The full-bridge DCDC output voltage does not satisfy Uout=2D*Uin*(N2/N1)? What is the reason?

jonny0811

The full-bridge DCDC output voltage does not satisfy Uout=2D*Uin*(N2/N1)? What is the reason? [Copy link]

 

Hello everyone. Recently I need to use a circuit to raise the 12V DC voltage to about 300V, and I plan to use a full-bridge DCDC. When I set up the power supply simulation in matlab, I found that it is not easy to converge... So this time I chose saber. After the model was built, I first took a very light load and worked in open loop. In the simulation: my duty cycle is 0.5, the input voltage is 20V, and the transformer ratio is P:S=2:20; then according to the full-bridge DCDC input-output DC transfer relationship Uout=2D*Uin*(N2/N1).Uout should be equal to 200V. But the voltage obtained is only about 130V. When I used matlab simulation to do other circuit topologies before, the simulation results should be almost the same as the theoretical calculations. What might be the reason?

maychang

Let's not talk about the output voltage. Is there no inductor filtering after your full-bridge output rectification? Do you use 1N4148 for 300V DC output?

jonny0811

maychang posted on 2017-1-14 23:09 Let's not talk about the output voltage. Is there no inductor filtering after your full-bridge output rectification? Do you use 1N4148 for 300V DC output?

4148 has a high frequency and a very small current. It is not used in practice. I didn't consider it so much in the simulation. I didn't use inductor filtering. I only learned about full-bridge DCDC before, and I haven't actually applied this topology before. If you use a capacitor to filter directly, it stands to reason that the voltage on the capacitor should also meet the relationship of Uout/Uin, but it doesn't. . . .

maychang

jonny0811 posted on 2017-1-14 23:18 4148 has a high frequency, and the current of 4148 is very small. It is not used in practice. I did not consider it so much in the simulation, and did not use inductor filtering...

The reverse withstand voltage of 1N4148 is only 100V peak. As for whether it breaks down above 100V in simulation, I don't know. In addition, the four MOS tubes of the full bridge seem to be driven by four independent pulse sources, and the phase of the four pulse sources cannot be seen. This is wrong.

maychang

jonny0811 posted on 2017-1-14 23:18 4148 has a high frequency and a very small current. It is not used in practice. I did not consider it so much in the simulation. I did not use inductor filtering...

"I have only learned full-bridge DCDC before, and have not actually applied this topology before." If you have learned full-bridge, you should know that it is impossible to do without inductor. Except for the flyback current, all other switching power supply circuits must use inductor filtering (actually energy storage).

maychang

jonny0811 posted on 2017-1-14 23:18 4148 has a high frequency and a small current. It is not used in practice. I did not consider it so much in the simulation and did not use inductor filtering...

"If a capacitor is used directly for filtering, it stands to reason that the voltage on the capacitor should also satisfy the relationship Uout/Uin." If a capacitor is used directly for filtering, it stands to reason that the relationship Uout=2D*Uin*(N2/N1) cannot be satisfied.

maychang

jonny0811 posted on 2017-1-14 23:18 4148 has a high frequency and a very small current. It is not used in practice. I did not consider this much in the simulation and did not use an inductor to filter...

The first post said "but the voltage obtained is only about 130V". How did you know that "the voltage obtained is only about 130V" during the simulation?

maychang

jonny0811 posted on 2017-1-14 23:18 4148 has a high frequency and a very small current. It is not used in practice. I did not consider it so much in the simulation. I did not use an inductor filter...

Since it is a simulation, calling an oscilloscope is just a matter of two clicks of the mouse. Why not use an oscilloscope? An oscilloscope is indispensable for switching power supplies. Switching power supplies do not have very high requirements for the bandwidth of the oscilloscope, 20MHz is about right. But it must be dual-trace, a single-trace oscilloscope is not enough. To some extent, the old analog oscilloscope may be more suitable. When simulating, use an oscilloscope to look at the transformer secondary voltage waveform, the waveform at both ends of the capacitor after rectification, the waveform at both ends of the transformer primary to ground... It is easy to find the error. The same is true after the actual circuit is built. The oscilloscope is still a powerful testing tool in the actual circuit.

maychang

jonny0811 posted on 2017-1-14 23:18 4148 has a high frequency and a very low current. It is not used in practice. I did not consider it so much in the simulation. I did not use an inductor filter...

Saying "the voltage on the capacitor should also satisfy the relationship of Uout/Uin" is enough to show that you have not yet mastered the working principle of the switching power supply. Nothing will be burned during simulation. If you set up a real circuit and power it on without mastering the working of the switching power supply, it is almost certain that it will either smoke or explode.

jonny0811

maychang posted on 2017-1-15 08:36 Saying "the voltage on the capacitor should also satisfy the relationship of Uout/Uin" is enough to show that you have not yet mastered the work of switching power supply...

