Electronics: free practice, theory and problems

A diode is a component that lets current through in (almost) only one direction, and a transistor lets a small current or voltage control a much larger current. Together they make it possible to rectify AC voltage, stabilize voltages, amplify signals and build digital switches – the foundation of power supplies, sensor interfaces and logic.

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Diodes and transistors
Operational amplifiers
Filters and signals

1. Diodes and transistors

What is it about?

Concepts and formulas

Silicon diode in forward bias: the voltage drop is roughly constant, $V_D \approx 0.7$ V, almost independent of the current.
Diode in reverse bias: it (almost) blocks all current, until the breakdown voltage is reached.
Zener diode: used backwards (in reverse bias) to hold a fixed voltage $V_Z$ . Simple voltage regulator: a series resistor $R_S$ between the source $V_{in}$ and the zener diode, $R_S = \dfrac{V_{in} - V_Z}{I_Z}$ , where $I_Z$ is the desired zener current.
LED: behaves like a diode with a larger voltage drop (often 1.8–3.3 V). Always needs a series resistor: $R = \dfrac{V_{supply} - V_{LED}}{I_{LED}}$ .
BJT: $I_C = \beta I_B$ in the active region. In cutoff, $I_B \approx 0$ and the transistor conducts no current. In saturation, $V_{CE}$ is very low (typically 0.1–0.3 V) regardless of $\beta$ – the transistor behaves like a closed switch, and $I_C$ is then set by the rest of the circuit, not by $\beta I_B$ .
MOSFET: controlled by the voltage $V_{GS}$ between gate and source. The gate draws almost no DC current.
Bridge rectifier: four diodes turn both half-cycles of an AC voltage into the same polarity; a smoothing capacitor after the bridge reduces the ripple voltage.

How to solve the problems

Identify whether the component is a diode, zener diode, LED or transistor, and which region it operates in (forward/reverse bias, active/saturation/cutoff).
For diodes: use $V_D \approx 0.7$ V (silicon) in forward bias, and find the current from Ohm's law for the rest of the circuit.
For zener/LED regulators: use $R = (V_{in} - V_{diode})/I$ , keeping all voltages and currents in the same unit.
For a BJT: check whether the transistor is in the active region ( $I_C = \beta I_B$ ) or in saturation (then $I_C$ is set by the load circuit).
Check the answer: are the current and voltage realistic (mA/V range, not A/kV)?

Example

A zener diode with $V_Z = 5.6$ V is to stabilize the voltage from a 12 V source. The desired zener current is 20 mA. What series resistance is needed?

Voltage that must drop across the resistor: $V_{in} - V_Z = 12 - 5.6 = 6.4$ V.
$R_S = \dfrac{6.4}{0.02} = 320$ Ω.

Answer: $R_S = 320$ Ω.

Common mistakes

Believing $I_C = \beta I_B$ holds in saturation – there $I_C$ is set by the load circuit, not by $\beta$ .
Forgetting the series resistor for an LED and connecting it directly to a voltage source (it is destroyed by excess current).
Using the wrong voltage drop: about 0.7 V for silicon diodes, 0.3 V for Schottky, 1.8–3.3 V for LEDs depending on color.
Mixing mA and A, or kΩ and Ω, in the same calculation.

A diode passes current in one direction with an almost constant voltage drop; a BJT is current-controlled (

I_C = \beta I_B

) in the active region, but behaves like a switch in saturation; a MOSFET is controlled by voltage.

Concepts in this part

Diode and RectifierA diode lets current through in only one direction, with an almost constant voltage drop while conducting. A s…Zener DiodeA zener diode is connected backward, in reverse bias, and holds a fixed voltage across itself while conducting…BJT TransistorA BJT is a current-controlled transistor: a small base current controls a much larger collector current. In th…

Practise diodes and transistors in the app →

2. Operational amplifiers

What is it about?

An operational amplifier (op-amp) is an amplifier with enormously high gain that is almost always used together with a few resistors and capacitors in feedback. It is the feedback network, not the op-amp's own (unpredictably high) open-loop gain, that decides what the circuit does. With a handful of basic circuits – inverting and non-inverting amplifier, follower, summing, difference, integrator and differentiator – you can amplify sensor signals, filter, add voltages and build comparators.

