Diodes - 中興電機cc.ee.nchu.edu.tw/~aiclab/teaching/Electronics1/ch3C_Diodes.pdf · Ching-Yuan Yang National Chung Hsing University Department of Electrical Engineering Diodes

Ching-Yuan Yang

National Chung Hsing UniversityDepartment of Electrical Engineering

Diodes

Read Chapter 3, Section 3.4-3.6, 3.9

Sedra/Smith’s Microelectronic Circuits

3C-1 Ching-Yuan Yang / EE, NCHUElectronics (I)

Zener diode – Operate in the reverse breakdown region

Called breakdown diode or Zener diode, it can be used as voltage regulator.

Breakdown voltage VZK and reverse IZK current at VZK

(also called knee current) is the turning point of the diode into the breakdown region.

The terminal voltage VZ at a specified test current IZT , and at this point we can define rz

as

ΔV = rz⋅ΔI.

rz is called incremental resistance at operating point Q.

(symbol)


Model for the zener diode

VZ0 , the equivalent voltage source, denotes the point at which the straight line of slope 1/rz intersects the voltage axis. Although it is slightly different from the knee voltage VZK, in practice their values are almost equal.

Model:

0Z Z z Z Z ZKV V r I I I= + > for .


Use of the zener as a shunt regulator

a) Find VO with no load and with V + at its nominally value.

b) Find the change in VO resulting from the ±1V change in V + . (Line regulation = ΔVO / ΔV + )

c) Find the change in VO resulting from conne-cting a load resistance RL that draws a current IL = 1 mA, and hence find the load regulation (ΔVO / ΔIL ).

d) Find the change in VO when RL = 2 kΩ.

e) Find the value of VO when RL = 0.5 kΩ.

f) What is the minimum value of RL for which the diode still operates in the break down region?

ExampleExample The 6.8-V zener diode in the circuit is specified to have VZ = 6.8 V at IZ = 5 mA, rZ = 20 Ω, and IZK = 0.2 mA. The supply voltage V + is nominally 10 V but can vary by ±1 V.


(a) With no load connected, the current through the zener is given by

(b) For ±1-V change in V +, the change in output voltage

(c) With IL = 1 mA, the zener current decreases by 1 mA.

SolutionSolution

0 0Z Z z Z Z Z z ZV V r I V V r I= + ⇒ = − = − × = 6.8 20 0.05 6.7 V

0ZZ

z

V VI I

R r

+ − −= = = =

+ +10 6.7

6.35 mA0.5 0.02

0O Z z ZV V r I= + = + × = 6.7 6.35 0.02 6.83 V

zO

z

rV V

R r+Δ = Δ = ± × = ±

+ +20

1 38.5 mV500 20

OV

V +

Δ≡ =Δ

Line regulation 38.5 mV/V

O z ZV r IΔ = Δ = × − = −20 ( 1) 20 mV

O

L

V

I

Δ≡ = −Δ

Load regulation 20 mV/mA


(d) When RL = 2 kΩ, IL = 6.8V/2kΩ = 3.4 mA. ΔIZ = −3.4 mA, and the corresponding change inzener voltage (output voltage) will thus be

(e) When RL = 0.5 kΩ, IL = 6.8V/0.5kΩ = 13.6 mA.This is not possible because the current I suppliedthrough R is only 6.4 mA (for V + = 10V). Therefore, the zener must be cut off. VO is determined by the voltage divider formed by RL and R,

(f) For the zener to be at the edge of the breakdown region, IZ = IZK = 0.2mA and VZ

≈ VZK ≈ 6.7V. At this point the lowest current supplied through R is (9 − 6.7)/0.5 = 4.6 mA, and thus the load current is 4.6 − 0.2 = 4.4 mA.

SolutionSolution

O z ZV r IΔ = Δ = × − = −20 ( 3.4) 68 mV

LO

L

RV V

R R+Δ = Δ = × =

+ +0.5

10 5 V0.5 0.5

Since the voltage is lower than the breakdown voltage of the zener, the diode is indeed not operating in the breakdown region.

