Analog circuits Test 1

When,

D₁ → ON; D₂ → OFF: R_AB = 10 kΩ

D₂ → ON; D₁ → OFF: R_AB = 10 kΩ

∴ The impedance by the circuit across the terminals A and B  of 10 kΩ

A) In graph ‘A’ the output follows input, but above certain positive input, output stays constant. The circuit is called positive clipper.

B) Similarly the graph ‘B’ is of negative clipper.

C) In this path positive and negative side are clipped, it is a double-ended clipper.

D) In this input output follows relation V_out = |V_in|. This is a full-wave rectifier circuit.

Concept:

For VDS ≥ VGS - VTN, the MOSFET operates in the saturation region, with the saturation current given by:

ID(sat) = Kn (VGS - VTN)2

\({K_n} = \frac{{W{\mu _n}{C_{ox}}}}{{2L}}\)

Application:

Since V_GS= V_DS for the given MOSFET, we observe that V_DS will always be greater than V_GS - V_TH. Hence the MOSFET is operating in saturation.

Given for V_GS = 4 V and V_TH = 3 V, the current is 3 mA, we can write:L

I_D = K [4 – 3]²

\(\frac{3m}{1} = K\)

K = 3 mA/V²

∴ For V_GS = 5 V, the drain current will be:

The ripple factor measures AC fluctuations in a rectifier’s output. A full-wave rectifier has a lower ripple factor than a half-wave rectifier because it rectifies both halves of the AC input, producing an output with twice the frequency (e.g., 120 Hz for a 60 Hz input). This higher frequency results in smaller voltage fluctuations between peaks, reducing the ripple factor.

Option A: Correct, as higher frequency reduces ripple magnitude.

Let the diode is ON

KCL at node B:

\(\frac{{{V_B} - 3}}{{1k{\rm{\Omega }}}} + \frac{{{V_B}}}{{5k{\rm{\Omega }}}} - 1 \times {10^{ - 3}} = 0\) 

\(\Rightarrow {V_B} = \frac{{10}}{3} = 3.33\;V\) 

\(\Rightarrow {I_B} = \frac{{3 - 3.33}}{{1 \times {{10}^3}}} =\) -ve not possible

⇒ Diode is OFF

So current passing through diode = 0

∴  voltage drop across ab will be zero volts.

The ripple factor for an inductor filter is

Diodes are essential components in electronic circuits, primarily used to manage voltage levels. They perform this function by behaving like a voltage source under certain conditions. Here are the key points regarding their operation:

• Clipping Voltage: Diodes can limit the maximum voltage in a circuit, preventing damage to other components.
• Forward Bias: When a diode is forward-biased, it allows current to flow, effectively acting as a voltage source.
• Reverse Bias: In reverse bias, the diode blocks current, protecting the circuit from excessive voltage.
• Applications: They are commonly used in power supply circuits and signal processing.

In summary, diodes play a crucial role in controlling voltage levels by functioning as a voltage source under specific conditions, making them indispensable in many electronic applications.

When a capacitor is connected across the output terminals of a rectifier, the output voltage:

• Essentially becomes a DC voltage.
• Undergoes minimal fluctuations, smoothing out the signal.
• Helps to store energy, providing a stable output.

This process is critical in converting the alternating current (AC) into a usable direct current (DC), which is essential for most electronic devices. The capacitor acts as a reservoir, filling in the gaps between the peaks of the rectified voltage and reducing the ripple effect.

 i = dQ/dt = 120/60 = 2A.

n = 10²⁰
Q = ne = e * 10²⁰ = 16.02 C.
The charge on the sphere will be positive.

• 100 = 65 + V2 ⇒ V2 = 35 V
• V3 – 30 = V2 ⇒ V3 = 65 V
• 105 – V3 + V4 – 65 = 0 ⇒ V4 = 25 V
• V4 + 15 – 55 + V1 = 0 ⇒ V1 = 15 V.

Req – 5 = 10(Req + 5)/(10 + 5 +Req).

Solving for Req we have

Req = 11.18 ohm.

Energy provided is equal to 0.5 CV² = 0.5x10^-6 x 100 x 100 = 0.05 J

The capacitor current is given as i=C*(dv/dt), where dv/dt is the derivative of voltage, dt=t2-t1 given as 10 sec and dv is the change in voltage which is given as 12 V.

So, we have C = i / (dv/dt)
C = 2mA/(12/10) = 2mA/(1.2).

Hence C = 1.67mF

There are two capacitors that arise between bases and emitter. One is C_je due to depletion region associated between base and emitter. C_b is another capacitor which arises due to the accumulation of electrons in the base which further results into the concentration gradient within the base of the transistor.

Only one capacitor up between the base and the collector. This is due to the depletion region present between the base and the collector region.

While observing the frequency response of a C.E. stage, we find that all the parasitic capacitances of the B.J.T. end up slowing the speed of the B.J.T. The frequency response of this stage is affected by all the parasitic capacitors.

The transit frequency is almost inversely proportional to the total capacitance between the base and the emitter of the B.J.T. Hence, the transit frequency will approximately reduce by 2 and the correct option becomes reduces by 2.

The miller effect results in a change in the capacitance seen between the base and the collector. This is why it affects the frequency response of the B.J.T. deeply by changing the poles and affecting the high frequency voltage gain stage.