- #31
canadiansmith
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Is this wrong? If so please help me. I have been working on this all day.
Engineering Electric circuit with negative reference voltages and diodes
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The discussion revolves around understanding negative reference voltages in a circuit with diodes. The user initially assumes all diodes are forward biased but struggles with calculating node voltages, particularly Va and Vb. After various calculations and clarifications, it becomes clear that Va cannot exceed the highest potential source of +5V, leading to the conclusion that diode D2 is off. The user ultimately realizes that negative potentials cannot exceed the original voltage, gaining a clearer understanding of the circuit dynamics. The conversation highlights the importance of systematic approaches in analyzing diode states within circuits.
- #32
gneill
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canadiansmith said:
Your formula for VA cannot yield 7.15V. It can't! VA must be less than 5V, since he current is flowing down through R1 (so there's a voltage drop across R1), and the top of R1 is at only +5V. It might be convenient to find the current in the branch first, then find the voltage drop on R1. +5V minus that voltage drop will give you VA.
The drop across D3 is 0.7V, so VB can only be 0.7V below ground potential. That makes VB -0.7V.
- #33
canadiansmith
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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
gneill
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canadiansmith said:
Show your calculation for Va if D2 is not conducting.
- #35
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gneill said:
Exactly what I was going to say
- #36
canadiansmith
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va would equal 5-VR1 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 = 5-7.15V = -2.15V is this right?
VR1 = 1.43mA*R1 = 7.15 V
and Va = 5-7.15V = -2.15V is this right?
- #37
gneill
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canadiansmith said:
It is. So in conclusion...?
- #38
canadiansmith
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In conclusion D2 is in fact off. Thank you very much. I got very confused with the negative potentials. But I realize now that they cannot make the potential at any point higher than the original potential.
- #39
gneill
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The important thing is that you got there in the end
- #40
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good work ... sounds like you learned something worth learning
- #41
canadiansmith
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Thank you both very much for your help. I didn't think I would ever see the light at the end of the tunnel. Thanks again
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