Electric circuit with negative reference voltages and diodes

[canadiansmith]

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

 

[gneill]

Ok I found Va by (R1/R1+R2)*(5-(-100-0.7) = 7.15V since the bottom of the left side is -10. Likewise for Vb i found 0-0.7-(-5) = 4.3 V

Your formula for V_A cannot yield 7.15V. It can't! V_A 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 V_A.

The drop across D3 is 0.7V, so V_B can only be 0.7V below ground potential. That makes V_B -0.7V.

 

[canadiansmith]

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

 

[gneill]

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

Show your calculation for Va if D2 is not conducting.

 

[phinds]

Show your calculation for Va if D2 is not conducting.

Exactly what I was going to say

 

[canadiansmith]

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?

 

[gneill]

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?

It is. So in conclusion...?

 

[canadiansmith]

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.

 

[gneill]

The important thing is that you got there in the end

 

[phinds]

good work ... sounds like you learned something worth learning

 

[canadiansmith]

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