Miscellaneous talks on rectification (VII)

maychang

Miscellaneous talks on rectification (VII) [Copy link]

 

　　I have posted a series of posts on the power supply section titled "Rectifier Talk". So far, six posts have been posted, mainly discussing the peak, average and effective values of the current in the rectifier circuit.
　　Although the voltage doubler rectifier circuit has been discussed in textbooks and there are quite a few posts discussing voltage doubler rectifier in this forum, every year there are netizens who raise questions about the voltage doubler rectifier circuit. So this post is going to discuss the details of voltage doubler rectifier.
　　Figure (01) is a half-wave rectifier capacitor input filter circuit.

Figure (01)
　　A double voltage rectifier circuit derived from Figure (01) is shown in Figure (02). The AC power supply is not shown in the figure. We assume that the capacitances of the two capacitors C1 and C2 are equal.

Figure (02)
　　For most circuits, their initial state must be determined when analyzing, that is, the voltage and current at each point in the circuit when time is zero. For the circuit in Figure (02), we assume that before power is turned on (that is, when no AC voltage is applied on the left), the voltage across C1 and C2 is zero, and of course the current is also zero.
　　We analyze the AC voltage input on the left half cycle by half cycle.

Figure (03)
　　starts from time zero, an AC voltage is applied on the left, positive at the top and negative at the bottom, with a peak value of Vp, as shown in Figure (03). The red arrow in the figure indicates the direction of the current in this "half cycle" (note that the current is not continuous within half a cycle, and the current is zero at the end of the half cycle, the same below). Ignoring the diode voltage drop, at the end of the first "half cycle", capacitor C1 is charged through the diode to a voltage of Vp across its two ends, positive on the left and negative on the right. Diode D2 is cut off, and the voltage across capacitor C2 is still zero.

Figure (04)
　　In the second "half cycle", the AC voltage on the left is reversed, with negative on the top and positive on the bottom. Diode D1 is cut off, and the current direction is as shown in Figure (04). The AC voltage on the left and the voltage across C1 are superimposed, and capacitor C2 is charged through diode D2. At the end of the second "half cycle", because the capacitance of capacitors C1 and C2 is equal, half of the power stored in capacitor C1 in the first "half cycle" will be consumed to charge C2, and capacitor C2 is charged to a voltage of 1.5Vp across the two ends, with negative on the top and positive on the bottom.

Figure (05)
　　In the third "half cycle", the voltage on the left becomes positive at the top and negative at the bottom. The voltage on the left charges capacitor C1 through diode D1, and the voltage across C1 is charged to Vp, with positive on the left and negative on the right. The current direction is as shown in Figure (05). Diode D2 is cut off, and the voltage across capacitor C2 remains unchanged, still 1.5Vp.

Figure (06)
　　In the fourth "half cycle", the AC voltage on the left is negative at the top and positive at the bottom, and diode D1 is cut off. The AC voltage on the left and the voltage across C1 are superimposed, and capacitor C2 is charged through diode D2. The direction of the current is as shown in Figure (06). At the end of the second "half cycle", because the capacitance of capacitors C1 and C2 is equal, one-fourth of the power stored in capacitor C1 in the first "half cycle" will be consumed to charge C2, and capacitor C2 is charged to a voltage of 1.75Vp across the two ends, with a negative voltage at the top and a positive voltage at the bottom.

In the fifth "half cycle" of Figure (07)
　　, the AC voltage on the left charges capacitor C1 through diode D1, and the current direction is as shown in Figure (07). Capacitor C1 is charged to the peak value Vp of the AC voltage.

In the sixth "half cycle" of Figure (08)
　　, the AC voltage on the left is still superimposed on the voltage across C1, and the capacitor C2 is charged through the diode D2. The current direction is as shown in Figure (08). The capacitor C2 is charged to a voltage of 1.875Vp across its two ends.

Figure (09)
　　The seventh "half cycle", as shown in Figure (09), the AC voltage on the left charges capacitor C1 through diode D1, and the current direction is shown in Figure (09). Capacitor C1 is charged to the peak value Vp of the AC voltage. In
　　the eighth "half cycle", it is obvious that the voltage Vp across C1 is superimposed on the AC voltage on the left, and capacitor C2 is charged through diode D2, so that the voltage across C2 is further increased. Since capacitor C1 is always charged to the peak value Vp of the AC voltage every time D1 is turned on, the negative half-cycle voltage of the AC voltage is superimposed on the voltage across C1, and the maximum is 2Vp. Therefore, after several AC cycles, the voltage across C2 will eventually rise to 2Vp.
　　Figure (02) Double voltage circuit, the common end of the right output end and the AC input is the output negative end. If you want to output a positive voltage to the common end, just reverse the direction of the two diodes. As shown in Figure (10).

