Low frequency power amplifier

Low frequency power amplifier
Power amplification is an energy conversion circuit. Under the effect of the input signal, the transistor converts the energy of the DC power supply into output power that changes with the input signal and sends it to the load.
(1) The output power should be large: To increase the output power of the amplifier, the transistor must be operated near the limit working area, which is determined by ICM, UCM and PCM.

Figure 1
(2) The efficiency η is higher: the efficiency η of the amplifier is defined as: η = AC output power / DC input power (3) The nonlinear distortion is within the allowable range: Because the power amplifier works under a large signal, nonlinear distortion is inevitable The problem is to control the distortion within the allowable range,
The power amplifier can be divided into the following types according to the working state and circuit form:
(1) Class A power amplifier: During the entire signal period, there is collector current;
(2) Class B power amplifier: within half a signal period, there is a collector current, which can be divided into:
1) Double-ended push-pull circuit (DEPP)
2) Single-ended push-pull circuit (SEPP)
3) Balanced transformerless circuit (BTL)
In practice, in order to overcome the crossover distortion, the push-pull Chang body tube circuit is working in Class A and B states.
1. Class A power amplifier

Figure 1 is a class A power amplifier, the load RL becomes the collector load RL = n through the impedance converter B RLo For DC, the primary DC resistance of transformer B and Re are very small, so the DC load line is close to a vertical line. See Figure 1 (b). To make the amplifier output more power, the AC load line can be placed at points a and b Position: Uce = UCM at point a, and operating point Q is at the midpoint of ab straight line, usually the saturation voltage drop and penetration current of the transistor are very small, in fact, Icmin = 0 and Ucemin = 0o Therefore, the current supplied to the load And voltage amplitude are:
Icm = IcM / 2, Ucem = UCM / 2 --------------------------------------- ---------Formula 1
The AC power (or amplifier output power) of the load is:
PL = (UceM / ) × (IcM / ) = (IcM / ) × (UcM / ) = (1/8) IcM × UcM ------ Equation 2
The collector current ICQ and voltage UceQ of operating point Q are:
ICQ = ICM / 2, UceQ = Ec = UCM / 2 ------------------------------------- -------- Form 3
Therefore, the input power of the DC power supply:
PD = IcQ × UceQ = (ICM / 2) × (UCM / 2) = 1 / 4IcMUcm ----------------------------- --Form 4
The efficiency of the Class A power amplifier is:
η = PL / PD = 50% ------------------------------------------ --------------- Form 5

It can be seen that: (1) The maximum collector voltage of the transistor is twice the power supply voltage EC.
(2) The static power consumption of the transistor is twice the output power.
(3) The maximum efficiency of Class A amplifier is only 50%.
2. Class B push-pull circuit
Figure 2 (a) is a class B push-pull circuit. Because the output uses a transformer, the transistor has two output terminals to ground. The circuit is completely symmetrical. When the input signal Us is a positive half wave, BG1 is turned off and BG2 is turned on. The voltage UL is a negative half-wave. Therefore, the two tubes are turned on in turn and work with one push and one pull, so it is called a push-pull circuit.
Since the two tubes work alternately, the output characteristics of the two tubes are stacked together in opposite directions. The AC load lines of the two tubes are connected in a straight line ab, and the operating point Q is at the midpoint of the straight line ab, as shown in Figure 2 (b). It can be seen from the figure that the relationship between the amounts of electricity:
(1) If the turns ratio of the primary and secondary windings of the output transformer is n, then the load resistance RL of each transistor is:
RL = (n / 2) RL = (n / 4) RL -------------------------------------------- Equation 6
The resistance RCC between the collector and the collector is
Rcc = n RL = 4RL ----------------------------------------------- ------ Form 7
(2) The voltage amplitude of the primary winding end of transformer B2 is:
Ucem = UceQ≈Ec --------------------------------------------- ------- Form 8
The primary winding current amplitude is:
Icm = IcM ----------------------------------------------- ----------- Form 9
So the power delivered to the primary winding is:
Ps = (Ucem / ) × (Icm / ) = (1/2) EcIcm ------------------------------ Equation 10
(3) The average value of the current through each transistor is:
Ico = IcM / π --------------------------------------------- ---------- Form 11
The power supplied by the DC power supply is
PD = (2Ico) Ec = 2 × (Icm / π) × Ec ---------------------------------- ---- Form 12
(4) The efficiency of the push-pull circuit is:
η = (Ps / PD) 100% = {(1/2 × Ec × Icm) / [2 × (Icm / π) × Ec]} 100% ≈78.5% ----- Eq. 13
Pay attention to when designing the push-pull circuit:
(1) To avoid crossover distortion, the transistor should have a certain bias current, but not too large, otherwise the circuit efficiency will be reduced.
(2) The maximum collector voltage of the transistor Ucm> 2Ec.
(3) The dissipated power of the transistor Pcm≥1.2Pc1, where Pc1 is the power that each transistor sends to the transformer B2 primary, that is, Pc1 = [(1/2) Pso]
(4) According to the requirements of Pc1 and Ec1, calculate the transistor load resistance PL and the turns ratio n of the output transformer.

figure 2
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