EE40 Lec 18EE40 Lec 18 Diode CircuitsDiode Circuitsee40/fa09/lectures/Lec_18.pdf · EE40 Lec 18EE40 Lec 18 Diode CircuitsDiode Circuits Reading Chap 10 of HambleReading: Chap. 10

EE40 Lec 18EE40 Lec 18

Diode CircuitsDiode Circuits

Reading Chap 10 of HambleReading: Chap. 10 of HambleySupplement Reading on Diode Circuits

http://www inst eecs berkeley edu/~ee40/fa09/handouts/EE40 MOS Circuit pdf

Slide 1EE40 Fall 2009 Prof. Cheung

http://www.inst.eecs.berkeley.edu/ ee40/fa09/handouts/EE40_MOS_Circuit.pdf

Diodes Circuits

–Load Line Analysisy–Analysis of Diode Circuits by

assumed statesassumed states–Diode Logic Circuits–Wave Shaping Circuits

Rectifying Circuits–Rectifying Circuits


SOLVING CIRCUITS WITH NONLINEAR ELEMENTS

Look at circuits with a nonlinear element like this:IL INL

Linear circuit

L+VL

Nonlinear element

NL+VNL

- -

A nonlinear element with its own I-V relationship, attached to a linear circuit with its own I-V relationship.

1. IL = fL(VL) (linear circuit I-V relationship)2. INL = gNL(VNL) (nonlinear element I-V relationship)3. INL = -IL4. VNL = VL


4. VNL VL

SOLVING CIRCUITS WITH NONLINEAR ELEMENTS

The 4 equations can be reduced to 2 equations in INL and VNL

INL = -fL(VNL) - the linear “loadline” INL = gNL(VNL)

which we can equate and solve for VNL,

or…graph the two equations and solve for the intersection.


EXAMPLE

+INL+

1 kΩ

+I Given : I0 = 10-15 A.+

- VNL

+

_2 V VL_

ILGiven : I0 10 A.Find VNL

1. IL = (VL- 2) / 1000

( )1e10I 026.0/NLV15NL −= −2.

3 I I Substitute 1 and 2 in 33. INL = -IL

4. VNL = VL( ) [ ]1000/)2V(1e10 NL

15 026.0/NLV −−=−−

Substitute 1 and 2 in 3


Solve by iteration, VNL ~ 0.725V

0.004linear

Graphical Solution

0.003

0.0035linear

nonlinear

0.002

0.0025 Loadline: I= - (V-2)/1000

0.001

0.0015

I_N

L

0

0.0005Diode I-V

0.725V

-0.001

-0.0005

0.725V


-1 -0.5 0 0.5 1

V_NL

Piecewise–linear Model of Nonlinear Devices

-5.5Vintercept

+1.5Vintercept

100/)5.1v(i:BSegment400/vi:ASegment−=

=


800/)5.5v(i:CSegment)(g

+=

Ideal Diode Model of PN Diode

ID (A)ID +ID

Circuit symbol I-V characteristic Switch model

forward biasVD

+VD–

reverse bias VD (V)–

Diode behaves like a switch: • closed in forward bias mode • open in reverse bias mode


•used when voltage of interest >> 0.6V

Piecewise Linear Model

ID + +ID

Circuit symbol I-V characteristic Switch modelID (A)

VDVDforward bias

+− VDon

VD (V)– –reverse bias

VDon

For a Si pn diode, VDon ≅ 0.6 V

Diode behaves like a voltage source in series with a switch: • closed in forward bias mode

i bi d


• open in reverse bias mode

Zener DiodeA Zener diode is designed to operate in the breakdown mode.

ID (A)(l k ) t ID (A)

V (V)

reverse (leakage) current

forward current

breakdown voltage VBD VD (V)

R tvs(t) >15V for all t

integratedi it

R

+v (t)

+v (t)


circuitVBD = 15V

vs(t)–

vo(t)–

Piecewise-linear Model of a Zener Diode


Diode Circuit Analysis by Assumed Diode States

•1) Specify Ideal Diode Model or Piecewise-Linear Diode Model

ID (A) ID (A)

reverse biasforward bias

VD (V)reverse bias

forward bias

•2) Each diode can be ON or OFF

VD (V)VDon

•3) Circuit containing n diodes will have 2n states•4) The combination of states that works for ALL di d ( i t t ith KVL d KCL) ill b th


diodes (consistent with KVL and KCL) will be the solution

Example Analysis by assumed Diode States

D1=on D2=on×1.75mA

0.5mA

D1=off D2=on +10 +3

D1=on D2=off

+30

×D1=off D2=off

+10×

√

D1=on D2=off +3+6


Transfer Function of Diode Circuits

Piecewise-Linear Model with 0.6V voltage drop


Diode Logic: AND Gate• AND gate

V Inputs A and B vary between 0


R

Vcc Inputs A and B vary between 0 Volts (“low”) and Vcc (“high”)Between what voltage levels does C vary with V =5VV

