Switching Power Supply Interest Group Task 08
[Copy link]
Question 07
1. If the current waveform in the switch tube in Figure (05) and Figure (06) of this article is the same as the waveform (c) and waveform (e) in Figure 2.1 of "Switching Power Supply Design 3rd Edition", do you think the current in the inductor L1 in the rectifier circuit in Figure (05) and Figure (06) of this article is in a continuous working state or a discontinuous working state? What is the reason?
The current in the inductor L1 in Figure (01)
must be in a continuous working state. If the current in the inductor L1 is intermittent, then the current waveforms of the switch tubes Q1 and Q2 in Figure (01) of this article, that is, Figure 2.1 of "Switching Power Supply Design Third Edition", will start from 0, that is, there is no initial current marked in red in the figure. In other words, the length of the red part is zero.
2. Please draw the current waveforms of the diodes D1 and D2 in Figure (08) or Figure (09) of this article based on the voltage waveform in Figure 2.9 of "Switching Power Supply Design Third Edition". Assume that the inductor L1 is very large, so that the current in it can be regarded as a constant current. Note: This is only a current waveform drawn based on the voltage waveform in Figure 2.9. The actual voltage waveform may be different from the waveform in Figure 2.9 of "Switching Power Supply Design Third Edition".
Figure (02)
Above the green horizontal line is the VI voltage waveform of Figure 2.9 of "Switching Power Supply Design 3rd Edition", and below is the current waveform of the two diodes D1 and D2.
We can see that: when the current in one diode is zero, the other diode passes the entire output current Io. This is consistent with the voltage waveform on the upper part of the green horizontal line. However, when the voltage waveform VI on the upper part of the green horizontal line is zero (actually a negative diode voltage drop), due to the existence of the filter inductor, the current must be continuous and still Io. At this time, the two diodes D1 and D2 each share half of the output current Io. Since the magnetic flux generated in the iron core by the two secondary windings where D1 and D2 are located is in opposite directions and cancels each other, the current of Io/2 in D1 and D2 does not affect the voltage across the two Ns windings to zero. For
the 08th event, please read Chapter 3 "Half-bridge and Full-bridge Converter Topologies" of "Switching Power Supply Design 3rd Edition".
Figure (03) in this article is reproduced from Figure 3.1 of "Switching Power Supply Design 3rd Edition". This is the so-called half-bridge converter topology.
Figure (03)
Figure (03) is Figure 3.1 of the third edition of Switching Power Supply Design. The feedback circuit for stabilizing the output voltage is not drawn therein.
In Figure (03), the first row is not a voltage or current waveform, but only represents the switching frequency of the switching power supply. The second and third rows are not waveforms either, but only represent the conduction and shutdown of the two switching tubes Q1 and Q2. High means conduction, and low means shutdown. It can be seen from the figure that Q1 and Q2 are alternately turned on, which is exactly the same as the push-pull circuit we discussed last time. Therefore, the working principles of the half-bridge circuit and the push-pull circuit are roughly the same.
Two left and right bidirectional arrows are drawn on the trailing edge of the conduction and shutdown of the second and third rows. These two left and right arrows represent the change in the duty cycle of the switching tube. In fact, when the PWM duty cycle decreases, the switching tube is turned off in advance, which is represented by the left arrow. However, whether the duty cycle increases or decreases, the duty cycles of the two switching tubes must change at the same time to ensure that the conduction time of the two switching tubes is the same when the switching power supply reaches a steady state.
As can be seen from Figure (03), the two switches are connected in series to the DC input power supply, so some books call this circuit a series push-pull circuit. However, it should be noted that the two switches are connected in parallel to the AC, that is, the primary winding Np of the transformer T1.
In the circuit of Figure (03), the rectifier filter circuit connected to the secondary of the transformer is no different from the push-pull switching power supply. It is still a full-wave rectifier or bridge rectifier inductor input filter circuit, and the current in the inductor is usually continuous, and the current discontinuous working mode is extremely rare.
Like the push-pull circuit, the half-bridge circuit absolutely does not allow the two switches to overlap in conduction time. If overlap occurs, it is "common conduction", and the two switches will almost short-circuit the 320V DC power supply, and a large current will flow through Q1 and Q2. Moreover, the energy represented by this large current is completely lost in the switch tube, causing the two switch tubes to heat up seriously. If the time is slightly longer (a few microseconds), the two switch tubes will burn out. In fact, all push-pull switching power supply circuits do not allow the conduction time of the two switch tubes that are alternately turned on to overlap.
In the half-bridge circuit, the transformer primary excitation current is reset by the anti-parallel diode on the other side of the switch tube that was once turned on and is now turned off.
