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# Gu Ee212 Midterm Essay

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
Vl=20V
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.

• Submitted by: VolbeatJohn
• on April 1, 2014
• Category: Science And Technology
• Length: 294 words
• Views: 108
• Popularity Rank: 527652
• 1 rating(s)

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