1

Equation Sheet

Switched Capacitors

Sum of an infinite geometric series 𝑆𝑘 =𝑎(1−𝑟𝑘)

1−𝑟

Universal Transformer Equation

𝐸𝑟𝑚𝑠 =2𝜋𝑓𝑁𝑎𝐵

√2≈ 4.44𝑓𝑁𝑎𝐵

Buck DC/DC Converter

Boost and Buck-Boost Converters

Thermal voltage across resistor 𝑣𝑛 = √4𝑘𝑇𝐵𝑅, 𝑘 = 1.38 × 10−23

Op-Amps: 𝑉𝑀𝜔 ≤ 𝑆𝑅

Definition of sensitivity 𝑆𝑥𝑦

=𝛿𝑦 𝑦⁄

𝛿𝑥 𝑥⁄=

𝑥

𝑦

𝛿𝑦

𝛿𝑥

𝐋𝐢𝐧𝐞 𝐑𝐞𝐠𝐮𝐥𝐚𝐭𝐢𝐨𝐧 = Δ𝑉𝑜

𝑉𝐼× 100 (%)

Diodes

LS

IOO

if

VVVL

)/1(L

Lo I

C

TItt

CC

QV

82222

11 OFFONIO DVV

IO VD

V

1

1IO V

D

DV

1

i

ooomin

V

VV

L

TI 1

2

ONsattIo

L tL

VVVi

OFF

DoL t

L

VVi

bi

R

j

bi

Rjj

V

V

C

V

VCC

1

10

2/1

0

1

eq

1

CfR

s

21

1

41

2 ESRCESRCCf

RRpump

SWO

LL Ii 2.0

2

2

22

1ESRC

CfIV

pump

ORIPPLE

1T

D

nV

v

SD eIi

2

h-Parameters

2121111 vhihv

2221212 vhihi

0

1

111 2

vi

vh

0

1

221 2

vi

ih 0

2

222 1

iv

ih

BJT CE Version

cerebiebe vhihv

ceoebfec vhihi

ℎ𝑖𝑒 = 𝑟𝑏 + 𝑟𝜋‖𝑟𝜇 ℎ𝑓𝑒 = 𝛽 ℎ𝑟𝑒 ≅𝑟𝜋

𝑟𝜇 ℎ𝑜𝑒 =

1 + 𝛽

𝑟𝜇+

1

𝑟𝑜

y-Parameters

2121111 vyvyi

2221212 vyvyi

0

1

111 2

vv

iy 0

2

112 1

vv

iy

0

1

221 2

vv

iy 0

2

222 1

vv

iy

0

2

112 1

iv

vh

3

RL, RC Series Parallel Transformations

𝑄𝑠 =|𝑋𝑠|

𝑅𝑠, 𝑄𝑃 =

𝑅𝑃

|𝑋𝑃|

4

Fourier Series

5

6

Second-Order Transfer Function Summary

Standard second-order form (s-plane)

𝐻(𝑠) = 𝜔02

𝑁(𝑠)

𝑠2 + 2𝜁𝜔0𝑠 + 𝜔02 =

𝑁(𝑠)

(𝑠

𝜔0)

2

+2𝜁𝜔0

𝑠 + 1= 𝜔0

2𝑁(𝑠)

𝑠2 +𝜔0

𝑄𝑠 + 𝜔0

2

Where 𝑄 = 1 2𝜁⁄ , and 𝐵 = 𝜔0 𝑄⁄ , and 𝜔0 is the undamped natural frequency (rad/s), and 𝜁 (Zeta) is the

damping ratio, and 𝑄 is the quality factor, and 𝐵 is the 3-dB bandwidth in (rad/s)

Case 𝜻 > 𝟏 (Overdamped): The poles are real and negative. The natural response consists of two

decaying exponentials.

Case 𝟎 < 𝜻 < 𝟏 (Underdamped): The poles are complex conjugate 𝑝1,2 = −𝜁ω0 ± 𝑗𝜔0√1 − 𝜁2

The natural response is a damped sinusoid (𝐴 is the residue at the upper pole):

𝑥0(𝑡) = 2|𝐴|𝑒−𝜁𝜔0𝑡 cos (𝜔0√1 − 𝜁2𝑡 + ∡𝐴)

Percentage overshoot (PO) with step response

𝑃𝑂 = 100𝑒−

𝜋𝜁

√1−𝜁2

Case ζ = 0 (No damping): The poles are complex conjugate: 𝑝1,2 = ±𝑗𝜔0. The natural response is a

sustained sinusoid with frequency ω0

Case ζ < 0 (Unstable): The poles lie in the right-hand plane causing a diverging response.

7

Phase-Locked Loop Equations

Phase Detectors

𝐾𝑑 Output With No Input?

XOR 𝑉𝐶𝐶 𝜋⁄ 𝑉𝑐𝑐 2⁄

Charge Pump 𝑉𝐶𝐶 4𝜋⁄ 0

Special Case: First-Order System

𝐻(𝑠) =𝜃𝑜(𝑠)

𝜃𝑖(𝑠)=

𝑇(𝑠)

1 + 𝑇(𝑠)=

𝐾𝑣𝐹(𝑠)

𝑠 + 𝐾𝑣𝐹(𝑠)

8

Special Case: Lag-Lead Filter

𝐻(𝑠) =𝜃𝑜(𝑠)

𝜃𝑖(𝑠)=

(2𝜁 − 𝜔𝑛 𝐾𝑣⁄ )(𝑠 𝜔𝑛⁄ ) + 1

(𝑠 𝜔𝑛⁄ )2 + 2𝜁(𝑠 𝜔𝑛⁄ ) + 1

𝜔𝑛 = √𝜔𝑝𝐾𝑣 𝜁 =𝜔𝑛

2𝜔𝑧(1 +

𝜔𝑧

𝐾𝑣)

𝜔−3dB = 𝜔𝑛 [1 ± 2𝜁2 + √1 + (1 ± 2𝜁2)2]

12

If 𝜔𝑧 = √𝜔𝑝𝐾𝑣 ⇒ 𝜃𝑚 = 45°

If 0.5 ≤ 𝜁 ≤ 1 then 𝜔𝑛 ≅ 1 𝜏⁄

The + sign is for high-gain loops, and the - sign is for low-gain loops. A high-gain loop is when

(𝐾𝑑𝐾𝑜) (𝑅2𝐶) ≫ 1⁄

𝑇(𝑠) = 𝐾𝑣

𝐹(𝑠)

𝑠

𝑇(𝑗𝜔) =1 + 𝑗𝜔 𝜔𝑧⁄

(𝑗𝜔 𝐾𝑣⁄ )(1 + 𝑗𝜔 𝜔𝑝⁄ )

𝐹(𝑠) =1 + 𝑠 𝜔𝑧⁄

1 + 𝑠 𝜔𝑝⁄

𝜔𝑧 = 1 𝑅2𝐶⁄ 𝜔𝑝 = 1 (𝑅1 + 𝑅2)𝐶⁄

𝑅1𝐶 determines settling time. Ratio 𝑅1 𝑅2⁄ determines the

damping ratio. Small values

overshoot, reduce phase margin.

Good place to start is 𝑅1 𝑅2⁄ ≅

0.1— 0.2

9

Wien Bridge Oscillator

To sustain oscillation, the loop gain must be 3. Loop gain with the two resistors is:

𝑅2

𝑅1+ 1

Also

𝑣𝑥 = 𝑣𝑦 =𝑣𝑜

3

Frequency of oscillation

𝜔𝑜 =1

𝑅𝐶 rad/s

3

ov

3

ov