In the above analysis of Figure (10)
　　, it is assumed that the output is open circuit, that is, no load. Only when the output is no load, the voltage across C2 can rise to 2Vp. If the load is not open circuit, then in the entire AC cycle, the capacitor C2 in Figure (09) must discharge to the load, which will reduce the average voltage across C2, that is, C2 cannot output such a high voltage as 2Vp. Moreover, the heavier the load (the larger the load current), the lower the output voltage of this double voltage rectifier circuit. In fact, no matter which double voltage rectifier circuit, the load capacity is not very strong and can only be used under low current conditions.
　　From this we can also know that the double voltage rectifier circuit shown in Figure (09) or Figure (10) is generally not afraid of output short circuit. From Figure (03) to Figure (09), we know that the charge of capacitor C1 in each AC cycle will not exceed C1*Vp, so even if the output is short-circuited, the short-circuit current will not exceed C1*Vp/T, where T is the cycle of the AC power. As long as the diode can withstand this current, the voltage doubler rectifier circuit will automatically limit the current even if the output is short-circuited.

Figure (11)
　　Based on the above analysis of the double voltage rectifier circuit in Figure (02), it is not difficult to construct a triple voltage and quadruple voltage rectifier circuit, as shown in Figure (11). If each diode and capacitor is considered a "section", such "sections" can be extended to the right to form a rectifier circuit with a higher voltage.
　　D1, C1, D2, and C2 in this circuit in Figure (11) are exactly the same as those in Figure (10). From the previous analysis, we can know that the voltage across C1 can reach Vp when the load is open, and the voltage across C2 can reach 2Vp when the load is open. According to the same analysis method as before, we can know that the voltage across capacitor C3 in Figure (11) can reach 2Vp when the load is open, and the voltage across capacitor C4 can reach 2Vp when the load is open. Therefore, the voltage between capacitors C1 and C3 (the left end of C1 and the right end of C3 in the figure) is 3Vp when the load is open, and the voltage between capacitors C2 and C4 (the left end of C2 and the right end of C4 in the figure) is 4Vp when the load is open. However, the larger the voltage doubler, the worse the load capacity, that is, the output voltage drops a lot when the output current is slightly larger.

This content is originally created by EEWORLD forum user maychang . If you want to reprint or use it for commercial purposes, you must obtain the author's consent and indicate the source

PowerAnts

Seeing 1.5->1.75->1.875... I think the kids behind will have a harder time. Why can't it be 1.6->1.8->1.9->1.95->... Building a ladder to carry things, 18 6 2 1 is the fairest, 8 4 2 1 is a no-brainer, 4 3 2 1 is the most economical, 2 2 2 2 or 4 4 4 4 is definitely the dumbest. I've never done multiplication before, so the above number series may not be good. Study it when you have time.

PowerAnts

First, determine the capacity of C4, the capacitor with the highest potential in the series capacitors: set it to 50Hz voltage doubler rectification, load impedance 100K, C4 bears a 50K load, and the half-wave rectification time constant is taken as 5 cycles, then C4=5*0.02/50000=2 (uF).

The capacity of C4, C3, C2, and C1 needs to be increased from right to left step by step, so as to gradually increase the charge carrying capacity under the appropriate pressure difference conditions, which is similar to the experience of the ancient military porters who delivered food. The closer to the source, the greater the conveying capacity, so that the design can ensure the farthest delivery.

maychang

PowerAnts posted on 2020-4-11 08:19 Seeing 1.5->1.75->1.875... I think the kids behind will have a harder time. Why can't it be 1.6->1.8->1.9->1.95->...Build a human ladder to resist...

The sequence of 1.5, 1.75, and 1.875 is calculated based on the fact that the capacitances of the two capacitors C1 and C2 are equal, the load is open circuit, and the AC power source charges C1 to Vp during each positive half cycle of the AC power supply, and the charges of the two capacitors are balanced during the negative half cycle.

Considering that most new netizens don’t like to look at calculation formulas, even a simple formula like Ohm’s law, I didn’t list the calculation process.