LR = = Ω6.7

1.5 k4.4

The minimum value of RL :


Temperature effects of the zener diode

VZ depends on temperature in term of its temperature coefficient TC (mV/°C).

TC depends on VZ , and TC varies with the operating current.

Design consideration:

For example, TC < 0 if VZ < 5 V; TC > 0 if VZ > 5 V; TC = 0 if VZ = 5 V.

TC = 0 can be designed for a 5-V reference voltage.

Another commonly used technique for obtaining a reference voltage with low TC is to connect a zenerdiode with a positive TC of about 2 mV/°C in series with a forward-conducting diode (ΔV = 0.7 V, TC = −2 mV/°C ).

VO = VZ + 0.7 with a TC of about zero.


Diode application circuits - Rectifier

Block diagram of a dc power supply

Half-wave rectifier: Using the battery-plus-resistance diode model, we have

In many applications

0 0 00, , O S D O S D S DD D

R Rv v V v v V v V

R r R r= < = − ≥

+ + ;

0.D O S Dr R v v V≈ − and



Discussion:The current handling capability required of the diode determined by the largest current flowing through the diode, should be considered.

The peak inverse voltage (PIV) that the diode must be able to sustain without breakdown, determined by the largest reverse voltage that is expected to appear across the diode, mush be taken into account in the design. For half wave rectifier, PIV = VS.

This kind of rectifier circuit does not function properly when the input signal is small.



Full-wave rectifier

Use center-tapped transformer

The full-wave rectifier obviously produces a more “efficient”waveform than that provided by the half-wave rectifier.

VO = VS − VD0

PIV = 2VS − VD0

In almost all rectifier applications, one opts for the full-wave type.



Bridge rectifier



Bridge rectifier

Compared to the previous rectifiers, this

one has

No need of center-tapped transformer

Halved PIV

Halved turns for the secondary wing of the transformer

VS > 2VD0

VS < 2VD0

0

0 0

2O S D

O D S D

V v V

V V v V

= −

= + = −

;

PIV



PIV OV Efficiency Transformer turns

Halfwave rectifier Sv

02 S Dv V−

0S Dv V−

0S Dv V−

0S Dv V−

02S Dv V−

~ 40%

~ 90%

~ 80%

1 2:n n

1 2: 2n n

1 2:n n

Center - tapped

Fullwave rectifier

Bridge

Fullwave rectifier

Comparison


Diode application circuits – Peak Rectifier

Peak Rectifier

With a filter capacitor connected to the load in parallel.

When vI > vO, diode conducts and the C charges up. vO follows vI .

After the peak, vI decreases; C holds vO; vI < vO; the diode is off.

Peak detector is a design example, such as an AM demodulator.


Diode application circuits - Peak Rectifier

Half-wave Peak Rectifier

During the diode-off interval, the C discharges through R and thus vO decays exponentially with a time constant of RC. The value of C must be selected such that RC >> T .


When the diode is conducting t ∈[t1, t2]:OI

D C L L L

vdvi i i C i i

dt R= + = + = where

Observations:Diode conducts for a brief interval, Δt.

t > t2, iD = 0:

During the diode-off interval, Cdischarges through R and thus vO

decays exponentially with time constant CR.

When CR >> T , Vr is small and vO is almost constant and equal to the peak of vI.

Similarly, iL is almost constant and its dc component IL = Vp /R.

