Example ( Analog Communication )

For a light ray which have the wavelength equal to 200 nm, calculate the frequency of the ray.

Solution: As given, in the problem,

Wavelength of the light ray = 200 nm

$\lambda = 200 \times 10^{-9} m$

Also speed of light,

C = $3 × 10^8 m s{-1}$

Substituting these values in the equation above, we get that,

$\lambda = \frac {c}{f}$

Rearranging the formula,

f = $\frac {c}{\lambda}$

= $\frac {3\times 10^8}{200 \times 10^{-9}}$

= $1.5 \times 10^{14} Hz.$

The frequency of the wave is equal to $1.5 \times 10^{14} Hz.$

Example ( Analog Communication )

In an experiment in physics, the wavelength of a photon particle was observed to be 500 nm. What will be the frequency of the wave?

Solution: As given here, the wavelength of the photon particle = 500 nm.

i.e. $\lambda = 500 nm$

i.e. $\lambda = 500 \times 10^{-9} m$

Also speed of light,

C = $3\times 10^8 m s{-1}$

In order to calculate the frequency of the photon particle, we will use the formulas:

$\lambda = \frac {c}{f}$

Rearranging the formula,

f = $\frac {c}{\lambda}$

= $\frac {3\times 10^8}{500 \times 10^{-9}}$

= $6 \times 10 ^{14} Hz$

Why Open Op-amp is unsuitable for many linear application?

Question:

Why Open Op-amp can not be used for amplification application? 

Answer: 

The open-loop gain of the op-amp is very high. ( Approx 2 lakh)

Only the smaller signal (in order of microvolts or less) having very low frequency maybe amplified accurately without distortion. 

However, signals this small are very susceptible to noise and are almost impossible to obtain in the laboratory. 

Besides being a large gain, it is also not constant. The voltage gain varies with change in temperature and power supply as well as mass production techniques. The variation in the voltage gain is relatively large in the open-loop op-amp in particular, which make the open-loop op-amp unsuitable for the many linear application.

Op - Amp Example 2

Example 2: Determine the output voltage for the inverting amplifier for the give case.

Assume that op-amp is a 741.

Answer:

As they said op-amp is 741, you should consider this by default.

Consider A=2,00,000 , R_i = 2Mohm, R₀=75 ohm, +V_cc= +15 V, -V_ee=-15 V and output voltage swing = ±14V.

So now for case 1:

  

This is the theoretical value. The actual value will be a negative saturation voltage of -14 V.

For case 2:

This means the output is a sine wave since it is less than the output voltage swing of ±14V.

Op-Amp Example 1

 Determine the output voltage in the following cases for open-loop differential amplifier.

Case 1:  V_in1 = 5uV dc and V_in2 = -7 uV dc

Case 2: V_in1 = 10mV rms and V_in2= 20 mV rms

Consider A=2,00,000 , R_i = 2Mohm, R₀=75 ohm, +V_cc= +15 V, -V_ee=-15 V and output voltage swing = ±14V.

Do comment on output.

 Answer:

Now as per the equation of Voltage gain

 

Now For Case 1

 

Now For Case 2

 

The theoretical value of output is -2000 V rms. However, the op-amp saturates at ±14 V. Therefore, the actual output waveform will be clipped.

This situation is unacceptable in amplifier applications. The normal solution to this problem is to use negative feedback.