GATE ECE Analog Circuits Formula Sheet 2027, With Traps

Q: What is the formula for gm of a BJT and of a MOSFET?

For a BJT, gₘ = IC/VT with VT ≈ 26 mV at room temperature, so the transconductance is proportional to collector current. For a MOSFET in saturation, gₘ = 2ID/Vov = √(2μₙ Cₒₓ(W/L)ID), so it grows only as the square root of drain current. Quadrupling the current quadruples the BJT value but only doubles the MOSFET value.

By MD ANISH AHAMADUpdated 4 Oct 20269 min read

GATE ECE Analog Circuits Formula Sheet 2027, With Traps

A GATE ECE analog circuits formula sheet needs a short list of standard results, not hundreds. Analog Circuits averaged 10.8 marks a paper across the five GATE EC papers counted question by question (2010, 2019, 2020, 2025 and 2026), and most of those marks came from a small set of formulas. This sheet gives each one with a short example of its use and the trap that costs the mark, plus the Electronic Devices formulas that analog questions lean on.

In this guide

Key takeaways
Notation used on this sheet
The diode equation and the thermal voltage
Reading an ideal-diode network
Rectifier averages, clampers and the doubler
Designing a Zener regulator at both corners
BJT small signal: transconductance and the four gains
MOSFET square law and small-signal transconductance
Current mirrors and the differential amplifier
Op-amp amplifiers and the integrator
Active filter cut-off: hertz against radians per second
Schmitt trigger thresholds and hysteresis
Where this sheet fits in your preparation
Quick revision

Key takeaways

Almost every analog question starts with a decision: is the diode on, is the transistor in its active region, does the op-amp have negative feedback.
For a BJT, $g_m = I_C/V_T$ is proportional to current; for a MOSFET, $g_m$ grows only as $\sqrt{I_D}$ .
Removing an emitter bypass capacitor lowers the voltage gain and raises the input resistance, and both directions are tested.
A Zener regulator is sized at one corner of the line and load range and checked at the opposite corner.
The single-ended output of a differential pair has half the gain of the differential output.
The virtual short holds only under negative feedback, so a Schmitt trigger is solved with a voltage divider.
An RC cut-off is $1/(2\pi RC)$ in hertz and $1/RC$ in radians per second, and the factor $2\pi$ is a ready-made wrong option.

Notation used on this sheet

The thermal voltage is $V_T = kT/q$ , about 26 mV at 300 K. The threshold voltage of a MOSFET is written $V_{th}$ here. Some texts write $V_T$ for threshold too, so read which one a question means before you substitute.

$V_m$ is the peak of a sinusoid. The overdrive is $V_{ov} = V_{GS} - V_{th}$ . The transconductance $g_m$ is the change in output current per volt of input. Small-signal means the small AC variation around the DC operating point, so DC sources become shorts and DC current sources become opens. $\beta$ is the BJT current gain and $\parallel$ means in parallel.

The diode equation and the thermal voltage

Analog questions borrow these from Electronic Devices, where the diode equation appeared in 2019 Q59, 2025 Q63 and 2026 Q25. The current rises exponentially with forward voltage, scaled by the ideality factor $\eta$ :

I = I_S\left(e^{V/\eta V_T} - 1\right), \qquad V_T = \frac{kT}{q}

Formula	One-line example	Watch out for
$V_T = kT/q \approx 26\text{ mV}$ at 300 K	At 360 K, $V_T = 0.026 \times 360/300 = 31.2\text{ mV}$	Rescale with temperature whenever the question is not at 300 K
One decade of current needs $\Delta V = 2.3\,\eta V_T$	$\eta = 1$ : about 60 mV; $\eta = 2$ : about 120 mV	The ideality factor is a factor of two hiding in a logarithm
$D/\mu = kT/q = V_T$ (Einstein relation)	$\mu = 1000\text{ cm}^2/\text{V s}$ gives $D = 26\text{ cm}^2/\text{s}$	cm²/s against m²/s is a factor of $10^4$

You will take natural logs and exponentials of these on the GATE virtual calculator, so practise the key sequence before the exam.

Reading an ideal-diode network

Ideal-diode networks appeared in four of the five counted papers (2019 Q34, 2020 Q27, 2025 Q57, 2026 Q65 among them). The circuit changes every year; the method does not.

