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schematic

simulate this circuit – Schematic created using CircuitLab

In order for a diode to allow current flow, the voltage on the anode must be greater than the voltage on the cathode. In the above circuit, one can easily calculate that when the diode is not conducting, the current on R1 is equal to (Vb - Vin) / R1 and when it is, 0. But I have trouble in determining the voltage Vin=V0, above which the diode is ON(I can guess that it is probably equal to Vb but why is that?). Thanks.

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  • \$\begingroup\$ As shown, it is impossible to add a Vdc across a 0 Ohm Vac generator. So put all in series with say Vdc = 600mV then apply 100mVp \$\endgroup\$ Jun 4 at 17:14
  • \$\begingroup\$ @Alema you shouldnt connect 2 voltage sources in parallel to each or 2 current sources in series with each other. \$\endgroup\$
    – Miss Mulan
    Jun 4 at 17:32
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As shown, it is impossible to add a Vdc across a 0 Ohm Vac generator.

So put all in series with say Vdc = 600mV then apply 100mVp or simulation or add a much bigger R in series with Vac then compute the attenuation of Vac into Idc to compute the incremental resistance of the diode. But a series model with one R is easier to compute.

The diode resistance will rise from Vf max according to an exponential value as Vf decreases and thus also mean current. When biasing Vbe for high gain , it is important to minimize variations of Vbe for linearity, so small fixed Re values are used similar to re or use negative feedback.

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When Vin goes positive D1 short circuits it. If these are ideal components Vin will drive an infinite current through D1.

In practice

  1. The AC source will burn up or
  2. The voltage will collapse to the forward voltage of D1 or
  3. The diode will burn up.
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