Compensating the forward voltage drop of a diode signal rectifier

Posted on Feb 7, 2014

It says that D1 compensates for D2`s forward drop by providing 0. 6V of bias. Is the +5V an external 5v source How does it compensate The R1, R3 and D1 basically creates a 0. 6V bias on the other side of the capacitor, so that a positive swing in the signal does not have to overcome a0. 6V hurdle. D_1 and R_3 form a shunt voltage regulator. The 0. 6V voltage is conveyed to D2 which is on the verge

Compensating the forward voltage drop of a diode signal rectifier
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of conducting, as a result. So only a small positive upswing from the input is required to bring it to conduction. Because the input is capacitively coupled, it is pure AC. Its swings are additively superimposed on top of the bias voltage that exists on the other side of the capacitor. The 5V source is just from somewhere in the rest of the circuit. There is nothing special about it. Perhaps you can get a different perspective by redrawing the circuit so that voltage falls from top to bottom. In this view, we highlight how the input is biased to 0. 6V, but the output is 0. 6V below that, down across the voltage drop of D1. So for instance suppose that the input creates a positive swing of 0. 1V. This becomes 0. 7V at the top of D2 (the whole point of the bias). At the bottom of D2, that swing is 0. 1V again. D2 lets through enough current so that R2 has 0. 1V across it. A negative swing of 0. 1V turns to 0. 5V. But this cannot create a -0. 1V output at the bottom of D2; that is nonsense because it is outside of our supply range. 0. 5V is not enough to forward bias D2, and so the output is at 0V, pulled to ground by R2, which has almost no current flowing across it to create any voltage. The purpose of R1 is to act as a flexible linkage to separate the reference 0. 6 voltage, which is quite stiff, from the point where the signal is injected, which must on the contrary be free to swing about 0. 6V. R1 also protects the diode...

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