Four Layer Diode or Schockley

 

Four Layer Diode or Schockley’s Diode 1. the capac capacitor itor of the ffigure igure ch changes anges fr from om 0.7 V to 9 V causi causing ng the four  layered diode to break over. In this example what is the current  through the 2 kohm just before the diode breaks over and when it is conducting? Solution:  Just before break-over the voltage across the capacitor and diode is slightly less than 9V. Therefore the current through the 2 kohm is:  I  =

VI  − Vb  R

=

15 − 9 2k Ω

= 3mA

Since the diode is off, all the current is charging the capacitor. When the diode is conducting, it has a voltage of 0.7V. Therefore the current  through the 2kohm is: 15 − 0.7

−

 I  = VI 

 R

Vb

=

2 k Ω

= 7.15mA

 

Silicon Controlled Rectifier  2. The SCR on the figure has Vt=0.75 It=7mA and Ih=6mA. What is the output voltage when the SCR is off? What is the input voltage that  triggers the SCR? If Vcc is decreased until SCR opens what is the value of Vcc. Solution: When the SC is not conducting there is no current through 100 ohms thus; Vi=Vo=15v  Since the trigger voltage is 0.75 and the trigger current is 7 mA, Required Vi for triggering: Vi

= 0.75V  + 7 * 10

−3

 A 1 * 10 3 Ω

= 7.75V 

Because the Ih is 6mA the V across the 100 ohms a current dropout is: Vi

= 6 * 10

−3

 A(100 Ω)

= 0.6V 

With the ideal approximation this means we have to decrease Vcc to slightly less than 0.6V to turn off the SCR. Therefore we need tod  decrease Vcc to slightly less than: Vcc = 0.75 v + 6 * 10  A(100 Ω) = 01 .35V  ≈1.3V  −3

 

Silicon-Controlled Switch 3.the trigger voltage is 0.7 in the figure what is the gate current? If the trigger current 5mA, what is the approximate current through the 68 ohms? Solution: In the second approximation we use 0.7V for the gate voltage is means the voltage across the 820 ohms V 

= Vg  − 0.7 = 9 − 0.7 = 8.3V 

 And the current through the 820 ohms is,  I  =

8.3V  820 Ω

= 10 .1mA

This is the gate current since the gate current is more than trigger  current, the SCR is conducting ideally it has zero voltage across its output terminal this mean the current through 68 ohms is, 25V   I  =

68 Ω

= 10 .1mA

 

Bidirectional Thyristors TRIAC 4. In the figure the switch is closed if the triac has fired what is the approximate current through the 22 ohms? Solution: Ideally the TRIAC has 0V across, when conducting therefore the current  through the 22 ohms is,  I  =

75V  22 Ω

= 3.41 A

Even if the TRIAC has 1 or 2V across it the current through the 22 ohms is still very close to 3.41 A because of the large supply voltage.

 

Unijunction Transistor (UJT) 5. Find the range of values of R1 in the figure that will ensure proper  turn-on and turn-off of the UJT. η=0.68, Vv=0.8V, Iv=15 mA, Ip=10 uA and Vp=10 V. Solution: VBB − Vp  Ip 12 −10 10 * 10

−6

>  R1 >

>  R1 >

VBB − Vv  Iv

12 − 0.8 15 * 10

−3

200 kohms >  R1 > 746 .67 ohms ←

Therefore the value of R1 must be from 200 kohms (maximum) to 746.67 ohms (minimum).

 

Technological Institute of the Philippines 938 A.Blvd Cubao, Quezon City 

College of Engineering and Architecture

Sample Problems on Thyristors  Assignment No.: 2

Orias, Jayson D.V. EC483L1/EC42FC1

Rating

Engr. Victoriano Dayto II Instructor 

 July 30, 2010 Date of Submission