Indeed, I am still a student. My current direction is motor control; I am working on sensorless FOC. I am just interested in power supply, so I tinkered with it myself. Oh, it seems that the water of switching power supply is very deep. . . .

jonny0811

maychang posted on 2017-1-15 08:22 The first post said "but the voltage obtained is only about 130V", how did you know "the voltage obtained is only about 130V" during the simulation?

During the simulation, I used an oscilloscope and saw that it was about 130V.

jonny0811

maychang posted on 2017-1-15 08:20 "If a capacitor is used for filtering directly, it stands to reason that the voltage on the capacitor should also satisfy the relationship of Uout/Uin." If a capacitor is used for filtering directly, it stands to reason that...

Oh, I am so ashamed. I have made a fool of myself. . . .

jonny0811

maychang posted on 2017-1-15 01:15 The reverse withstand voltage of 1N4148 is only 100V peak. As for whether it will break down when it is over 100V during simulation, I don’t know. In addition, the four MOS tubes of the full bridge seem to...

The driving signal should be generated by comparing the triangle wave with the ref, but I don’t know how to use saber. However, the convergence of matlab simulation of this model with magnetic components is not good.

PowerAnts

maychang posted on 2017-1-15 08:17 "I have only learned about full-bridge DCDC before, and have never actually applied this topology before." If you have learned about full-bridge, you should know that it is absolutely impossible to do without inductors...

There is a special feature: the DC boost part of the inverter, the secondary side rectifier basically has no filter inductor, and is directly a "DC electronic transformer". Take 12V battery input and 220V AC output as an example. The battery discharge voltage range is about 11-14V, and the DC conversion output voltage is 310-400V. After the SWPM closed loop, a stable 220V AC is output. If it is a modified sine wave, The DC conversion output is about 280-350V, and then the duty cycle of the H bridge is adjusted to output a 50Hz rectangular wave (duty cycle is less than 0.5, non-square wave is called modified sine wave :pleased:). In order to save costs, this inverter generally performs average sampling rather than effective value sampling. The waveform factor of a duty cycle of 0.5 is 0.45, and the half-wave average value is about 99V. Here the waveform factor is smaller. Generally, the output of a half-wave average value of 90V is considered to be an effective value of 220V.

maychang

jonny0811 posted on 2017-1-15 10:19 Indeed, I am still a student. My current direction is motor control; I am working on sensorless FOC. I am just interested in power supply, so I tinker with it myself. ...

Motor control is currently achieved through switching power supply. Switching power supply is not just DC-DC conversion, there are many branches of switching power supply. Of course, the most used is small and medium power DC-DC conversion.

maychang

jonny0811 posted on 2017-1-15 10:20 When I simulated, I used an oscilloscope and saw that it was about 130V.

Then we need to look at the current of the four power tubes of the full bridge. Not only the total current, but also the current waveform of each power tube.

maychang

PowerAnts posted on 2017-1-15 10:51 There is a special case: the DC boost part of the inverter, the secondary side rectifier basically has no filter inductor, it is directly a "DC electronic transformer" ...

This, I have never seen it before. In this case, doesn't it mean that the peak current of the power tube is very large? Only the transformer leakage inductance works. Obviously, it is not good for the work of the power tube.

PowerAnts

maychang posted on 2017-1-15 11:31 I have never seen this before. In this case, wouldn't the peak current of the power tube be very large? Only the transformer leakage inductance works. Obviously, the power tube works...

This type of inverter has always been made in this way. I have helped others make several models. This picture is a project that Bu Yixin worked on when he helped me in my studio. This picture was also drawn by Bu Yixin. We connected a current sensing resistor in series at the source stage of the MOS to limit the current overshoot when powering on. Fortunately, it usually takes 20 cycles to fully charge the energy storage capacitor, which takes about several hundred uS. Calculated by the transient thermal resistance method, the junction temperature is controlled at 100 degrees.

jonny0811

PowerAnts posted on 2017-1-15 12:05 This type of inverter has always been made in this way. I have helped others make several models. This picture is a project that Bu Yixin worked on when he helped me in my studio. ...

Hello, moderator. Later I thought about it. In fact, the question I raised before; is it like this: this 2D*(N2/N1) formula can only provide a transformation ratio. This formula is obtained by calculating the power average value through the integral median, and the number of turns of N1 and N2 must be calculated instead of N2/N1=x calculated by me, and then I just set the number of turns of P and S (satisfying S/P=X). The correct way is to calculate the transformation ratio first, and then calculate N1 and N2 according to the formula. The result Uin Uout obtained in this way satisfies the formula I mentioned before. :)

jonny0811

maychang posted on 2017-1-15 11:31 I have never seen this before. In this case, doesn't it mean that the peak current of the power tube is very large? Only the leakage inductance of the transformer works. Obviously, the operation of the power tube is...

Hello, moderator, please take a look at my reply under the reply of the power moderator. Do I understand it well this time? Moreover, the model that Saber imitated at the beginning had a large voltage drop at the IRF end, about 2V. The MOS driver voltage I gave was 10V. For IRF830, this driver is enough.