![]() ![]() Here, the common gate amplifier has less input resistance, which can be given as Rin = 1/gm. ![]() If, the applied voltage is ‘Vgs’ & the current at the source is ‘Vgs*gm’, then: Here in the ‘T’ model, the gate current is always zero. The small-signal model and T model of a common-gate amplifier equivalent circuit are shown below. The common gate amplifier configuration is mainly used to provide high isolation in between i/p & o/p to prevent oscillation or less input impedance. In the CG configuration, the source terminal (S) of the transistor works like the input whereas the drain terminal works like the output & the gate terminal is connected to the ground (G). Common-Gate (CG) AmplifierĪ common-gate (CG) amplifier is normally used as a voltage amplifier or current buffer. A CS MOSFET amplifier suffers from a poor high-frequency performance like most of the transistor amplifiers do. The output resistance can be reduced by decreasing the RD but also the voltage gain can also be decreased. Thus, the CS MOSFET amplifiers have infinite i/p impedance, high o/p resistance & high voltage gain. So, the voltage gain (Gv) is the similar as the voltage gain accurate (Av), ![]() The load resistor (RL) is connected to the o/p across RD, then the terminal voltage gain through the voltage divider formula can be expressed as Īv = Avo (R L/R L + Ro) = −gm (R DR L/R L + R D) = −gm(R D||RL)įrom the information that Rin = ∞, after that vi = vsig. And through the test-current technique, the o/p resistance is To verify the Norton equivalence resistance, set vi = 0, so that the circuit will be an open circuit, so there is no current flow. In this case, using the Norton equivalence is more convenient. One can replace a linear circuit driven by a source by its Th´evenin equivalence.įrom the small-signal circuit, one can change the output fraction in the circuit by a Norton’s or Thevenin’s equivalence. ![]() So, the current induced within the o/p port is i = −g mv gs as specified through the current source. This small-signal circuit can be replaced by the hybrid-π model which is shown in the following figure. In the following small-signal CS MOSFET amplifier, the ‘RD’ resistor measures the resistance in between the drain (D) & the ground (G). The small-signal and hybrid π model of a common source MOSFET amplifier is shown below. This is very popular due to high gain and larger signal amplification can be achieved. The common-source MOSFET amplifier is related to the CE (common-emitter) amplifier of BJT. In this configuration, the source terminal acts as a common terminal in between the i/p and o/p. Common Source MOSFET AmplifierĬommon source amplifier can be defined as when the i/p signal is given at both the terminals of the gate (G) & source (S), the o/p voltage can be amplified & attained across the resistor at the load within the drain (D) terminal. MOSFET amplifiers are available in three types like common source (CS), common gate (CG), and common drain (CD), where each type along with its configuration is discussed below. In the above equation, sign “-” comes from the fact that the MOSFET amplifier inverts the o/p signal in equivalence with the BJT CE Amplifier. After that simplification, the equation will become Av = – RD/Rs=1/gm The voltage gain (AV) is the ratio of input voltage and output voltage. Vout = – RD x ID = -gmVGS RD Voltage Gain The o/p voltage (Vout) is simply given through the voltage drop across the drain resistor (RD) So, ID = gm×VGS & the input voltage (Vin) can be factored by VGS like the following. The voltage drop across the RS resistor can be given by RS×ID.Īccording to the transconductance (gm) definition, the ratio of ID (drain current) to VGS (gate-source voltage) once a constant drain-source voltage is applied. The input voltage (Vin) can be given through the gate (G) to source (S) voltage like VGS. To make it simpler, we need to consider that there is no load is connected with the drain branch in parallel. ![]()
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