Analog Circuits - Online Test

Q1. A good current buffer has
Answer : Option B
Explaination / Solution:
No Explaination.


Q2. In the ac equivalent circuit shown in the figure, if iin is the input current and Rf is very larger, the type of feedback is

Answer : Option B
Explaination / Solution:

From the circuit, we observe that output is Vout (Voltage). Feedback is current through resistance Rf , which is added to input current iin . Thus, the configuration is voltage-current feedback.

Q3. In the circuit shown, the op-amp has finite input impedance, infinite voltage gain and zero input offset voltage. The output voltage Vout is

Answer : Option C
Explaination / Solution:

Given that the op-amp has infinite voltage gain, i.e.

AOL

and zero input offset voltage

VIO = 0

So, we redraw the op-amp circuit as


Hence, the current I1 is drawn through resistance R2. So, the output voltage is

Vout = I1R2


Q4.
In the circuit shown below what is the output voltage  (vout)if a silicon transistor Q and an ideal op-amp are used?


Answer : Option B
Explaination / Solution:

For the given ideal op-amp, negative terminal will be also ground (at zero voltage) and so, the collector terminal of the BJT will be at zero voltage.
i.e., VC = 0 volt
The current in 1 k resistor is given by
This current will flow completely through the BJT since, no current will flow into the ideal op-amp (I/P resistance of ideal op-amp is infinity). So, for BJT we have 

i.e.,the base collector junction is reverse biased (zero voltage) therefore, the collector current (IC ) can have a value only if base-emitter is forward biased.
Hence,



Q5. The circuit below implements a filter between the input current ii and the output voltage v0. Assume that the opamp is ideal. The filter implemented is a

Answer : Option D
Explaination / Solution:

From diagram we can write 

Transfer function


Q6. Consider the oscillator circuit shown in the figure. The function of the network (shown in dotted lines) consisting of the 100kΩ resistor in series with the two diodes connected back-toback is to:

Answer : Option A
Explaination / Solution:

When the output voltage is positive the diode D1 is turned on making 100kΩ resistor to become parallel to 22.1kΩ. So the gain is reduced. When the output voltage becomes negative the diode D2 is turned on thereby again 100kΩ resistor to become parallel to 22.1kΩ. So the gain is reduced. With the use of diodes, the non ideal OP-Amp is made stable to produce steady

Q7. In the circuit shown below the op-amps are ideal. Then, Vout in Volts is

Answer : Option C
Explaination / Solution:
No Explaination.