As shown in Figure (01), there are two batteries and a resistor in series. In this circuit, does battery E1 charge battery E2, or does battery E2 charge battery E1?
Figure (01)
Let's measure voltage U1 and voltage U2. If voltage U1 is greater than voltage U2, we can determine that the potential at the left end of resistor R is higher than the potential at the right end. Obviously, the direction of the current in the resistor is from left to right, as shown in Figure (02). At this point, we can determine that battery E1 is charging battery E2. Of course, if the results of measuring U1 and U2 are opposite, then E2 is charging E1.
E1 charging E2 means that energy flows out of E1 and flows into E2.
Figure (02)
From Figure (02), we can see that: when the voltage direction is opposite to the current direction (such as the left side of the figure above), energy flows out of E1, and when the voltage direction is the same as the current direction (such as the right side of the figure above), energy flows into E2.
Is this a law that can be universally applied? According to the definition of voltage, this is indeed a law that can be universally applied. This is true
not only in DC circuits, but also in AC circuits.
Figure (03) In a simple AC circuit, there is only an AC power source E and a resistor R as a load. We assume that the positive direction of the voltage and current in the circuit is as shown in the figure.
Figure (03)
We know that the current in a resistor has the same phase as the voltage across it. The waveform of the relationship between the voltage u across the resistor R and the current i therein is shown in Figure (04).
Figure (04)
Figure (04) waveform, in the first half of the cycle, the voltage u across the resistor and the current i through the resistor are both in the positive direction. As mentioned above, energy flows into the resistor (converted into heat). In the second half of the cycle, the voltage across the resistor is negative, and the current through the resistor is also negative. The two are still in the same direction, and energy flows into the resistor and is converted into heat.
Only at the moment when the AC voltage passes through zero, the voltage and current are both zero at the same time, and their product is also zero, and there is no energy transmission, but at other times, energy flows into the resistor.
However, if we replace the resistor in Figure (03) with an inductor as shown in Figure (05), the situation is very different. For simplicity, we will not consider the various losses in the inductor for the time being, and assume that the inductor L in Figure (05) is an ideal inductor. In addition, we only consider the simplest sinusoidal voltage waveform.
Figure (05)
We all know that the voltage u across the inductor leads the current i in the inductor. For an ideal inductor, the leading angle is 90°. The voltage and current waveforms in the circuit in Figure (05) are shown in Figure (06). The blue waveform in the figure is the voltage waveform across the inductor, and the red waveform is the current waveform through the inductor.
Figure (06)
The voltage and current waveforms in Figure (06) sometimes have the same direction and sometimes have opposite directions. To analyze the direction of energy flow in more detail, we divide one cycle of AC into several stages in Figure (07).
Figure (07)
In Figure (07), we start the analysis from the moment when the current changes from negative to positive. The quarter cycle after this moment is recorded as stage 1. In stage 1, although the voltage and current are constantly changing, the voltage is decreasing and the current is increasing, but both the voltage and current are positive (positive means the same direction as marked in Figure (05)). It can be concluded that in stage 1, energy flows out of the AC power supply E and flows into the inductor L.
The quarter cycle after stage 1 is recorded as stage 2. In stage 2, the voltage and current are also constantly changing, but the voltage is negative and the current is positive. The voltage and current are in opposite directions, so it can be concluded that in stage 2, energy flows out of the inductor L and flows into the AC power supply E.
The quarter cycle after stage 2 is recorded as stage 3. In stage 3, the voltage and current are still changing, but it can be seen from Figure (07) that in this stage, the voltage and current are both negative (opposite to the direction marked in Figure (05)). Both are negative and in the same direction, so it can be concluded that in stage 3, energy flows out of the AC power source E and flows into the inductor L.
The quarter cycle after stage 3 is recorded as stage 4. In stage 4, the voltage is positive and the current is negative, and the two directions are opposite. Therefore, it can be concluded that in stage 4, energy flows out of the inductor L and flows into the AC power source E.
Stage 5 is exactly the same as stage 1 and will not be repeated.
As mentioned above, for an ideal inductor, in each cycle of AC, there are two times (one quarter cycle each) when energy flows from the AC power source to the inductor, and two times (one quarter cycle each) when energy flows from the inductor to the AC power source. Moreover, due to the symmetry of the sine wave, the energy flowing from the AC power source into the inductor in stage 1 must be equal in quantity to the energy flowing from the inductor to the AC power source in stage 2.
