Electronics

A. Models, biasing, and performance of discrete devices

Discrete Device Models and Biasing

Pick the right diode model, place a BJT or MOSFET Q-point with a load line, and extract the small-signal parameters every amplifier problem then reuses.

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B. Amplifiers

Transistor Amplifiers

Turn a biased transistor into a gain stage: compute common-emitter/source voltage gain, input and output impedance, gain in dB, bandwidth, and feedback.

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C. Operational amplifiers

Operational Amplifiers

Use the two golden rules of the ideal op-amp to solve inverting, noninverting, summing, difference, integrator, and differentiator circuits, then bound them with gain-bandwidth and slew rate.

The operational amplifier is the highest-yield analog topic on the FE because almost every problem reduces to two assumptions applied with discipline. An ideal op-amp has infinite open-loop gain, infinite input impedance, and zero output impedance; the FE Reference Handbook states it directly under Operational Amplifiers as

v_o=A(v_1-v_2)

with

A

large and, in the linear region,

v_2-v_1=0

. From that single line every closed-loop formula follows. Points are lost not on the algebra but on sign and on which input the feedback returns to — so this concept drills the method, not just the results.

The two golden rules

When an ideal op-amp is in linear operation with negative feedback, two rules hold and solve nearly everything. First, no current flows into either input (infinite input impedance). Second, the two input voltages are equal — the virtual short — because infinite gain drives the differential input to zero. When the noninverting input is grounded, the inverting input is held at

0\,\text{V}

: a virtual ground. Apply KCL at the inverting node using these two rules and the circuit falls out in one equation.

i_+ = i_- = 0,\qquad v_+ = v_-\quad(\text{virtual short})

Inverting amplifier

Ground the noninverting input, drive through

R_1

into the inverting node, and feed back through

R_f

. The inverting node is a virtual ground, so the input current

v_{in}/R_1

must flow entirely through

R_f

(none enters the op-amp). That forces the output negative, giving an inverting gain set purely by the resistor ratio. The input resistance seen by the source is just

R_1

— a drawback of this topology.

v_o=-\frac{R_f}{R_1}\,v_{in}

Noninverting amplifier

Drive the signal into the noninverting input and feed

R_f

/

R_1

from the output back to the inverting input, with

R_1

to ground. The virtual short forces the inverting node to follow

v_{in}

, and the

R_f

–

R_1

divider sets the gain. The gain is always at least 1 and the input impedance is ideally infinite (the source sees only the op-amp's noninverting input). Setting

R_f=0

(or

R_1=\infty

) gives a unity-gain buffer/voltage follower, used for impedance isolation.

v_o=\left(1+\frac{R_f}{R_1}\right)v_{in}

Summing and difference amplifiers

An inverting summer ties several inputs through their own resistors into the virtual-ground node; by KCL each contributes independently, producing a weighted, inverted sum. A difference (subtractor) amplifier feeds one signal to the inverting side and one to the noninverting side; with the matched-resistor condition

R_f/R_1

on both sides it outputs a scaled difference — the basis of the instrumentation amplifier and the practical realization of common-mode rejection.

v_o=-R_f\!\left(\frac{v_1}{R_1}+\frac{v_2}{R_2}+\cdots\right),\qquad v_o=\frac{R_f}{R_1}\,(v_b-v_a)

Integrator and differentiator

Replace the inverting amplifier's feedback resistor with a capacitor and you get an integrator: the virtual-ground current

v_{in}/R

charges

C

, so the output is the (inverted) time integral of the input. Swap the roles — capacitor in, resistor feedback — and you get a differentiator, whose output is proportional to the rate of change of the input. Integrators are common in filters and analog computers; differentiators are noise-sensitive and usually tamed with a series resistor.

v_o(t)=-\frac{1}{RC}\int v_{in}\,dt\quad(\text{integrator}),\qquad v_o(t)=-RC\,\frac{dv_{in}}{dt}\quad(\text{differentiator})

Gain-bandwidth product

A real op-amp's open-loop gain rolls off at

-20\,\text{dB/decade}

above a low dominant pole, so the product of closed-loop gain and bandwidth is a constant — the gain-bandwidth product (GBW), equal to the unity-gain frequency

f_t

. A circuit with a higher closed-loop gain therefore has proportionally less bandwidth. This is the first thing to check when an op-amp circuit fails to pass a high frequency.

