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I have been trying to analyze this circuit and calculate the maximum voltage across the capacitor. Here is the diagram:

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And, here are the graphs for various inputs and the resultant outputs:

VG1(peak): 300mV
VM2: 157mV

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Even though the diode is not conducting(in cutoff region), the capacitor is getting charged. I don't know why.

VG1(peak): 1V
VM2: 769mV

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VG1(peak): 30V
VM2: 29.61V

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What decides the maximum voltage on the capacitor?

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  • \$\begingroup\$ "Even though the diode is not conducting(in cutoff region), the capacitor is getting charged" - how did you come to that conclusion? \$\endgroup\$ – Nick Johnson Jul 22 '15 at 13:37
  • \$\begingroup\$ @NickJohnson: I mean the supply voltage does not cross diode threshold voltage of 750mV. In that sense I assumed it to be operating in cutoff region. Maybe I am wrong here. \$\endgroup\$ – om sai Jul 22 '15 at 13:41
  • \$\begingroup\$ I've updated my answer to address that point. \$\endgroup\$ – Nick Johnson Jul 22 '15 at 13:44
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The maximum voltage on the capacitor will be equal to the maximum input voltage, less the diode's voltage drop. Whenever the input voltage is more than a diode drop above the capacitor voltage, the diode conducts and charges up the capacitor further. Whenever the diode is reverse biased, no current flows.

It's worth noting that a diode's forward voltage is dependent on current: it's not a perfect 'elbow'. A diode will allow a small amount of current to flow even at very low forward voltages, and with high currents, the forward voltage will be substantially higher than the 'typical' forward voltage.

The capacitor will continuously discharge (slowly) through its parasitic series resistance, so in reality it won't remain fully charged forever.

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  • \$\begingroup\$ Okay. Now I understand that the forward voltage drop depends on the current drawn. But how to calculate the maximum voltage across the capacitor? For instance, in 3rd case, why is the capacitor charged to 29.61V? How to know that forward voltage drop is 400mV and not 750mV. Can it be known through datasheet only? \$\endgroup\$ – om sai Jul 22 '15 at 13:57
  • \$\begingroup\$ @omsai Sure, look at, eg, chart 2 in the 1n4148 datasheet, which shows forward voltage with current. \$\endgroup\$ – Nick Johnson Jul 22 '15 at 14:01
  • \$\begingroup\$ Thanks for the datasheet. I did notice the voltage drop of around 400mV at the charging current of 1mA. What about the first case when there is no question of forward drop, why is the capacitor voltage 157mV? How do I calculate that? \$\endgroup\$ – om sai Jul 22 '15 at 14:10
  • \$\begingroup\$ @omsai The datasheet's current curve doesn't go below 1mA; in your first example it's that low because some current can still flow. Which could be accurate, or it could just be an artifact of the model. \$\endgroup\$ – Nick Johnson Jul 22 '15 at 14:17

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