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
Akira Posted on September 16, 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.
Taya Posted on September 16, 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.
Aya Posted on September 16, 2020
But the pulse signal will be smoothed after passing the capacitor? There is a charge and discharge curve. So my understanding is that the curve at VF2 is: the capacitor charge and discharge curve is superimposed on the 8v DC voltage. Is this understanding? Thank you!
