curs numai pentru incepatori audio amp

8/7/2019 Curs numai pentru incepatori Audio Amp

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and also the peak value of output cuurent as:

and, of course, we will need equally that much in the positive rail, which means th

This very simplified calculation of the (conditionally taken) a standard-basi

For someone this might not appear logical, but in my opinion it is best to be

To calculate the required supply voltage we have to assess the additional v

Taking into account all that we have calculated, we can choose the appropr


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to be on a safe side weshould add at least another30%, and therefore Uceshould be chosen to be atleast 90V.

Another important parameteris determining what is thedissipation of outputtransistors. It is calculatedas follows:

and this is the dissipation of each output transistor. Itmust not be forgotten thatbecause of operation in AB

class, output devices arepassing some quiescentcurrent too, and since it israrely more than about100mA and the voltage is36V, we get additionaldissipation from thatquiescent current of some3,6 W to be added to the 16.2W and we can round total of about 20W of dissipation pertransistor. With a largeselection of transistors thatare available today, itshouldn't be difficult to find asuitable output pair.

To determine the parameters of driver stage transistors T3 and T4 we must


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and to be safe, we should usetransistors with at least twotimes higher collectorcurrent, let's say 250mA, andeven better to choose evenhigher value which will notbe a problem for moderndriver transistors. Formaximum collector-emitter

voltage of the drivers weshould choose the same asfor the output transistors, ieabout 90V. Because they areworking on the same voltageas the outputs, theirdissipation will be smallerthan dissipation of theoutputs transistor fordifference in currents theyare passing, that meanreduced by the current gainof output transistors.Therefore:

Realistic, and inpractice frequently foundusual value of current gain of the driver transistors is atleast 50, and knowing thatwe can calculate basecurrents of T3 and T4, whichis:


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This informationallows us to determine thecollector current of transistorT5, which has the function of a linear voltage amplifierwho works in class A. Classof operation dictate us to seta collector current of T5 so

that it is always in a linearmode. To ensure this, andhaving in mind the basecurrent of T3 and T4, we willadopt 2-5 times highervalue... say 10mA. Thishigher current is needed formany reasons and especiallydue to the fact that driversand output transistors,together as a system, are notthe constant load for thisstage since the current gain

of the transistor changeswith the change of currentthrough them, and thathappens all the time whenamplifier is operating.

Collector current of T5 is comming from a constant current source with tr

To enable T6 to work as a constant current source we have to keep its bas

To assure constant voltage at their ends, in most low-power Zener diodes

Voltage across R4 will be 36V-2,1V = 33.9 V and with the current of 5mA,


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to have just sufficientvoltage drop that is requiredto start "opening" theoutputs. Value of thatvoltage we can estimateknowing that silicontransistors needs about 0.7 Vhigher (for NPN's) or lower(for PNP's) potential than

their emitters. Becauseemitters are connected to thesame point, outputtransistors will have the totalvoltage difference betweentheir bases of 2 x 0,7 V, i.e.~1.4 V to begin to conduct.In doing so we canpractically ignore the emittervoltage drop over emitterresistors R1 as the voltagedrop on them with just aquiescent current passingwill be very small (about10..30 mV) due to their lowresistance.So, knowing that across R3should be around 1.4 V andtaking into account quiescentcurrent of the driver pairsuch as 4mA and all that isgoing through R3, we cancalculate its value as:

Its value is rounded to the nearest standard value of 330ohms which will n

And now let us deal with differential input stage with T7 and T8. The large

To achieve the best possible performance, we should not go with too smal

Taking into account that the collector currents of T7 and T8 are 1.2 mA eac

To calculate the resistance R8 starting point is the fact that it passes thro

In better - improved constructions, instead of R9 you will most often find


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Base currents of T7 and T8 are very small but still shouldn t produce any

The result of 1.15 k you can freely round to the nearest standard value whi

Capacitors C1 and C1a have a double role. For DC, they are infinitely large

If you pay attention to the scheme, you will see that the ground of the inp

Now, let us see the role of transistor T9 and the related elements around i

If we want to be able to completely stop any current flow through the out

From this, we can now easily compute the value of R12= 5,7k-1,5k = 4,2k.


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Dr. Borivoje Jagodi

significantly reduce thepossibility for oscillation andinstabilities when amp isworking with reactive loads.Their values are usually 10ohms and 100nF and as suchare in most schemes but it isnot because it is some kindof a fashion, but becausethese values are very realand proven to be optimal inpractice. Coil "L" is used toreduce or prevent negativeinfluence of Capacitive Loadsand its value in practiceranges from about 1uH tomaximally 6uH. It should bedone with enameled copperwire for transformers andoptimal thickness of about1,2-1,5 mm. Quality factor(Q-factor) of that coil should

be minimized as much aspossible to prevent it to formsome kind of resonant circuitwith various capacitances(speaker cable capacitance,wiring between the boards ...and so on). Therefore, itshould always have attacheda resistor in parallel whoseresistance should not begreater than few ohms (1-5ohms).

