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The circuit below shows a Darlington pair that is used to amplify an AM signal picked up by an LC tank. As you can see in the input and output waveforms below, the signal is amplified but not decoupled from DC, despite using a 1u capacitor on the transistor output. Why is this the case?

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    \$\begingroup\$ There's no current through the capacitor. \$\endgroup\$ – immibis Dec 10 '17 at 0:56
  • \$\begingroup\$ I found the problem -- Use Initial Conditions (UIC) option on the simulation command string. My understanding is that this enabling this option skips the DC operating point calculation. I'm not sure why one would actually ever need to do that. \$\endgroup\$ – Qubit1028 Dec 10 '17 at 15:29
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Since you don't have a DC return path to ground on the right side of the capacitor, it simply retains its initial charge of 0V throughout the simulation.

Try connecting a 1MΩ resistor between Vout and ground.

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A series cap is a HPF with a Load R, use 500 Ohms 1uF*500R= 500us=T
AM is > 500kHz with a cycle time of 2us. L1C1 has a Zo=1k reactance so Q depends on series 120k feedback // Zbe so use hFE>200.

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