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
Dewi Posted on January 22, 2021
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.
Marylou Posted on January 22, 2021
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.
Brylin Posted on January 22, 2021
Why doesn't capacitor C1 charge and discharge VG1? I mean the output voltage of VF2 is: capacitor charge and discharge curve superimposed on DC bias 8V. Please guide, thanks!
Jina Posted on January 22, 2021
Don't forget: the capacitor is [through pass through].
