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ece240: finish diodes
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@ -73,3 +73,18 @@ $$r_d=\left(\frac{\partial i_D}{\partial v_D}\right)^{-1} = \frac{V_T}{I_D}$$
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$$i_D(t)=I_D+\frac{1}{r_d}V_d(t)$$
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### Signal analysis
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1. Analyse DC signals
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- assume blocking capacitors are open circuits
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- turn off AC sources
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2. Analyse AC signals
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- assume blocking capacitors are shorts
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- turn off DC sources
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- replace diode with effective resistor (the differential resistor)
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!!! tip
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Most $R$s in the circuit can be assumed to be significantly greater than $r_d$, so $r_d$ can be removed in series or $R$ can be removed in parallel.
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!!! warning
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Oftentimes, turning off a DC source to nowhere is actually a short to ground.
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