『原创』菜鸟课堂2--快速稳定你的回路

1.Review the feedback-loop path transfer function includes the TL431 and the op-couple.
The feedback-loop path looks like fig 1:
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The fellow equation can be driven from fig 1 directly:
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CTR: The current transfer ratio of the op-couple.

The equivalent circuit of TL431 can be drawn as fig 2:
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Insert (3) and (2) into (1):

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2. The original parameters.
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   In this condition, the system is unstable. Fig 3 and fig 4 show the Vctrl, Vds and 5v output   ripple waveforms.
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From fig 3, the Vds waveform(blue one) is very chaos. Some times it has 3 valleys. Some times it has 5 valleys. And it also has 1 valley some times. The pink one is the VCtrl waveform. It is a sine-wave and the peak-peak voltage is about 192mv. This VCtrl waveform can not be good regulate the PSU.
Fig 4 is the 5V output ripple waveform. It looks awful and the peak-peak voltage is almost 100mv.

Then used the network analyzer to analyse the whole loop, the op-couple and the modulator-power stage bode diagrams. They are showed in fig5~fig7.


To be continued......

顶,顺便问一下兄弟最近跳了否

顶上学习

楼主继续,顶上

蛮详细的,也挺实际的~
再接再励

要快速稳定回路,简单的方法就是:1,降低增益.这样子的做法是降低了你的带宽,可能会引起诸如 输出纹波变大,源 负载效应变差等不良后果,但带来的好处是可以很快的看到回路稳定了,提高对做好电路的信心.2,降低零点位置.降低零点的位置,也许可以提升更大的相位.但不总是一定的,有时候降低零点位置反而会降低相位(PS:今天有人问我这个怎么理解,很多时候在降低零点位置的时候,极点位置也跟着降低,所以很有可能会引起相位提升的降低...额,不知道清楚了没有).但不管怎么说,当不知道怎么下手的时候,改变一下零点的位置是值得尝试的一个好办法.

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PS:Please pay attention to fig 7. Do you find anything interesting?

今天晚上网络好慢啊 :(.明天再继续吧...

^_^,换了个地方...但估计不会很长久....:(

写的真详细,顶一下

楼主幸苦了~
期待后续,看看怎样由现在的状态调整至稳定.

From fig 5, the crossover frequency is 14.97kHz. And from fig 7, the RHP-zero is about 40kHz. The crossover frequency is too high. It is close to the RHP-zero. So the response is too fast. It controls the IC to do the regulation in one cycle. So everything is wrong.

3.Improve the parameters.
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Due to express (4), the new R1,C1 value make the gain smaller than the old one. And provide a little higher zero. It will boost more phase.
   This time, the system is stable. Fig 8 and fig 9 show the Vctrl, Vds and 5v output   ripple waveforms.

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From fig 8, the Vds and the Vctrl waveforms are the exactly waveforms what are expected. And the 5V output ripple is much smaller than the original one. This one is the natural one.
Fig 10~fig 12 show the whole loop, the op-couple and the modulator-power stage bode diagrams.

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From fig 10, the crossover frequency is 5.01kHz and the phase margin is about 82.6 degrees. From fig 12, the RHP-zero is about 40 kHz. The crossover frequency is about   of the RHP-zero.

THE END

谢谢了!

收益.
楼主,问几个问题,后面有点没看懂
1.Fig 10~fig 12 show the whole loop, the op-couple and the modulator-power stage bode diagrams.
这句话请帮忙用汉语翻译一下.能不能解释一下Fig 10~fig 12这几张图,看不懂哦
2.the crossover frequency is 14.97kHz. And from fig 7, the RHP-zero is about 40kHz. The crossover frequency is too high. It is close to the RHP-zero. So the response is too fast. It controls the IC to do the regulation in one cycle. So everything is wrong.
中crossover frequency 交叉频率是含义不是很明白啊,还有RHP-zero也不知道什么意思
crossover frequency =14.97Hz,怎么会be close to the RHP-zero?
40-15似乎不是很close
确实不懂,大胆说出来了,还望能解释一下

1.Fig 10~fig 12 show the whole loop, the op-couple and the modulator-power stage bode diagrams.
图十到图十二,分别展示了整个回路,光偶的反馈回路,以及主电路部分的bode图.
crossover frequency  穿越频率
rhp-zero: right half plane zero.
15k只有40k的不到1/3很接近,很接近了...

good, ths, it is very visual ,continue concerning

i am very interesting in the measure process about the bode diagram , such as you said: the whole loop, the op-couple and the modulator-power stage . can you draw some circuit to express?

Buy a loop analyzer (usually it it very expensive, more than 10k usd. Please find some day that your boss looks great, then tell him. "A loop analyzer can improve our product. I think we need one."). Good luck to you!
There will be a instruction. Then you can do these.

什么书上有介绍bode图的?楼主,这只能上电源网,所以不能去google

From fig 10, the crossover frequency is 5.01kHz and the phase margin is about 82.6 degrees. From fig 12, the RHP-zero is about 40 kHz. The crossover frequency is about   of the RHP-zero.
1.楼主,除了bode图,怎么样分析crossover frequency,the phase margin,RHP-zero ?
2.The crossover frequency is about   of the RHP-zero.
该怎么理解?

ding

谢谢楼主...

RC值怎么确定?

楼主,
1.你前面说当参数变为R=1.5Kohms,C=470nF时,降低了环路增益,那么根据你开始将的,肯定降低了环路带宽.
G(s)=-R1*Rb1*C1(S+1/R1*Rb1*C1)/Rb1*C1*S,
那么C由47nF增加到470nF,G(s)肯定减小了.f=1/2*3.14*R1C1,RC乘积没有变,带宽怎么会减小了呢?
2.如你讲" provide a little higher zero",你的意思是提高了零点的位置吗? 零点s=-1/R1*Rb1*C1,RC乘积没变,零点位置为什么升高了呢?
3.穿越频率,右半平面零点和R,C的变化又有什么关系?
还请楼主解释一下啊,想了2天了, 太痛苦了,问题可能幼稚,不过不提出来我睡不着啊

零点你写错了,应该是(1+s(r1+rb1)c1)
增益和1/c1成比例,当c1变大10倍,增益下降20db,但是由于rb1的存在(如果r1比rb1大的话),当r1减少到1/10之一的时候,零点提供的增益增加将不会到20db,所以穿越点将会提前.  不知道说清楚了没有...
provide a little higher zero, 这个意思应该是把零点的位置提前了...没表达清楚
3.穿越频率,右半平面零点和R,C的变化又有什么关系?  额,这个好复杂.本来在dcm的模式下,是不会存在右半平面零点的,但测出来的就有这个东西,说明在实际电路中,还有一些理论上没覆盖到的东西吧...穿越频率是你自己来确定的....R,c会影响到你的穿越频率的点

i am not care about the price,i have measured the bode diagram, but there are big different in magnitude and phase between measured and calculated.so if you can show me the measured process, such as the access and output point in the circuit, i will very appreciate.

I will draw a draft and scan it next week. I'll be out of office tomorrow.

G(s)=-(1/Rb1) *[(1/C1S)+R1]=-[(1+R1C1S)/Rb1C1S]零点怎么是(1+s(r1+rb1)c1) ,还请指教