ECET311Electronic Devices and Circuits

TheoryDIODE APPLICATIONS

(part 1)

DIODE APPROXIMATIONS

Equivalent circuit – a combination of elements properly chosen to best represent the actual terminal characteristics of the device, system or such in a particular operating region.

To simplify diode networks analysis, three diode approximations are used:

1. Piecewise-linear equivalent circuit

2. Simplified equivalent circuit

3. Ideal equivalent circuit

Piecewise-linear equivalent circuit

Characteristic curve

- The exact analysis for diodes

Simplified equivalent circuit

Ideal diode

ID

ID

VD0.7 V for Si0.3 for Ge

Characteristic curve

- Use this model only if R network >> rav

0.7 V for Si0.3 for Ge

Ideal equivalent circuit

ID

Characteristic curve

ID

VD

- Use this model only if R network >> ravE network >>VT

LOAD-LINE ANALYSISLOAD-LINE ANALYSIS

The load line plots all possible combinations of diode current (ID) and voltage (VD) for a given circuit. The maximum ID equals E/R, and the maximum VD equals E.

The point where the load line and the characteristic curve intersect is the Q-point, which identifies ID and VD for a particular diode in a given circuit.

EXAMPLE 1

For the series diode configuration and employing the diode characteristics below, determine:

(a)VDQ and IDQ

(b)VR

Si

10 VE

R1k

ID

10ID(mA)

VD(V)0.8

ANSWER:

(a) Max VD = E = 10V; because the diode is forward biased, there is no voltage drop in the diode.

Max ID = VD/ R = 10 V/ 1 k = 10 mA

(b) VR = IRR = IDQR = (9.25 mA)(1k) = 9.25 V

Or VR = E – VD = 10 – 0.78 V = 9.22 V

10V

IDQ = 9.25 mA

VDQ = 0.78 V

Q point

EXAMPLE 2

For the series diode configuration and employing the diode characteristics below, determine:

(a)VDQ and IDQ

(b)VR

Si

10 VE

R1k

ID

10ID(mA)

VD(V)0.7

ANSWER:

(a) Max VD = E = 10V; because the diode is forward biased, there is no voltage drop in the diode.

Max ID = VD/ R = 10 V/ 1 k = 10 mA

(b) VR = IRR = IDQR = (9.25 mA)(1k) = 9.25 V

Or VR = E – VD = 10 – 0.7 V = 9.3 V

10V

IDQ = 9.25 mA

VDQ = 0.78 V

Q pointQ point

EXAMPLE 3

For the series diode configuration and employing the diode characteristics below, determine:

(a)VDQ and IDQ

(b)VR

Si

10 VE

R1k

ID

10ID(mA)

VD(V)0 V

ANSWER:

(a) Max VD = E = 10V; because the diode is forward biased, there is no voltage drop in the diode.

Max ID = VD/ R = 10 V/ 1 k = 10 mA

(b) VR = IRR = IDQR = (10 mA)(1k) = 10V

Or VR = E – VD = 10 – 0V = 10 V

10V

IDQ = 10 mA

VDQ = 0.78 V

Q point

VDQ = 0V

SERIES DIODE CONFIGURATIONS

Constants• Silicon Diode: VD = 0.7 V• Germanium Diode: VD = 0.3 V

Analysis (for silicon)• VD = 0.7 V (or VD = E if E < 0.7 V)• VR = E – VD

• ID = IR = IT = VR / R

Forward BiasForward Bias

Diodes ideally behave as open circuits

Analysis• VD = E• VR = 0 V• ID = 0 A

Reverse BiasReverse Bias

EXAMPLE 4:

For the series diode configuration of the figure below,

Determine:

(a) VD

(b) ID

(c) and VR.

Si

8 VE

R2.2k

ID

Note: In the preceding discussions, we will be utilizing the simplified equivalent circuit of a diode

ANSWER:

VD = 0.7VR = E – VD = 8 – 0.7 = 7.3 VID = IR = VD/ RD = VR/IR = 7.3 V/ 2.2 k = 3.32 mA

EXAMPLE 5:

For the series diode configuration of the figure below,

Determine:

(a) VD

(b) ID

(c) and VR.

