# Circuit operation explanation of astable multivibrator

Posted on Feb 5, 2014

It doesn`t break if being little even if it crosses the value of the input/output electric current characteristic. It breaks when short-circuiting and doing the extreme thing in the output. It is explained by the following circuit operation but because the input electric current must flow little at the circuit this time, 74HC04 can not be used. First, I assume that the input (the 9th pin) of the inverter A is the low level.

At this time, the output (the 8th pin) of the inverter A becomes high level. At first, the condition of the inverter B is unsettled but because the output of the inverter B is not totally in the high level condition, above-mentioned electric current flows. When the electric charge begins to store up in C5, C5 can be seen like the short circuit condition. Because it is, the input of the inverter B becomes the high level condition. The output of the inverter B becomes the low level and above-mentioned electric current route is totally formed by this operation. When the electric charge is stored up in C5, the electric current decreases. Input voltage of the inverter B, too, goes down with it and approaches the threshold voltage of the inverter B. Because the output of the inverter B is the low level (being 0V almost), the electric current alwayses fall through more through R2, D2 and the input voltage of the inverter B becomes the threshold voltage (VT). The output of the inverter B is conveyed to the input of the inverter B through R2 when there is not D2. Then, it repeats the operation that the input of the inverter B becomes high level again and that the output becomes the low level and the inverter B has worked in the oscillation. The high level output of the inverter B makes the input of the inverter A the high level condition through C6. At this time, the electric current flows with the inverter B output When the input...

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