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Analog Electronics Solved Problems with Worked Examples

In short: analog electronics deals with continuously varying signals and the devices that amplify, filter and condition them. This page works the standard numerical problems through in full: BJT bias, small-signal gain, op-amp gain and filter cutoff. FLN Tutors (Freelancer Network) also offers one-to-one help with the same topics. Matched to a verified specialist who knows your course, your university's marking criteria, and your deadline.

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FLN offers one-to-one online Analog Electronics tutoring with verified subject specialists. Sessions focus on calculations, design problems and the theory behind them, including Operational Amplifiers, Diodes, Transistors and BJT, using your own course material, so you understand the subject and can apply it yourself.

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What is analog electronics?

Analog electronics works with signals that vary continuously, rather than the discrete levels of digital circuits. It covers the devices that do the work, diodes, bipolar transistors, field effect transistors and operational amplifiers, and the circuits built from them to amplify, filter, rectify and condition signals.

It stays essential because the physical world is analog. Every digital system meets reality through analog circuitry: a sensor, an amplifier, a filter and a converter.

Diodes and rectification

A diode conducts in one direction once forward biased, with a forward drop of roughly 0.7 V for silicon and about 0.3 V for germanium.

CircuitOutputRipple frequency from 50 Hz
Half-waveOne half cycle50 Hz
Full-wave centre-tappedBoth half cycles100 Hz
Full-wave bridgeBoth half cycles100 Hz

The bridge is usually preferred because it uses the whole secondary winding without a centre tap, though two diode drops appear in series rather than one. A reservoir capacitor then smooths the output, and ripple falls as capacitance and load resistance rise.

For circuit analysis, decide first whether to use the ideal model, the constant-drop model or the full exponential model. Using the exponential model where a 0.7 V drop would do wastes time; using the ideal model where the drop matters loses accuracy.

Worked example: BJT voltage divider bias

A common-emitter stage uses VCC = 12 V, R1 = 47 kΩ, R2 = 10 kΩ, RC = 2.2 kΩ, RE = 1 kΩ, with β = 150.

Base voltage, from the divider:

VB = 12 × 10 ÷ (47 + 10) = 12 × 0.1754 = 2.105 V

Emitter voltage, allowing 0.7 V across the base-emitter junction:

VE = 2.105 − 0.7 = 1.405 V

Emitter current:

IE = 1.405 ÷ 1,000 = 1.405 mA

Taking IC ≈ IE, the collector voltage is:

VC = 12 − (1.405 mA × 2.2 kΩ) = 12 − 3.092 = 8.908 V

So VCE = 8.908 − 1.405 = 7.503 V

That sits comfortably in the active region, roughly mid-rail, which is what you want for maximum undistorted swing.

Why this bias arrangement. The emitter resistor provides negative feedback: if IC rises, VE rises, which reduces the base-emitter voltage and pushes IC back down. That makes the operating point far less sensitive to β, which varies widely between devices and with temperature. A design whose Q-point depends on β is not manufacturable, and saying so is often worth more than the arithmetic.

Worked example: small-signal gain

Continuing the stage above, the intrinsic emitter resistance at room temperature is approximately:

re = 26 mV ÷ IE = 26 ÷ 1.405 = 18.5 Ω

With the emitter resistor fully bypassed by a capacitor, the voltage gain is:

Av ≈ −RC ÷ re = −2,200 ÷ 18.5 = −119

The minus sign means inversion, a 180° phase shift.

Without the bypass capacitor, the gain becomes approximately −RC ÷ (re + RE) = −2,200 ÷ 1,018.5 = −2.16. The gain collapses, but it is now set almost entirely by the ratio of two resistors, so it is stable and predictable. That trade between gain and stability is the point of the comparison.

Operational amplifiers

An ideal op-amp has infinite gain, infinite input impedance, zero output impedance and infinite bandwidth. Real devices do not, but the ideal model plus two rules solves most circuits:

  1. No current flows into the inputs.
  2. With negative feedback, the op-amp drives its output until the two inputs are at the same voltage, the virtual short.
ConfigurationGain
Inverting−Rf ÷ Rin
Non-inverting1 + Rf ÷ Rin
Voltage follower1
SummingWeighted sum of inputs
DifferenceAmplifies the difference between inputs
Integrator, differentiatorSet by R and C

Worked example. With Rf = 100 kΩ and Rin = 10 kΩ, the inverting configuration gives −100 ÷ 10 = −10, and the non-inverting configuration gives 1 + 100 ÷ 10 = +11.

