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A practical approach to amplifier output stage design There are many requirements for a successful amplifier design. One of the least understood is how to design the output and driver stages to remain within their safe operating areas. Faulty design here can be more than just expensive; the smoking remains are down right embarrassing. By DAVID EATHER Many cope with this task by copying from other designs. This is a limiting approach and almost invariably leads to designs with a certain sameness about them. Also, bad design is perpetuated. There is a better way. This article shows a practical approach to amplifier output stage design and covers the calculation of power supply voltages, output load . lines, derating transistor SOAR curves and heatsink selection. You get nothing for nothing so before you start there is a fair swag of calculations to be done. You also need access to a power transistor data book. The payoff is reliable amplifiers with output. power levels customised for your needs. There are no definitive answers as to exactly how many or what type of transistors you have to use in your 14 SILICON CHIP design; a lot depends on taste. At the same time, there are certain design rules that should not be transgressed. My method is a simplified approach the overall aim has been to achieve a reliable design without too much pain. What has been shaved off one area is generally compensated for elsewhere. For illustration, I will be designing the output and driver stage of a general purpose amplifier capable of 25 watts into 8Q. The circuit is a simple 10-transistor design as shown in Fig.l. After a general discussion on each step, I will provide some specific results, so you can check your understanding of the principles being discussed. Assuming you already know how much power output you want, the first step begins with calculations to find the peak voltage and peak current delivered to the load. Use the following formulas. Vmax load = ✓ (2 x P x Z) lmax load = ✓ (2 x P/Z) For my amplifier this works out as: Vmax load= ✓ (2 x P x Z) = ✓ (2 X 25 X 8) = 20 volts Imax load= ✓ (2 x P/Z) = ✓ (2 X 25/8) = 2.5 amps Emitter resistors At this point, I will give a quick mention of the emitter resistors, Rl and R3, on Fig. l. These resistors help provide thermal stability of the output stage bias current and in designs with output transistors in parallel they help to ensure equal current sharing. The higher the resistance the better the thermal stability and current sharing but the more power they waste. The final value is a compromise. As a guide, you would normally try for about 0.6 volts across the emitter resistors at lmax load. For this design, 0.22Q should prove adequate. The resistance is a little low but I would not expect any problems for the following two reasons: (1) there is only one output transistor for each rail so there is no current sharing; and (2) I don't intend to set a high quiescent current. The next step is to work out the required supply voltage (±Vee). You must consider the requirements of Vmax load, the voltage drops caused by the driver and biasing circuitry and the voltage drop caused by the emitter resistors in the output transistors, and lastly the ripple voltage (hum and audio signal) on the supply rails (±Vee). At this stage, refer to back to Fig, 1. , - - - -.....- - - - - - - - - - - - . - - - - - - . . . . . - - - + V C C R1D R9 01 ? 09 BC547 R17 07 BC549 R9=R10=2R8 C4 06 -YCC+1.4Y BC639 RB ...__ _ _----,.__ _ _ _ _...__ _ _ _ ___.____ _ vcc Fig.1: the circuit for a general purpose 10-transistor audio amplifier capable of delivering about 25W into 8Q. Note that the design uses split supply rails and complementary output stages. Notice that the bias voltages applied to the bases of Q5 and Q6 are equal. Notice also that the outputs, drivers and pre-drivers are mirror images of each other (R5, Q5, Q2 and Ql vs R6 , Q6, Q4 and Q3). This is a common situation and allows calculation for ±Vee by considering just the positive side. In cases where the quiescent biasing varies for the positive and negative sides, the ±Vcc rails are worked out separately. Depending on the design, use the larger value for both supply rails. In this example (see Fig.1 again), there is 1.4 volts across the base-emitter junction of Q5 and R3. We will assume that at full power, the same voltage of 1.4 volts appears between the collector of Q5 and +Vee, 0.55 volt peak (Imax load x Rl) across Rl, and 1.6 volts across the base-emitter junctions of Ql and Q3. This gives a total overhead of 3.55 volts. Power supply ripple Next, you have to make an estimate for the ripple on the power