Switching Power Supply Interest Group Task 06
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This post was last edited by maychang on 2020-9-11 11:27
Question 05:
1. We already know from Question 04: In an ideal Boost circuit, the voltage across the two ends is equal to the output voltage when the switch is turned off. In an ideal Buck circuit, the voltage across the two ends is equal to the input voltage when the switch is turned off. So, for an ideal Buck/Boost circuit, what is the maximum voltage across the two ends when the switch is turned off? In an ideal
Boost circuit, the maximum voltage across the two ends when the switch is turned off is equal to the sum of the absolute values of the input voltage and the output voltage. In a Boost circuit, at the moment the switch is turned off, the voltage across the inductor L is equal to the output voltage. At this time, the voltage across the two ends of the switch is equal to the superposition of the input voltage and the voltage across the two ends of the inductor L.
2. In the half-wave rectifier circuit of the industrial frequency transformer, the negative pole of the rectifier is usually connected to a capacitor to the ground, that is, one end of the capacitor is connected to the rectifier, and the other end is connected to the negative end of the output. Why can't the Buck circuit be connected like C2 in the figure below? Wouldn't this filter be more thorough?
The Buck circuit cannot work if connected like this, or in other words, it is not a Buck circuit if connected like this.
The original intention of the Buck circuit is to control the increase in the energy of the inductor during the conduction period of the switch tube by controlling the conduction time of the switch tube. When the switch tube is turned off, the energy stored in the inductor is released to the load to control the output voltage. If a capacitor is connected between the output end of the switch tube and the ground as shown in C2 in the figure above, then when the switch tube is turned on, the capacitor will be charged to the voltage across the capacitor C1, and the freewheeling diode D will never be turned on, and the voltage on the load cannot be controlled.
3. We have discussed three types of non-isolated switching power supplies, namely Buck circuit, Boost circuit and Buck/Boost circuit, as shown in the figure below:
We can see that the only difference between these three different circuits is the different positions of the switch tube, diode and inductor.
However, there are six possibilities for three different objects placed in three different positions (the full permutation of 3, i.e. the factorial of 3, 3!).
Can you find three other non-isolated switching power supply circuits with different positions of the switch tube, diode and inductor? Note: There must be a common terminal between the input and output.
The other three are the Buck, Boost and Buck/Boost circuits with the load on the right side of the above figure as the power supply and the power supply on the left side of the above figure as the load. In the above figure, energy is transferred from the left to the right, while in the other three, energy is transferred from the right to the left. In other words, in the figure below, RL is used as a DC power supply and the battery V is used as a load, which is the other three circuits.
Therefore, there are only three basic non-isolated switching power supply topologies.
In the book "Bidirectional DC Converter", it is believed that there are six types of non-isolated single-tube DC converters. In addition to the three types of Buck, Boost and Buck/Boost listed above, there are three types of Cuk, Zeta and Sepic that use capacitors to transfer energy. Different opinions have different views, and it is enough for us to know that they are different. For
the 06th event, please read Chapter 2, Section 2.1 "Introduction" and Section 2.3 "Forward Converter Topology" in "Switching Power Supply Design Third Edition".
Why don't we read Chapter 2, Section 2.2 "Push-Pull Topology" first? This is because: the single-ended forward circuit is the basic circuit, and the single-ended forward circuit derives the dual-tube forward, staggered parallel forward, push-pull, half-bridge, full-bridge... (Of course, this is just my personal opinion). So we read the single-ended forward first, and then read the push-pull. As for the dual-tube forward and staggered parallel forward, they are rarely used, so it's better not to read them.
If you still don't understand the operation of the forward converter after reading the forward converter section in "Switching Power Supply Design 3rd Edition", please read the detailed explanation below carefully.
In "Switching Power Supply Design 3rd Edition", the forward converter circuit and waveform are shown in Figure (01) below.
Figure (01)
We draw the electrical schematic diagram in a simpler way, as shown in Figure (02). In Figure (02), the feedback control part is omitted, and two of the three windings and their rectifier and filter parts are also omitted.
In Figure (02),
we can see that the secondary of the forward converter transformer plus the rectifier and filter parts is a Buck circuit, as shown in the red box in Figure (03).
