Match the following : Codes :List – I List – II a. h-parameters i. O/P voltage varies as the slope of i/p voltage b. differentiator ii. Noise division c. half-wave rectifier iii. Function of a Q point d. integrator iv. series diode clipper
a-iii, b-i, c-iv, d-ii
Two pairings are certain and they select the code by themselves: a-iii, b-i, c-iv, d-ii — option 1.
| Item | Match |
|---|---|
| a. h-parameters | iii. Function of a Q point |
| b. Differentiator | i. Output varies as the slope of the input |
| c. Half-wave rectifier | iv. Series diode clipper |
| d. Integrator | ii. Noise division |
b is the surest anchor. "Slope" is the derivative, and a differentiator's defining equation is
\(V_{o}=-RC\dfrac{dV_{i}}{dt}\)
so its output is proportional to the rate of change — the slope — of the input. Nothing else in the list could claim that description.
a is the second anchor. The hybrid parameters are small-signal quantities, obtained by linearising the transistor's curves about its bias point. Since the slope of a curve differs from place to place, \(h_{ie}\) and \(h_{fe}\) depend on where the device is biased — for instance
\(h_{ie}\approx\dfrac{\eta V_{T}}{I_{B}}\)
which is inversely proportional to the quiescent base current. A transistor's data sheet accordingly quotes h-parameters at a stated collector current and collector-emitter voltage.
c — the half-wave rectifier as a series clipper. A single diode in series with a load passes one polarity and blocks the other, which is precisely the description of a series clipper set to remove everything below zero. Rectification and clipping are the same operation seen with different intent: the rectifier is used to obtain DC, the clipper to limit a waveform.
d — the integrator and "noise division". The remaining pairing is by elimination, and the phrase is awkward, but it is defensible. An integrator's gain falls with frequency as
\(|A|=\dfrac{1}{\omega RC}\)
so it attenuates — "divides" — high-frequency noise, acting as a low-pass filter. This is the exact opposite of the differentiator, whose gain rises with frequency and which therefore amplifies noise. It is why integrators are preferred wherever a choice exists between the two.
The answer is flagged only because item ii's wording is unclear; the other three pairings are unambiguous, and no other code accommodates them.
Hence, the correct code is a-iii, b-i, c-iv, d-ii.
Similar Questions
Assertion (A) : An Op-Amp is a direct coupled high gain amplifier.
Reason (R) : It consists of one or more differential amplifiers and usually followed by a level translator and push pull stage.
The input to a differentiator is –5 V. Its output will be
Assertion (A) : Op-Amp is used for sensor circuit.
Reason (R) : A small signal amplifier amplify weak measured signals.
Consider the following circuit, the switch S1 allows the output to switch between two ranges of amplitudes from 0-0.1 V and 0-1 V. Arrange these values of R1, R2 and R3 in increasing order.
(A) Value of R1
(B) Value of R2
(C) Value of R3Choose the most appropriate answer from the options given below :
The given operational amplifier circuit corresponds to which electronic circuit application ?
Statements in connection to Op-Amp applications are :
A. If we use a square wave generator followed by integrator circuit we get a triangular wave at the output
B. The logarithmic amplifier called a log-amplifier or a logger, is basically a current to voltage converter.
C.
is a first order high pass filter with voltage followerD. If we use a square wave generator followed by a clipping circuit then we get a saw-tooth wave generator.
Choose the correct answer from the options given below:
Match List I with List II
LIST I LIST II A. Single polarity positive voltage clipper circuit I. 
B. Negative clamping circuit II. 
C. Differentiator circuit III. 
D. Logarithmic Amplifier IV. 
Choose the correct answer from the options given below:
The correct flow in the basic block diagram of op-Amp is
A. Complementary symmetry push pull amplifier block.
B. Dual input unbalanced output differential amplifier block
C. Emitter follower with constant current source block
D. Dual-Input balanced output differential amplifier block
Choose the correct answer from the options given below :
Match the following :
List - I List - II a) 
i) 
b) 
ii) 
c) 
iii) 
d) 
iv) 
Codes :
\(\dfrac{d^{2}y}{dt^{2}}+y=0\) can be solved using an Operational Amplifier. Which one is preferred ?
Important Questions from Op-Amp and Its Applications
In the circuit shown below, $R$ is $1\text{ k}\Omega$ and $R_f$ is $10\text{ k}\Omega$. If $V_{\text{in}}$ is $100\text{ mV}$ and the op-amps have supply voltages of $\pm 15\text{ V}$, then $V_{\text{out}}$ is _______ V.
For IC 741 circuit shown below, the maximum input frequency above which the output will be distorted is ____________.
Assume that an op-amp 741 connected as a unit gain inverting amplifier is applied with an input change of 10 V. Determine the time taken for the output to change by 10 V. [Slew rate for op-amp 741 = 0.5 V/µs]
What is the value of output voltage $V_o$ in the given circuit? Assume op-amp to be ideal and MOSFET operating in saturation region.
In an ideal op-amp, the common mode gain is _____________.