Si

8 VE

R2.2k

ID

ANSWER:

VD = 8 VVR = E – VD = 8 – 8 = 0 VID = IR = VD/ RD = VR/IR = 0V/ 2.2 k = 0 A

EXAMPLE 6:

For the series diode configuration of the figure below,

Determine:

(a) VD

(b) ID

(c) and VR.

Si

0.5 VE

R1.2 k

ID

ANSWER

VD = 0.5 V; although the diode is forward biased, the forward bias voltage is not enough to make the diode conduct, and so the diode acts like an open switch.

VR = E – VD = 0.5 – 0.5 = 0 V

ID = IR = VD/ RD = VR/IR = 0V/ 1.2 k = 0 A

EXAMPLE 7

Determine Vo and ID for the series circuit below:

12 V Vo

Si red R680

Color Construction

Typical Voltage

(V)

Amber AlInGaP 2.1

Blue GaN 5.0

Green GaP 2.2

Orange GaAsP 2.0

Red GaAsP 1.8

White GaN 4.1

Yellow AlinGaP 2.1

Note:Typical Forward Voltages for LEDs

ANSWER

Red LED, Forward voltage = 1.8 V

12 – 0.7 -1.8 – 680IR = 0

ID = IR = (12–0.7 – 1.8)/680

= 13.97 mA

Vo = ID R = 9.500 V

or Vo = 12-0.7-1.8

= 9.500 V

12 V Vo

Si red R680

EXAMPLE 8

Determine ID, VD2 and Vo for the circuit below:

EXAMPLE 8

Determine ID, VD2 and Vo for the circuit below

R5.6 k

ANSWER

IR = 0, because the circuit is open at diode 2.

VD2 = 12 V

The KVL:

12 – 0 – 12 – Vo = 0

12 – 0 – 12 – (5.6k)IR = 0

Vo = 0V

EXAMPLE 9

Determine I, V1, V2 and Vo for the series dc configuration below.

I

+ V1 -

+

V2

-

ANSWER

10 – (4.7k)I – 0.7 – (2.2k)I –(-5) = 0

10 – (4.7k)I – 0.7 – (2.2k)I + 5 = 0

I = (10 – 0.7 + 5)/(4.7k + 2.2k)

I = 2.07 mA

V1= IR1= 9.73 V

V2 = IR2 = 4.55 V

Vo – V2 – E2 = 0

Vo = V2 + E2 = 4.55 – 5V = - 0.45 V

PARALLEL CONFIGURATIONS

mA 142

mA 28

D2I

D1I

mA 28.33kΩ

V .7V 10

R

DVE

RI

V 9.3R

V

V 0.7O

VD2

VD1

V

V 0.7D

V

EXAMPLE 10:

Determine the current I for the network below

R2.2 k

E1=20V E2=4V

I

Si

Si

D1

D2

ANSWER:

Given:

-E1 +IR+VD1+E2 =0

I = (E1 - VD1 - E2)/R

I = 6.95 mA

D1

D2

EXAMPLE 11

Determine the voltage Vo for the network below12 V

Vo

R2.2 kΩ

Si Ge

ANSWER

Given:

-12 +VGE + Vo = 0

Vo = 12 -VGE = 12V – 0.3V

Vo = 11.7 V

EXAMPLE 12

Determine the currents I1, I2 and ID2 for the network below:

R13.3 kΩ

R25.6 kΩ

I1

I2 ID2

E = 20V

Si

Si

D1

D2

ANSWER:

Given

-E +VD1 +VD2 +I2R2 = 0 -E + VD1 + I1R1 + I2R2 = 0

I2 = (E -VD1 -VD2 )/= R2 I1 = (E - VD1 - I2R2)/R1

I2 = 3.321 mA I1 = 0.213 mA

I2 = ID2 + I1; ID2 = I2 – I1

ID2 = 3.108 mA

1

2

D1

D2

END