When the switch tube Q1 turns from off to on, the voltage across the primary winding Np of the transformer T is positive at the top and negative at the bottom. From the same-name terminal (the end with a dot), we know that the voltage across the secondary winding Ns of the transformer is also positive at the top and negative at the bottom. At this time, the diode D2 is turned on, and D3 is turned off due to the reverse voltage. The voltage across Ns "charges" the inductor L1, and the current in the inductor L1 increases linearly. The switch tube Q1 turns from on to off, and the voltage across Ns also reverses, the inductor "discharges", and the diode D3 turns on to continue the current for the inductor L1. Ideally, the voltage across the load R1 is the average value of the voltage across D3. This is exactly the working state of the Buck circuit. So we can think that the forward converter circuit is developed from the Buck circuit.
Figure (03)
is different from the Buck circuit in that the forward converter circuit adds a transformer T to isolate the DC power supply Vdc from the load.
But it is this transformer T that makes the forward converter circuit much more complicated than the Buck circuit. This is because the transformer is never an ideal component, but a component that deviates a lot from the ideal. For such a component that deviates a lot from the ideal, we cannot use an ideal component to replace it and analyze the operation of the forward converter circuit. This requires us to have a relatively full understanding of the actual operation of the transformer in order to correctly analyze the operation of the forward converter circuit.
The use of the ideal transformer model is subject to certain conditions. In the forward converter circuit, there is no way to use the ideal transformer model.
The ideal transformer does not consider the excitation current, which actually assumes that the primary inductance of the transformer is infinite. The ideal transformer also does not consider the leakage inductance, which actually assumes that the magnetic flux generated by the current in the primary winding is completely coupled with the secondary winding, and there is no leakage flux.
Figure (04)
Of course, the primary inductance of an actual transformer will not be infinite, and the current in the primary winding cannot be completely coupled with the secondary winding. There is always leakage flux.
Therefore, in Section 2.3 of Chapter 2 of Professor Zhao Xiuke's book "Magnetic Components in Switching Power Supplies", the equivalent circuits of the transformer no-load working state, the transformer loaded working state and the actual transformer are given. Figure (04) in this article is the equivalent circuit of the actual transformer copied from the book "Magnetic Components in Switching Power Supplies".
First, the primary inductance of the transformer is not infinite, so in the equivalent circuit, a finite value inductor is connected in parallel with the primary of the ideal transformer, and the current flowing through the finite value inductor is used to equal the primary excitation current. Therefore, the current flowing through the primary of the actual transformer is equal to the current reflected from the secondary winding to the primary plus the excitation current.
Secondly, leakage flux always exists. This part of the flux that is not coupled to the secondary will also generate an induced electromotive force in the primary winding. Therefore, it is equivalent to connecting an inductor in series with the primary of the ideal transformer, that is, the leakage inductance of the primary to the secondary. At the same time, part of the magnetic flux generated by the secondary current is not coupled to the primary, so it is equivalent to connecting an inductor in series with the secondary, that is, the leakage inductance of the secondary to the primary.
Finally, considering that the winding always has a certain resistance, the iron core always has a certain loss, and there is always a certain capacitance between the transformer windings. This requires adding corresponding resistance and capacitance to the equivalent circuit to represent it.
Figure (05)
Figure (05) uses the actual transformer equivalent circuit in the red box to replace the circuit of transformer T in Figure (02). Because the winding resistance and distributed capacitance are not important in the current discussion, they are not drawn in Figure (05). Only the excitation inductance Lm, the leakage inductance of the primary to the secondary Lp, and the leakage inductance of the secondary to the primary Ls are drawn.
For beginners who have been exposed to non-isolated switching power supply circuits such as Buck, Boost and Buck/Boost, but have not been exposed to forward-pulse circuits with transformer isolation, they often have such confusion: Isn’t there a diode D3 in Figure (03) as a freewheeling current? Why do we need to add a winding Nf to the transformer and a diode D1 for freewheeling current?
Figure (06)
Another problem that is easy to cause confusion is that in Buck, Boost and Buck/Boost circuits, the current when the freewheeling diode is turned on is exactly equal to the current in the switch tube, as shown in Figure (06). The left side is the continuous current working state. The distance between the two red lines below is the current when the diode starts to conduct, and the distance between the two red lines above is the current when the switch tube is turned off. These two currents are exactly equal. The right side is the intermittent current working state. The distance between the two red lines below is also equal to the distance between the two red lines above, that is, the current when the diode starts to conduct is exactly equal to the current when the switch tube is turned off. However, in the waveform part of Figure (01), as shown in Figure (07), the current when D1 starts to conduct, that is, the distance between the two red lines below in Figure (07), is less than the current when the switch tube is turned off, that is, the distance between the two red lines above. The two currents are different, so where does the difference between the two currents go?
