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
Femke Posted on August 25, 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.
Veasna Posted on August 25, 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.
Jinny Posted on August 25, 2020
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!
Pocahontas Posted on August 25, 2020
Capacitance is [through pass through]