As for the fact that the capacitance increases gradually from right to left, that calculation is more complicated. It is best to avoid it when discussing the most basic principles, but when discussing in depth, we should imagine the benefits of different capacitances of capacitors.

PowerAnts

maychang published on 2020-4-11 17:20 The sequence of 1.5, 1.75, 1.875 is based on the fact that the capacitance of the two capacitors C1 and C2 is equal, the load is open circuit, and the AC power supply charges C1 every time the AC positive half cycle...

Yes, this voltage sequence is equal capacity voltage multiplication. I propose to start with a higher rate sequence, considering that the capacity is preferably negatively correlated with the position gradient. This is conducive to improving the output voltage and load capacity when the gradient is higher and the total capacity is equal.

PowerAnts

The 5th floor has an extra 0

zhaoyanhao

Teacher, I am really confused when I see 1.5, 1.75, 1.875. Is the calculation process complicated?

maychang

zhaoyanhao posted on 2020-4-14 15:11 Teacher, I was really confused when I saw 1.5, 1.75, and 1.875. Is the calculation process complicated?

It is not very complicated and only requires elementary algebra knowledge.

In fact, I have already said it on the 4th floor: the sequence of 1.5, 1.75, and 1.875 is calculated based on the fact that the capacitances of the two capacitors C1 and C2 are equal, the load is open circuit, and the AC power supply charges C1 to Vp in each positive half cycle of the AC power supply, and the charges of the two capacitors are balanced in the negative half cycle.

maychang

zhaoyanhao posted on 2020-4-14 15:11 Teacher, I was really confused when I saw 1.5, 1.75, and 1.875. Is the calculation process complicated?

Here we will only talk about how 1.5Vp comes from.

At the end of the first "half cycle", capacitor C1 is charged to a voltage of Vp across its terminals, and the voltage across capacitor C2 is zero. In the second "half cycle", as shown in Figure (04), capacitors C1 and C2 are connected in series, and C1 and C2 are equal. The power supply voltage Vp is superimposed on the voltage Vp across C1, which is 2Vp, charging C2. During the charging process, the charge on C1 decreases. In the end, the charge on C1 decreases from C1*Vp to 0.5C1*Vp, while the charge on C2 increases from 0 to 1.5C2*Vp, of which 0.5C2*Vp is transferred from C1, and 1C2*Vp is from the power supply charging C2.

zhaoyanhao

maychang published on 2020-4-14 17:11 Here I will only talk about how 1.5Vp comes from. At the end of the first "half cycle", capacitor C1 is charged to a voltage of Vp across both ends, and capacitor C2 is charged to a voltage of Vp across both ends.

Thank you teacher

zhaoyanhao

maychang published on 2020-4-14 17:11 Here I will only talk about how 1.5Vp comes from. At the end of the first "half cycle", capacitor C1 is charged to a voltage of Vp across both ends, and capacitor C2 is charged to a voltage of Vp across both ends.

"1C2*Vp is the power supply charging C2", I know, but what I don't understand is why C1 charges C2 with 0.5Vp, and why C1 charges C2 with a quarter or an eighth of the power in the subsequent cycles.

maychang

zhaoyanhao published on 2020-4-15 09:46 "1C2*Vp comes from the power supply charging C2", I know, but what I don't understand is why C1 charges C2 with 0.5Vp, and...

The voltage across C1 is Vp at the beginning of this "half cycle", and the power supply voltage is added. The power supply voltage plus the voltage of C1 charges C2 from zero, while C1 discharges. C1 discharges to 0.5Vp, and the power supply voltage Vp is added, and C2 is charged to 1.5Vp.

他们逼我做卧底

Thank you teacher for sharing ~ I am a newbie and will learn from you!

lqsgg

Nice article! Thanks for sharing! Learn more!

yhhhy

maychang published on 2020-4-11 17:20 The sequence of 1.5, 1.75, 1.875 is based on the fact that the capacitance of the two capacitors C1 and C2 is equal, the load is open circuit, and the AC power supply charges C1 every time the AC positive half cycle...

Teacher, can you give me the formula to calculate the charge balance of the two capacitors in the negative half cycle?

maychang

yhhhy posted on 2023-5-28 14:59 Teacher, can you give the balance formula for calculating the charge of two capacitors in the negative half cycle?

Are you referring to the circuits in Figures (02) to (09)? The voltages to which capacitor C2 is charged in each negative half cycle are listed.