Figure (07)
These problems can be explained by the actual transformer model in Figure (05).
In Figure (05), the current flowing through the switch tube Q1 when it is turned on is the sum of two currents. One is the current flowing through the primary winding Np of the ideal transformer, and the other is the current flowing through the excitation inductance Lm. The current flowing through the excitation inductance Lm is to generate the necessary magnetic flux in the iron core. If the magnetic flux in the iron core does not change, no induced electromotive force will be generated in each winding, so the current change in the excitation inductance Lm is a necessary condition for each winding to generate electromotive force. The
current flowing through the primary winding Np of the ideal transformer is the current reflected from the secondary winding Ns of the ideal transformer to the primary, and the current flowing through Ns, as shown in the red box in Figure (03), is equivalent to the current in a Buck circuit switch tube. This has been mentioned when explaining the circuit in Figure (03).
Since the current in Ns is equivalent to the current flowing through the switch tube in the Buck circuit, the current in Ns must rise linearly when Q1 is turned on. Therefore, the current in Np also rises linearly when Q1 is turned on.
As for the current in the excitation inductor Lm, during the conduction period of the switch tube Q1, if the core is not saturated, it will of course also rise linearly.
Therefore, during the conduction period of the switch tube Q1, the current passing through Q1 is the sum of two linearly rising currents. One is the current flowing through the diode D2 in Figure (05), that is, the current in the inductor L1 (the diode D3 is turned off during the conduction period of Q1) reflected to the primary of the transformer, and the other is the current in the excitation inductor Lm.
The sum of the two linearly rising currents in Figure (08)
is still a linearly rising current. We use the red part in Figure (08) to represent the linearly rising part of the current reflected to the primary of the transformer by the inductor L1 in Figure (05), and the green part to represent the current in the excitation inductor Lm. The
red part, that is, the rising part of the current reflected to the primary of the transformer, represents the increase in energy that does not need to be continued in the primary of the transformer, because the energy carried by this part of the current increment has been transmitted to the inductor L1 in Figure (05), and the continued flow is completed by the diode D3 in Figure (05). The green part, that is, the current in the excitation inductor Lm, must be continued in the primary of the transformer to release the energy carried by this part of the current. Otherwise, the magnetic flux in the transformer core cannot be reset to the value before Q1 starts to conduct, and the magnetic flux in the transformer core will continue to increase in one cycle after another until the core is saturated, making the transformer unable to work.
The release of the magnetic energy stored in the transformer core depends on the transformer winding Nf and the diode D1.
In Figure (03), the winding Nf and the primary winding Np are usually wound in parallel with two wires, so the number of turns is equal.
When Q1 is turned on in Figure (03), the voltage direction across winding Np is positive at the top and negative at the bottom, so the voltage across winding Nf is also positive at the top and negative at the bottom, and diode D1 is turned off. When Q1 turns from on to off, the excitation current in Np must be continued by Nf through D1. In other words, when Q1 turns from on to off, the current in winding Nf is equal to the green part in Figure (08). This results in unequal distances between the upper and lower red lines in Figure (07), because the current increment in winding Nf is only a part of the current increment of Q1, that is, the green part in Figure (08).
We already know: when Q1 turns from on to off, the excitation current in Np must be continued by Nf through D1. But what is the voltage across winding Nf? What is the voltage across Q1?
It can be seen from Figure (05) that when Q1 turns from on to off, the voltage across winding Np changes from positive at the top and negative at the bottom to negative at the top and positive at the bottom. Of course, the voltage across winding Nf also changes from positive at the top and negative at the bottom to negative at the top and positive at the bottom, and diode D1 is turned on. If the voltage drop of D1 is ignored, the upper end of winding Nf is grounded and the lower end is connected to the DC power supply Vdc, so the voltage across winding Nf is equal to the DC power supply voltage Vdc. Since Nf and Np are tightly coupled, the voltage across winding Np is also the DC power supply voltage Vdc, and the upper end is negative and the lower end is positive. If the induced electromotive force in the leakage inductance Lp is ignored, then the voltage across Np and the DC power supply voltage are superimposed, and the voltage across Q1 is 2Vdc, which is twice the DC power supply voltage. This can be seen clearly in Figure (01), Figure (07) and Figure (08).
