Circuit Simulator - Common-Source Amplifier
Why should I learn to use the circuit simulator to design Common-Source Amplifier circuits?
Learn how to use circuit simulator software to design your own Common-Source Amplifier circuits.
Where can I get a Common-Source Amplifier circuit diagram with an explanation?
IndiaBIX provides numerous Common-Source Amplifier circuit diagrams with detailed explanations and working principles.
How do I design a Common-Source Amplifier circuit with this circuit simulator?
You can easily design Common-Source Amplifier circuit diagrams by practising with the given circuit simulator. With this online circuit simulator, you can design and simulate your own electronic circuits.
Common-Source Amplifier
Circuit Description:
This is a common-source amplifier, which amplifies the input voltage about 30 times.
The gain of this amplifier is determined partly the transconductance of the MOSFET. This depends on the bias point in the circuit; here it averages about 9 mA/V. This means that a change in the gate voltage causes a change in the drain current that is 9 mA/V times the gate voltage change.
The drain current goes through a 4k ohm resistor. The capacitor connected to the source is supposed to act as a short circuit at the input frequency, so it and the source resistor can be ignored. A change in input of 50 mV causes a change in drain voltage of 9 mA/V * 50 mV * 4000 ohms, or about 1.8 V. This means the predicted output gain is 36. Here it is less because the transconductance changes over the input cycle (the amplifier isn't quite linear), we are ignoring the output resistor, and the source capacitor isn't quite a short circuit like we supposed.
The gain of this amplifier is determined partly the transconductance of the MOSFET. This depends on the bias point in the circuit; here it averages about 9 mA/V. This means that a change in the gate voltage causes a change in the drain current that is 9 mA/V times the gate voltage change.
The drain current goes through a 4k ohm resistor. The capacitor connected to the source is supposed to act as a short circuit at the input frequency, so it and the source resistor can be ignored. A change in input of 50 mV causes a change in drain voltage of 9 mA/V * 50 mV * 4000 ohms, or about 1.8 V. This means the predicted output gain is 36. Here it is less because the transconductance changes over the input cycle (the amplifier isn't quite linear), we are ignoring the output resistor, and the source capacitor isn't quite a short circuit like we supposed.
Discussion:
1 comments Page 1 of 1.
Azad said:
1 decade ago
Hi, can you more information about it, to any student like come read it can get some information to use it in the discussion of lab experiment.
Thanks.
Thanks.
(7)
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