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Electric circuit with negative reference voltages and diodes 
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#20
Oct1511, 08:15 PM

P: 36

OK. So if d2 is on does this change my value for va and vb? I don't think it would.



#21
Oct1511, 08:33 PM

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P: 6,505




#22
Oct1511, 09:05 PM

P: 36




#24
Oct1511, 09:20 PM

P: 36

oh no. I see what you are saying now. So is everything that i have done wrong now? Like the way I calculated Va and Vb is this still valid.



#25
Oct1511, 09:36 PM

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Suppose that you remove D2 from the circuit (excise it... open circuit it). What would be the potential V_{A}? What would be the potential V_{B}? What's V_{A}  V_{B}?



#26
Oct1511, 09:38 PM

P: 36

VaVb would be 7.154.3 = 2.85V



#27
Oct1511, 09:46 PM

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#28
Oct1511, 09:48 PM

P: 36

That would be the potential with respect to ground.



#29
Oct1511, 09:53 PM

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#30
Oct1511, 09:56 PM

P: 36




#31
Oct1511, 10:02 PM

P: 36

Is this wrong? If so please help me. I have been working on this all day.



#32
Oct1511, 11:12 PM

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The drop across D3 is 0.7V, so V_{B} can only be 0.7V below ground potential. That makes V_{B} 0.7V. 


#33
Oct1611, 02:02 PM

P: 36

ok what you are saying makes perfect sense but the voltage at Va will still be positive and Vb will be negative so that means D2 has the wrong polarity to turn it off



#34
Oct1611, 02:13 PM

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#36
Oct1611, 02:34 PM

P: 36

va would equal 5VR1 the voltage drop across R1 would be the current * R1. OH! VR1 would be the current induced by both the 10 draw and the +5 push. therefore the total current in the left side would be 5+10.7 =14.3V/R1+R2 = 1.43mA
VR1 = 1.43mA*R1 = 7.15 V and Va = 57.15V = 2.15V is this right? 


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