How to Avoid OCL Crossover Distortion
01 Composition of OCL schema and how it works
To eliminate crossover distortion created by basic OCL circuit, it is necessary to set an appropriate static operating point so that both amplifying transistors operate in a critical state of conduction or microconduction. An OCL circuit that can eliminate crossover distortion is shown in Figure 1.
Figure (1) OCL Circuit for Eliminating Crossover Distortion
In above figure, in a static state, a DC current flows from +12V through R5, R6, D1, D2, R7, R8 to GND, and voltage generated between two bases Q1 and Q2 is Ub1b2=UR6. +UD1+UD2+UR7, so that Ub1b2 is slightly larger than sum of emitter junction turn-on voltages of Q1 and Q2, so that both tubes are in a weakly conductive state, that is, there is a small current base, Ib1 and Ib2, respectively. Adjusting R6 and R7 can make static emitter potential Vout equal to 0V, that is, output voltage Vo is 0V.
When applied signal Ui changes according to law of forward rotation, since dynamic resistances of diodes D1 and D2 are very small, and resistance values R6 and R7 are also relatively small, we can assume that potential change in base of tube Q1 is same as that at base of tube Q2. The pole potentials are approximately equal, that is, Ub1≈Ub2≈Ui.
That is, we can assume that potential difference between bases of two tubes is basically a constant value, and potentials of two bases change in same way with Ui. Thus, with Ui > 2.5 V and a gradual increase in Vbe1 gradually increases, base current Ib1 of transistor Q1 increases accordingly, emitter current Ie1 inevitably increases, and load resistance RL receives current in positive direction; at same time, an increase in Ui leads to a decrease in Veb2, and when it decreases to a certain value, tube Q2 stops. In same way, when Vi<2.5v and gradually decreases, Veb2 gradually increases, base current Ib2 of transistor Q2 increases accordingly, and emitter current Ie2 will inevitably increase, and load resistor RL current in negative direction; and At same time, a decrease in Ui leads to a decrease in Vbe1, and when it decreases to a certain value, tube Q1 is cut off. Thus, even if Ui is very small, at least one crystal transistor can always be guaranteed to be on, which eliminates crosstalk. A graphical analysis of input characteristics of lamps Q1 and Q2 under influence of Ui is shown in Figure 3:
Figure (1.1) Graphical analysis of input features
Below static operating point of Q1 and Q2, input signal gets larger and smaller until Q1 stops, Ui has no signal, and Q1 returns to static operating point current.
The smaller input signal, larger it is. When Q2 stops and Ui is very small (no signal), Q2 returns to static current operating point.
So, positive half cycle of input signal is mainly driven by Q1 tube emitter, and negative half cycle is mainly driven by Q2 tube emitter, and conduction time of two tubes is longer than half input signal cycle. , so they work in class A and B state.
It is worth noting that if static operating point is not adjusted, for example, any component in R6, D1, D2 and R7 is soldered, then from 12V through emitter junction R5 and Q1, R9R10, and emitter junction from Q2 to R8 forms path to GND, with large base currents Ib1 and Ib2 flowing, resulting in large collector currents Ic1 and Ic2 for Q1 and Q2, and maximum voltage drop across VCE tube of each tube is approximately 12V. so tubes Q1 and Q2 can be damaged due to excessive power consumption. Therefore, role of R9 and R10 is very important, which can divide voltage drop of VCE into Q1Q2.
02 Output power and efficiency of OCL circuit
The most important technical indicators of a power amplifier circuit are maximum output power Pom and efficiency η (Ita) of circuit. To solve Pom, you must first find maximum output voltage amplitude that can be obtained at load. When input voltage is large enough and there is no saturation distortion, circuit analysis is shown in Fig. 3.2.
Region I in figure is Q1 output characteristic and region II is Q2 output characteristic. Since quiescent current of two lamps is very small, it can be considered that operating quiescent point is on horizontal axis, as noted in figure above, so maximum output voltage amplitude is equal to power supply voltage minus saturation. transistor voltage drop, i.e. (Vcc-Vces1).