supply. For this I like to use what I call Eatber's rule of thumb. Stated as a formula it looks like this: Vripple = 6300 X lmax load/Cwhere Vripple is the peak to peak voltage ripple on the power supply and C is the filter capacitor size in microfarads. The capacitors should be a minimum of 100-200µF per watt of output power for a class B amplifier with a full wave rectifier. (Actually, Eather's rule of thumb is not just a whim of mine but is a condensation of the maths for capacitor-input power supplies). For this example, I have elected to use two 2500µF capacitors for each power supply rail: Vripple = 6300 X lmax load/C = 6300 X 2.5/2500 = 3.15 volts The value for ±Vee is: ±Vee= Vmax load+ Vripple + circuit overhead For my design this becomes: ±Vee= Vmax load+ Vripple + circuit overhead = 20 + 3.15 + 3.55 = 26.7 volts This can be safely rounded off to ±27 volts. Transistor load lines OK so far? The next step is to figure out the transistor load lines. We are not going to bother with the load lines for resistive loads. These are straight lines and not really the problem for amplifiers. We are concerned with reactive load lines. These show the instantaneous voltage and current flowing through the transistors when driving a complex load impedance such as a speaker. To do this you need the output power, Imax load and ±Vee. You also need the value of the emitter resistors in the output stage (Rl, R2 in my case), the power output of the amplifier and the load impedance, Zl. Before leaping into the computations , we need to make an estimate of the maximum phase shift caused by the inductive portion of the speaker load. 45° seems to be the accepted standard in many electronics magazines and is the value we shall use FEBRUARY1991 15 20 " "- .... le = lmax load x sin(wt - 0) Write down each result in turn for the value of wt. Next is Vee using the more complex formula: Vee = Vee - Imax load X Zl X sin(wt) - le x RE The column for Ppk(W) is calculated by multiplying the collector emitter voltage Vee by the collector current le: Ppk(W) = V ce X le Table 1 shows the results for the amplifier under discussion. I'\ 10 .... " I'-- ~ 1 ", 1" I Sn LINE "-.... \ -........I'- V I '\ I \ \ \ \ Load variations I I I 0.5 I 0.3 0.2 I \ \ \ \ \ I I \ I 3 I 20 10 30 60 Vee (VI Fig.2: the 4Q & an load lines for the output transistors of the amplifier. These curves were plotted using the data shown in Tables 1 & 2 respectively. Note that the load lines should be fully enclosed by the DC SOAR curve of the selected transistor as shown here. here. However, if you intend to use your amplifier with highly reactive loads such as electrostatic speakers or line transformers, 60° would be a better choice. Drawing up a table Now we draw up a table with five columns and 13 rows. The columns are labelled: (wt - 0), wt , le, Vee and Ppk(W). Theta (0) is the electrical phase shift caused by the speaker. The term "wt" is the instantaneous phase of the signal frequency and is expressed in degrees. le is the instantaneous current through the collector of th e output transistor. Vee is the instantaneous voltage across the output transistor. Ppk(W) is the instantaneous power dissipated by the output transistor. The (wt - 0) column starts at 0 and steps up to 180° in 15° increments. Down the wt column write the corresponding value of wt. This is the same as adding the selected value of 0 (45° in our case) to the adjacent value of (wt - 0). This leaves the wt column with values starting at 45° and ending at 225 °. Start the calculations with le, using the formula: 16 SILICON CHIP Now take a deep breath. A general purpose amplifier could drive all sorts of speakers, some with only a very nominal 8Q impedance. For amplifiers in this situation, it is normal to design the amplifier so that it can safely drive into half the nominal load impedance. This may not be necessary if the amplifier is to drive a known speaker impedance or if using electronic limiting. If electronic limiting is not done carefully though, the amplifier may produce objectionable distortion if pushed hard into a nonresistive load. The rule of thumb for estimating power output into half the nominal load impedance is that the amplifier will produce about 50% more power. This won't apply if the amplifier has a well regulated power supply and large filter capacitors, in which case the power output will be closer to double. Conversely, if the power supply has poor regulation and small filter capacitors, the amplifier may only deliver a few percent more power into half its nominal load impedance. First, assume your amplifier will deliver 50% more power. Then you have to check that your amplifier will really deliver