\text{GBW}=A_{CL}\cdot f_{-3\text{dB}}=f_t\ \Rightarrow\ f_{-3\text{dB}}=\frac{f_t}{A_{CL}}

Slew rate and full-power bandwidth

GBW is a small-signal limit; slew rate is a large-signal one. The output cannot change faster than the slew rate

SR

(in

\text{V}/\mu\text{s}

), set by how fast the internal compensation capacitor can charge. For a sinusoid

v_o=V_p\sin(2\pi f t)

the peak rate of change is

2\pi f V_p

; requiring that to stay below

SR

gives the full-power bandwidth. Exceed it and the sine wave distorts into a triangle. The exam wants you to distinguish: small signals are limited by GBW, large signals by slew rate — whichever bites first.

SR\ge 2\pi f\,V_p\ \Rightarrow\ f_{\max}=\frac{SR}{2\pi V_p}\ \text{(full-power bandwidth)}

Exam strategy

For any op-amp problem: (1) note whether the input enters the inverting or noninverting terminal — that fixes the sign and base formula; (2) if you are unsure, write KCL at the inverting node using

i=0

and

v_+=v_-

from scratch — it always works and protects against memorized-formula slips; (3) for a difference amp, check the matched-resistor condition before using the simple form; (4) decide whether the limit is gain-bandwidth (small-signal) or slew rate (large-signal swing) and apply the right one. Inverting input impedance is

R_1

(finite); noninverting is ideally infinite — a frequent comparison question. Keep

SR

in consistent units (

\text{V}/\mu\text{s}

means multiply by

10^6

for

\text{V/s}

).

Key equations

Ideal op-amp golden rules

i_+=i_-=0,\qquad v_+=v_-

Linear region with negative feedback: no input current, zero differential input (virtual short). Solve via KCL at the inverting node.

Open-loop relation

v_o=A(v_1-v_2),\quad A\to\infty

Defines the device; finite differential input × huge gain ⇒ $v_1\approx v_2$ when not saturated. $v_1$ = noninverting, $v_2$ = inverting.

Inverting amplifier

v_o=-\frac{R_f}{R_1}\,v_{in}

Noninverting input grounded; gain set by resistor ratio, negative sign. Input resistance to source = $R_1$ .

Noninverting amplifier

v_o=\left(1+\frac{R_f}{R_1}\right)v_{in}

Signal at noninverting input; gain ≥ 1, ideally infinite input impedance. $R_f=0$ gives a unity buffer.

Inverting summer

v_o=-R_f\left(\frac{v_1}{R_1}+\frac{v_2}{R_2}+\cdots\right)

Weighted inverted sum; each input scaled by $R_f/R_k$ . Inputs are isolated by the virtual ground.

Difference amplifier

v_o=\frac{R_f}{R_1}\,(v_b-v_a)

Valid with matched ratios ( $R_f/R_1$ on both inverting and noninverting sides). Rejects common-mode signal.

Integrator

v_o(t)=-\frac{1}{RC}\int v_{in}\,dt + v_o(0)

Capacitor in feedback. Output = inverted time-integral; $RC$ sets the time constant.

Differentiator

v_o(t)=-RC\,\frac{dv_{in}}{dt}

Capacitor at input, resistor feedback. Output ∝ rate of change; noise-prone, usually damped.

Gain-bandwidth product

\text{GBW}=A_{CL}\,f_{-3\text{dB}}=f_t

Constant equal to unity-gain frequency $f_t$ . Bandwidth $=f_t/A_{CL}$ ; higher gain ⇒ less bandwidth (small-signal).