In Figure (03), four diodes are used to form a bridge rectifier at the AC mains input, but a switch S1 is added at the connection of D3D4, so that the rectifier circuit can be used for both AC 220V and AC 120V. When switch S1 is disconnected, D1~D4 is an ordinary bridge rectifier circuit, and C1 and C2 are connected in series and used as energy storage filtering. In this case, the maximum voltage across the two ends of C1 and C2 after being connected in series is the peak value of the AC input voltage. When switch S1 is closed, diodes D3 and D4 are connected in anti-parallel with C1 and C2 respectively and do not work. Diodes D1 and D2 and capacitors C1 and C2 form a full-wave voltage doubler rectifier circuit. In this case, the maximum voltage across the two ends of C1 and C2 after being connected in series is twice the peak value of the AC input voltage.
In the circuit of Figure (03), a capacitor Cb is connected in series with the primary winding Np of the transformer. It is this capacitor Cb that makes the half-bridge circuit very resistant to imbalance. When discussing the push-pull circuit last time, the author of Sections 2.2.5 to 2.2.8 of "Switching Power Supply Design, Third Edition" once said: If the conduction time of the two switch tubes in the push-pull circuit is slightly different due to the dispersion of the tube parameters, the transformer core will generate DC flux, which is the so-called "flux imbalance", causing the core to enter magnetic saturation, and for this reason, "using current mode topology" was proposed. However, this problem does not exist in the half-bridge circuit.
Assume that the two switch tubes in Figure (03) are slightly different, and the conduction time of Q1 is always slightly longer than that of Q2. Then this difference in conduction time will cause a DC voltage to appear across the capacitor Cb. When the conduction time of Q1 is slightly longer, the voltage across Cb is negative on the left and positive on the right. Thus, when Q1 is turned on, the voltage across C1 and the voltage across Cb are in a subtractive relationship, and the voltage obtained on winding Np is slightly smaller than the voltage across C1. When Q2 is turned on, the voltage across C2 and the voltage across Cb are in an additive relationship, and the voltage obtained on winding Np is slightly larger than the voltage across C2. Therefore, the voltage waveform across winding Np is shown in Figure (04). The rectangular wave above the horizontal axis lasts slightly longer but has a slightly smaller amplitude, while the rectangular wave below the horizontal axis lasts slightly shorter but has a slightly larger amplitude. The area of the upper rectangle is equal to that of the lower rectangle. The area of this rectangle is exactly the "volt-second product" we mentioned earlier. Therefore, due to the effect of capacitor Cb, the half-bridge circuit transformer will not have flux imbalance caused by unequal volt-second products, and will not cause core saturation due to flux imbalance.
Figure (04)
Netizens who are new to switching power supplies often do not understand this. They think that the filter capacitor has been divided into two parts, C1 and C2, in the power frequency rectifier circuit. Even if Cb is short-circuited, there will be no DC component in the primary of the transformer. Why should Cb be connected in series? This is because the capacity of C1 and C2 may not be completely equal. In a short period of time after the AC power is turned on, the half-bridge switching power supply is in an unstable state. Especially when the AC power is double-voltage rectified (switch S1 is closed), the two capacitors C1 and C2 must be charged first and then charged later. The capacitor that is charged first can be charged close to the peak value of the AC power input, while the voltage across the capacitor that is charged later is close to zero. The voltages of the two capacitors are not equal, and the power supply for the half-bridge is asymmetric. Moreover, this state in which one capacitor is charged and the other is not charged can last for half a cycle of the AC power, that is, 10ms. In 10ms, the half-bridge has already worked for hundreds of switching cycles. In this case, it is obvious that the excitation current of the primary of the transformer cannot be fully continued, and the magnetic flux in the iron core cannot be reset. During the period when the circuit has not reached stability, the primary current of the transformer may have a considerable DC component. This DC component may cause the transformer core to enter saturation, and the saturation of the transformer core is likely to damage the switch tube. To avoid this danger, it is better to connect a capacitor Cb in series with the primary of the transformer. This capacitor cannot use a polar electrolytic capacitor like C1C2, but must use a non-polar capacitor, and Cb must be able to withstand the high voltage at the operating frequency of the switching power supply (the voltage across the two ends can exceed half of the DC power supply voltage), so the loss of this capacitor should be relatively small. The capacity of this capacitor usually requires that the voltage drop on it should not exceed 10% of the primary voltage of the transformer. The capacity of this capacitor Cb is relatively large for non-polar capacitors. Note: The voltage drop on Cb cannot be too small, for example, the voltage drop on Cb is only one thousandth of the primary voltage of the transformer. Because the voltage drop on Cb is very small, the capacity of Cb must be very large. This will not only increase the cost, but also make the circuit load change or the AC mains voltage change cause the "flux imbalance" for a long time, and the flux imbalance time is too long will cause the core to enter saturation, resulting in damage to the power switch tube in the switching power supply.