Assume a state for each diode: on (a short) or off (an open).
Solve the linear circuit that results.
Check: an on diode must carry forward current, and an off diode must see a reverse or zero voltage. If either check fails, flip that diode and solve again.

Example: a 5 V source, an ideal diode and a 2 V battery opposing it, in series with 1 kΩ. Assume on: $I = (5 - 2)/1\text{ k}\Omega = 3\text{ mA}$ . The current is positive, so the assumption holds.

Rectifier averages, clampers and the doubler

Formula	One-line example ( $V_m = 10$ V)	Watch out for
Half-wave: $V_{avg} = V_m/\pi$ , $V_{rms} = V_m/2$	$3.18$ V and $5$ V	$V_m$ is the peak after any diode drops the question specifies
Full-wave: $V_{avg} = 2V_m/\pi$ , $V_{rms} = V_m/\sqrt{2}$	$6.37$ V and $7.07$ V	A clipper output is not a rectifier: integrate the actual waveform
Clamper: DC shift so one peak sits at the clamp level	Clamp the top peak at 0 V: output swings 0 to $-20$ V	Peak-to-peak is unchanged at $2V_m$
Voltage doubler: $V_o \approx 2V_m$	$V_o \approx 20$ V	Use the peak, not the rms value of the mains

Designing a Zener regulator at both corners

Zener-referenced regulators appeared in 2019 Q48, 2020 Q54 and 2025 Q47. The Zener stays in breakdown only while it carries at least $I_{Z,\min}$ . The hardest moment is lowest input with heaviest load, so the series resistor is sized there:

R_s = \frac{V_{in,\min} - V_Z}{I_{Z,\min} + I_{L,\max}}, \qquad I_{Z,\max} = \frac{V_{in,\max} - V_Z}{R_s} - I_{L,\min}

Example: $V_{in}$ from 15 to 20 V, $V_Z = 10$ V, $I_{Z,\min} = 2$ mA, load from 0 to 18 mA. Then $R_s = 5/20\text{ mA} = 250\ \Omega$ . At the opposite corner, $I_{Z,\max} = 10/250 - 0 = 40$ mA, so the Zener dissipates $10 \times 0.04 = 0.4$ W.

Trap: The resistor is sized at one corner and the Zener power is checked at the opposite corner. A question that asks for both and gets only one is still a lost mark.

BJT small signal: transconductance and the four gains

BJT single-stage amplifiers appeared in four of the five counted papers (2010 Q9, 2020 Q45, 2025 Q21, 2026 Q50 among them). Everything follows from $g_m$ . In the examples, $I_C = 1.3$ mA, $\beta = 100$ , $R_C = 4\text{ k}\Omega$ and $R_E = 1\text{ k}\Omega$ .

Formula	One-line example	Watch out for
$g_m = I_C/V_T$	$1.3/26 = 50\text{ mA/V}$	Use $I_C$ at the bias point, not the signal current
$r_\pi = \beta/g_m$	$100/0.05 = 2\text{ k}\Omega$	$r_\pi$ is in the base; $r_e = V_T/I_E$ is in the emitter
CE, bypassed: $A_v = -g_m R_C$	$-200$	Include any load: $R_C \parallel R_L$
CE, unbypassed $R_E$ : $\displaystyle A_v = \dfrac{-g_m R_C}{1 + g_m R_E}$	$-200/51 \approx -3.9$ , close to $-R_C/R_E$	Gain falls and input resistance rises together
$R_{in} = r_\pi + (\beta + 1)R_E$	$2 + 101 = 103\text{ k}\Omega$	Bias resistors appear in parallel with this
CC (emitter follower): $A_v \approx +1$	Low output resistance	No inversion
CB: $A_v = +g_m R_C$ , $R_{in} \approx 1/g_m$	$+200$ and $20\ \Omega$	Same magnitude as CE but non-inverting

Remember: Removing the emitter bypass capacitor lowers the gain and raises the input resistance. Both directions, or the mark is gone.

