#### introduction

In electronics once the basic behavior of a device is understood it’s function and response in an infinite variety of configurations can be examined. In previous page we discussed about energy band theory, construction of diodes but now we learn how to use them.

#### load-line analysis

The figure(a) given above is used define characteristics of diode in figure(b) the blue line is characteristic curve of diode. In the figure we can see that +ve terminal is connected anode and -ve terminal is connected with cathode and current is same in direction as the arrow in the diode which states that diode is in ‘on’ state means diode is the forward-bias state resulting in a voltage across the diode in the neighborhood of 0.7V and a current on the order of 10mA or more. Applying KVL on figure(a)

– E + Vd + Vr = 0  -(1)

E = Vd + Vr        -(2)

we can also write Vr = Id . R

E = Vd + Id . R -(3)

When Vd=0 equation 3 take form

E = Id.R

Id = E / R

When Id=0 then eq. 3 take form

E = Vd

plotting there graph

after joining these two points we got a straight line called load line. The point of intersection between load line and characteristic curve is called quiescent-point abbreviated as (Q-point). The solution obtained at the intersection of two curves is the same as obtained by mathematical solution of

Id = E / R – Vd / R

#### Example

Find Idq, Vdq and Rd in the given figure

we can see that Vdq=0.78V approx. and Idq=18.5mA approx. doted lines representing them from this we can calculate Rd

Rd = Vdq / Idq

Rd = 0.78V / 18.5mA

so Rd is equal to 42.16 OHM.

### Series diode configurations

For conduction the difference between silicon and ideal diode is the vertical shift in characteristics in ideal case it is at 0V and in silicon it is at 0.7V. The approximate models will now be used to investigate a number of series diode configurations with dc inputs.Before starting analysis of any circuit first we configure state of diode either it’s ‘on’ or ‘off’. For voltage less than 0.7V for silicon diode or 0V for ideal diode the resistance is so high compared to other elements that it’s equivalent is the open circuit.

In general, a diode is in the “on” state if the current established by the applied sources is such that its direction matches that of the arrow in the diode symbol, and VD >= 0.7 V for silicon, VD >= 0.3 V for germanium, and VD >= 1.2 V for gallium arsenide.

Note:If the direction of current is same as diode symbol representing then diode is in ‘on’ state.

#### example: Determine the following

1)Vd

2)Vr

3)Id

so diode is in ‘on’ state Vd = 0.7 V. Applying KVL on above figure

Vr = E – Vd

Vr = 8 – 0.7

Vr = 7.3 V

Id = Ir = Vr / R

Id = 7.3 V / 2.2k

Id = 3.32 mA

Note: An open circuit can have any voltage across its terminals, but the current is always 0 A. A short circuit has a 0-V drop across its terminals, but the current is limited only by the surrounding network.

#### Example: for a series circuit determine following

1)Vd

2)Vr

3)Id

current is in direction of diode symbol but it is insufficient to on diode so Id = 0A so it is treated as open circuit.

Vr = (0)(1.2k) = 0V

Vd = E = +0.5V

#### Example: Determine the following

1)Id

2)Vd2

3)Vo

we can see that Si1 and Id have same direction so that one is in ‘on’ state but Si2 and Id have opposite direction so it is considered ‘off ‘

so Id=0 and Vd=0 and Vo = Id.R = 0.R = 0V , Vd2 = applied source = 20V

### parallel & series-parallel configurations

Let’s take an example to better understand this concept

#### example: Determine the following

1)Vo

2) I1

3)Id1 and Id2

we know that dc voltage for silicon diode is 0.7V so replacing diodes with 0.7 voltage source so figure take form

so it is clearly give that Vo = 0.7V the current is

I1 = Vr / R

I1 = E – Vd / R = 10 – 0.7 / 0.33k = 28.18 mA

Assuming diodes of similar characteristics

Id1 = Id2 = I1 / 2 = 28.18 mA / 2 = 14.04 mA

#### example: determine the currents

1)I1

2)I2

3)Id2

current is in the direction of both diodes so both are in ‘on’ state

I1 = Vk2 / R1

I1 = 0.7 / 3.3k

I1 = 0.212mA

Applying KVL in first loop so the equation take form

– E + Vk1 + Vk2 + V2 = 0 -(1)

V2 = 20 – 0.7 – 0.7

V2 = 18.6V

I2 = V2 / R2 = 18.6 / 5.6k = 3.32mA

At the bottom node applying KCL

Id2 + I1 = i2

Id2 = 3.32mA – 0.212mA

Id2 = 3.11mA

### Rectification

Conversion of alternating current into direct current is called rectification. Semiconductor diodes are extensively used for this purpose. There are two very common rectifications

1)Half-wave rectification

2)Full-wave rectification

#### Half-wave rectification

An alternating voltage of period T called input voltage is applied to a diode which is connected in series with a load resistance R.

During the +ve half cycle of the input alternating voltage i.e during the interval 0 to T/2 the diode is forward biased , so it offers very low resistance and current flows through R. The flow of current through R causes a potential drop across it which varies in accordance with the alternating input.

During the -ve half cycle i.e during the period T/2 to T the diode is reverse biased. Now it offers very high resistance so practically no current flows through R and potential drop across it is almost zero. The same event repeat during the next cycle and so on. The voltage which appears across the load resistance R is known as output voltage.

#### Full-wave rectification

In half-wave rectification only unidirectional current is used while in full-wave rectification both halves can be utilized.It’s circuit is consists of 4 diodes connected in bridge type structure. Do not forget that diode conduct current when it is forward bias.

during the time 0 – T/2 the point of intersection between D4 and D1 is positive with respect to its other point of intersection D3 and D2. Now the diodes D1 and D3 become forward bias and conduct.

During the negative half cycle i.e during the time interval T/2 – T now the diodes D4 and D2 conduct and flows current through the circuit. It can be seen that direction of current flow through the load resistance R is the same in both halves of the cycle. The output is not smooth but pulsating.

Previous page

-Semiconductor Diodes

https://www.electricalengineering4u.com/electronic-devices-and-circuit/semiconductor-diodes/