The amplifiers $X, Y$ and $Z$ are connected in series. If the voltage gains of $X, Y$ and $Z$ are $10, 20$ and $30$ respectively,and the input signal is $1 \, mV$ peak value,then what is the output signal voltage (peak value) $(i)$ if $DC$ supply voltage is $10 \, V$? $(ii)$ if $DC$ supply voltage is $5 \, V$?
(A) The total voltage gain of amplifiers connected in series is the product of individual gains: $A_v = (A_v)_1 \times (A_v)_2 \times (A_v)_3 = 10 \times 20 \times 30 = 6000$.
The output voltage is given by $v_0 = A_v \times v_i$.
Given input signal $v_i = 1 \, mV = 10^{-3} \, V$.
Therefore,the theoretical output voltage is $v_0 = 6000 \times 10^{-3} \, V = 6 \, V$.
$(i)$ If the $DC$ supply voltage is $10 \, V$,the amplifier can support an output of $6 \, V$ because $6 \, V < 10 \, V$. Thus,the output is $6 \, V$.
$(ii)$ If the $DC$ supply voltage is $5 \, V$,the amplifier cannot produce an output higher than the supply voltage. Since the required output $6 \, V$ exceeds the supply $5 \, V$,the output is clipped at the supply voltage limit. Thus,the output is $5 \, V$.
The output voltage is given by $v_0 = A_v \times v_i$.
Given input signal $v_i = 1 \, mV = 10^{-3} \, V$.
Therefore,the theoretical output voltage is $v_0 = 6000 \times 10^{-3} \, V = 6 \, V$.
$(i)$ If the $DC$ supply voltage is $10 \, V$,the amplifier can support an output of $6 \, V$ because $6 \, V < 10 \, V$. Thus,the output is $6 \, V$.
$(ii)$ If the $DC$ supply voltage is $5 \, V$,the amplifier cannot produce an output higher than the supply voltage. Since the required output $6 \, V$ exceeds the supply $5 \, V$,the output is clipped at the supply voltage limit. Thus,the output is $5 \, V$.
Explore More
Similar Questions
When a potential difference is applied across,the current passing through:
MediumIIT 1999
View SolutionThe frequency response curve of an $RC$ coupled amplifier is shown in the figure. The bandwidth of the amplifier will be
Medium
View SolutionThe correct curve between voltage gain $({A_v})$ and load resistance $({R_L})$ is
Medium
View SolutionThe box in the circuit below has two inputs marked $V_{+}$ and $V_{-}$ and a single output marked $V_{o}$. The output obeys the following relation:
$V_{o} = \begin{cases} +10 \, V & \text{if } V_{+} > V_{-} \\ -10 \, V & \text{if } V_{+} < V_{-} \end{cases}$
Which of the following graphs represents the output $V_{o}$ as a function of time $t$?
$V_{o} = \begin{cases} +10 \, V & \text{if } V_{+} > V_{-} \\ -10 \, V & \text{if } V_{+} < V_{-} \end{cases}$
Which of the following graphs represents the output $V_{o}$ as a function of time $t$?
Consider the following statements:
Statement $(A)$: The resistance of an ideal diode in forward biased condition is zero.
Statement $(B)$: In a half-wave rectifier,the load current flows only for every half cycle of the input signals.
Statement $(C)$: In the breakdown region,a Zener diode behaves as a constant voltage source.
Statement $(A)$: The resistance of an ideal diode in forward biased condition is zero.
Statement $(B)$: In a half-wave rectifier,the load current flows only for every half cycle of the input signals.
Statement $(C)$: In the breakdown region,a Zener diode behaves as a constant voltage source.
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo