Download EENG 270 HW2: Diodes and Voltage Drops and more Assignments Electronics in PDF only on Docsity!
EENG 270
HW
Carlyn Annunziata Q1: a)Practical model, ๐๐ท drop across each diode in series: ๐๐ท = ๐๐ = 3๐๐ท = 2.1V b) ๐๐ท = 5 ๐โ 2. 1 ๐
- 5 ๐ฮฉ = 1.16mA; in series, current is the same across all diodes. c)For ๐๐= 2.61V: ๐๐
= ๐ 1 โ ๐๐ท= 2.39V, where ๐๐ is the voltage drop across one diode, ๐๐=
- 39 ๐ 3 = 0.8V^. For ๐๐
= 0.25W: ๐๐
= 0.25W = ๐๐
^2 ๐
=^ ( 2. 39 ๐)^2 ๐
, Solving for R: ๐
= ( 2. 39 ๐)^2
- 25 ๐
For ๐๐ท = 0.125W: ๐๐ท =0.125W= ๐๐๐๐ Solving for ๐๐: ๐๐= (0.00116)(0. 125 W)= 0.145mV. Yes, diodes and resistor can survive. Q2: a)๐ท 1 is 5 .1V Zener diode in Reverse bias and regular diode in Forward bias. ๐ท 2 is a PN junction diode, 0.7V when forward bias and OFF when Reverse bias. In the condition when both diodes are ON, there is a contradiction. When both ON, the ๐๐ for analysis of ๐ท 1 is 5.1V, and for ๐ท 2 , ๐๐ is 0.7V.
If Zener diode is ON, and ๐ท 2 is OFF: Cathode side is 10V and anode side is 5.1V. Then: ๐๐ท = 10V-5.1V= 4.9V, Which is less than the cut-in voltage required. Current flowing to the Zener diode is 0mA. When ๐ท 2 ON and Zener diode, ๐ท 1 , is OFF: ๐๐ = 0.7V The current through the PN junction diode is: ๐๐ = ๐๐ โ๐ 0 ๐
10 ๐โ 0. 7 ๐ 1 ๐ฮฉ = 9 .3mA. b)In Figure 2.2, when both ๐ท 1 and ๐ท 2 are ON, by voltage division: ๐๐ = ๐
2 ๐
1 +๐
2 (๐๐ )^ =^ 6.5V. For the case when both ๐ท 1 and ๐ท 2 are OFF, the diodes act as an open circuit: ๐๐ = 10V, ๐๐ 1 = ๐๐ 2 =0mA. ๐ = 10 ๐
- 1 ๐ฮฉ = 0.99mA. For ๐๐ท 1 : ๐๐ท 1 = ๐๐
1 =0.99mA(.1kฮฉ) = 0.099V, Which is sufficient, and this diode is OFF. For ๐๐ท 2 : ๐๐ท 2 = ๐๐
1 =0.99mA( 5 kฮฉ) = 4. 95 V Which is insufficient cut-in voltage for the Zener diode. The diode, ๐ท 2 , is also OFF. Q3.1&3.2:
For the half wave rectified circuit, the secondary output voltrage is found from the expression: ๐ 2 ๐ 1 =^ ๐๐ 2 ๐๐ 1 , Where N 1 :N 2 is given to be 10:1, and ๐๐ 1 is 170V. Solving for VS 2 : ๐๐ 2 = 17V, The peak to peak output voltage is given by the expression: ๐๐๐ = ๐๐ 2 - (๐ฃ๐ท)= 17V-0.7V = 16.3V, With diode of 0.7V on for each half wave. ๐ผ๐ = ๐๐๐ ๐
=^
- 3 ๐ 1 ๐ฮฉ = 16.3mA. Using peak resistor current, given to be 7mA, the minimum resistor voltage: ๐๐
๐๐๐ = (7mA)(1kฮฉ) = 7V, Ripple voltage, ๐๐, is: ๐๐ = 16.3V-7V = 9.3V, From this, the capactitance is found using the approximated form for ripple voltage in the half wave rectified case: ๐๐ = ๐๐๐๐ ๐
๐ถ
Where T = 1/f. Solving for capacitance: C = 2 9 ๐F.
0 0 1 1 2 2 3 3 4 4 5 5 6 6 7 7 A B C D E F G V 5V D 1N
0 0 1 1 2 2 3 3 4 4 5 5 6 6 7 7 A B C D E F G V 10V R 1kฮฉ D D 1N4733A
EENG 270 HW 2 q2.
Printing Time:Tuesday, October 26, 2021, 3:24:18 PM
DC Operating Point Analysis
Variable Operating point value 1 2 I(D2[ID]) 9.28726 m V(vout) 712.74144 m
EENG HW2 q2 b
Printing Time:Tuesday, October 26, 2021, 4:13:12 PM
DC Operating Point Analysis
Variable Operating point value 1 2 V(vo)
I(R1) 990.09902 u
DC Operating Point Analysis
Variable Operating point value 1 2 V(vd2)
I(R2) 990.09901 u
0 0 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 A^ A B^ B C^ C D^ D E^ E F^ F G G D 1N R120k ฮฉ C110.7ฮผF Vout V118Vpk50Hz0ยฐ XSC A^ B Ext Trig
_ _^ +^ _
0 0 1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 A^ A B^ B C^ C D^ D E^ E F^ F G G D 1N R11k ฮฉ C129ฮผF Vout V117Vpk60Hz0ยฐ XSC A^ B Ext Trig
_ _^ +^ _
