RB RL ⎟ ⎜ + 2 2 2 2 C C B A ⎜R R ( ) +R R ( ) +R R RB RL ( B ) + RA RB ( B ) + RA RL ⎟⎠ B L A B A ⎝ A L A (29) (24) ⎧⎪ ⎫⎪ RL RA ZCOMP = RB || Z B = {RB } || ⎨ 2⎬ 2 ⎪⎩ RL ( AB ) + RA ( CB ) ⎪⎭ ⎛ 2 I MAX = ⎜⎜ 4 ⎝ The two primary voltages (vA and vB) can be derived from vS based on the respective turns ratios, as follows: ⎞ ⎟× ⎟ ⎠ ⎞ ⎛ RA RB RL ( CA ) RA RB RL ( CB ) ⎟ ⎜ + 2 2 2 C 2 A ⎜ R R ( B ) + R R (C ) + R R RB RL ( B ) + RA RB ( B ) + RA RL ⎟⎠ B L A B A ⎝ A L A (28) Rev. 0 | Page 10 of 12 (32) ⎛ 2 I MAX v NORM = vCOMP = ⎜⎜ 8 ⎝ ⎛ 2 I MAX vS = ⎜⎜ 2 ⎝ (31) ⎞ ⎞⎛ RO RL ⎟ ⎟⎜ ⎟⎜ R + 2R N 2 ⎟ O ⎠⎝ L ⎠ ⎞⎛ NRO RL ⎞ ⎟ ⎟⎜ ⎟⎜ R + 2R N 2 ⎟ O ⎠⎝ L ⎠ (33) (34) AN-912 NOTES Rev. 0 | Page 11 of 12 AN-912 NOTES ©2007 Analog Devices, Inc. All rights reserved. Trademarks and registered trademarks are the property of their respective owners. AN06737-0-5/07(0) Rev. 0 | Page 12 of 12

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