How the audio amplifier works

Introduction

Good afternoon, dear habrayuzer, I want to tell you about the basics of building audio frequency amplifiers. I think this article will be interesting to you if you have never worked on electronics, and of course it will be funny to those who do not part with the soldering iron. And so I will try to tell about this topic as easily as possible and unfortunately omitting some of the nuances.

An audio amplifier or a low frequency amplifier, in order to figure out how it all works and why there are so many transistors, resistors and capacitors, you need to understand how each element works and try to find out how these elements are arranged. In order to assemble a primitive amplifier, we need three types of electronic elements: resistors, capacitors, and of course transistors.

Resistor

So, our resistors are characterized by resistance to electric current and this resistance is measured in Ohms. Each electrically conductive metal or metal alloy has its own specific resistance . If we take a wire of a certain length with a large resistivity, then we get a real wire resistor. In order for the resistor to be compact, the wire can be wound on the frame. Thus, we will have a wire resistor, but it has several disadvantages, so the resistors are usually made of a cermet material. This is how the resistors on electrical circuits are designated:

The upper version of the designation adopted in the US, lower in Russia and in Europe.

Capacitor

A capacitor consists of two metal plates separated by a dielectric . If we apply a constant voltage to these plates, an electric field will appear, which, after turning off the power, will maintain positive and negative charges on the plates, respectively.

The basis of the design of the capacitor - two conductive plates, between which there is a dielectric

Thus, the capacitor is able to accumulate electric charge. This ability to accumulate electrical charge is called electrical capacitance , which is the main parameter of the capacitor. Electrical capacity is measured in Farad. What else is characteristic is that when we charge or discharge a capacitor, an electric current flows through it. But as soon as the capacitor is charged, it ceases to pass an electric current, and this is because the capacitor took charge of the power source, that is, the potential of the capacitor and power source are the same, and if there is no potential difference (voltage), there is no electric current. Thus, a charged capacitor does not pass a direct electric current, but passes an alternating current, since when it is connected to an alternating electric current, it will be constantly charged and discharged. On electrical circuits it is designated as:

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Transistor

In our amplifier, we will use the most simple bipolar transistors . The transistor is made of a semiconductor material . The property we need for this material is the presence of free carriers of both positive and negative charges in them. Depending on what charges more, semiconductors are distinguished into two types according to conductivity: n- type and p- type (n-negative, p-positive). Negative charges are electrons released from the outer shells of the atoms of the crystal lattice, and positive charges are the so-called holes. Holes are vacancies that remain in the electron shells after electrons leave them. Conventionally, we denote atoms with an electron on the outer orbit with a blue circle with a minus sign, and atoms with a vacant place - an empty circle:

Each bipolar transistor consists of three zones of such semiconductors, these zones are called the base, emitter and collector.

Consider the example of the transistor. To do this, we connect to the transistor two batteries for 1.5 and 5 volts, plus to the emitter, and minus to the base and the collector, respectively (see the figure):


At the contact of the base and the emitter, an electromagnetic field will appear that literally pulls electrons from the outer orbit of the base atoms and transfers them to the emitter. Free electrons leave holes behind them, and occupy vacant places already in the emitter. The same electromagnetic field has the same effect on the collector atoms, and since the base in the transistor is quite thin relative to the emitter and collector, the collector electrons pass easily through it into the emitter, and in much larger quantities than from the base.

If we disconnect the voltage from the base, then there will be no electromagnetic field, but the base will act as a dielectric, and the transistor will be closed. Thus, when a sufficiently low voltage is applied to the base, we can control the larger applied voltage to the emitter and collector.

The pnp type transistor we considered is because it has two p- zones and one n- zone. There are also npn transistors, the principle of operation in them is the same, but the electric current flows in them in the opposite direction than in the transistor considered by us. This is how bipolar transistors are indicated on electrical circuits, the arrow indicates the direction of current:

ULF

Well, let's try to design from this a low-frequency amplifier. First, we need a signal that we will amplify, it can be a computer sound card or any other audio device with a linear output. Suppose our signal with a maximum amplitude of about 0.5 volts at a current of 0.2 A, something like this:



And to make the simplest 4-ohm 10-watt loudspeaker work, we need to increase the signal amplitude to 6 volts, with a current strength of I = U / R = 6/4 = 1.5 A.

So, let's try to connect our signal to the transistor. Remember our circuit with a transistor and two batteries, now instead of a 1.5 volt battery we have a line-out signal. Resistor R1 serves as a load, so that there is no short circuit and our transistor is not burned.



But then there are two problems at once, firstly our npn- type transistor, and opens only with a positive half-wave, and with a negative closes.



Secondly, a transistor, like any semiconductor device, has non-linear characteristics with respect to voltage and current, and the lower the current and voltage values, the stronger these distortions are:



Not only did the half-wave remain of our signal, it will also be distorted:


This is the so-called stair type distortion.

To get rid of these problems, we need to shift our signal to the working area of ​​the transistor, where the entire sine wave of the signal will fit and the nonlinear distortion will be negligible. To this end, a bias voltage, say 1 volt, is applied to the base using a voltage divider composed of two resistors R2 and R3.



And our signal entering the transistor will look like this:



Now we need to remove our useful signal from the collector of the transistor. To do this, install the capacitor C1:



As we remember, the capacitor passes an alternating current and does not pass a constant current, therefore it will serve us as a filter passing only our useful signal - our sinusoid. A constant component is not passed through the capacitor will be scattered on the resistor R1. The alternating current, our useful signal, will tend to pass through a capacitor, so the resistance of the capacitor for it is negligible compared with the resistor R1.