MOSFET square law and small-signal transconductance

In saturation the drain current follows the square law, valid while $V_{DS} \ge V_{GS} - V_{th}$ :

I_D = \frac{1}{2}\mu_n C_{ox}\frac{W}{L}\,(V_{GS} - V_{th})^2, \qquad g_m = \frac{2I_D}{V_{ov}} = \sqrt{2\mu_n C_{ox}\frac{W}{L}\,I_D}

Example: $\mu_n C_{ox}(W/L) = 2\text{ mA/V}^2$ , $V_{GS} = 1.5$ V, $V_{th} = 0.5$ V. Then $\displaystyle I_D = \tfrac{1}{2} \times 2 \times 1^2 = 1$ mA, valid if $V_{DS} \ge 1$ V, and $g_m = 2 \times 1/1 = 2\text{ mA/V}$ . Check whether a question's $k$ already contains the $\displaystyle \tfrac{1}{2}$ .

Device	$g_m$ depends on current as	Current quadrupled
BJT	$g_m \propto I_C$	$g_m$ quadruples
MOSFET	$g_m \propto \sqrt{I_D}$	$g_m$ doubles

Carrying the BJT intuition across to a MOSFET is the error the options are written for. It matters for GATE 2027 in particular. In 2019, the one counted paper set by IIT Madras, the analog section had no op-amp and no BJT question, and both amplifier items (Q64 and Q65) were MOS. One paper is a signal, not a law, but make the MOS side as automatic as the BJT side.

The book's last-minute revision sheet carries every Analog Circuits and Electronic Devices formula in this form, with its trap and the counted questions behind it. The Analog Circuits chapter adds 35 questions with worked solutions, among 925 across the GATE ECE 2027 book.

Current mirrors and the differential amplifier

Mirrors and differential pairs appeared in 2010 Q8, 2019 Q65 and 2025 Q43. In the examples, $g_m = 2\text{ mA/V}$ , $R_C = 10\text{ k}\Omega$ and the tail resistance $R_{EE} = 50\text{ k}\Omega$ .

Formula	One-line example	Watch out for
Matched mirror: $I_{out} = I_{ref}$ ; MOS ratioed: $\displaystyle I_{out} = I_{ref}\dfrac{(W/L)_2}{(W/L)_1}$	$100\ \mu\text{A}$ with three times the $W/L$ gives $300\ \mu\text{A}$	For a BJT mirror, emitter-area ratio plays the role of $W/L$
Differential output: $A_d = g_m R_C$	$20$	Output taken between the two collectors
Single-ended output: $A_d = g_m R_C/2$	$10$	The factor of two is the trap
$A_{cm} \approx -R_C/(2R_{EE})$ (single-ended)	$-10/100 = -0.1$	Larger tail resistance, smaller common-mode gain
$\text{CMRR} = \vert A_d/A_{cm} \vert$	$10/0.1 = 100$ , or 40 dB	Compare single-ended with single-ended

Op-amp amplifiers and the integrator

Op-amp linear circuits appeared in four of the five counted papers, three times in 2026 alone (Q17, Q53, Q62). With negative feedback, the virtual short makes $V_+ = V_-$ and the input currents zero.

Formula	One-line example ( $R_1 = 10\text{ k}\Omega$ , $R_f = 47\text{ k}\Omega$ )	Watch out for
Inverting: $A_v = -R_f/R_1$	$-4.7$	The input resistance is $R_1$ , not infinite
Non-inverting: $A_v = 1 + R_f/R_1$	$5.7$	No minus sign, and never below 1
Summer: $V_o = -R_f \sum_k V_k/R_k$	1 V and 2 V through 10 kΩ each, $R_f = 10\text{ k}\Omega$ : $-3$ V	Each input sees virtual ground
Integrator: $\displaystyle V_o = -\dfrac{1}{RC}\displaystyle\int V_{in}\,dt$	$RC = 10$ ms, 1 V in: output ramps at $-100\text{ V/s}$	Add any initial capacitor voltage

In one line: Check for negative feedback first; only then write $V_+ = V_-$ .

Active filter cut-off: hertz against radians per second

A first-order active filter has its −3 dB cut-off where the reactance of $C$ equals $R$ . This was asked in 2020 Q51 and 2025 Q34.

f_c = \frac{1}{2\pi RC}\ \text{Hz}, \qquad \omega_c = \frac{1}{RC}\ \text{rad/s}

Example: $R = 10\text{ k}\Omega$ and $C = 10$ nF give $\omega_c = 10^4$ rad/s and $f_c \approx 1.59$ kHz. Read the unit the answer box asks for, because the other value is exactly $2\pi$ away.