The demagnetized winding Nf passes through diode D1, and the excitation current is reversely "injected" into the DC power supply Vdc. Therefore, the energy of the excitation current is not lost, and is not converted into heat, but returned to the DC power supply.
According to the theorem that the inductor flux resets and the inductor withstands the volt-second product must be zero, the voltage that winding Np withstands during Q1 conduction is Vdc, with the upper end being positive, and the voltage that winding Nf withstands during Q1 off is also Vdc, with the lower end being positive. From this we can infer that the conduction time of diode D1 must be equal to the conduction time of Q1. Only when the two times are equal, the flux in transformer T will be reset to the state before Q1 conduction. This is clearly seen in Figure (07).
From this we can also infer that if the demagnetization method is used to wind winding Nf and winding Np in parallel (with equal number of turns), the duty cycle of the forward converter circuit cannot exceed 50%. If the duty cycle of this forward converter circuit exceeds 50%, then when Q1 switches from off to on, there will be some flux in the iron core that has not been released. In the next cycle, there will be more flux in the iron core. After several cycles, the flux in the iron core will reach saturation, and the forward switching power supply will be damaged due to excessive current.
If you are not clear about the question that the inductor flux reset inductor must have a volt-second product of zero, please refer to my previous post "Ampere-second product of capacitor and volt-second product of inductor" .
Another question: Can the induced electromotive force generated by leakage inductance Lp and Ls when the current changes be ignored?
Let's first look at the leakage inductance Ls of the secondary to the primary. When Q1 turns from on to off, the electromotive force generated by the leakage inductance Ls in Figure (05) is positive on the left and negative on the right, while the voltage direction of the two ends of the secondary winding Ns is negative at the top and positive at the bottom, and the two directions are opposite. Therefore, when Q1 turns from on to off, the leakage inductance Ls will not cause the two ends of the rectifier diode D2 to bear a larger reverse voltage. When Q1 turns from off to on, the role of the leakage inductance Ls is just to slightly delay the time for D2 to turn on, which has no great effect on the operation of the circuit.
Let's look at the leakage inductance Lp of the primary to the secondary. As mentioned earlier: In order to reduce the leakage inductance between each other, the windings Nf and Np are usually wound in parallel with two wires. The distributed capacitance between the two windings of the bifilar winding is the largest among all winding methods. When Q1 switches from on to off, the current generated by the induced electromotive force at both ends of Lp will flow through the distributed capacitance between the two windings, so the voltage at both ends of Lp will not be very high. The peak voltage Q1 bears when it is turned off will not be very high. However, the LC loop formed by Lp and the distributed capacitance will produce attenuated oscillation, so the "leakage inductance spike" in Figure (01) often appears as a rapidly decaying oscillation rather than a single spike.
There is another question. We need to talk about the connection method of the demagnetization winding Nf and the diode D1 in Figures (01) and (02). Figure (02) is originally a simplification of Figure (01), and we only need to look at Figure (02). In
Figure (02), the lower end of the demagnetization winding Nf is connected to the positive DC power supply, and the lower end is grounded through a diode. When Q1 is on, the potential of the lower end of Nf is the positive DC power supply, and the potential of the upper end is negative Vdc. The potential of the lower end of Np is zero, and the upper end is the positive DC power supply. The voltage between the lower end of Nf and the lower end of Np is Vdc, and the potential of Nf is higher than that of Np.
However, after Q1 is turned off, the potential of the lower end of Nf is still the positive DC power supply, while the lower end of Np becomes 2Vdc. The voltage between the two windings is Vdc, and the potential of Nf is lower than that of Np. In other words, when Q1 switches from on to off, the voltage between the two windings changes by 2Vdc. As we have said before, Nf and Np are often wound in parallel with two wires to reduce leakage inductance. With this winding method, the distributed capacitance between the two windings is very large. Since the voltage between the two windings of Nf and Np changes greatly in each switching cycle, and there is a large distributed capacitance between the two windings, a considerable current must flow through the distributed capacitance. This current flowing through the distributed capacitance will not damage the switching power supply, but it will increase the burden on the switch tube, because this current is ultimately provided by the switch tube.
