LÝ THUYẾT DIODE (MẠCH điện tử SLIDE)

Properties of Diodes The Shockley EquationKristin Ackerson, Virginia Tech EE Spring 2002 • The transconductance curve on the previous slide is characterized by the following equation: •

 anode

 maximum forward current = dòng

thuận cực đại

3-8 Cách tính điện trở khối 3-9 Điện trở DC của diode

3-10 Đường tải

3-11 Diode dán bề mặt

Properties of Diodes The Shockley Equation

Kristin Ackerson, Virginia Tech EE Spring 2002

• The transconductance curve on the previous slide is characterized by

the following equation:

• As described in the last slide, I D is the current through the diode, I S is

the saturation current and V D is the applied biasing voltage.

• V T is the thermal equivalent voltage and is approximately 26 mV at room temperature The equation to find V T at various temperatures is:

at higher currents

Diode Circuit Models

Kristin Ackerson, Virginia Tech EE Spring 2002

The Ideal Diode

Model

The diode is designed to allow current to flow in only one direction The perfect diode would be a perfect conductor in one direction (forward bias) and a perfect insulator in the other direction

(reverse bias) In many situations, using the ideal diode approximation is acceptable.

Example: Assume the diode in the circuit below is ideal Determine the value of I D if a) V A = 5 volts (forward bias) and b) V A = -5 volts (reverse bias)

+

_

V A

I D

R S = 50  a) With V A > 0 the diode is in forward bias

and is acting like a perfect conductor so:

Diode Circuit Models

Kristin Ackerson, Virginia Tech EE Spring 2002

The Ideal Diode with

Barrier Potential

This model is more accurate than the simple ideal diode model because it includes the approximate barrier potential voltage

Remember the barrier potential voltage is the voltage at which appreciable current starts to flow.

Example: To be more accurate than just using the ideal diode model

include the barrier potential Assume V = 0.3 volts (typical for a

germanium diode) Determine the value of I D if V A = 5 volts (forward bias).

V+

Diode Circuit Models

The Ideal Diode

resistance) value is pretty constant For low-power germanium and silicon diodes the R F value is usually in the

2 to 5 ohms range, while higher power diodes have a R F value closer to 1 ohm.

Linear Portion of transconductance

Diode Circuit Models

The Ideal Diode

Diode Circuit Models

Kristin Ackerson, Virginia Tech EE Spring 2002

Values of ID for the Three Different Diode Circuit Models

Ideal Diode Model

Ideal Diode Model with Barrier Potential Voltage

Ideal Diode Model with Barrier Potential and Linear Forward Resistance

These are the values found in the examples on previous

slides where the applied voltage was 5 volts, the barrier

potential was 0.3 volts and the linear forward resistance

value was assumed to be 5 ohms.

Reverse

breakdown

Forward Bias Region

Reverse Bias Region

D T

V ηV

 An ideal diode is a two-terminal device

defined by the following non-linear

(currentvoltage) iv-characteristic:

R s is inevitable series resistance of a real device structure Current

controlled current source represents ideal exponential behavior of diode Capacitor specification includes

depletion-layer capacitance for

reverse-bias region as well as

diffusion capacitance associated with junction under forward bias

Typical default values: Saturation

current = 10 fA, R s = 0, Transit

time = 0 seconds

V and R may represent Thévenin

equivalent of a more complex

2-terminal network.Objective of

diode circuit analysis is to find

quiescent operating point for

diode, consisting of dc current and

voltage that define diode’s i-v

characteristic

Loop equation for given circuit is:

This is also called the load line for

the diode Solution to this equation can be found by:

• Graphical analysis using load-line method

• Analysis with diode’s mathematical model

• Simplified analysis with ideal diode model

• Simplified analysis using constant voltage drop model

D

I

Problem: Find Q-point

These points and the resulting

load line are plotted.Q-point is given by intersection of load line and diode characteristic:



D

I

Problem: Find Q-point for given diode

This equation is transcendental, so it

has no closed form solution

D S

D

V V

V nV

V I

exp 10

10

10

1 40

exp 10

1 exp

13 4

13

By iteratively guessing values for V D and increasing or decreasing V D until right side of equation equals 10:

Q-point = ( 0.943 mA, 0.574 V)

2 or 3 significant digits for V D is

usually plenty since I S , n, V T , and R

are rarely know to better precision.Typically, we use SPICE if we want

to use full math model

If diode is forward-biased, voltage across diode

is zero If diode is reverse-biased, current

through diode is zero

v D =0 for i D >0 and i D =0 for v D < 0

Thus diode is assumed to be either on or off

Analysis is conducted in following steps:

• Guess diode’s region of operation from circuit

• Analyze circuit using diode model appropriate for assumed operation region

• Check results to check consistency with

assumptions

Assume diode is on (since it looks like

anode will be at high voltage than

10

V)010(

(our assumption is right)

Assume diode is off Hence I D =0 Loop equation is:

Q-point is (0, -10 V)

V10

010

V

I V

our assumption is right

10

V)6.010(

k10

V)10

Ideal diode model is CVD model with V on = 0V

Analysis: Ideal diode model is chosen Since

15V source is forcing positive current through

D1 and D2 and -10V source is forcing positive

current through D2, assume both diodes are on

Since voltage at node D is zero due to short

circuit of ideal diode D1,

Q-points are (-0.5 mA, 0 V) and (2.0 mA, 0 V)

But, ID1 <0 is not allowed by diode, so try again

mA5

1k

10

V)015

Since current in D1 is zero, ID2 = I1,

Q-points are D1 : (0 mA, -1.67 V):off

D2 : (1.67 mA, 0 V) :on

Analysis: Since current in

D2 but that in D1 is invalid,

the second guess is D1 off

and D2 on

V67.17

.1615k

1015

mA67

115k

V25

0)10(k

5k

1015

1 1

1

2 1

I

I I

D

D

The Q Point

Kristin Ackerson, Virginia Tech EE Spring 2002

The operating point or Q point of the diode is the quiescent or

no-signal condition The Q point is obtained graphically and is really only needed when the applied voltage is very close to the diode’s barrier potential voltage The example 3 below that is continued on the next

slide, shows how the Q point is determined using the

transconductance curve and the load line.

I D = V A – V 

R S Using V  values of 0 volts and 1.4 volts we obtain

I D values of 6 mA and 4.6 mA respectively Next

we will draw the line connecting these two points

on the graph with the transconductance curve

This line is the load line.

The Q Point

ID (mA)

VD (Volts) 2

Q point in this example is located

Q Point: The intersection of the

load line and the transconductance curve.

Dynamic Resistance

Kristin Ackerson, Virginia Tech EE Spring 2002

The dynamic resistance of the diode is mathematically

determined as the inverse of the slope of the transconductance

curve Therefore, the equation for dynamic resistance is:

ID

The dynamic resistance is used in determining the voltage drop

across the diode in the situation where a voltage source is

supplying a sinusoidal signal with a dc offset.

The ac component of the diode voltage is found using the

following equation:

vF = vac rF

The voltage drop through the diode is a combination of the ac and

dc components and is equal to:

Dynamic Resistance

Kristin Ackerson, Virginia Tech EE Spring 2002

Example: Use the same circuit used for the Q point example but

change the voltage source so it is an ac source with a dc offset The

source voltage is now, v in = 6 + sin(wt) Volts It is a silicon diode so the barrier potential voltage is still 0.7 volts.