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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500) {this.resized=true; this.width=500; this.alt='這是一張縮略圖，點擊可放大。\n按住CTRL，滾動鼠標滾輪可自由縮放';this.style.cursor='hand'}" onclick="if(!this.resized) {return true;} else {window.open('http://u.dianyuan.com/bbs/u/70/253371225029187.jpg');}" onmousewheel="return imgzoom(this);">    (2)
CTR: The current transfer ratio of the op-couple.

The equivalent circuit of TL431 can be drawn as fig 2:
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500) {this.resized=true; this.width=500; this.alt='這是一張縮略圖，點擊可放大。\n按住CTRL，滾動鼠標滾輪可自由縮放';this.style.cursor='hand'}" onclick="if(!this.resized) {return true;} else {window.open('http://u.dianyuan.com/bbs/u/70/253371225029256.jpg');}" onmousewheel="return imgzoom(this);">   (3)

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.
500) {this.resized=true; this.width=500; this.alt='這是一張縮略圖，點擊可放大。\n按住CTRL，滾動鼠標滾輪可自由縮放';this.style.cursor='hand'}" onclick="if(!this.resized) {return true;} else {window.open('http://u.dianyuan.com/bbs/u/70/253371225029608.jpg');}" onmousewheel="return imgzoom(this);">   fig 3


500) {this.resized=true; this.width=500; this.alt='這是一張縮略圖，點擊可放大。\n按住CTRL，滾動鼠標滾輪可自由縮放';this.style.cursor='hand'}" onclick="if(!this.resized) {return true;} else {window.open('http://u.dianyuan.com/bbs/u/70/253371225029624.jpg');}" onmousewheel="return imgzoom(this);">   fig 4

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......

頂,順便問一下兄弟最近跳了否

頂上學習

樓主繼續(xù),頂上

蠻詳細的,也挺實際的~
再接再勵

要快速穩(wěn)定回路,簡單的方法就是:1,降低增益.這樣子的做法是降低了你的帶寬,可能會引起諸如 輸出紋波變大,源 負載效應變差等不良后果,但帶來的好處是可以很快的看到回路穩(wěn)定了,提高對做好電路的信心.2,降低零點位置.降低零點的位置,也許可以提升更大的相位.但不總是一定的,有時候降低零點位置反而會降低相位(PS:今天有人問我這個怎么理解,很多時候在降低零點位置的時候,極點位置也跟著降低,所以很有可能會引起相位提升的降低...額,不知道清楚了沒有).但不管怎么說,當不知道怎么下手的時候,改變一下零點的位置是值得嘗試的一個好辦法.

500) {this.resized=true; this.width=500; this.alt='這是一張縮略圖，點擊可放大。\n按住CTRL，滾動鼠標滾輪可自由縮放';this.style.cursor='hand'}" onclick="if(!this.resized) {return true;} else {window.open('http://u.dianyuan.com/bbs/u/70/253371225110617.jpg');}" onmousewheel="return imgzoom(this);">   fig 5

500) {this.resized=true; this.width=500; this.alt='這是一張縮略圖，點擊可放大。\n按住CTRL，滾動鼠標滾輪可自由縮放';this.style.cursor='hand'}" onclick="if(!this.resized) {return true;} else {window.open('http://u.dianyuan.com/bbs/u/70/253371225110649.jpg');}" onmousewheel="return imgzoom(this);">    fig 6

500) {this.resized=true; this.width=500; this.alt='這是一張縮略圖，點擊可放大。\n按住CTRL，滾動鼠標滾輪可自由縮放';this.style.cursor='hand'}" onclick="if(!this.resized) {return true;} else {window.open('http://u.dianyuan.com/bbs/u/70/253371225110662.jpg');}" onmousewheel="return imgzoom(this);">    fig 7

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

今天晚上網(wǎng)絡好慢啊 :(.明天再繼續(xù)吧...

^_^,換了個地方...但估計不會很長久....:(

寫的真詳細,頂一下

樓主幸苦了~
期待后續(xù),看看怎樣由現(xiàn)在的狀態(tài)調整至穩(wěn)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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500) {this.resized=true; this.width=500; this.alt='這是一張縮略圖，點擊可放大。\n按住CTRL，滾動鼠標滾輪可自由縮放';this.style.cursor='hand'}" onclick="if(!this.resized) {return true;} else {window.open('http://u.dianyuan.com/bbs/u/70/253371225199805.jpg');}" onmousewheel="return imgzoom(this);">     fig 11

500) {this.resized=true; this.width=500; this.alt='這是一張縮略圖，點擊可放大。\n按住CTRL，滾動鼠標滾輪可自由縮放';this.style.cursor='hand'}" onclick="if(!this.resized) {return true;} else {window.open('http://u.dianyuan.com/bbs/u/70/253371225199824.jpg');}" onmousewheel="return imgzoom(this);">    fig 12

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

謝謝了!

收益.
樓主,問幾個問題,后面有點沒看懂
1.Fig 10~fig 12 show the whole loop, the op-couple and the modulator-power stage bode diagrams.
這句話請幫忙用漢語翻譯一下.能不能解釋一下Fig 10~fig 12這幾張圖,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 交叉頻率是含義不是很明白啊,還有RHP-zero也不知道什么意思
crossover frequency =14.97Hz,怎么會be close to the RHP-zero?
40-15似乎不是很close
確實不懂,大膽說出來了,還望能解釋一下

1.Fig 10~fig 12 show the whole loop, the op-couple and the modulator-power stage bode diagrams.
圖十到圖十二,分別展示了整個回路,光偶的反饋回路,以及主電路部分的bode圖.
crossover frequency  穿越頻率
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.

什么書上有介紹bode圖的?樓主,這只能上電源網(wǎng),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.樓主,除了bode圖,怎么樣分析crossover frequency,the phase margin,RHP-zero ?
2.The crossover frequency is about   of the RHP-zero.
該怎么理解?

ding

謝謝樓主...

RC值怎么確定?

樓主,
1.你前面說當參數(shù)變?yōu)镽=1.5Kohms,C=470nF時,降低了環(huán)路增益,那么根據(jù)你開始將的,肯定降低了環(huán)路帶寬.
G(s)=-R1*Rb1*C1(S+1/R1*Rb1*C1)/Rb1*C1*S,
那么C由47nF增加到470nF,G(s)肯定減小了.f=1/2*3.14*R1C1,RC乘積沒有變,帶寬怎么會減小了呢?
2.如你講" provide a little higher zero",你的意思是提高了零點的位置嗎? 零點s=-1/R1*Rb1*C1,RC乘積沒變,零點位置為什么升高了呢?
3.穿越頻率,右半平面零點和R,C的變化又有什么關系?
還請樓主解釋一下啊,想了2天了, 太痛苦了,問題可能幼稚,不過不提出來我睡不著啊

零點你寫錯了,應該是(1+s(r1+rb1)c1)
增益和1/c1成比例,當c1變大10倍,增益下降20db,但是由于rb1的存在(如果r1比rb1大的話),當r1減少到1/10之一的時候,零點提供的增益增加將不會到20db,所以穿越點將會提前.  不知道說清楚了沒有...
provide a little higher zero, 這個意思應該是把零點的位置提前了...沒表達清楚
3.穿越頻率,右半平面零點和R,C的變化又有什么關系?  額,這個好復雜.本來在dcm的模式下,是不會存在右半平面零點的,但測出來的就有這個東西,說明在實際電路中,還有一些理論上沒覆蓋到的東西吧...穿越頻率是你自己來確定的....R,c會影響到你的穿越頻率的點

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]零點怎么是(1+s(r1+rb1)c1) ,還請指教