Slew rate limit

\left.\frac{dv_o}{dt}\right|_{\max}=SR

Maximum output rate of change (V/μs). A large-signal limit independent of GBW; exceeding it distorts the waveform.

Full-power bandwidth

f_{\max}=\frac{SR}{2\pi V_p}

Highest frequency at which a sinusoid of peak $V_p$ can be output without slew distortion.

CMRR

\text{CMRR}=\left|\frac{A_d}{A_{cm}}\right|,\quad \text{CMRR}_{dB}=20\log_{10}\left|\frac{A_d}{A_{cm}}\right|

Ratio of differential to common-mode gain; infinite for an ideal op-amp ( $A_{cm}=0$ ). Measures rejection of common-mode input.

Worked examples

Inverting amplifier output

Problem. An inverting amplifier has $R_f=100\,\text{k}\Omega$ and $R_1=10\,\text{k}\Omega$ , with $v_{in}=0.5\,\text{V}$ . Find the gain and output voltage.

Solution. Gain:

A_v=-R_f/R_1=-100\text{k}/10\text{k}=-10\,\text{V/V}

. Output:

v_o=A_v\,v_{in}=(-10)(0.5)=-5.00\,\text{V}

. Sanity: the output is the inverted, ×10 amplified input; the inverting node sat at

0\,\text{V}

(virtual ground) so

i=0.5\text{V}/10\text{k}=50\,\mu\text{A}

flowed through

R_f

, dropping

100\text{k}\times50\,\mu\text{A}=5\,\text{V}

✓. Answer:

v_o=-5.00\,\text{V}

.

v_o=-\frac{100\,\text{k}\Omega}{10\,\text{k}\Omega}(0.5\,\text{V})=-5.00\ \text{V}

Inverting summer and a noninverting stage

Problem. (a) A summer has $R_f=40\,\text{k}\Omega$ , $R_1=10\,\text{k}\Omega$ with $v_1=0.5\,\text{V}$ , and $R_2=20\,\text{k}\Omega$ with $v_2=1.0\,\text{V}$ . (b) A separate noninverting stage has $R_f=90\,\text{k}\Omega$ , $R_1=10\,\text{k}\Omega$ , $v_{in}=0.2\,\text{V}$ . Find both outputs.

Solution. (a)

v_o=-R_f\left(\dfrac{v_1}{R_1}+\dfrac{v_2}{R_2}\right)=-40\text{k}\left(\dfrac{0.5}{10\text{k}}+\dfrac{1.0}{20\text{k}}\right)=-40\text{k}(50\mu+50\mu)=-40\text{k}(100\,\mu\text{A})=-4.00\,\text{V}

. (b)

A_v=1+R_f/R_1=1+90\text{k}/10\text{k}=10\,\text{V/V}

;

v_o=10(0.2)=2.00\,\text{V}

. Sanity: the summer weights

v_1

by

40/10=4

and

v_2

by

40/20=2

:

4(0.5)+2(1.0)=4.0

, inverted ✓. The noninverting gain exceeds 1 with the same sign as input ✓. Answers:

v_o=-4.00\,\text{V}

;

v_o=+2.00\,\text{V}

.

v_o=-40\,\text{k}\!\left(\frac{0.5}{10\,\text{k}}+\frac{1.0}{20\,\text{k}}\right)=-4.00\ \text{V}

Difference amplifier

Problem. A matched difference amplifier has $R_f=50\,\text{k}\Omega$ and $R_1=10\,\text{k}\Omega$ on both sides, with $v_a=2\,\text{V}$ at the inverting input and $v_b=3\,\text{V}$ at the noninverting input. Find the output.