An important feature of the half-bridge circuit is that the induced electromotive force generated by the transformer primary leakage inductance when the primary current changes can be eliminated without special measures. The half-bridge circuit relies on the diode D5D6 connected in anti-parallel with the switch tube Q1Q2 to continue the excitation current. However, while continuing the excitation current, the induced electromotive force generated by the transformer primary leakage inductance is also clamped by the diode D5D6, and the energy is returned to the DC power supply (actually charging the capacitor C1C2). Because of this, the half-bridge circuit is not like a push-pull circuit, such as RsDsCs in Figure 2.7 of "Switching Power Supply Design Third Edition", which also requires a clamping circuit specifically for the induced electromotive force of the leakage inductance. This is a feature of the half-bridge circuit and the full-bridge circuit.
Figure (05) in this article is reproduced from Figure 3.3 of "Switching Power Supply Design Third Edition". This is the so-called full-bridge converter topology.
Figure (05)
Figure (05) is Figure 3.3 of the third edition of "Switching Power Supply Design". The feedback circuit for stabilizing the output voltage is not drawn therein. In addition, please note that the directions of the two diodes D2 and D4 in Figure (05) are reversed . If connected in this way, smoke will appear immediately after power is turned on, or even the tube will explode. Please correct it in your mind.
The function of the switch S1 at the AC mains input in Figure (05) is exactly the same as that in Figure (03).
In Figure (05), the two switch tubes (Q1 and Q4, Q2 and Q3) on the diagonal of the full bridge are turned on and off at the same time. Therefore, the peak voltage across the primary of the transformer is not half of the DC power supply voltage, but the entire DC power supply voltage. Compared with the half bridge, the primary voltage of the transformer is doubled. Under the same current condition, the power transmitted to the secondary is also doubled. Of course, under the same transformer core cross-sectional area, the number of primary winding turns should also be doubled.
In fact, the full-bridge circuit can be seen as two half-bridge circuits working in opposite phases, or a combination of two half-bridges.
Obviously, the transformer excitation current must continue to flow when the two switches on a certain diagonal are turned off, and the continuous flow is completed by the diode connected in anti-parallel to the switch on the other diagonal. The induced electromotive force generated by the transformer primary leakage inductance is also clamped to the power supply voltage by the diode connected in anti-parallel to the switch on the other diagonal.
Like the half-bridge circuit, a capacitor Cb should also be connected in series to the primary of the full-bridge circuit transformer to avoid the problem of DC in the transformer primary in a short time due to unequal voltages on the two capacitors C1 and C2 or unequal capacitances of the two capacitors, thus causing the iron core to enter magnetic saturation.
The full-bridge circuit also does not have the problem of clamping the induced electromotive force generated by the transformer primary leakage inductance. The induced electromotive force generated by the leakage inductance is also clamped to the power supply voltage by the diode connected in anti-parallel to the power switch.
Both the half-bridge and full-bridge circuits transfer energy to the secondary through the transformer during the conduction of the switch, so like the push-pull circuit, they also belong to the forward working mode. After understanding the operation of the forward circuit, it should not be difficult to understand the operation of the push-pull, half-bridge and full-bridge circuits.
Question 08
1. This article says that in the full-bridge circuit of Figure (05), Q1 and Q4 are turned on and off at the same time, and Q2 and Q3 are turned on and off at the same time. If Q1 and Q4 have a slightly longer conduction time than Q4 due to the dispersion of tube parameters (but there is still a "dead zone", that is, the conduction time of Q1 and Q2 does not overlap), what is the impact on the operation of the full-bridge switching power supply? If the conduction time of Q1 and Q4 is equal, and the conduction time of Q2 and Q3 is also equal, but the conduction time of Q1 is slightly longer than the conduction time of Q2?
2. If the full-bridge circuit in Figure (05) of this article has four switch tubes Q1~Q4 in the switching state that Q1Q2 is turned on alternately, and Q3Q4 is also turned on alternately, but the left and right sides of the four switch tubes of the full-bridge work simultaneously, that is, Q1 and Q3 are turned on and off at the same time, and Q2 and Q4 are turned on and off at the same time, what kind of waveform should the voltage across the primary of the transformer be? As
mentioned earlier, the full-bridge circuit in Figure (05) can be regarded as the left and right half-bridge circuits working in reverse phase, and the working state in which Q1 and Q3 are turned on and off at the same time, and Q2 and Q4 are turned on and off at the same time can be regarded as the left and right half-bridge circuits working in the same phase.
Between the two working states of the left and right half-bridges working in reverse phase and in phase (the phases are neither the same nor opposite, but have a certain phase difference), a new control method for the full-bridge circuit is generated: phase-shifted full-bridge. Phase-shifted full-bridge has many characteristics. But this is a later story. For the time being, it is enough for everyone to understand the left and right half-bridges working in reverse phase and in phase.
|
提升卡
变色卡
千斤顶
; in fact, the time difference is not much, and the charging and discharging process may not be obvious or difficult to observe.
京公网安备 11010802033920号
