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# Problems 7 and 8 - PowerPoint PPT Presentation

Problems 7 and 8. Assume V out,max = V out, min = power supply rail 7. A V (V + - V - ) = V out V out = 10000 * (6 – 5) = 50 kV > 10 V = V max V out,max = 10V Diode off 8. V - = 12 [3.3k/(2.2k + 3.3k)] = 7.2V V out = I LED (R) + V LED = 10mA (200 W ) + 2V = 4V

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## PowerPoint Slideshow about ' Problems 7 and 8' - taro

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Assume Vout,max = Vout, min = power supply rail

7. AV (V+ - V-) = Vout

Vout = 10000 * (6 – 5) = 50 kV > 10 V = Vmax

Vout,max = 10V

Diode off

8. V- = 12 [3.3k/(2.2k + 3.3k)] = 7.2V

Vout = ILED(R) + VLED = 10mA (200W) + 2V = 4V

Av (V+ - V-) = Vout = 10000 (Vin – 7.2V) = 4V  Vin = 7.2004 V

In practice, setting Vin to this level is impossible. Vin will be set to drive Vout

to the rail Iout = (15-2)/200 = 6.5mA

EET 109 Midterm

Assume Vout,max = Vout,min = power supply rail

15c) LTP = -12 [22k / (3.3k + 2.2k)] = -4.8

UTP = +12[22k / (3.3k + 2.2k)] = +4.8

15d) Each trip point is Voffset + VR1.

UTP = 1 + [(12 – 1)/(3.3k + 2.2k)](2.2k) = 5.4

VR1

LTP = 1+ [(-12 -1)/(3.3k + 2.2k)](2.2k) = -4.2

EET 109 Midterm