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555定時(shí)器非穩(wěn)態(tài)(或自由運(yùn)行)多諧振蕩器 原理圖及電路分析

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  電容 C 通過(guò) R1 和 R2 充電,當(dāng)電壓上升到 2/3 的電源電壓時(shí),六腳檢測(cè)到此狀態(tài),并且 7 腳 連到地。電容通過(guò) R2 放電,直到它的電壓小于
1/3 的電源電壓,并且 2 腳檢測(cè)到該狀態(tài),關(guān)閉 七腳,重復(fù)以上的循環(huán)。 上面的電阻是為了防止 7 腳被損壞,因?yàn)楫?dāng)六 腳檢測(cè)到 2/3 的電源電壓時(shí)它短接到 0V。它的 阻值相對(duì)于 R2 來(lái)說(shuō)很小,并且不會(huì)干擾到振蕩 周期的計(jì)算。


下圖適用于非穩(wěn)態(tài)電路:


Using the graph:(計(jì)算部分,不敢隨便翻譯)
Suppose R1 = 1k, R2 = 10k and C = 0.1u (100n).
Using the formula on the graph, the total resistance = 1 + 10 + 10 = 21k
The scales on the graph are logarithmic so that 21k is approximately near the "1" on the 10k. Draw a line parallel to the lines on the graph and where it crosses the 0.1u line, is the answer. The result is approx 900Hz.
Suppose R1 = 10k, R2 = 100k and C = 1u
Using the formula on the graph, the total resistance = 10 + 100 + 100 = 210k The scales on the graph are logarithmic so that 210k is approximately near the first "0" on the 100k. Draw a line parallel to the lines on the graph and where it crosses the 1u line, is the answer. The result is approx 9Hz.
The frequency of an astable circuit can also be worked out from the following formula:


文章摘自:50個(gè)555定時(shí)器電路分享(一個(gè)國(guó)外優(yōu)秀文檔的翻譯,個(gè)人覺(jué)得是對(duì)于掌握555定時(shí)器電路最有幫助的資料)每個(gè)電路都有詳細(xì)的解說(shuō),全部?jī)?nèi)容請(qǐng)下載:http://www.torrancerestoration.com/bbs/dpj-80748-1.html
作者:MCU起航

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