So we got the first transistor cascade of our amplifier. But there are two more little nuances:

We do not know 100% what signal enters the amplifier, all of a sudden the signal source is faulty, anything happens, again static electricity or a constant voltage passes with the useful signal. This may cause the transistor to not work properly or even cause it to break. To do this, we will install a capacitor C2, it will, like a capacitor C1, block a constant electric current, and the limited capacitance of the capacitor will not let through peaks of large amplitude that could damage the transistor. Such voltage surges usually occur when the device is turned on or off.



And the second thing, any signal source requires a certain specific load (resistance). Therefore, the input resistance of the cascade is important for us. To adjust the input resistance, add a resistor R4 to the emitter circuit:



Now we know the purpose of each resistor and capacitor in the transistor cascade. Let's now try to calculate which element values ​​to use for it.

Initial data:

• U = 12 V - supply voltage;
• U be ~ 1 V - Voltage emitter-base of the working point of the transistor;

Choosing a transistor, suitable for us npn transistor 2N2712

• P max = 200 mW - maximum power dissipation;
• I max = 100 mA - maximum constant current collector;
• U max = 18 V - maximum allowable voltage collector-base / collector-emitter (We have a supply voltage of 12 V, so that is enough with a margin);
• U EB = 5 V - maximum allowable voltage emitter-base (our voltage is 1 volt ± 0.5 volts);
• h21 = 75-225 - base current gain, the minimum value is 75;

1. We calculate the maximum static power of the transistor, it takes 20% less than the maximum power dissipation, so that our transistor does not work at the limit of its capabilities:
   
   P st.max = 0.8 * P max = 0.8 * 200mW = 160 mW;
   
   
2. Let us determine the collector current in the static mode (without a signal), despite the fact that no voltage is applied to the base through the transistor anyway, electric current still flows to a small extent.
   
   I k0 = P Art .max / U ke , where U ke is the collector-emitter junction voltage. The transistor dissipates half the supply voltage, the second half will dissipate on the resistors:
   
   U ke = U / 2;
   
   I 0 = P st.max / ( U / 2) = 160 mW / (12V / 2) = 26.7 mA;
   
   
3. Now let's calculate the load resistance, initially we had one resistor R1, which fulfilled this role, but since we added the resistor R4 to increase the input resistance of the cascade, now the load resistance will be added from R1 and R4:
   
   R n = R1 + R4 , where R n - the total resistance of the load;
   
   The ratio between R1 and R4 is usually taken 1 to 10:
   
   R1 = R4 * 10;
   
   Calculate the load resistance:
   
   R1 + R4 = ( U / 2) / I k0 = (12V / 2) / 26.7 mA = (12V / 2) / 0.0267 A = 224.7 Ohm;
   
   The nearest resistors are 200 and 27 ohms. R1 = 200 Ω, and R4 = 27 Ω.
   
   
4. Now we find the voltage on the collector of the transistor without a signal:
   
   U 0 = ( U ke0 + I 0 * R4 ) = ( U - I 0 * R1 ) = (12 –0.0267 * 200 Ohm) = 6.7 V;
   
   
5. Transistor control base current:
   
   I b = I to / h21 , where I to - collector current;
   
   I to = ( U / R n );
   
   I b = ( U / Rn ) / h21 = (12V / (200 Ohm + 27 Ohm)) / 75 = 0.0007 A = 0.07 mA;
   
   
6. The total base current is determined by the bias voltage on the base, which is set by a divider R2 and R3 . The current set by the divider should be 5-10 times the base control current ( I b ), so that the base control current itself does not affect the bias voltage. Thus, for the value of the current divider ( I cases ), we take 0.7 mA and calculate R2 and R3 :
   
   R2 + R3 = U / I = 12V / 0.007 = 1714.3 Ω
   
   
7. Now we calculate the voltage at the emitter in the resting state of the transistor ( U e ):
   
   U e = I K0 * R4 = 0.0267 A * 27 Ohms = 0.72 V
   
   Yes, I k0 is the quiescent current of the collector, but this same current also passes through the emitter, so that I k0 is considered the quiescent current of the entire transistor.
   
   
8. Calculate the total voltage on the base ( U b ) taking into account the bias voltage ( U cm = 1V):
   
   U b = U e + U cm = 0.72 + 1 = 1.72 V
   
   Now, using the voltage divider formula, we find the values ​​of resistors R2 and R3 :
   
   R3 = ( R2 + R3 ) * U b / U = 1714.3 Ω * 1.72 V / 12 V = 245.7 Ω;
   
   The nearest nominal resistor is 250 Ohm;
   
   R2 = ( R2 + R3 ) - R3 = 1714.3 Ohm - 250 Ohm = 1464.3 Ohm;
   
   The nominal value of the resistor is chosen in the direction of decrease, the nearest R2 = 1.3 kΩ.
   
   
9. Capacitors C1 and C2 usually set at least 5 microfarads. Capacity is chosen so that the capacitor does not have time to recharge.
   

Conclusion

At the output of the cascade, we receive a proportional amplified signal both in current and voltage, that is, in power. But one cascade is not enough for the required gain, so we have to add the next and the next ... And so on.

The considered calculation is rather superficial and such a gain scheme is of course not used in the structure of amplifiers, we should not forget about the range of transmitted frequencies, distortions and much more.