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Electric circuit with negative reference voltages and diodes

by canadiansmith
Tags: circuit, electric
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phinds
#19
Oct15-11, 08:04 PM
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Quote Quote by canadiansmith View Post
no. If the voltage is less than the "on" voltage then the diode will still be off.
Good.

Also, note the post I just made about D2 being on.
canadiansmith
#20
Oct15-11, 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.
phinds
#21
Oct15-11, 08:33 PM
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OK. So if d2 is on does this change my value for va and vb? I don't think it would.
I think you've missed the point. My post "about D2 being on" was to the effect that it is NOT on.
canadiansmith
#22
Oct15-11, 09:05 PM
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Quote Quote by phinds View Post
I think you've missed the point. My post "about D2 being on" was to the effect that it is NOT on.
What? I am confused. What post about D2 being on? Can you please explain.
phinds
#23
Oct15-11, 09:14 PM
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What? I am confused. What post about D2 being on? Can you please explain.
Read through the posts again in order. You'll get it.
canadiansmith
#24
Oct15-11, 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.
gneill
#25
Oct15-11, 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 VA? What would be the potential VB? What's VA - VB?
canadiansmith
#26
Oct15-11, 09:38 PM
P: 36
Va-Vb would be 7.15-4.3 = 2.85V
gneill
#27
Oct15-11, 09:46 PM
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Quote Quote by canadiansmith View Post
Va-Vb would be 7.15-4.3 = 2.85V
Where would a 7.15V potential come from? The highest potential source I can see is +5V.
canadiansmith
#28
Oct15-11, 09:48 PM
P: 36
That would be the potential with respect to ground.
gneill
#29
Oct15-11, 09:53 PM
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Quote Quote by canadiansmith View Post
That would be the potential with respect to ground.
What would be the potential with respect to ground? All potentials are with respect to ground. The highest potential source (with respect to ground) shown in the circuit is +5V.
canadiansmith
#30
Oct15-11, 09:56 PM
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Quote Quote by gneill View Post
What would be the potential with respect to ground? All potentials are with respect to ground. The highest potential source (with respect to ground) shown in the circuit is +5V.
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
canadiansmith
#31
Oct15-11, 10:02 PM
P: 36
Is this wrong? If so please help me. I have been working on this all day.
gneill
#32
Oct15-11, 11:12 PM
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Quote Quote by canadiansmith View Post
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 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.
canadiansmith
#33
Oct16-11, 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
gneill
#34
Oct16-11, 02:13 PM
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Quote Quote by canadiansmith View Post
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
#35
Oct16-11, 02:15 PM
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Quote Quote by gneill View Post
Show your calculation for Va if D2 is not conducting.
Exactly what I was going to say
canadiansmith
#36
Oct16-11, 02:34 PM
P: 36
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?


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