Silicon ChipBalanced/Unbalanced Converter For Audio Signals - June 2008 SILICON CHIP
  1. Outer Front Cover
  2. Contents
  3. Publisher's Letter: New Zealanders can legally do their own wiring - why can't Australians?
  4. Feature: DIY Electrical Work: Are Aussies Dumber Than Kiwis? by Ross Tester
  5. Feature: A Look At Crash Test Dummies by Peter Holtham
  6. Project: DSP Musicolour Light Show by Mauro Grassi
  7. Project: PIC-Based Flexitimer Mk.4 by Jim Rowe
  8. Project: USB Power Injector For External Hard Drives by Greg Swain
  9. Project: Balanced/Unbalanced Converter For Audio Signals by John Clarke
  10. Review: Altitude 3500-SS Stereo Valve Amplifier by Leo Simpson
  11. Project: A Quick’n’Easy Digital Slide Scanner by Brian Coulson
  12. Vintage Radio: The Pye TRP-1 portable HF transceiver by Rodney Champness
  13. Book Store
  14. Advertising Index
  15. Outer Back Cover

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Articles in this series:
  • DSP Musicolour Light Show (June 2008)
  • DSP Musicolour Light Show (June 2008)
  • DSP Musicolour Light Show; Pt.2 (July 2008)
  • DSP Musicolour Light Show; Pt.2 (July 2008)
  • DSP Musicolour Light Show; Pt.3 (August 2008)
  • DSP Musicolour Light Show; Pt.3 (August 2008)
  • DSP Musicolour Light Show; Pt.4 (September 2008)
  • DSP Musicolour Light Show; Pt.4 (September 2008)
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Items relevant to "Balanced/Unbalanced Converter For Audio Signals":
  • Balanced/Unbalanced Converter PCB [01106081] (AUD $5.00)
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  • Balanced-Unbalanced Converter PCB pattern (PDF download) [01106081] (Free)

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Balanced/unbalanced converter for audio work By JOHN CLARKE If you work in the professional audio field, you need to use balanced lines for long signal runs to prevent hum and noise pickup. This Balanced/Unbalanced Converter is really two projects in one. It can convert an unbalanced input to balanced outputs and vice versa. P ROFESSIONAL AUDIO GEAR invariably has balanced inputs and outputs. However, what if you want to connect standard audio equipment that has unbalanced outputs to equipment that has balanced inputs? Alternatively, what if you want to connect a balanced output signal to an unbalanced input? Either way this Balanced/Unbalanced Converter project can do the job. The reason professional audio equip­ment utilises balanced inputs and outputs is quite simple. It’s done so that audio connections can be made over quite long distances without add68  Silicon Chip ing extra noise to the signal. These balanced connections use 3-pin XLR plugs and sockets and screened twincore cable. Fig.1 shows the basic arrangement. Basically, the audio output signal is coupled to two separate amplifiers and these drive the two signal leads in the cable in anti-phase (ie, the signals have opposite phases). In this case, Amplifier 1 has an output signal that’s in phase with the input, while Amplifier 2 has an output that’s opposite in phase with the input. The output impedance of each amplifier is the same and the twin-core cable carries the signal to the equipment at the other end. However, in some cheaper balanced line drivers, one core does not carry any signal but is grounded instead. So in this case, Amplifier 2 is left out and the lefthand side of resistor R2 is grounded. In operation, there will be some noise and hum pickup over the length of the cable even though the cable is shielded. However, because the cores in the cable are close together, any signal that is picked up will be common to both. At the receiving end, the signal in each of the two cores is subtracted to produce the original audio signal. At the same time, this also removes most of the noise and hum that was picked up in the leads, since the same noise signal is present in both. If one of the cores is grounded, as in the cheaper type of balanced driver, then the signal level after subtraction will be the same as the signal in the main core. Alternatively, if anti-phase signals are applied to both cores, the subtraction process produces an audio signal level that’s twice the level in the individual cores. As well as the increased signal level at the receiving end, using two antiphase signals gives a better result than using a balanced line driver with an earthed line. There are several reasons for this. First, when using two anti-phase signals, the two amplifiers that drive them are similar and basically follow the same impedance