When I make a DC analysis for the following circuit the capacitors are open circuit.
But what if C1 does not exist, what is the effect in DC analysis?
Does the signal generator still exist or should I short it?
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1\$\begingroup\$ It's either oc or sc - which one? \$\endgroup\$– Andy akaCommented Mar 25, 2022 at 11:58
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\$\begingroup\$ exactly , i dont know how the circuit would be like. \$\endgroup\$– alawnehCommented Mar 25, 2022 at 12:18
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1 Answer
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When C1 is present, you replace it with an open-circuit for DC analysis, meaning that the gate is driven by the R1/R2 voltage divider.
If C1 were absent and the signal generator were connected directly to the input, you would need to use the generator's steady-state DC voltage when solving for the operating point of the circuit.
answered Mar 25, 2022 at 14:31
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\$\begingroup\$ that mean i have to short the signal generator cause it has a zero value at steady state ? \$\endgroup\$– alawnehCommented Mar 25, 2022 at 16:15
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\$\begingroup\$ @alawneh This particular design would not make much sense in DC analysis if the generator had a zero value at steady state, and C1 were replaced with a wire. At the steady state of 0 V, the transistor would not be turned on. With the capacitor still present, you ignore the generator because C1 is an open circuit, and the transistor is biased by R1, R2. \$\endgroup\$ Commented Mar 25, 2022 at 16:49