Solution. Matched form:

v_o=\dfrac{R_f}{R_1}(v_b-v_a)=\dfrac{50\text{k}}{10\text{k}}(3-2)=5(1)=5.00\,\text{V}

. Sanity: a common-mode shift (add 1 V to both) leaves

v_b-v_a

unchanged, so

v_o

stays

5\,\text{V}

— the circuit rejects common mode, as it should ✓. Answer:

v_o=5.00\,\text{V}

.

v_o=\frac{50\,\text{k}\Omega}{10\,\text{k}\Omega}(3-2)=5.00\ \text{V}

Gain-bandwidth and slew-rate limits

Problem. An op-amp has $\text{GBW}=1\,\text{MHz}$ and slew rate $SR=0.5\,\text{V}/\mu\text{s}$ . (a) Find the closed-loop bandwidth at a gain of 10. (b) Find the full-power bandwidth for a $5\,\text{V}$ peak output. (c) What slew rate is needed for a $10\,\text{V}$ peak, $20\,\text{kHz}$ sine?

Solution. (a)

f_{-3\text{dB}}=\text{GBW}/A_{CL}=1\text{MHz}/10=100\,\text{kHz}

. (b)

f_{\max}=\dfrac{SR}{2\pi V_p}=\dfrac{0.5\times10^{6}\,\text{V/s}}{2\pi(5)}=15.9\,\text{kHz}

. (c)

SR_{req}=2\pi f V_p=2\pi(20{,}000)(10)=1.26\times10^{6}\,\text{V/s}=1.26\,\text{V}/\mu\text{s}

. Sanity: the

0.5\,\text{V}/\mu\text{s}

part cannot do (c) (

1.26>0.5

), so the

10\,\text{V}

/

20\,\text{kHz}

sine would slew-distort — large-signal limit bites first. Answers:

100\,\text{kHz}

;

15.9\,\text{kHz}

;

1.26\,\text{V}/\mu\text{s}

required.

f_{\max}=\frac{0.5\times10^{6}}{2\pi(5)}=15.9\ \text{kHz}

Common pitfalls

•Sign errors: the inverting amp is $-R_f/R_1$ , the noninverting is $+(1+R_f/R_1)$ . Identify which terminal the signal enters before writing any formula.
•Treating inverting and noninverting input impedance as the same. The inverting amp's source sees only $R_1$ (finite); the noninverting input is ideally infinite — a classic comparison question.
•Forgetting the '+1' in the noninverting gain. It is $1+R_f/R_1$ , not $R_f/R_1$ ; the minimum gain is 1, never 0.
•Applying the simple difference-amp formula without the matched-resistor condition. If the ratios are not equal on both sides, common mode leaks through and you must use the full two-source expression.
•Confusing gain-bandwidth (small-signal) with slew rate (large-signal). Small fast signals are GBW-limited; large swings are slew-limited; check which constraint is violated first.
•Unit slips on slew rate: $SR$ in $\text{V}/\mu\text{s}$ must be converted to $\text{V/s}$ (× $10^6$ ) before using $f_{\max}=SR/(2\pi V_p)$ .
•Assuming the virtual short when the op-amp is saturated or has no negative feedback (e.g., a comparator). The golden rules apply only in the linear region with negative feedback.

References

NCEES FE Reference Handbook — Electrical and Computer Engineering: Operational Amplifiers — Ideal op-amp definition, two-source output, inverting/noninverting forms, CMRR.
NCEES FE Reference Handbook — Electrical and Computer Engineering: Electronics — Surrounding device and amplifier context.
Sedra & Smith, Microelectronic Circuits — Virtual-short method, integrator/differentiator, GBW and slew-rate derivations.

D. Instrumentation

Instrumentation

Turn a sensor's tiny resistance or voltage change into a clean, amplified, digitized number using bridges, instrumentation amplifiers, CMRR, and ADC resolution.

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E. Power electronics

Power Electronics

Switch, rectify, filter, and convert power: diode/SCR/MOSFET/IGBT devices, half/full-wave Vdc, ripple, buck/boost gains, and PWM inverter output.

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