Anti Essays offers essay examples to help students with their essay writing.

Sign Up

Gu Ee212 Midterm Essay

Open Document

Below is an essay on "Gu Ee212 Midterm" from Anti Essays, your source for research papers, essays, and term paper examples.

3-4: Il=Vs/R, 20V/1kOhm, Il=20mA
Pl=(Vl)(Il), (20V)(20mA), Pl=400mW
Pd=Vd(Id), (0V)(20mA), Pd=0W
Ptot=Pl+Pd, 400*10-3+0, Ptot=400mW
3-8: Id=0A, Vd=12V
4-4: Vp=Vrms/0.707, 15V/0.707, Vp(in)=-21.22
Vp(out)=Vp(in)+0.7V, -21.22V+0.7V, Vp(out)=-20.52V
Vavg=0.318Vp(out), 0.318(-20.52V), Vavg=-6.52V, Vdc=-6.52V
4-12: Vp(out)=170/8=21.25V, Vavg=(2*21.25) /pi=13.5, Vdc=(2*21.25)/pi=13.5

5-4: Vl=(Rl/(Rs+Rl))Vs, (1.5kOhm/(470Ohm+1.5kOhm))(24V), (0.76142(24V), Vl=18.27V

5-16: Pz=Vz*Iz, (10V)(20*10-3A), 200*10-3, Pz=0.2W
6-4: Ib=Ic/Bdc, 100mA/65, 1.54mA, Ie=Ib+Ic, 1.54mA+100mA, Ie=101.54mA
6-8: Vce=Vcc-Ic*Rc, 20V-(6mA)(1.5kOhm), 20V-9V, Vce=11V

6-12: Ib=Vbb/Rb, 12V/680kOhm, Ib=17.65uA
Ic=Bdc*Ib, (175)(17.65uA), Ic=3.1mA
Vce=Vcc-Ic*Rc, 12V-(3.1mA)(1.5kOhm), Vce=7.35V
Pd=Vce*Ic, (7.35V)(3.1mA), Pd=22.79mW
Ib=Vbb-Vbe/Rb, (12V-0.7V)/680kOhm, Ib=16.62uA
Ic=Bdc*Ib, (175)(16.62uA), Ic=2.91mA
Vce=Vcc-Ic*Rc, 12V-(2.91mA)(1.5kOhm), Vce=7.64V
Pd=Vce*Ic, (7.64)(2.91mA), Pd=22.23mW
6-16: DeltaT=65C-25C, DeltaT=40C
DeltaP=(5mW/C)(40C), DeltaP=200mW
Pd(max)=625mW-200mW, Pd(max)=425mW
The transistor would likely be destroyed since the maximum power dissipation is 425mW, but
the dissipation in the circuit is 625mW.

Show More


MLA Citation

"Gu Ee212 Midterm". Anti Essays. 9 Dec. 2018


APA Citation

Gu Ee212 Midterm. Anti Essays. Retrieved December 9, 2018, from the World Wide Web: http://trenmayamx.com/free-essays/Gu-Ee212-Midterm-602780.html