The non-inverting gain cannot be less than one, which is why an inverting stage is used where attenuation is needed.

Real op-amp limitations

LimitationEffect
Finite gain-bandwidth productGain falls as frequency rises
Slew rateLimits how fast a large output can change
Input offset voltageA DC error at the output
Input bias currentError across high source resistances
Finite CMRRCommon-mode signal appears at the output
Output swing limitsOutput cannot reach the supply rails

Gain and bandwidth trade directly: with a gain-bandwidth product of 1 MHz, a gain of 100 leaves about 10 kHz of bandwidth. A design asking for high gain and wide bandwidth from one stage usually needs to be split into two.

Worked example: filter cutoff

A first-order RC low-pass filter has R = 1.6 kΩ and C = 100 nF.

fc = 1 ÷ (2πRC) = 1 ÷ (2π × 1,600 × 100 × 10⁻⁹) = 995 Hz

At the cutoff frequency the output is 3 dB down, which corresponds to about 70.7% of the input amplitude, with 45° of phase shift. A first-order filter rolls off at 20 dB per decade; each additional order adds another 20 dB per decade.

For a sharper transition you raise the order, and the standard responses trade differently: Butterworth gives the flattest passband, Chebyshev a sharper transition at the cost of passband ripple, and Bessel the best phase linearity. Name the response you chose and the reason.

Feedback

Negative feedback trades gain for everything else that matters: it stabilises gain against device variation, reduces distortion, extends bandwidth and lets input and output impedance be set by design.

The closed-loop gain approaches 1/β, the reciprocal of the feedback fraction, once loop gain is large, which is why a circuit's behaviour comes to depend on passive components rather than on the transistor.

Stability is the limit. Feedback that arrives with excessive phase shift becomes positive at some frequency, and if the loop gain is still at or above unity there the circuit oscillates. That is what gain and phase margin measure, and it is why a high-gain amplifier may need compensation.

Typical exam and viva questions

These are the forms the standard questions take in university exams, GATE-style papers and lab vivas. Each one is answered somewhere on this page.

Numerical - Find the Q-point of a given voltage-divider-biased stage. - Find the voltage gain with and without the emitter bypass capacitor. - Find the gain of an inverting or non-inverting op-amp stage. - Find the cutoff frequency of an RC filter, and the output at cutoff. - Find the bandwidth left at a given gain from the gain-bandwidth product.

Conceptual and viva - Why is voltage divider bias preferred over fixed bias? - Why does an emitter resistor stabilise the operating point? - What happens to gain if the bypass capacitor is removed, and why? - What is the virtual short, and when may you assume it? - Why can a non-inverting amplifier not have a gain below one? - What is the difference between slew rate and bandwidth? - Why does negative feedback reduce distortion but risk oscillation?

For a viva, be ready to state the assumption behind each formula, not just the formula. Being asked "where does 26 mV come from?" after quoting re is extremely common.

Where students lose marks

  • Designing a bias circuit whose operating point depends on β.
  • Forgetting the 0.7 V base-emitter drop in a bias calculation.
  • Omitting the minus sign, and so the phase inversion, on an inverting stage.
  • Using re without stating the operating current it was derived at.
  • Claiming a non-inverting gain below one.
  • Asking one op-amp stage for gain and bandwidth beyond its gain-bandwidth product.
  • Choosing a filter order with no stated attenuation requirement.
  • Treating negative feedback as free, with no mention of stability.

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Analog Electronics tutoring: frequently asked questions

What is analog electronics?

Analog electronics deals with continuously varying signals and the devices that process them: diodes, bipolar and field effect transistors, and operational amplifiers. It covers amplification, filtering, rectification and signal conditioning, and it remains essential because sensors and the physical world are analog.

Where can I find analog electronics solved problems?

This page works four of the standard problems through in full with numbers: the Q-point of a voltage-divider-biased stage, small-signal gain with and without the bypass capacitor, inverting and non-inverting op-amp gain, and RC filter cutoff. Each shows the method rather than only the answer.