This tutorial isintended to be simple asmuch as possible, adapted tothe level of knowledge andunderstanding of young orless experienced DIYbuilders, and I hope it willgive them a sufficient basisfor the first steps alone.


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at the power supply on its output should provide + - 36V DC voltage with no load.

c modern transistor amplifier as is commonly seen today, originated from the desire to s

gin the process starting from the output stage and working all the way to input. Some th

ltage drop at R1, T2, T4, T5 and R5. For R1, T2 and T4 we can count to overall voltage d

iate output transistors. Knowing that the peak collector current is 3.5Amp, to be on a saf


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proceed from assumptions that the current gain of output transistors is 30 which is quit


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nsistor T6 and adjacent components. That current passes through the collector-emitter

e on a constant voltage which will be independ of variation of supply voltage Ub, and tha

you have to pass at least about 5mA or more and that current passes also through the R

it will dissipate 0.17 W and it is advisable to use at least a 1/2W resistor. T5 collector cu


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ot significantly disrupt anything except that the quiescent current of the drivers will be l

st part (but not all!) distortions as well as "slew rate" parameter of this type of amplifie

l collector currents of T7 and T8. Often you will find designs where there are no resistors

, we can calculate the value of R9 as:

gh the collector current of T7 which we set to 1.2 mA, and that current should create ac

some type of constant current source (similar to a circuit with T6, Dz, R2 and R4) and ins


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significant voltage drop across R6 and R10. These two resistors should have the same v

ch is 1.1k or 1.2 k for it in practice will make no difference. R11 should have the same v

resistance and virtually insulating the lower end of R7 as if it s not connected anywher

ut signal, the lower end of the R10 and the lower ends of C1 and C1a are all connected t

t. This circuit is a characteristic of the transistor output stages and in foreign literature y

ut transistors by turning RV preset, we have to bring voltage higher than some 0,6-0,7

We'll take the nearest standard value of 4.3 k. If we ever move Rv slider all the way to t


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mehow help the young and less experienced DIY enthusiasts. The construction of such uni

ings we have to assume as pre-seted, so let say we decided to build one amplifier working

op of about 2V. At the maximum negative amplitude of the collector voltages of T5 and R5

e side we will use transistors that have peak collector current at least two times higher, an


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realistic and even a modest value for today's output transistors. Assuming this value, we


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unction of T6 transistor and therefore it must pass through its emitter resistor R2 on whic

t we are achieving with Zener diode in its base. Voltage of the Zener diode must be for the

4 and its value we can get as:

rrent is also passing through its emitter resistor R5 and we will chose so that it creates a v


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ittle higher and that could only benefit from that.

, depends to a great extent of this stage, so it pays to devote enough attention to it. Ampli

Re in emitters and a collector current "set" in the 300-500uA range. It's much better to in

ross R8 a voltage drop equal to the sum of the voltage on the R5 and the base-emitter volt

tead of R8 there will be so-called "current mirror", but those are further developments wit


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lues and we take the arbitrary value of 22k for both, i.e. R6 = R10 = 22k. Ratio of the resi

lue to void to significantly disturbe overall balance of differential pair. Input capacitor Cin

e, and therefore for DC the amplifier has unity gain. For audio signal that is AC in nature, th

the same point. Truth is also that the lower end of the R4 and Cs as well as both electrolyt

ou'll find it under the name "Vbe multiplier". It serves as a "sensor" for the temperature o

to the base of T9. To be safe, let it be 0.75V. For that voltage to appear on the base of T9,

he end connected to base of T9, we will actually make a short circuit between the base and


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t is not a "mystery", although it might seem like that for beginners. Amplifier model that

in class AB and producing 50W average power on 8 ohm loads. You need to first determine

we can count with about 1.5 V on each of them, i.e. together that is approximately 3V. Als

d there will suit those which have that parameter in the range of 7...10A. Maximum voltag


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can calculate the maximum collector current of the drivers as:


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h it creates a voltage drop of 1.5 V, and value of R2 we can get from

base-emitter voltage of T6 higher than the voltage over R2 and is calculated as:

oltage drop of 1.2 V and compute its value from:


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ier, which as the input stage have a differential pair (or "Long Tail pair" as you will find in

crease the currents of T7 and T8 to for example 1.2 mA each, and introduce the so-called "

ge of T5 and we can count R8 as:

h more complex solutions that really made improvements but are beyond the scope of this


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tance of R6 and R7 determines the voltage gain of the amplifier and we choose it to be aro

serves primarily to prevent the introduction of any DC voltage that may exist on the output

ey have relatively little resistance, but which gradually rise up as the frequency goes lowe

ic in the positive and negative rail of the power supply are also going to ground, BUT THOS

the output transistors and the purpose is to compensate-regulate quiescent current (BIAS

we have to turn Rv slider completely on one end that is connected to the collector of the T

emitter of T9, which will stop its conduction and its collector-emitter junction will behave


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as used as an example for this explanation is a "stripped" basic concept that is commonly

the peak output voltage on the speaker as:

, do not forget that most amplifiers are powered from unstabilised power supplies which

between collector and emitter of output transistor we have at the moment when the com


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the literature on the Internet) work mostly with the use of voltage feedback, which is actu

mitter degeneration" by inserting the resistor Re in their emitters whose practical value is

tutorial.