Output dc voltage:

O Odv vC

dt R= +0

12O p rV V V= −


During the diode-off interval

At the end of the discharge interval

O Odv vC

dt R= +0 /t CR

O pv V e−=

/T CRp r pV V V e−− ≈

/ 1 / .T CRpCR T V e T CR−⇒ ≈ −

pr p

VTV V

CR fCR≈ =

/r p L pV V I V R⇒ ≈ constant


During the diode-on interval

Small ωΔt and

Find the average diode current:

Find the peak diode current:

212

cos( )

cos( ) 1 ( )

p p rV t V V

t t

ω

ω ω

Δ = −

Δ ≈ − Δ

and

2 /r pt V VωΔ ≈

r pV V :

supplied lostCav rQ i t Q CV= Δ = and

supplied lostQ Q=Since ,

( )1 2 /Dav L p ri I V Vπ= +

( )max 1 2 2 /D L p ri I V Vπ= +

max 2r p D DavV V i i≈For ,


Half-wave Peak Rectifier

Find C :

Conduction angle:

Diode conducting cycle: (0.2/2π) 100% = 3.18 %

Average diode current:

Peak diode current:

ExampleExample

Consider a peak rectifier fed by a 60-Hz sinusoid having a peak value Vp = 100 V. Let the load resistance R = 10 kΩ. Find the value of the capacitance C that will result in a peak-to-peak ripple of 2 V. Also calculate the fraction of the cycle during which the diode is considering, and the average and peak values of the diode current.

3

10083.3 F

2 60 10 10p

r

VC

V fRμ= = =

× × ×

SolutionSolution

2 / 2 2 /100 0.2 radr pt V VωΔ = = × =

( )1 2 / 324 mA / 10 mA.Dav L p r L pi I V V I V Rπ= + = = =where

( )max 1 2 2 / 638 mAD L p ri I V Vπ= + =


Diode application circuits - Peak Rectifier

Full-wave Peak Rectifier

The ripple frequency is twice that of the input.

For this case:

Need a capacitor half the size of that required in the half-wave rectifier.

Each diode current is approximately half that in the half-wave circuit.

2p

r

VV

fCR= ( )1 / 2Dav L p ri I V Vπ= + ( )max 1 2 / 2D L p ri I V Vπ= +


Diode application circuits – Half-wave Rectifier

Precision half-wave rectifier – the super diode

Almost identical to the ideal characteristic of a half-wave rectifier:

The op amp has non-ideal effects.

0

0 0O I I

O I

v v v

v v

= ≥

= <

if

if (No offset)


Diode application circuits – Limiting (Clamping) circuits

Double Limiter

(hard limiter)


Diode application circuits – Limiting (Clamping) circuits

Double Limiter(soft limiter)

Soft limiting is characterized bysmoother transitions between the linear region and the saturation regiona slop greater than zero in the saturation region.


Diode application circuits – A variety of basic limiters


Diode application circuits – A variety of basic limiters


Diode application circuits – Clamping circuits

Clamping shifts an entire signal by a dc level: (clamped capacitor)

Assume Ideal diode.

The output voltage has its lowest peak clamped to 0V, which is why the circuit is called clamping circuit. Reversing the diode polarity will provide an output waveform whose highest peak is clamped to 0V.

6 V 0 V 6 V

4 V 6 V 10 VI O C

I C O I C

v v v

v v v v v

= − ⇒ = =

= + ⇒ = = + =

diode conducts, and

diode turns off, and


Diode application circuits – Clamping circuits

Clamped capacitor with a load resistance R:

During [t0, t1], the output voltage falls exponential with time constant CR.

At t1 the input decreases by Va, and the output attempts to follows. This causes the diode to conduct heavily and to quickly charge the capacitor.

At the end of the interval t1 to t2, vO would normally be ~(−0.5) V.

At t1 the input rises by Va, and the output attempts to follows.


Diode application circuits – Clamping & Rectifying circuits

Voltage doubler

Two section in cascade:

A clamp (C1 & D1)

A peak rectifier (D2 & C2)


SPICE diode model


SPICE zener diode model


Design a 5-V dc power supplyHomeworkHomework


SPICE format (txt) :


Diode model parameters:

Documents

Diodes - 中興電機cc.ee.nchu.edu.tw/~aiclab/teaching/Electronics1/ch3C_Diodes.pdf · Ching-Yuan Yang National Chung Hsing University Department of Electrical Engineering Diodes