Therefore, in the circuit composed of the AC power source and the ideal inductor as shown in Figure (05), energy flows back and forth between the power source and the inductor, and the reciprocating frequency is twice the frequency of the AC power. In one cycle of the AC power, the average value of the work done by the power source on the load is zero. Because of this, the product of the effective value of the AC voltage u and the effective value of the AC current i in Figure (05) does not represent the power of the power supply. This product is called apparent power. When the voltage and current phase difference in Figure (05) is 90°, the power emitted by the AC power supply is completely reactive power.
If the inductor in Figure (05) is replaced with an ideal capacitor, when the circuit reaches a steady state, the AC current will lead the voltage by 90°. The energy flow direction deduced from Figures (06) and (07) is still applicable in the capacitor circuit, that is, in one cycle of AC, energy flows from the power supply to the capacitor twice, and energy returns from the capacitor to the power supply twice. In one cycle of AC, the average value of the work done by the power supply on the capacitor is zero.
However, the ideal situation does not exist. The wire always has a certain resistance, and the actual inductor also always has a certain resistance. Such a load is shown in Figure (08).
In this case, the angle of current lagging behind voltage is less than 90°, but a value between 0 and 90°. Figure ( 09
) shows the voltage and current waveforms in this case.
Figure (09)
We can also consider a cycle of AC in several stages.
The time starts from the time when the current goes from negative to positive and crosses zero, and ends at stage 1 when the voltage goes from positive to negative and crosses zero. Stage 2 ends when the current goes from positive to negative and crosses zero. Stage 3 ends when the voltage goes from negative to positive and crosses zero. Stage 4 ends when the current goes from negative to positive and crosses zero.
In Figure (09), in stage 1, the direction of the voltage and current are the same, and the power supply E does work on the inductor L and the resistor R. In stage 2, the direction of the voltage and current are opposite, and the energy stored in the inductor L returns to the power supply, while a part of the energy is consumed in the resistor R. In stage 3, the direction of the voltage and current are the same, and the power supply E does work on the inductor L and the resistor R. In stage 4, the direction of the voltage and current are opposite, and the energy stored in the inductor L returns to the power supply, while a part of the energy is consumed in the resistor R.
We can see that for an inductive load containing a resistor, a part of the energy still returns to the power supply during a cycle of AC. However, unlike the ideal situation in Figure (05), the energy returned to the power supply in one cycle is only a small part of the power supply output energy, not all of it.
The question is: does the AC power supply allow energy to return to the power supply? This is not determined by default. Some power supplies allow energy to return, but others do not.
Figure (10)
Figure (10) is a very common audio push-pull power amplifier. T1 and T2 are two complementary triodes, SP is a speaker, and E1 and E2 are DC power supplies. For simplicity, the driving circuit diagram of T1 and T2 is not drawn.
The structure of a dynamic speaker is that a voice coil wound with copper or aluminum wire is placed in the magnetic field gap of a magnet. When current passes through the voice coil, it will be affected by the magnetic field force, pushing the diaphragm of the speaker to move and produce sound. A dynamic speaker is a typical load that contains both inductance and resistance.
We know that the two tubes of a push-pull power amplifier can work in Class A, Class B, or Class A and B.
Push-pull working in Class A means that there is always current in any tube in a cycle and it will not be interrupted, that is, there will not be a period of time when the current is zero in a cycle of AC. In other words, the current I1 in tube T1 (indicated by green) and the current I2 in tube T2 (indicated by blue) in Figure (11) will never be interrupted at any time. In Figure (11), we can see that the current I in the load is the difference between the current I1 in T1 and the current I2 in T2.
Push-pull operation in Class B means that the two tubes are turned on alternately, and at any time, there is always a zero current in one tube. In other words, the current I1 in T1 (indicated by green) and the current I2 in T2 (indicated by blue) in Figure (11) will never appear at the same time.
Push-pull operation in Class A and B means that the current I1 in T1 (indicated by green) and the current I2 in T2 (indicated by blue) in Figure (11) are sometimes interrupted and sometimes appear at the same time.
