Analog circuits miscellaneous
- A hysteresis type TTL inverter is used to realize an oscillator in the circuit shown in the figure.
If the lower and upper trigger level voltages are 0.9 V and 1.7 V, the period (in ms), for which output is LOW, is __________.
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Given LTP = 0.9
UTP = 1.7
VC (t) = Vmax + (Vinitial – Vmax )e – t / RC
LTP = 0 + (1.7 – 0)e – t / RC = 0.9
⇒ t = 0.635 ms (Given R = 10k, C = 0.1 μF).Correct Option: A
Given LTP = 0.9
UTP = 1.7
VC (t) = Vmax + (Vinitial – Vmax )e – t / RC
LTP = 0 + (1.7 – 0)e – t / RC = 0.9
⇒ t = 0.635 ms (Given R = 10k, C = 0.1 μF).
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The sinusoidal ac source in the figure has an rms value of 20 V . Considering all √2
possible values of RL, the minimum value of Rs in Ω to avoid burnout of the Zener diode is _______.
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To avoid burn-out IS ≥ IZ + IL
For RL = ∞ [considering all possible values]
IL = 0then , IS ≥ IZ > 20 - VZ RS Since , 5.IS = 1 4 ⇒ IS = 1 A 20 Then , 1 ≥ 20 - 5 20 RS
⇒ RS ≥ 300 Ω
Minimum value of RS = 300 Ω
Correct Option: D
To avoid burn-out IS ≥ IZ + IL
For RL = ∞ [considering all possible values]
IL = 0then , IS ≥ IZ > 20 - VZ RS Since , 5.IS = 1 4 ⇒ IS = 1 A 20 Then , 1 ≥ 20 - 5 20 RS
⇒ RS ≥ 300 Ω
Minimum value of RS = 300 Ω
- The transistor in the given circuit should always be in active region. Take VCE (sat) = 0.2 V. VBE = 0.7V. The maximum value of Rc in Ω which can be used, is ______
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In B-E loop, using KVL, we have
5 – 2KIB – VBE = 0⇒ IB = 5 - 0.7 = 2.15 mA 2K
and IC = βIB = 100 × 2.15 mA = 215 mA
Using KVL in C-E loop, we get
VCE = 5 – 0.215 RC
For active region, VCE > 0.2 V
⇒ 0.215 RC < 5 – 0.2⇒ RC = 4.8 ≅ 22.32 Ω 0.215
Correct Option: A
In B-E loop, using KVL, we have
5 – 2KIB – VBE = 0⇒ IB = 5 - 0.7 = 2.15 mA 2K
and IC = βIB = 100 × 2.15 mA = 215 mA
Using KVL in C-E loop, we get
VCE = 5 – 0.215 RC
For active region, VCE > 0.2 V
⇒ 0.215 RC < 5 – 0.2⇒ RC = 4.8 ≅ 22.32 Ω 0.215
- In the following circuit, the transistor is in active mode and VC = 2V. To get VC = 2V. To get VC = 4V, we replace RC with R'C. Then the ratio R' C / RC is ______.
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we have Vc = 2V;
Now, Ic Rc = 10 – 2 = 8 ...(i)
We have Vc = 4V
and Ic Rc = 10 – 4 = 6... (ii)Ic Rc = 6 Ic Rc 8 ∴ Rc = 3 = 0.75 Rc 4
Correct Option: C
we have Vc = 2V;
Now, Ic Rc = 10 – 2 = 8 ...(i)
We have Vc = 4V
and Ic Rc = 10 – 4 = 6... (ii)Ic Rc = 6 Ic Rc 8 ∴ Rc = 3 = 0.75 Rc 4
- A(0– 50A) moving coil ammet er has a volt age dr op of 0.1V across its terminals at full scale deflection. The external shunt resistance (in milliohms) needed to extend its range to (0 – 500A) is ______.
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Voltage drop across its terminals at full scale deflection = 0.1 volt
450 × Rsh = 0.1Rsh = 0.1 = 0.22 mΩ 450
Required shunt resistance, Rsh = 0.22m Ω
Correct Option: A
Voltage drop across its terminals at full scale deflection = 0.1 volt
450 × Rsh = 0.1Rsh = 0.1 = 0.22 mΩ 450
Required shunt resistance, Rsh = 0.22m Ω