Hey guys please help me
The text says:
"When Vin is high the diode is FB and capacitor charges up to peak. When Vin is low the diode is RB and the capacitor tends to discharge. The reason given for this is the diode leakage currents.."
But i don't understand one thing that is that as we know that the leakage current of diode is the current that the diode will leak when a reverse voltage is applied to it and the leakage current is entire function of the reverse bias voltage applied to it...so why the text tells us that this leakage current of diode is responsible for the discharge of the capacitor...??
or if i am interpreting it right then it might be inferred as that the diode due to its leakage current provides a path for the capacitor current to discharge and get dumped in the op amp....??
Please correct me.....?
In the second figure during when Vin is high both diodes are FB and capacitor charges ....and when Vin is low then Vout which is equal to capacitor voltage will be feedback to node x via R1 thus causing zero potential across diode D2 hence eliminating leakage current of D2 ..but what happen to the leakage current of diode D1.The text say it is isolated from the capacitor but suppose if I artificially induce same Vout at node x and feedback resistor is absent then...does the leakage current of diode D1 interfere with the capacitor...??
The text says:
"When Vin is high the diode is FB and capacitor charges up to peak. When Vin is low the diode is RB and the capacitor tends to discharge. The reason given for this is the diode leakage currents.."
But i don't understand one thing that is that as we know that the leakage current of diode is the current that the diode will leak when a reverse voltage is applied to it and the leakage current is entire function of the reverse bias voltage applied to it...so why the text tells us that this leakage current of diode is responsible for the discharge of the capacitor...??
or if i am interpreting it right then it might be inferred as that the diode due to its leakage current provides a path for the capacitor current to discharge and get dumped in the op amp....??
Please correct me.....?
In the second figure during when Vin is high both diodes are FB and capacitor charges ....and when Vin is low then Vout which is equal to capacitor voltage will be feedback to node x via R1 thus causing zero potential across diode D2 hence eliminating leakage current of D2 ..but what happen to the leakage current of diode D1.The text say it is isolated from the capacitor but suppose if I artificially induce same Vout at node x and feedback resistor is absent then...does the leakage current of diode D1 interfere with the capacitor...??
Attachments
-
18.1 KB Views: 16
-
21.4 KB Views: 23
Last edited: