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Active Analogue Circuits Year 2 B. Todd Huffman

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Page 1: Active Analogue Circuits Year 2 - University of Oxford ...huffman/MPhys/Circuits/AADy2... · • Actually; start from Ebers-Moll equation • If V BE ≈ 0.6 V or more I C starts

Active Analogue Circuits Year 2

B. Todd Huffman

Page 2: Active Analogue Circuits Year 2 - University of Oxford ...huffman/MPhys/Circuits/AADy2... · • Actually; start from Ebers-Moll equation • If V BE ≈ 0.6 V or more I C starts

Circuit Theory Reminders

Basics, Kirchoff’s laws, Thevenin and Norton’s theorem, Capacitors, Inductors

AC theory, complex notation, LCR circuits

• Passive Sign Convention

• What is “Passive Sign Convention”?

• Good Texts:

• Electronics Course Manual for 2nd year lab.

• “Art of Electronics” by Horowitz and Hill

October 2016 Todd Huffman

Page 3: Active Analogue Circuits Year 2 - University of Oxford ...huffman/MPhys/Circuits/AADy2... · • Actually; start from Ebers-Moll equation • If V BE ≈ 0.6 V or more I C starts

V0

I R1

R2

R3

-V0+IR1+IR2+IR3=0

0Vn

+

+

+

+

–V0

+IR1

+IR2

+IR1

I1

I3

I2

I4

I1+I2–I3–I4=0

0In

Kirchoff’s laws

KCL

KVL

Page 4: Active Analogue Circuits Year 2 - University of Oxford ...huffman/MPhys/Circuits/AADy2... · • Actually; start from Ebers-Moll equation • If V BE ≈ 0.6 V or more I C starts

AC circuit theory

• Voltage represented by complex exponential

• Impedance relates current and voltage V=ZI in complex notation:

Resistance R Inductance jL Capacitance 1/(jC)

and combinations thereof

• Impedance has magnitude and phase

0V V cos t represented by real component of j t

0V V e

easily shown from

dI

V Ldt

jZ Z e

Q=VC

Page 5: Active Analogue Circuits Year 2 - University of Oxford ...huffman/MPhys/Circuits/AADy2... · • Actually; start from Ebers-Moll equation • If V BE ≈ 0.6 V or more I C starts

• Current is given by

• So |Z| gives the ratio of magnitudes of V and I, and give the phase difference by which current lags voltage

• Notice that the time dependent part is a common factor – So ejt can be removed and is “understood” to be present when

returning to the time domain.

– WARNING!!! This is only true for circuits with Linear behaviour!

j tj t0 0

j

V V e VI e

Z Z e Z

Page 6: Active Analogue Circuits Year 2 - University of Oxford ...huffman/MPhys/Circuits/AADy2... · • Actually; start from Ebers-Moll equation • If V BE ≈ 0.6 V or more I C starts

Op-amps Gain is very large (A)

Inputs draw no current (ZIN=)

Feedback v+=v–

VOUT +

v+

v–

VIN

R1 R2

Non-Inverting Amplifier Circuit

+ VOUT

VIN R1

Inverting Amplifier Circuit

R2

– v–

v+ i

i

Page 7: Active Analogue Circuits Year 2 - University of Oxford ...huffman/MPhys/Circuits/AADy2... · • Actually; start from Ebers-Moll equation • If V BE ≈ 0.6 V or more I C starts

First Non-ideal model

+

-

A()dV

+

dV

Instead of infinite gain, the device

has finite, and frequency dep. Gain.

V0 -

+

Page 8: Active Analogue Circuits Year 2 - University of Oxford ...huffman/MPhys/Circuits/AADy2... · • Actually; start from Ebers-Moll equation • If V BE ≈ 0.6 V or more I C starts

A() behaves like an RC filter.

Magnitude |A()|

Phase A()

With a gain factor of over a million; and a roll-off around = 1 rad/s

A() ≈ 106/(1+j)

Page 9: Active Analogue Circuits Year 2 - University of Oxford ...huffman/MPhys/Circuits/AADy2... · • Actually; start from Ebers-Moll equation • If V BE ≈ 0.6 V or more I C starts

Model of this non-ideal gain curve

• Vx = A0V1

• KCL • (V2 – Vx)/R + jCV2 = 0

– Substitute expression for Vx above and some algebra

• V2(1 + jCR) = A0V1

• V2/V1 = A0/(1 + jCR) ≡ A()

A0

R

C V1

V2

Vx

Note:

Also Draw filter

on Blackboard

Page 10: Active Analogue Circuits Year 2 - University of Oxford ...huffman/MPhys/Circuits/AADy2... · • Actually; start from Ebers-Moll equation • If V BE ≈ 0.6 V or more I C starts

How does this effect our negative feedback circuits?