What topics come up most in analog electronics exams?

Transistor biasing and Q-point calculation, small-signal gain, op-amp configurations and their limitations, diode rectifier circuits, filter cutoff and roll-off, and feedback with its effect on gain, bandwidth and stability. Numerical questions concentrate heavily on biasing and on gain calculations, while viva questions tend to probe the assumptions behind each formula.

What is the difference between analog and digital electronics?

Analog circuits handle signals that vary continuously over a range, so amplitude carries information and noise adds directly to it. Digital circuits use discrete levels, which gives noise immunity and easy storage, but every digital system still needs analog circuitry at its interface.

What is the forward voltage drop of a diode?

Roughly 0.7 V for a silicon diode and about 0.3 V for germanium, once conducting. For many circuits this constant-drop model is accurate enough; the exponential model is needed only where the precise current-voltage relationship matters, such as in temperature-sensitive or logarithmic circuits.

What is the difference between half-wave and full-wave rectification?

Half-wave rectification passes only one half of each input cycle, so ripple is at the supply frequency. Full-wave rectification uses both halves, doubling the ripple frequency and making smoothing easier, which is why full-wave bridges are standard in power supplies.

Why is voltage divider bias preferred for BJT circuits?

Because the emitter resistor provides negative feedback that holds the operating point steady: a rise in collector current raises the emitter voltage, reducing the base-emitter voltage and opposing the change. The Q-point then depends on resistor ratios rather than on the transistor's beta.

Why does beta variation matter in biasing?

Because beta varies widely between nominally identical devices and changes with temperature and collector current. A bias design whose operating point depends on beta will drift and will not reproduce across a production batch, so good designs deliberately make the Q-point independent of it.

How do you calculate the Q-point of a common-emitter stage?

Find the base voltage from the divider, subtract the base-emitter drop to get the emitter voltage, divide by the emitter resistor for the emitter current, take collector current as approximately equal, then compute the collector voltage and subtract the emitter voltage for the collector-emitter voltage.

What is re in small-signal analysis?

It is the intrinsic emitter resistance seen by small signals, approximately the thermal voltage of about 26 mV at room temperature divided by the emitter current. Because it depends on operating current, any gain figure derived from it must state the bias point used.

Why does bypassing the emitter resistor increase gain?

Because the capacitor shorts the emitter resistor at signal frequencies, removing the negative feedback that it provides to AC signals while leaving it in place for DC bias. Gain rises substantially, but it becomes dependent on the operating point rather than on a resistor ratio.

What are the ideal op-amp assumptions?

Infinite open-loop gain, infinite input impedance, zero output impedance, infinite bandwidth and zero offset. With negative feedback these lead to two working rules: no current enters the inputs, and the two inputs sit at the same voltage, known as the virtual short.

What is the gain of an inverting op-amp amplifier?

The negative ratio of the feedback resistor to the input resistor. The minus sign indicates inversion, a 180 degree phase shift, and omitting it is a common error. Unlike the non-inverting configuration, this arrangement can provide gains of less than one.

What is gain-bandwidth product?

It is the roughly constant product of closed-loop gain and bandwidth for an op-amp. With a one megahertz product, a gain of one hundred leaves about ten kilohertz of bandwidth. A requirement for both high gain and wide bandwidth usually means splitting the design across two stages.

What is slew rate?

Slew rate is the maximum rate at which an op-amp output can change, usually quoted in volts per microsecond. It limits large-signal performance independently of bandwidth, so a circuit can have adequate small-signal bandwidth yet still distort a fast, large-amplitude waveform.

How do you calculate the cutoff frequency of an RC filter?

Cutoff equals one divided by two pi times resistance times capacitance. At that frequency the output is three decibels below the passband, about seventy per cent of the input amplitude, with forty-five degrees of phase shift and a twenty decibel per decade roll-off beyond it.

What are the benefits and risks of negative feedback?

It stabilises gain against device variation, reduces distortion, extends bandwidth and allows input and output impedances to be set by design, all at the cost of gain. The risk is instability: if feedback becomes positive while loop gain is at least unity, the circuit oscillates.

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