**7. Determine the Miller input and output capacitances for the amplifier in Figure 10–57.****Circuit Diagram Description:**The circuit features an N-channel JFET amplifier configured as follows:- The input signal $ V_{\text{in}} $ is applied through a coupling capacitor of 0.001 µF.- A 50 Ω resistor is connected to $ V_{\text{in}} $.- The gate of the JFET is connected to the input capacitor and a 10 MΩ resistor which leads to the ground.- The drain is connected to a +10 V supply through a 1.0 kΩ resistor.- A capacitor and resistor, both 1.0 kΩ, are connected in parallel between the source and ground.- The output is taken across the 10 kΩ load resistor connected to the drain through a 0.001 µF capacitor.- A 0.1 µF capacitor connects the source to the ground in parallel to the 1.0 kΩ source resistor.**Parameters:**- $ C_{\text{iss}} = 10 \, \text{pF} $- $ C_{\text{rss}} = 3 \, \text{pF} $- $ I_{\text{GSS}} = 18 \, \text{nA} \, @ \, V_{\text{DS}} = -10 \, \text{V} $- $ V_{\text{GS(off)}} = -8 \, \text{V} $- $ I_{\text{DSS}} = 10 \, \text{mA} $This setup aims to provide a clear understanding of how Miller capacitance affects the input and output capacitive characteristics in this amplifier circuit.

Miller Capacitance - When reading about guitar amplifier circuit design, the term "Miller capacitance" comes up frequently. The plate-to-grid capacitance in a triode tube (or transistor) is effectively multiplied by the gain of the amplifying stage. A tube must work against the plate-to-grid capacitance when amplifying a signal, charging and discharging it as the signal varies. Because the grid has a high impedance and cannot sink or source current, the charging current must be sourced or sunk through the prior stage's driving source resistance. This results in a lowpass filter with a corner frequency dictated by the previous stage's source resistance and input capacitance. Because Miller capacitance may easily ruin the frequency response, there are a few factors to keep in mind when designing guitar amplifiers. If you want to reduce the frequency response effect of the Miller capacitance, you can do the following: Reduce the previous stage's output impedance. Reduce the value of the plate load resistor, use a tube with a lower internal plate resistance, or reduce the value of any series or shunt attenuation resistors to achieve this. Obviously, all of these factors will have an impact on stage gain, so this must be considered as well.Reduce the stage's gain.The miller effect is named after John Milton Miller's work. The capacitance of the equivalent circuit of an inverting voltage amplifier can be increased using the miller theorem by adding more impedance between the input and output terminals of the circuit. According to Miller's theorem, a circuit with an impedance (Z) connecting two nodes with voltage values V1 and V2 has an impedance (Z). When this impedance is replaced with two different impedance values and linked to the same input and output terminals to the ground, the frequency response of the amplifier can be analyzed as well as the input capacitance can be increased. The Miller effect is the name for this type of effect. The following data is given along with the circuit diagram shown in fig .1 Ciss= 10 pFCrss= 3 pFIGSS =18 nA @ VDS = -10 VVGS(off)=-8 VIDSS = 10mAGate to source voltage - When a gate-source voltage greater than the rated gate threshold voltage Vth is applied, the MOSFET switches on, as illustrated in the diagram. When the gate voltage is near the threshold voltage, however, the drain-source on-state resistance is higher than the specified value because the channel is not sufficiently formed. Calculate the value of the gate−source voltage : VGS =VG-Vs where VG=IGR and Vs=IDRs But as no gate current flow through reverse biased gate source IG=0 As the gate current is Zero, the gate voltage is VG = 0 Now the gate −source voltage reduces to VGS=-Vs =-IDRs Calculate the value of the gate-source voltage Calculate the equation of drain current ID = IDSS1-VGSVGS(off)2 =(10m)1-(-IDx1.0 k)(-8)2 =(10x10-3)1+ID2(1x106)64-ID(1x103)4 =10x10-3+ID2(10x103)64-10ID4Th above equation can be simplified and represented as 156.25ID2-3.5ID+10x10-3=0---(1) Solving equation 1 we get ID =0.019 A and ID=3.36mA But the value of ID will be less than the value of I Therefore the value of drain current is ID =3.36mA Calculate the value of gate source voltage VGS=-IDRS =(-3.3x10-3)(1x103) Therefor the gate source voltage is VGS=−3.36 V Transconductance - Certain electrical components have the property of transconductance. Transconductance is the ratio of the current change at the output port to the voltage change at the input port; conductance is the reciprocal of resistance. gm=-21DSSVGSS(off)1-VGSVGS(off) =(-2)(10x10-3)(-8)1--3.36-8 =1.45 mS Therefore the value of transconductance is gm=1.45 mS Calculate the voltage gain for the FE amplifier Av = -gm(RD∥RL) =(1.45m)(1.0k∥10k) =1.32 The voltage gain is Av= 1.32 Calculate the value of gate −drain capacitance Cgd=VRss =3 pF Therefore the gate −drain capacitance is Cg=3pF Calculate the miller capacitance Cinmiller=Cgd(Av+1) =(3pF)(1.32+1) =6.95 pF Therofore the input capacitance is 6.95 pF According to the output ripple voltage, which is centred on the output voltage, the output capacitor is repeatedly charged and discharged. After that, we'll talk about output capacitors. When choosing an output capacitor, keep the following three things in mind. Cout= Cgd(Av+1Av) =(3pF)(1.32+11.32) =5.28pF Therefore the output capacitance is 5.28 pF