variations over siliconchip.com.au BALANCED OUTPUTS AMPLIFIER 1 A = +1 R1 BALANCED INPUT 2 TWIN CORE SHIELDED CABLE 2 1 SIGNAL INPUT 3 AMPLIFIER 2 (UNBALANCED) A = –1 R2 3 SHIELD Fig.1: the basic arrangement for converting an unbalanced audio input signal to a balanced signal and back again. the audio frequency range. Second, with the full anti-phase (or differential) lines, the electromagnetic field due to the signal in each is theoretically zero and so crosstalk into adjacent cables is minimised. And third, the cable will still supply signal should one of the cores be shorted due to a wiring fault (or damage). How it works Refer now to Fig.2 for the circuit details. As can be seen, it’s based on three LM833 op amps (IC1-IC3). IC1a, IC1b & IC2a make up the “Balanced Input To Unbalanced Output Converter” section. As shown, the balanced input signal is fed in via pins 3 & 2 of the XLR socket. These inputs are each tied to ground using a 100kW resistor to prevent the signal lines from “floating” with no input connected. From there, the audio signals are coupled via 10mF non-polarised (NP) electrolytic capacitors to pins 3 & 5 of op amps IC1a & IC1b respectively. The 220pF capacitor between the two inputs and the 100pF capacitors at pins 3 & 5 are included to filter RF (radio frequency) signals. In addition, pins 3 & 5 are each tied to ground via a 10kW resistor to set the DC bias for IC1a and IC1b. These 10kW resistors either connect to the signal ground or to a half-supply ground, depending on the power supply configuration used. IC1a & IC1b both operate as noninverting amplifiers with a gain of 1, as set by their 10kW feedback resistors and resistor R1 (20kW). A 100pF capacitor across each 10kW feedback resistor rolls off high-frequency signals above about 160kHz. The outputs from IC1a and IC1b apsiliconchip.com.au SIGNAL OUTPUT (UNBALANCED) 1 pear at pins 1 & 7 respectively and are summed in differential amplifier stage IC2a. For signals from IC1a, IC2a functions as an inverting amplifier – ie, it operates with a gain of -1. Conversely, for signals on its pin 3 input, it operates as a non-inverting amplifier with a gain of 2. Because of this, the signals from IC1b are divided by two using a 10kW resistive divider before being fed to IC2a. This means that each signal path has overall unity gain through IC2a. However, IC2a inverts the signals from IC1a so that they are now in-phase with the signals from IC1b. as a result, both signals add to provide an overall gain of 2. The resulting unbalanced signal appears at pin 1 of IC2a and is AC-coupled to the output via a 22mF NP capacitor and a 150W resistor. The 100kW resistor from the 22mF capacitor to ground ensures that the output signal swings above and below ground with no DC bias. Unbalanced to balanced stage A single LM833 dual op amp (IC3) is used for the “Unbalanced Input To Balanced Output” stage. As shown, the audio input signal is AC-coupled via a 10mF NP capacitor to the noninverting input (pin 3) of IC3a. A 100pF capacitor shunts any RF signal Parts List 1 PC board, code 01106081, 103 x 85mm 1 2.5mm PC-mount DC socket 2 3-way screw terminal blocks (5.08mm or 5mm spacing) 4 2-way screw terminal blocks (5.08mm or 5mm spacing) 4 M3 x 6.3mm tapped standoffs 4 M3 x 6mm screws 2 2-way pin headers (2.54mm spacing) 1 3-way pin header (2.54mm spacing) 3 jumper shunts 1 60mm length of 0.8mm tinned copper wire Semiconductors 3 LM833 dual op amps (IC1-IC3) 2 15V 1W zener diodes (ZD1,ZD2) 2 IN4004 1A diodes (D1,D2) Capacitors 2 470mF 25V PC electrolytic 1 100mF 25V PC electrolytic 3 22mF NP electrolytic 1 10mF 16V PC electrolytic 3 10mF NP electrolytic 3 100nF MKT polyester 1 220pF ceramic 7 100pF ceramic Resistors (0.25W, 1%) 6 100kW 1 4.7kW 1 20kW 4 150W 13 10kW 2 33W Specifications • • • Signal to noise ratio: -100dB with respect to 1V output, 4.7kW input load. • Signal handling: supply dependent; requires 30VDC or ±15V for 9V RMS signal handling. Frequency response: -3dB at 2Hz and 200kHz. Total harmonic distortion: less than .001% from 20Hz to 20kHz with a 1V input. June 2008  69 Table 1: Link Configurations SUPPLY LK1 ±9-15V DC OUT IN LK2 LK3 OUT OUT + 0V – 7-12V AC OUT IN OUT IN 9-30V DC IN OUT IN LK4 POWER INPUT + 0V OUT + 0V Note: install LK4 for an AC supply only Fig.2: the circuit can be split into three sections: (1) a balanced input to unbalanced output converter (top); (2) an unbalanced input to balanced output converter (centre); and (3) the power supply circuitry (bottom). to ground, while the associated 10kW resistor sets the DC bias for