In fact, same conclusion can be reached without plotting. It can be imagined that in positive half-cycle of sinusoid, Ui gradually increases from 0, and output voltage gradually increases accordingly. The voltage drop across CE tube on Q1 tube should decrease gradually. When voltage drop across tube drops to saturation voltage drop. The output voltage reaches its maximum value, and its value is (Vcc-Vces1), so effective value of maximum undistorted output voltage is:
Uom=(Vcc-Vces1)/ , assuming that parameters of transistor are same, saturation voltage drop is also same, that is, Vces1=-Vces2=Vces.
Maximum output power: Pom=Uom^2/R=(Vcc-Vces)^2/2RL+2R9 or loop resistance R10 2RL+2R9 in one cycle
In case of ignoring base loop current, current provided by Vcc power supply:
When load receives maximum AC power, average power consumed by power supply is equal to product of average current and power supply voltage,
Pv=1/Π (Vcc-Vces)/RL sinwt*Vcc dwt=2/Π*Vcc (Vcc-Vces)/RL
Received after completion, conversion efficiency
In an ideal situation, i.e. voltage drop of saturated tube is negligible, and resistors R9 and R10 are relatively small and negligible (negative emitter feedback resistance of Q1 and Q2), in case
Here, it should be noted that pressure drop of a saturated high power tube is 2-3v, so pressure drop of a saturated tube cannot be neglected at all, that is, above three formulas cannot be used.
03 Selection of transistors in OCL circuits
In a power amplifier circuit, transistor should be selected according to maximum lamp voltage drop Vces, maximum collector current Icm, and maximum power consumption.
1. Maximum pressure drop in a pipe
From analysis of principle of operation of OCL circuit, it can be seen that among two lamps of power amplifier, lamp with Q1 and Q2 in off state will carry a greater pressure drop in tube. Assuming that input voltage Ui is a positive half-cycle, Q1 is on and Q2 is off, As Ui increases from 0 to peak value, emitter potential Ve of lamps Q1 and Q2 gradually increases to (VCC-Vces1) because voltage drop across tube tube Q2 The value of Vec2 is Vec2=(Ve-0)=Ve, Vce2max=Vcc-Vces1. Since average current Ie is relatively small and resistance R9R10 is relatively small, voltage drop generated by these two resistors is initially ignored. .Using same parse method for parsing, we can get:
When Ui is value of second half cycle, Q1 tube withstands maximum tube pressure drop, and value is VCC-Vces2. Therefore, it is considered that a certain margin and maximum pressure drop is reserved, which tubular bears /Vcemax/=Vcc.
2. Maximum Collector Current
From analysis of maximum output power of circuit, it can be seen that emitter current of transistor is equal to load current, and maximum voltage across load resistor is Vcc-Vces1, so maximum value of collector current is:
Consider leaving some stock
3. Maximum Collector Power
In power amplifier circuit, power provided by power supply, excluding converted output power, is mainly consumed by power transistors Q1 and Q2. The power lost by transistor can be considered to be Pq=Pv-Po. When drive signal input voltage is 2.5V, that is, when output power is smallest, tube loss is very small because collector current is very small; losses are also very small, it can be seen that largest losses in lamp will not occur even whenthe lowest voltage, nor highest input voltage. The relationship between collector power consumption Pq of transistor and peak output voltage Vom is given below, and then expression of voltage drop across lamp and instantaneous value of collector current is calculated for Vom:
The power consumption Pq is average power consumed by power amplifier tubes Q1 and Q2, so expression for collector power consumption of each transistor is as follows:
Instantaneous maximum tube voltage drop*instantaneous current then average
Pq=1/2Π (Vcc-Vomsin wt)*Vom/RL*sin wt*dwt
Assuming dPq/dVom=0, we can get Vom=2/Π*Vcc≈0.6Vcc.
The above analysis shows that when Vom≈0.6 Vcc, Pq=Pqmax. Substitute Uom into Pq==1/RL(VccVom/Π-Vom^2/4)
When Vces=0, according to Pom=Vom^2/RL=Vcc^2/2RL
It can be seen that maximum collector power consumption of transistor is only one-fifth of maximum output power in an ideal state (saturation voltage drop is 0).
When referring to a transistor selection guide, use limit parameters
Pcm>0.2Pom/Vces=0, collector power consumption Pcm
Here it is also necessary to emphasize that when choosing a transistor, its limiting parameters, especially Pcm, must have a certain margin, and strictly follow manual layout of printed circuit board or install a radiator.
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