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Capacitor charging and discharging problems.

Hardware design
10月 30, 2020 by Leighton 481

As shown in the figure, 12v DC and 0-3V square wave pass through a simple resistance capacitor circuit to output a 6.5-9.5V square wave signal.

And the output signal amplitude can be adjusted by adjusting the lower resistance.

I want to know, what is the principle of intermediate output? Thank you all

TIM截图20180330104309.png

所有評論

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Carl 發表於 October 30, 2020

The value of VF2 is the divided voltage of R1 and R2 to 12V DC VS1, and then the square wave of VG1 is superimposed on the DC blocking capacitor.

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Orlando 發表於 October 30, 2020

6.5+9.5=16/2=8. That is, R1 and R2 are divided into 8V DC, and the upper and lower 3/2=1.5 amplitude square wave fluctuations.

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Bryant 發表於 October 30, 2020

Don't forget: the capacitor is [through pass through].

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