this power into the new load. Why? Because if it can, it will have to dissipate a lot more power and we need to know that the transistors can stand this extra stress. Work out the required value for Vee for the increased power output. This means going through the same procedure you did before, finding the required voltage across the load, the amplifier overhead and the power supply ripple using the new load impedance. A required value for Vee much larger than that available from your power supply means that the amp1ifier won't be able to deliver the extra 50% power, except maybe for short peaks. If the required value for Vcc is less than the actual supply, the amplifier will deliver a bit more than an extra 50%. Most times, allowing 50% gives a close estimate of what will actually happen. If, in your case, the value you came up with for Vee was very different, adjust your estimate of output power and go through the checking procedure again. For my amplifier, half the load impedance equals 4Q and I expect the amplifier to deliver about 37 watts. The calculations for Vmax load and lmax load give 17.3 volts and 4.33 amps respectively. The amplifier overhead goes up by 0.4 volts to 3.95 volts due to the higher current through the emitter resistors. The ripple on the supply also increases to 5.45 volts. So the Vcc needed is still about 27 volts. The next step is to calculate another load line for the new load impedance. Use the new values for Zl and lmax load. My results for output into a 4Q load are tabulated in Table 2. We can now draw some conclusions about the possible output transistors. The output devices must have an le rated higher than Imax load. The Vce must be twice Vcc and the power rating should be at least 50% greater than the largest value for Ppk(W) for most designs. In suggesting 50% more for the power rating of the output transistors, I am assuming power dissipation is the limiting factor and not secondary breakdown. It is just a ballpark figure and may need adjustment. Output transistors For your design, you should be able to make an educated guess about what output transistors or combination of transistors you will need to use. For my job, I will need the output transistors to have an le of more than 5 amps, a Vee of 60 volts or more, and a power rating of around 100 watts. I c;an now select some possible devices. For my design, transistor pairs such as the MJE3055/MJE2955, TIP3055/ TIP2955, 2N3055/MJ2955 or MJ15003/MJ15004 could all be suitable. I won't use the TIP and MJE pairs because the packages are not pin compatible even though they are of- HEAVY DUTY TV/SPEAKER WALL-CEILING BRACKETS Table 1 rot-0 0 15 30 45 60 75 90 105 120 135 150 165 180 rot 45 60 75 90 105 120 135 150 165 180 195 210 225 Vee le 12.858 9.537 7.406 6.611 7.205 9.148 12.308 16.469 21.347 26.611 31.901 36.858 41 .142 0.000 0.647 1.250 1.768 2.165 2.415 2.500 2.415 2.165 1.768 1.250 0.647 0.000 Ppk(W) 0.000 6.171 9.258 11 .687 15.600 22.091 30.770 39.769 46.218 47.042 39.877 23.848 0.000 rot 45 60 75 90 105 120 135 150 165 180 195 210 225 Vee le Ppk(W) 14.753 11.754 9.794 9.006 9.445 11.080 13.800 17.420 21 .692 26.326 31.007 35.414 39.247 0.000 1.121 2.165 3.062 3.750 4.182 4.330 4.182 3.750 3.062 2.165 1.121 3 0.000 0.000 13.172 21.204 27.576 35.418 46.343 59.755 72.858 81.344 80.605 67.129 9.687 0.000 ten sold as being interchangeable. The mounting tab is on opposite sides when placed into a PC board. This has the potential for mistakes during construction or repairs. Also, the power rating for these transistors is a bit low (only 90 watts). They could work but I will look around for something else. The 2N3055/MJ2955 pairs have a higher power rating for only a few cents more. This higher power rating could-lead to cost and size savings by enabling the use of a smaller heatsink. The MJ-15003 /15004 pairs are nice but relatively expensive. Compared to the 2N3055/MJ2955 pair, they do offer a better current gain-bandwidth product (ft) and would give slightly berotated 360 degrees ~:~ ~1H i1 li1 _, 1 as well as being swivelled up or down to any viewing or listening angle. The metal platforms have predrilled holes for mounting and are easily adjusted with a large alien key supplied with the unit. ~~ Imparted 1111d distributed by: Table 2 rot-0 0 15 30 45 60 75 90 105 120 135 150 165 180 The M83 and MB5 are heavy duty two platform mounting brackets designed for securing small TV's and speakers to walls, ceilings, desks or bench tops. When mounted both units can lower distortion. Also the much higher power rating means I could possibly get away with quite a small heats ink. Overall, the 2N3055/MJ2955 pairs should perform satisfactorily in my circuit so they are my first choice. At this stage, the choice of output transistors is no more than an educated guess. It is possible that the transistors may not be suitable. The graph of the load lines, transistor SOAR curves and the heatsink calculations will confirm the final choice. This is the time to check your transistor data book closely. Enlarge or redraw the safe operating area (SOAR) graphs for the transistors until they are a convenient size. Then plot the ~ ELECTRONICS Available through the fol/owing retailers: David J Reid All Electronic Electronics. Components. 