To avoid the current in this distributed capacitance, just swap the Nf winding and the diode D1, as shown in Figure (09).
Figure (09) is copied from Professor Zhao Xiu Ke's book "Magnetic Components in Switching Power Supplies". Compared with Figure (02), Figure (09) only swaps the positions of the demagnetizing winding and the diode. In Figure (09), the same-name end (the end with a "dot") of the demagnetizing winding N3 is grounded, and the other end is connected to the DC power supply through a diode. When the switch tube is turned on, the same-name end of the transformer primary winding is at the positive end of the DC power supply, the same-name end of the demagnetizing winding is at the ground potential, and the voltage between the two is the DC power supply voltage. When the switch tube is turned off, the same-name end of the transformer primary winding is still at the DC power supply voltage, and the same-name end of the demagnetizing winding is still at the ground potential. The voltage between the two windings remains unchanged within one switching cycle and is a DC voltage. No matter how large the distributed capacitance between the two windings is, the DC voltage will not generate current in the distributed capacitance between the two windings. This reduces the burden on the switch tube and reduces the current capacity requirements of the switch tube.
Figure (09)
Therefore, although the connection method of the demagnetizing winding and the diode D1 in Figure (01) copied from the "Switching Power Supply Design 3rd Edition" cannot be said to be wrong, it is not as good as the connection method in Figure (09) where the demagnetizing winding and the diode are swapped.
Finally, some netizens may raise the following question: When a single transistor drives a relay winding, isn't it as simple as connecting a diode in parallel at both ends of the relay winding as shown in Figure
(10) to prevent the transistor from being broken down due to the high induced electromotive force of the relay winding? Why can't a diode be connected in parallel to the primary winding of the transformer in Figure (02) to demagnetize? Must a winding Nf be added to the transformer?
Simply connecting a diode in parallel as shown in Figure (10) is feasible in the circuit of a transistor driving a relay, but it is not feasible in a switching power supply.
When a diode is connected in parallel as shown in Figure (10), the diode D will indeed continue the current for the relay winding L after the transistor Q is turned off. However, during the continuous current process, the voltage across the winding L is only the forward voltage drop of the diode (in fact, the voltage drop on the DC resistance of the current flowing through the winding must be added). It has been said before that the magnetic flux reset should be zero volt-second product. During the reset process of the circuit in Figure (10), the voltage across the winding is very small, so the magnetic flux reset time will inevitably be very long. For relays, a reset time of tens of ms or even hundreds of ms is not a problem, because the speed of relay action is not very fast. Switching power supplies are different. They must be reset very quickly after the switch tube is turned off, and the longest time cannot exceed the switch tube conduction time (otherwise the duty cycle will be reduced to less than 50%). Therefore, simply connecting a diode in parallel as shown in Figure (10) does not work in the forward circuit. The forward circuit must add another winding to the transformer to reset the magnetic flux, and this winding must be tightly coupled with the primary winding, preferably with two wires in parallel.
Doing so is of course very troublesome, but we have to do so.
Question 06
1. We know that the Buck circuit has two working states: continuous inductor current and discontinuous inductor current. We also know that the forward circuit is developed from the Buck circuit, as shown in Figure (03) of this article. If Figure (07) of this article is the working waveform of the circuit in Figure (03), please ask whether the Buck circuit in the red box in Figure (03) works in the continuous inductor current state or the discontinuous inductor current state? Please explain why you judge it as continuous current or discontinuous current.
2. The top row of waveforms in Figure (07) is the voltage waveform across the switch tube. After Q1 is turned off, except for a very short period of leakage inductance spikes, the voltage Q1 bears is twice the power supply voltage, i.e., 2Vdc. But after a period of time, the voltage Q1 bears suddenly drops to one times the power supply voltage, i.e., Vdc, until Q1 turns on again. Why does Q1 bear one times the power supply voltage after a period of time? How long does it bear two times the power supply voltage?
3. In the top row of the waveforms in Figures (01) and (07), two areas A1 and A2 are marked with slashes, and it is visually observed that the areas of A1 and A2 are equal. What does it mean that the areas of A1 and A2 are equal?
4. There is an error in Figure 2.11 of "Switching Power Supply Design, Third Edition". Please find the error.
5. There is a printing error from the 4th to the 3rd from the last line on page 47 of "Switching Power Supply Design, Third Edition". Please find the error.
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