Concepts and formulas

Ideal op-amp with negative feedback: no current flows into the inputs, and the two inputs have the same voltage (virtual short).
Inverting amplifier: $A = -R_f/R_{in}$ .
Non-inverting amplifier: $A = 1 + R_f/R_g$ (always $\ge 1$ ).
Voltage follower (buffer): $A=1$ , high input impedance, low output impedance.
Inverting summing amplifier: $V_o = -R_f\left(\dfrac{V_1}{R_1}+\dfrac{V_2}{R_2}+\dots\right)$ .
Difference amplifier with matched resistor pairs ( $R_3=R_1$ , $R_4=R_2$ ): $V_o = \dfrac{R_2}{R_1}(V_2-V_1)$ .
Integrator: $V_o = -\dfrac{1}{RC}\displaystyle\int V_{in}\,dt$ . Differentiator: $V_o = -RC\,\dfrac{dV_{in}}{dt}$ .
Saturation: the output can never go past the supply rails; rail-to-rail op-amps get very close to them.
Gain-bandwidth product (GBW): for most op-amps, (closed-loop gain) × (upper cutoff frequency) $\approx$ a constant $=GBW$ . Higher gain therefore gives lower bandwidth: $f_{-3dB} \approx GBW/A$ .
Slew rate ( $SR$ ): the maximum rate the output can change, in V/µs. The time for a voltage step $\Delta V$ is at least $t = \Delta V/SR$ .

How to solve the problems

Identify which basic circuit it is (inverting, non-inverting, follower, sum, difference, integrator/differentiator), and use the matching formula.
For gain/bandwidth problems: find the gain $A$ first, then use $f_{-3dB} = GBW/A$ .
For slew rate: find the voltage step $\Delta V$ and divide by the slew rate.
Always check whether the answer is within what the op-amp can actually deliver (saturation at the supply rails).
Watch the sign: inverting circuits give an output with the opposite sign of the input.

Example

An op-amp has $GBW = 2$ MHz and is used as a non-inverting amplifier with $R_f = 47$ kΩ and $R_g = 1$ kΩ. What is the upper cutoff frequency?

Gain: $A = 1 + 47/1 = 48$ .
$f_{-3dB} = \dfrac{GBW}{A} = \dfrac{2\text{ MHz}}{48} \approx 41.7$ kHz.

Answer: about 41.7 kHz.

Common mistakes

Assuming the gain stays the same at all frequencies – in practice the GBW limits the bandwidth.
Using $A=1+R_f/R_g$ for an inverting amplifier, or the reverse.
Forgetting that the output saturates at the supply rails even though the ideal formula gives a larger number.
Mixing up $R_f$ and $R_{in}$ in the inverting formula.

The feedback network sets the gain, not the op-amp's own open-loop gain. But neither the gain nor the speed is infinite: GBW limits the bandwidth at high gain, and slew rate limits how fast the output can change.

Concepts in this part

Op-Amp Golden RulesAn ideal operational amplifier with negative feedback follows two simple rules: no current flows into the inpu…Inverting AmplifierIn an inverting amplifier the signal enters through a resistor to the negative input, while the feedback resis…Non-Inverting AmplifierIn a non-inverting amplifier the signal enters directly at the positive input, while and form a feedback divid…

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3. Filters and signals

What is it about?

A filter shapes how a circuit responds to different frequencies: a low-pass filter passes low frequencies and attenuates high ones, a high-pass filter is the opposite, and a band-pass filter passes a specific range of frequencies. Filters are everywhere – removing noise, preventing aliasing before an ADC, blocking a DC level (AC coupling), and shaping audio signals. The key to describing gain over many decades of frequency is the decibel (dB) scale, a logarithmic scale that lets you add instead of multiply.