Schmitt trigger thresholds and hysteresis

A Schmitt trigger feeds the output back to the non-inverting input, so the output sits at $\pm V_{sat}$ and switches when the input crosses a threshold. For the inverting form, with $R_1$ from the non-inverting input to ground and $R_2$ from the output to that input:

V_{TH} = \pm\frac{R_1}{R_1 + R_2}V_{sat}, \qquad \text{hysteresis} = V_{UT} - V_{LT}

Example: $V_{sat} = 12$ V, $R_1 = 10\text{ k}\Omega$ , $R_2 = 20\text{ k}\Omega$ . The thresholds are $\pm 4$ V and the hysteresis is 8 V. This circuit last appeared in 2020 Q18. Do not write a node equation with $V_+ = V_-$ here.

Where this sheet fits in your preparation

Analog Circuits is the third-largest technical section by counted marks, behind Networks and Engineering Mathematics, as the GATE ECE subject-wise weightage shows. The GATE ECE important topics guide ranks the recurring concepts across every section. Many analog items are MCQs, and the marking scheme, which is common to every GATE paper, deducts marks only for wrong MCQ answers. Read this sheet daily in the last two months, attaching the trap to each formula as you go.

More formula sheets: all of GATE ECE · Communications · Electromagnetics · Signals and Systems

Quick revision

Decide every diode state first, then check it against current and voltage signs.
Half-wave average $V_m/\pi$ , full-wave $2V_m/\pi$ ; a clamper keeps peak-to-peak unchanged.
Size the Zener resistor at minimum input and maximum load, then check power at the opposite corner.
$g_m = I_C/V_T$ , $r_\pi = \beta/g_m$ , CE gain $-g_m R_C$ , divided by $1 + g_m R_E$ when unbypassed.
MOSFET $g_m = 2I_D/V_{ov}$ grows as $\sqrt{I_D}$ ; BJT $g_m$ grows as $I_C$ .
Differential pair: $g_m R_C$ differential output, $g_m R_C/2$ single-ended.
Inverting $-R_f/R_1$ , non-inverting $1 + R_f/R_1$ , but only with negative feedback.
Cut-off $1/(2\pi RC)$ in Hz, $1/RC$ in rad/s; Schmitt thresholds $\pm R_1 V_{sat}/(R_1 + R_2)$ .

Frequently asked questions

What is the formula for gm of a BJT and of a MOSFET?

For a BJT, $g_m = I_C/V_T$ with $V_T \approx 26$ mV at room temperature, so the transconductance is proportional to collector current. For a MOSFET in saturation, $g_m = 2I_D/V_{ov} = \sqrt{2\mu_n C_{ox}(W/L)I_D}$ , so it grows only as the square root of drain current. Quadrupling the current quadruples the BJT value but only doubles the MOSFET value.

What happens to a CE amplifier when the emitter bypass capacitor is removed?

The emitter resistor is no longer shorted for signals, so it degenerates the gain. The voltage gain falls from $-g_m R_C$ to $-g_m R_C/(1 + g_m R_E)$ , close to $-R_C/R_E$ when $g_m R_E$ is large. The input resistance rises from $r_\pi$ to $r_\pi + (\beta + 1)R_E$ . Questions on this circuit test both directions, so state both changes.

How do you find the series resistor of a Zener regulator?

Size it at the worst corner for keeping the Zener in breakdown, which is minimum input voltage with maximum load current. That gives $R_s = (V_{in,\min} - V_Z)/(I_{Z,\min} + I_{L,\max})$ . Then check the opposite corner, maximum input with minimum load, where the Zener current and power are largest. Solving only one corner leaves half the question unanswered.

Why is single-ended differential amplifier gain half the differential gain?

A differential input splits equally between the two transistors, so each collector swings by only half the full output. Taking the output between the two collectors gives $A_d = g_m R_C$ . Taking it from one collector to ground gives $g_m R_C/2$ . When you compute CMRR, use the single-ended differential gain against the single-ended common-mode gain, so both refer to the same output.

Does the virtual short apply to a Schmitt trigger?

No. The virtual short holds only when the op-amp has negative feedback. A Schmitt trigger uses positive feedback, so its output sits at one saturation level and the two inputs are not equal. Find the thresholds from the voltage divider on the non-inverting input instead. The switching points are the values of input that make the two op-amp inputs equal at that instant.

Sources

Dates, fees and the syllabus are set by the GATE 2027 organising institute and can change. Always confirm at gate2027.iitm.ac.in.