• KVL • VR1 – dV – Vin = 0

• VR2 + dV + V0 = 0

• KCL • VR1/R1 = VR2/R2

• And also the Gain relationship

• dVA() = V0

Solve on board

+ VOUT

VIN R1

Inverting Amplifier Circuit

R2

– v–

v+ i

i

𝑉0𝑉𝑖𝑛

=−𝑅2

𝑅1 +𝑅1 + 𝑅2 1 + 𝑗𝜔

106

Page 11: Active Analogue Circuits Year 2 - University of Oxford ...huffman/MPhys/Circuits/AADy2... · • Actually; start from Ebers-Moll equation • If V BE ≈ 0.6 V or more I C starts

The Transistor!

Page 12: Active Analogue Circuits Year 2 - University of Oxford ...huffman/MPhys/Circuits/AADy2... · • Actually; start from Ebers-Moll equation • If V BE ≈ 0.6 V or more I C starts

• Silicon (Si) is a semiconductor

– Also Ge

• Atoms in diamond lattice

• Doping

• P-type

• N-type

N N P

base

emmitter

collector Coll.

emmitter

base

Page 13: Active Analogue Circuits Year 2 - University of Oxford ...huffman/MPhys/Circuits/AADy2... · • Actually; start from Ebers-Moll equation • If V BE ≈ 0.6 V or more I C starts

Simple Transistor Model

• It can be a “switch”

– Flow is “on” one way

– Flow is “off” the other way

• It can be an amplifier

– The flow is proportional to the amount you turn the valve.

– If you turn the valve fast enough you can communicate in Morse-Code-litres

Page 14: Active Analogue Circuits Year 2 - University of Oxford ...huffman/MPhys/Circuits/AADy2... · • Actually; start from Ebers-Moll equation • If V BE ≈ 0.6 V or more I C starts

Bipolar Junction Transistor curves

Page 15: Active Analogue Circuits Year 2 - University of Oxford ...huffman/MPhys/Circuits/AADy2... · • Actually; start from Ebers-Moll equation • If V BE ≈ 0.6 V or more I C starts

BJT – How to approach this?!

Assume it is working as expected

• Find an “operating point” using DC

parameters (check assumptions!)

• Use some kind of “equivalent circuit”

which is linear

• Solve linear circuit for “small signals”

• Check consistency

Page 16: Active Analogue Circuits Year 2 - University of Oxford ...huffman/MPhys/Circuits/AADy2... · • Actually; start from Ebers-Moll equation • If V BE ≈ 0.6 V or more I C starts

How to use graphs?

• Actually; start from Ebers-Moll equation

• If VBE ≈ 0.6 V or more IC starts to blow up. If IC changes by 10x, VBE still ~0.6 V

1. Assume VBE ≈ 0.6 V is true!

2. Assume VCE is ≥ 1 V (transistor is “active”)

3. Assume b = 100

4. Solve and rethink assumptions if inconsistency is found.

CmVq

kTII e kT

qV

C

BE

0.27@2510

Page 17: Active Analogue Circuits Year 2 - University of Oxford ...huffman/MPhys/Circuits/AADy2... · • Actually; start from Ebers-Moll equation • If V BE ≈ 0.6 V or more I C starts

Biasing the BJT –

“active” operation • Will use npn transistors

(but pnp are simple complements of npn)

• If “Active” the Base-Emitter voltage drop is a diode drop ~0.6 V.

– Assume this is the case

• Use KVL and KCL and Ohm’s law on rest of circuit to find the DC collector current, IC.

N N P

base

emmitter

collector

0.6v

+

Page 18: Active Analogue Circuits Year 2 - University of Oxford ...huffman/MPhys/Circuits/AADy2... · • Actually; start from Ebers-Moll equation • If V BE ≈ 0.6 V or more I C starts

vBE

Model works for npn and pnp (follow passive sign conv. on resistor)

CB

Cm

II

IkT

qg

b

+ –

First Transistor Small Signal Model

gmvBE b/gm

base collector

emitter

Typical npn form shown

b is related to details of trans.

Construction: b 100 good start

Page 19: Active Analogue Circuits Year 2 - University of Oxford ...huffman/MPhys/Circuits/AADy2... · • Actually; start from Ebers-Moll equation • If V BE ≈ 0.6 V or more I C starts

Our First Transistor Circuit

VOUT

VIN

RB RC

+

+

How is it Biased?

What does it do?

Page 20: Active Analogue Circuits Year 2 - University of Oxford ...huffman/MPhys/Circuits/AADy2... · • Actually; start from Ebers-Moll equation • If V BE ≈ 0.6 V or more I C starts

The 741 op-amp’s actual diagram