a)                                                                   b)

c)vi= 2 sin 4tV    d) vi= 4tV

We know that

v0 = - 1/RC + v0(0)

Here R= 500K Ω   and C = 1µF

1/RC = 1/ 500 x 1000 x 1 x 10-6  

= 1/RC = 2

= -2          ; vi =- 5V

= -2

= 10t V

Which means it is a ramp voltage with positive slope.

When vi=5V

-2          ; vi = 5V

= -2

= -10t V

Which means it is a ramp voltage with negative slope.

When vi= 2 sin4t V

Vo =  -2

      = -4 [-cos 4t]  /4   = cos 4t

When vi=4t V

Vo =

            = -8 [t2/2] =   - 4V

A4)

 Vout = 0V

Va = 166V

Vout = 5V

Va = 333V

A5) Rectifier circuit gives average value of input signal; but in practice we need peak value of input signal. This is achieved by peak detector circuit. The following figure shows a simple peak detector circuit using diode and capacitor.

Fig 1 Peak detector

In the positive half cycle, diode D is forward biased and capacitor C starts charging. When input reaches its peak value, capacitor gets charged to positive peak value.

In negative half cycle, as input decreases, diode D is reversed biased and capacitor is isolated and holds the peak value of previous cycle. Hence called as peak detector.

But in practice, output is taken across some load RL, so when input voltage decreases capacitor discharges through load RL. To avoid this, select RL of very large value so that capacitor discharges very slowly hence almost hold the charge. Whatever charge it lost through RL is gets back in next half cycle.

A6) As the name says it is an amplifier which produces the output proportional to logarithmic of the applied input. The log amplifier using op-amp is shown below. The input is applied through the inverting end of op-amp. As the non-inverting end has voltage zero then by virtual ground concept the voltage at inverting terminal also becomes zero.

Fig 2 Log Amplifier

The equation for input voltage will be

+If = 0

If =

The current flowing through diode is given as

Where:

Is = Saturation Current

Vf = Voltage drop across diode in forward bias

VT = Thermal equivalent voltage

For feedback loop the KVL equation will be

0-Vf -V0 = 0

Vf = -V0

Substituting Vf in above equation of If

Equating both equations of If

Taking natural log of both sides we get

The above equations shows that the output is natural log of the applied input.

A7) This device produces the output proportional to antilog of input. The inverting op-amp is used in this case as well. The figure below shows an antilog amplifier with its inverting terminal connected to the input end with diode and the non-inverting terminal is grounded.

Fig 3 Antilog Amplifier

Applying KCL at input terminal we get

The current flowing through diode is given as

Substituting If in above voltage equation we get

At inverting terminal applying KVL we get

Substituting Vf in equation of V0 we get

The above equations shows that the output is natural antilog of the applied input.

A9) The Instrumentation amplifiers consist of three op-amps. In this circuit, a non-inverting amplifier is connected to each input of the differential amplifier.

This instrumentation amplifier provides high input impedance for exact measurement of input data from transducers. The circuit diagram of an instrumentation amplifier is as shown in the figure below.

Fig 3 Instrumentation Amplifier

The op-amps 1 & 2 are non-inverting amplifiers and together form an input stage of the instrumentation amplifier. The op-amp 3 is a difference amplifier that forms the output stage of the instrumentation amplifier.

Working

The output stage of the instrumentation amplifier is a difference amplifier, whose output Vout is the amplified difference of the input signals applied to its input terminals.

If the outputs of op-amp 1 and op-amp 2 are Vo1 and Vo2 respectively, then the output of the difference amplifier is given by,

Vout = (R3/R2)(Vo1-Vo2)

The expressions for Vo1 and Vo2 can be found in terms of the input voltages and resistances.

Consider the input stage of the instrumentation amplifier as shown in the figure below.

Fig 4 Input Stage of Instrumentation Amplifier

The potential at node A is the input voltage V1. Hence the potential at node B is also V1, from the virtual short concept. Thus, the potential at node G is also V1.

The potential at node D is the input voltage V2. Hence the potential at node C is also V2, from the virtual short. Thus, the potential at node H is also V2.

Ideally the current to the input stage op-amps is zero. Therefore, the current I through the resistors R1, Rgain and R1 remains the same.

Applying Ohm’s law between the nodes E and F,

I = (Vo1-Vo2)/(R1+Rgain+R1) ——————— 1

I = (Vo1-Vo2)/(2R1+Rgain)

Since no current is flowing to the input of the op-amps 1 & 2, the current I between the nodes G and H can be given as,

I = (VG-VH)/Rgain = (V1-V2)/Rgain ————————- 2

Equating equations 1 and 2,

(Vo1-Vo2)/(2R1+Rgain) =  (V1-V2)/Rgain 

(Vo1-Vo2) = (2R1+Rgain)(V1-V2)/Rgain —————— 3

The output of the difference amplifier is given as,

Vout = (R3/R2) (Vo1-Vo2)

Therefore, (Vo1 – Vo2) =  (R2/R3)Vout

Substituting (Vo1 – Vo2) value in the equation 3, we get

(R2/R3)Vout = (2R1+Rgain)(V1-V2)/Rgain

i.e. Vout = (R3/R2){(2R1+Rgain)/Rgain}(V1-V2)

The above equation gives the output voltage of an instrumentation amplifier. The overall gain of the amplifier is given by the term (R3/R2){(2R1+Rgain)/Rgain}.