IC3a. Note that this 10kW resistor either 70  Silicon Chip connects to the signal ground or to a half-supply ground, depending on the power supply configuration used (this is the reason for the different earth symbol at the bottom of this resistor). The 100kW resistor at the input ties the siliconchip.com.au Power supply Power for the circuit can come from a 9-30V DC source, a ±9-15V DC source or a 7-20VAC source. The current requirements are quite modest at just 30mA. The simplest supply arrangement is to use a ±9-15V DC source (this type of supply can often be found in existing equipment). The positive rail is simply connected to the “+” supply input, the negative rail to the “–“ input and the ground to 0V. Diodes D1 & D2 provide reverse polarity protection, while two 470mF capacitors filter the supply rails. Zener diodes ZD1 & ZD2 protect the op amps by conducting if the input voltage rails exceed ±15V. A 33W resistor in series with each supply line limits the current through ZD1 and ZD2 when they conduct but note that voltages above ±18V may destroy these zener diodes. With this supply arrangement, the two different grounds on the circuit are tied together using link LK2 (see siliconchip.com.au + SIG 100nF LK3 NP 22 F 0V LK1 150 150 100k 10k NP 0V 100pF 0V 33 33 ZD2 470 F 10 F 10k 10 F NP NP – – 100k 10k IC3 LM833 100 F 220pF 100pF 100pF D2 LK4 4.7k 100k 10k 10k D1 ZD1 10k 100nF 20k 22 F 100pF 100k 10k 150 10k 10k IC1 LM833 10k 22 F NP 100nF NP 100pF 10 F 10k IC2 LM833 10k 10k + 150 100pF 100k 0V LK2 100pF 10 F – POWER INPUT 0V DC SOCKET BALANCED OUT UNBALANCED OUT 100k input line to ground when no signal is connected. IC3a is wired as a unity gain buffer stage and so its pin 1 output follows the signal input. The non-inverting (+) component for the balanced signal is then AC-coupled via a 22mF NP capacitor and a 150W resistor to pin 2 of the XLR output socket. The 150W resistor isolates IC3a’s output from external capacitive loads, to ensure stability. The 100kW resistor on the output side of the 22mF capacitor ensures that the signal swings symmetrically above and below ground. The out-of-phase signal is derived using IC3b. This stage is also fed from pin 1 of IC3a and functions as an inverting amplifier with a gain of -1 as set by its 10kW feedback resistor. As before, a 100pF capacitor across the feedback resistor shunts any frequencies above 160kHz to prevent amplifier oscillation. IC3b’s output at pin 7 is inverted compared to IC3a’s output. It drives pin 3 of the XLR socket via another 22mF capacitor and 150W resistor combination. Note that the pin assignments on the XLR socket follow standard practice. Pin 1 is the ground, while pin 2 is for the “hot” or non-inverted (+) signal and pin 3 is for the “cold” or inverted signal. + BALANCED IN 470 F /DE C NALA B DE C NALA B NU RETREV N O C 18060110 SIG 0V UNBALANCED IN Fig.3: install the parts on the PC board as shown in this parts layout diagram. Table 1 (facing page) shows how to install links LK1-LK4 to suit the selected power supply. Table 1). This biases the op amp inputs at 0V so that the signal swings above and below ground. AC supply A 7-12V AC supply can also be used to derive positive and negative supply rails. In this case, the “+” and “-” inputs are connected together using link LK4 and the supply is connected between either of these two inputs and the 0V (ground) terminal. With this supply configuration, diodes D1 & D2 function as half-wave rectifiers, with filtering provided by two 470mF capacitors. D1 conducts on the positive half-cycles to derive the positive rail, while D2 conducts on the negative half-cycles to derive the negative rail. As before, the two grounds are connected using link LK2. 9-30V DC supply The circuit is a little more complicated for a 9-30V DC supply. That’s because the signal can no longer swing below the 0V rail, since there’s no negative supply. As a result, the op amps must be biased to a mid-supply voltage, so that the signal can swing symmetrically about this voltage. This mid-supply voltage is produced using a voltage divider consist- ing of two 10kW resistors between the V+ rail and ground. A 100mF capacitor filters this half-supply rail which is then fed to IC2b. IC2b is wired as a unity gain buffer stage. Its pin 7 output drives a 10mF capacitor via a 150W decoupling resistor to produce the Vcc/2 half-supply rail to bias the op amps in the converter stages. In this case, links LK1 & LK3 are installed. Link LK1 connects the Vcc/2 rail to the junction of the 10kW bias resistors on the pin 