127 York Street. 118 Lonsdale St. Sydney. 2000. Melbourne. 3000. NSW. (02) 267-1385. Vic. (03) 662 3506. SOUND AUSTRALIA Your P.A. Accessory Specialist Electronic Component Specials 555 Timer 4001 4007 4011 4013 -15volt regulator LED's 5mm Red, Green, Orange, the same price V28 7MHz $0 .1 0ea $0.15ea $0.15ea $0.15ea $0.15ea $0 .15ea Yellow All $0 .1 0ea $7.95ea HEAVY DUTY CARPETED ROAD CASE 4 Unit $100.00 6 Unit $115.00 8 Unit $135.00 10 Unit $155.00 12 Unit $175.00 Rack cases can be made to order in any size and any carpet colour. HANDLES 6" Strap $1.95 s· Strap 10• Strap $4.95 s· Nylon Briefcase Style Chest Handle Metal Recessed Spring Handle [Z] $ 2.45 $ 2.95 $ 2.95 $ 5.95 $12.95 - Please Call for a catalogue SOUND AUSTRALIA 28 Walker St. Dandenong, VIC, 3175 Telephone: (03) 791 1622 FEBRUARY1991 17 10 wt-e wt 0 15 30 45 60 75 90 105 120 135 150 165 180 45 60 75 90 105 120 135 150 165 180 195 210 225 Table 3 Vee 12.858 9.537 7.406 6.611 7.205 9.148 12.308 16.469 21 .347 26.611 31 .901 36.858 41 .142 I le Ppk(W) 0.000 0.032 0.063 0.088 0.108 0.121 0.125 0.121 0.108 0.088 0.062 0.032 0.000 0.000 0.309 0.463 0.584 0.780 1.105 1.538 1.988 2.311 2.352 1.994 1.192 0.000 '- '\ '\ \ ~ 0.1 ~ \ '\\ ' I I\ I I \ .01, 1 \\ 10 100 Vee (V) wt-e wt 0 15 $0 45 60 75 90 105 120 135 150 165 180 45 60 75 90 105 120 135 150 165 180 195 210 225 Table 4 Vee 14.781 11 .789 9.834 9.048 9.486 11.117 13.831 17.442 21.705 26.328 30.997 35.394 39.219 load lines for the amplifier using the points you calculated above. Fig.2 is my plot for the 4Q and 8Q loads using the data in Tables 1 & 2. You may be used to seeing these load lines in text books on graphs with a linear scale and think mine look a bit strange. Don't worry, they are the same type of graph only the scales arf) logarithmic. Notice how the 2N3055 SOAR curve fully encloses the load lines. If this were not the case, then the amplifier may die on the first occasion it is required to give a big burst of power. Make sure the load line~ are fully enclosed by the DC SOAR curve of your transistor. The fully enclosed load lines show that the selected 18 SILICON CHIP le Ppk(W) 0.000 0.056 0.108 0.153 0.187 0.209 0.216 0.209 0.187 0.153 0.108 0.056 0.000 0.000 0.659 1.062 1.382 1.774 2.319 2.987 3.639 4.060 4.021 3.348 1.979 0.000 power transistor can work in the design. Driver transistors Now we come to the driver transistors. The first step is to calculate the load impedance presented to the driver transistors. In my case, it is simply the speaker impedance multiplied by the minimum beta of the output transistor over the range of currents of interest (0 to lmax load). For the 2N3055, the minimum beta is 20. Calculate lmax for the driver by dividing Imax load by the beta of the output transistor. If your circuit calls for it, make sure you add in any Fig.3: these curves show the load lines for the driver transistors & were plotted using the data shown in Tables 3 & 4. As with the output devices, the load lines must be fully enclosed by the DC SOAR curve of the transistor. other currents the driver transistor must supply. Now calculate points for the load lines for the driver transistors in the same way as for the output transistors . Remember to do this for half the nominal load impedance if applicable. My results are tabulated in Tables 3 and 4 while Fig.3 shows the plotted load lines. Note that, as with the output devices, these load lines must be fully enclosed by the DC SOAR curve of the selected transistor. Go through the same selection procedure as you did for the output transistors. For my case, a look down the tables on this page shows a peak power dissipation of 4.06 watts and a maximum current of 216mA. Notice that even at these modest power levels, small signal transistors like the BC546/7 /8 aren't able to cope. Be wary of designs that suggest they will. I will try a BD139/BD140 pair for the driver transistors. Next month, we will see if a pair of BD139/BD140 transistors is up to the task of being driver transistors in the circuit of Fig.1. I think that they will but we'll find out for sure, next month. See you then. SC