Concepts and formulas

Cutoff frequency of an RC low-pass: $f_c = \dfrac{1}{2\pi RC}$ . Same formula for an RC high-pass.
Amplitude ratio of a first-order low-pass: $|H(f)| = \dfrac{1}{\sqrt{1+(f/f_c)^2}}$ . At $f=f_c$ , $|H|=1/\sqrt2 \approx 0.707$ (−3 dB).
Decibels for voltage/current (gain $A = V_{out}/V_{in}$ ): $dB = 20\log_{10}A$ . For power: $dB = 10\log_{10}(P_{out}/P_{in})$ (the factor is 20 for voltage/current because power $\propto V^2$ ).
Attenuation in dB for a first-order filter: $dB(f) = 20\log_{10}|H(f)| = -20\log_{10}\sqrt{1+(f/f_c)^2}$ .
Phase of a first-order low-pass: $\varphi(f) = -\arctan(f/f_c)$ . At $f=f_c$ , $\varphi=-45°$ .
Roll-off (slope far from $f_c$ ): 20 dB/decade per order. A second-order filter (two RC stages) falls 40 dB/decade.
Band-pass filter: high-pass (cutoff $f_1$ ) followed by low-pass (cutoff $f_2 > f_1$ ); bandwidth $BW=f_2-f_1$ .
SNR (signal-to-noise ratio): $SNR_{dB}=10\log_{10}(P_s/P_n)$ .

How to solve the problems

Identify the filter type (low-, high- or band-pass) and find the cutoff frequency (or frequencies) from $f_c=1/(2\pi RC)$ .
To find the attenuation in dB at a given frequency: compute $|H|$ first, then take $20\log_{10}|H|$ .
To find $R$ or $C$ for a desired $f_c$ : solve the formula for the unknown, e.g. $C=1/(2\pi Rf_c)$ .
For a band-pass filter: find $f_1$ and $f_2$ separately, and subtract to get the bandwidth.
Remember that $\log_{10}$ of a ratio below 1 gives a negative number (attenuation), and above 1 gives a positive number (gain).

Example

An amplifier has $V_{in}=50$ mV and $V_{out}=2$ V. What is the gain in dB?

Voltage gain: $A = V_{out}/V_{in} = 2/0.05 = 40$ .
$dB = 20\log_{10}40 \approx 32.0$ dB.

Answer: about 32.0 dB.

Common mistakes

Using $10\log_{10}$ for a voltage or current gain instead of $20\log_{10}$ .
Thinking that −3 dB means the signal is gone – it means roughly half the power (71% amplitude).
Mixing up the sign: attenuation gives negative dB, gain gives positive dB.
Using the raw frequency $f$ instead of the ratio $f/f_c$ in the amplitude and phase formulas.

Decibels are a logarithmic scale:

20\log_{10}A

for voltage/current,

10\log_{10}

for power. At the cutoff frequency, the attenuation is always −3 dB and the phase is always −45°, no matter what

R

and

C

are used.

Concepts in this part

RC Low-Pass FilterAn RC low-pass filter passes low frequencies and attenuates high ones. At the corner frequency the amplitude h…RC High-Pass FilterAn RC high-pass filter passes high frequencies and attenuates low ones, the opposite of the low-pass filter, a…Decibels (dB)The decibel is a logarithmic way of stating the ratio between two voltages (or other quantities), used because…

Practise filters and signals in the app →

Example problems with solutions

Here are some of the problems in electronics. In the app, calculation problems get new numbers every time, so you can practise until it sticks – and take a graded practice exam before the real one.

Diodes and transistors: Roughly how large is the voltage drop across a forward-biased silicon diode?

Answer: 0.7 V

Schottky diodes have around 0.3 V, and LEDs 1.8–3.3 V.

Operational amplifiers: Which two rules apply to an ideal op-amp with negative feedback?

Answer: No current flows into the inputs, and the two inputs are at the same voltage

The second rule is called the virtual short.

Filters and signals: What is the cutoff frequency of an RC low-pass filter?

Answer: $f_c = \dfrac{1}{2\pi RC}$

At $f_c$ the gain is $1/\sqrt2$ , i.e. −3 dB.

Diodes and transistors: What is a Zener diode typically used for?

Answer: Holding a stable reference voltage in reverse bias

It conducts at the Zener voltage in reverse bias.

Practise all the problems →

Matches these university courses

The content covers the syllabus found in engineering degrees, for example:

TTT4203 (NTNU)
FYS235 (NMBU)

Electronics: free practice, theory and problems

Contents

1. Diodes and transistors

What is it about?

Concepts and formulas

How to solve the problems

Example

Common mistakes

Concepts in this part

2. Operational amplifiers

What is it about?

Concepts and formulas

How to solve the problems

Example

Common mistakes

Concepts in this part

3. Filters and signals

What is it about?

Concepts and formulas

How to solve the problems

Example

Common mistakes

Concepts in this part

Example problems with solutions

Matches these university courses