If the audio push-pull power amplifier in Figure (10) operates in Class A, then the dynamic speaker load SP, which contains both resistance and inductance, can return the energy in the inductor to the amplifier in stage 2 of Figure (09). In phase 1 of Figure (09), the voltage at point A in Figure (11) is positive, but at the beginning of phase 2, the voltage at point A turns negative. From Figure (09), we can see that the current in the speaker continues to flow in the positive direction. For a push-pull power amplifier operating in class A, this is not a problem: as long as the current I1 is slightly greater than I2, the current in the speaker SP can flow in the original direction (phase 1). In other words, for a class A push-pull power amplifier, the load is allowed to return energy to the power supply (push-pull power amplifier).
Figure (11)
But for class B push-pull power amplifiers, it is different. In class B push-pull power amplifiers, current I1 (green) and current I2 (blue) never appear at the same time. When I2 is generated, I1 must be zero. In stage 2 in Figure (09), transistor T1 in Figure (11) has been turned off, transistor T2 is turned on, and the current in SP still has to flow in the direction indicated by the red arrow in the figure. But this is impossible, because transistor T1 has been turned off, although T2 is turned on, T2 does not allow current to flow in the opposite direction, and can only flow from T2 emitter to collector. If the current in SP continues to flow in the direction of the red arrow in Figure (11), the transistor will be broken down.
However, most of the early transistor radios used push-pull power amplifier circuits and dynamic speakers as shown in Figure (10). Why didn't the transistors get damaged?
That's because the inductance of the first speaker is not strong, it is mainly resistive. If the nominal impedance of an ordinary dynamic speaker is 4 ohms, the resistance is about 3.2 ohms when measured with a multimeter. It can be seen that the dynamic speaker is still mainly resistive. The inductance is not strong, so the energy returned to the power amplifier will not be very large. Second, the triode is not necessarily damaged by breakdown, and it will only be damaged if the power loss exceeds the allowable value. The voltage regulator tube we commonly use works in the breakdown state for a long time, and it will not be damaged as long as the power loss is within the rated value. Third, this type of push-pull power amplifier works in Class A and Class B state, that is, when the current in T2 increases to a relatively small value, T1 is completely turned off. When there is current in both T1 and T2, the energy in the speaker is allowed to return to the amplifier.
However, "even if it is broken down, it will not be damaged" is limited to portable radios with an audio power of more than 100 mW, at most 1 to 2W. If the amplifier output power is large (of course, the matching speaker is also large), the energy returned from the speaker to the power amplifier is relatively large, and the tube in the power amplifier may be damaged. In particular, this type of power amplifier always makes the circuit work as close to Class B as possible to improve efficiency.
In fact, it can be seen from Figures (10) and (11) that to protect T1 and T2 from breakdown and damage, just connect a diode in anti-parallel with T1 and T2. After the diodes are connected in anti-parallel, the energy stored in the speaker can be returned to the DC power supply through the diodes, without causing the voltage across T1 and T2 to increase too much and break down and damage.
From the typical application circuits in the manuals of early audio power amplifier chips, the above-mentioned problem of speaker energy return can be seen. Figure (12) is the application circuit in the manual of the typical 18W output audio power amplifier chip TDA2030A. As can be seen in the figure, the amplifier output end (pin 4) is connected to the positive and negative ends of the power supply with an ordinary diode 1N4001. The reason for connecting these two diodes is to prevent the energy stored in the speaker from breaking down the power tube when it flows back to the push-pull power tube, because the power supply voltage of the audio amplifier chip with higher power often reaches the limit value, with not much surplus, and the speaker power is larger, so the energy that can be stored is also larger.
Figure (12)
So, why don't all audio power amplifier chips require these two diodes to be connected externally? This is because most audio power amplifier chips have these two diodes built into the chip, which is not difficult. For example, the LM12 chip with an output power of 80W has these two diodes built into the chip, as shown in the red circle in Figure (13).
Figure (13)
Figure (13) is the internal circuit of the audio power amplifier chip LM12. Q14 and Q15 are set to prevent the energy stored in the output load from flowing back into the amplifier and causing Q12 or Q13 to break down. In fact, Q14 and Q15 use their emitter junctions and connect the collector of the tube to the power supply.
提升卡
变色卡
千斤顶
京公网安备 11010802033920号