3 & pin 5 inputs of IC1a & IC1b. It also connects the Vcc/2 rail to the pin 3 input of IC3b via another 10kW resistor. Link LK3 connects the negative supply pins for the op amps to ground. Finally, the AC coupling capacitors at the inputs and outputs of the various op amps remove any DC component from the signal. Building it The assembly is straightforward with all the parts installed on a PC board coded 01106081. This board also carries screw terminal blocks for the audio input and output connections, plus a DC socket for the power supply connections (depending on the supply used). Fig.3 shows the parts layout. Begin June 2008  71 Table 3: Capacitor Codes Value 100nF 220pF 100pF A 9-30V DC supply can be connected either via the DC socket or via the “+” and 0V terminals on the “Power Input” screw terminal block. An AC supply is connected in exactly the same manner (ie, via the DC socket or between the “+” and 0V terminals). For the ±9-15V DC supply option, connect the positive lead to the “+” terminal, the negative lead to the “-” terminal and the 0V lead to the 0V terminal. Again, make sure the links are correct – see Table 1. Apply power and check that close to the supply voltage appears between pins 8 & 4 the ICs. If the supply is 12V DC, for example, then the pin 8 to pin 4 voltage should be close to 10.3V (after allowing for a 1.7V drop across D1 and its series 33W resistor). The Vcc/2 supply, as measured at pin 6 of IC2b and at the pin 1 & pin 7 outputs of the other op amps, should be close to 10.3V/2 or 5.15V. For an AC supply, the pin 8 voltage should be positive with respect to ground and the pin 4 voltage negative. The actual voltages should be about 1.414 times the AC voltage minus about 1.7V for the diode and resistor drop. Thus, for a 9VAC supply, the voltage should be about 12.7V - 1.7V = 11V DC. This means that there should be +11V with respect to ground on pin 8 of each IC and -11V on pin 4 of each IC. Finally, for a ±9-15V DC supply, the pin 8 and pin 4 voltages should be about 1.7V less than the input voltages. For example, if the supply is ±12V DC, there should be about +10.3V on pin 8 of each IC and -10.3V on pin 4 SC of each IC. This view shows the fully-assembled PC board. Take care to ensure that the semiconductors and electrolytic capacitors are correctly installed. by checking the board for any defects such as shorted tracks or breaks in the tracks. Check also that the hole sizes for the DC socket and screw terminal blocks are correct by test fitting these parts and check that the four corner holes are drilled to 3mm. Install the links first, followed by the resistors. Table 2 shows the resistor colour codes but you should also check each resistor using a DMM before soldering it in place, as some colours can be difficult to decipher. The diodes and zener diodes can go in next, followed by the three ICs. Take care to ensure that these parts are all oriented correctly and be sure to use the correct diode at each location. We used IC sockets on the prototype but this is not really necessary and you can simply solder the ICs straight in. The capacitors are next on the list. Take care with the electrolytic types, as they must all be fitted with the correct polarity. The two 470mF capaci- mF Code IEC Code EIA Code 0.1mF 100n 104   NA 220p 221   NA 100p 101 tors are mounted on their sides, with their leads bent down through 90° so that they pass through the holes in the board. Finally, install the pin headers (for the links), the DC socket and the screw terminal blocks. The 4-way screw terminal blocks are made by sliding two 2-way terminals together, using the dovetail mouldings on either side. Installation As mentioned earlier, there are several supply options for the Balanced/ Unbalanced Converter. The current requirements are quite low at 30mA maximum when each output is driving a 1V signal into 600W. Installation is basically a matter of deciding which type of supply you want to use and then choosing the linking options – see Table 1. Note that link LK4 is installed only for an AC supply. Table 2: Resistor Colour Codes o o o o o o o No. 6 1 13 1 4 2 72  Silicon Chip Value 100kW 20kW 10kW 4.7kW 150W 33W 4-Band Code (1%) brown black yellow brown red black orange brown brown black orange brown yellow violet red brown brown green brown brown orange orange black brown 5-Band Code (1%) brown black black orange brown red black black red brown brown black black red brown yellow violet black brown brown brown green black black brown orange orange black gold brown siliconchip.com.au