A

RAND

C

does C vary with VCC=5VVOUT

5

BSlope =1

EOC

O t t lt C i hi h Shift 0.7V UpOutput voltage C is high only if both A and B are high


VIN0 5

0both A and B are high

Diode Logic: OR Gate• OR gate

Inputs A and B vary between 0 V lt (“l ”) d V (“hi h”)


A

B C

Volts (“low”) and Vcc (“high”)Between what voltage levels does C vary with VCC=5V?

BROR

CVOUT

5

Output voltage C is high if

EOC

Output voltage C is high if either (or both) A and B are high

Slope =1Shift 0.7V Down


VIN0 5

0

0.7V

Diode Logic: Incompatibility and Decay

AND gate OR gate

Signal Decays with each stage (Not regenerative)

output voltage is high only if both A and B are high

output voltage is high if either (or both) A and B are high

Vcc

RAND

A

B CA CAND

BROR

COR

B0.6V drop


Clipper Circuits

Assume forward diode has 0 voltage drophas 0 voltage drop


Peak Detector Circuit

Vi(t)

Assume the ideal (perfect rectifier) model.

+

VC(t)Vi(t)+C Vi+−

−

VC(t)i( )−

t

VC(t) VCIdea: The capacitor h d tcharges due to one

way current behavior of the diode.


Peak Detector with Load Resister


Level Shift Circuit

- VC + VIN VC

VOUT

+

VIN

+

C t

--

V = V + VVOUT 1 3

VOUT = VC+ VINt

21) Diode =open, VC=0, VOUT = VIN2) Diode =short, VC= -VIN , VOUT=03) Diode =open V = V (min) V = V +V


3) Diode =open, VC= -VIN (min) , VOUT= VIN+VC,

Clamp Circuit (level shifter)

Max of vin(t)=5 sin(ωt)is shifted by -5V b h di d lby the diode-voltage source combination


Voltage Doubler Circuit- VC1 +

+ VC

21

R2VOUT

+

VIN

+

C2R1

VOUT

+

VIN

+

C1-

--

--

Level Shift Peak Detect

See Homework problemOutput is the peak to peak voltage of the input


Output is the peak to peak voltage of the input.

Half Wave Rectifier Equivalent circuit

V >0.6V, diode = short circuitV = VI - 0 6Vo VI 0.6

V < 0.6V, diode = open circuitpVo =0


Adding a capacitor: what does it do?

+

Vm sin (ωt)R V0

C0

-


Half-Wave Rectifier

Current charging up capacitor


Full Wave Rectifier


Small –Signal Linear Equivalent CircuitSuppose the nonlinear device has the functional dependence I = i(v) is biased with a DC voltage vG at the Q-point (quiescent point) A small differential voltage ∆v ispoint (quiescent point). A small differential voltage ∆v is added on top of vG. Using Taylor series expansion

.........vdi)v(i)vv(i QQ +∆•+=∆+ .........vdv

)v(i)vv(iQv

QQ +∆+∆+

We can define a dynamic resistance r at the Q point

di1r ≡

i Tangent line

vi ∆∆

Qvdv∆i


ri ≅∆

vvG

∆v

Small –Signal Model of Diode

Q2∆i

Q1

∆i

rvi ∆

≅∆

∆i


∆v ∆v

Small –Signal Model Example

VC and RCDetermines rd atdQ point of diode


Small –Signal Model Example

The large capacitors and DC bias source are effective shortsfor the ac signal in small-signal circuits


for the ac signal in small signal circuits* See Hambley for an application of voltage controlled Attenuator

Documents

EE40 Lec 18EE40 Lec 18 Diode CircuitsDiode Circuitsee40/fa09/lectures/Lec_18.pdf · EE40 Lec 18EE40 Lec 18 Diode CircuitsDiode Circuits Reading Chap 10 of HambleReading: Chap. 10