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enter image description here

Hello, I have some question about the circuit i posted as photo file. I have there a AC source 4V amplitude. R= 100k ohm. Uv=1.5v. RB=680k ohm. I know that diode flows current at one direction. And when u have AC source you have either a only positive voltage through a diode or a only a negetive voltage. But I dont understand the circuit when you have a DC source of 1.5V for first part and then 4.5 V for second part.

I did the simulation of the two parts: The green one is AC signal and the blue one is de voltage through de Resistor 680K ohm. But I dont understand the working of this circuit? Can someone please explain me this circuit. enter image description here

enter image description here

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The diode only conducts when Anode is more positive then the Cathode. Since the Anode is fixed at 1.5V, the Cathode could not be lower than 0.8V during AC swing of the voltage source. Therefore Vx varies between ~0.8V and \$ \frac{R_B}{R_B+R} \$.4V =3.4V.

The voltage at the cathode actually follows the supply voltage change. For example, if the voltage source decreases, then does so the cathode's. The problem is that the cathode cannot be brought less than 0.8V due to the anode being held fixed at 1.5V. So during the negative cycle of the voltage source the voltage at the cathode, or across Rb, stays fixed at about 0.8V. In this case, the diode is forward biased.

But during the positive cycle of the voltage source the diode is reverse-biased because the cathode is being more positive than the anode. This shuts down the diode and no current would go through the middle branch. The circuit then reduces to a simple resistive divider. /end

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  • \$\begingroup\$ Thank you for the answer. If i look at the simulation i have there no negative wave. But there is a DC horizontal voltage. What is the reason for this? \$\endgroup\$ Dec 10 '17 at 14:57
  • \$\begingroup\$ @RoshanTimsina see the update. \$\endgroup\$
    – dirac16
    Dec 10 '17 at 15:42

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