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und 20 and from that the value of R7 is calculated as:

of some pre-stage which preceded this amp, because any additional DC would completely

. Since they are in series with R7, whose resistance determines the gain by the ratio to the

E TWO GROUNDS ARE NOT THE SAME POINT AND SAME COPPER TRACK ON THE PCB! The fi

) through them, that would vary significantly during the operation if there is no circuit wit

. At that position Rv behaves as an ordinary resistor and knowing that it form a divider wit

s a very high resistance. Because of that, voltage between the bases of T3 and T4 will be


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seen today. It consists of the input differential pair, voltage amplification stage (VAS) in cl

ill (even when they are very generously dimensioned) have output voltage falling by at lea

lementary transistor is inactive, is then:


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isrupt DC operating points of the entire amplifier. You should use highest quality capacito

resistance of R6, for C1 you should use a value large enough to prevent drop of the gain of

rst is the so-called "input signal ground and it is connected with separate wire to the cen

T9. Bipolar transistors have so-called "positive temperature coefficient," which simply me

h R12, and total resistance of divider is 5.7 k, value of Rv we can calculate as:

uch higher than necessary 2,8-2,9 V, which will in turn produce that the output transistor


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ass A with a constant current source, a complementary pair driver stage and complementa

st 2-2.5 V at full load. The sum of these voltages will give us the required negative half of t


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differential pair transistors) and this circuit is actually performing a comparison of input si

nation of purpose and calculation of these resistors is beyond the scope of this simplified t


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there (preferably NOT an electrolytic) and its value shall not be less than 470nF to avoid

whole amplifier on those low frequencies. Therefore, the C1 shouldn t have a value of les

ral grounding point called star ground , which is usually between the two main large el

ans that their collector current increases with warming, and that leads to further increase

immediately begin to "draw" very high current, and probably blow of immediately! To pre


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y output stage in the emitter-follower configuration. Once again I have to strongly highlig

he supply voltage (on standby), which will amount to:


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nals and scaled down but faithful sample of output signals from amplifier speaker output

torials, and it is enough to know that their insertion in emitters significantly linearise ope


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loss of the lowest frequencies.

s than some 47-100uF in general, but in practice there is no need to go over 470-1000uF. C

ectrolytic capacitors in the power supply unit. The "other" ground of R4, Cs and so on, is al

f the heat and that will very quickly lead to overheating and failure of the output devices.

vent that, it is better to use lower resistance value for Rv, for example something like 250


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t that the calculation is very simplified, but for the beginning... for "getting started" it is q


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nd it practically attempts to correct the differences that exist between those two signals.

ration of this stage and also significantly reduce distortion and additionally, differences in


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apacity of C1a (usually around 100nF) does not contribute much to the overall capacity of

o routed to the same point but with another-separate wire. This will reduce to a minimum

In order to "sense" the change of temperature, the transistor T9 is mounted so that it is in

r 500 ohms, and in series with Rv (from his top side to the basis of T9) insert another resi


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practical construction are done correctly. That won't be a "High End" design, but honestly


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ll stages during the process of amplification because each one is adding a little "dirt" on th


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apacitor, which can affect the gain at high frequencies.

llector and emitter of T9 are connected to the bases of driver transistors and there should


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speaking, NO ONE SHOULD START THE FIRST STEPS IN AUDIO ELECTRONIC WITH High-E


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e main signal. Therefore, the differential input stage is also the error amplifier at the same


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d! (Frankly, most people would like to make ultimate amplifier immediately ... but in that c


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time, so it is important to make sure that at least that stage contributes as little distortion


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rs are sufficiently "open" to start conducting a quiescent current. As the main heatsink is h


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ase they will most likely start with disappointments, because for High-End one needs a lot


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eating more and more during the operation of the amp, that change in temperature will be


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of knowledge and experience!).


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transferred on T9 too, causing its internal resistance to drop which would cause a reductio


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the collector and emitter, T9 we have to set up a proper operating condition for that transi


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stor, and that is done by insertion of preset Rv and resistor R12. Serial connection of the t


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o resistors is situated between the bases of T3 and T4 and we have seen that voltage bet


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een these two points should be about 2,8-2,9 V. In this circuit transistor T9, preset Rv and


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R12 passes all current that is going through transistor T5 which we have already said that


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it is 10mA. Through a serial link Rv and R12 should not pass more than about 1/20 of that


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total current of T5 which means about 0.5 mA and the total resistance Rv + R12 can be cal


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culated as:


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