# Negative voltage

Discussion in 'Homework Help' started by killerfish, Feb 14, 2010.

1. ### killerfish Thread Starter Member

Feb 27, 2009
17
4
Hi,

i understand there is a similar thread was discussed very long time ago but it was locked. I'm still unclear to this qns. If a resistor is connected between negative voltage and ground, the current flow from negative to ground(0v)? since i think ground is a reference between positive and negative voltage, so it is the lower potential???

Quote from n9xv
Thanks.

Last edited: Feb 14, 2010

Feb 17, 2009
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3. ### killerfish Thread Starter Member

Feb 27, 2009
17
4

I attached a image of my problem. I do not understand why Diode 1 on part a is not conducting(the ans) whereas the diode 1 on the part is conducting, why is that so? On the 2 problems, there is a change of source of voltage, but the current flow from aren't the same for both, from the high potential to lower?

the ans:
(a) D1: -2.13V, 0A
D2: 0.75V, 0.808mA
(b) D1: 0.75V, 0.283mA
D2: 0.75V, 0.667mA

4. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,991
1,115
To determine whether D1 conducts or nor we need to do some calculations.
First we determine the voltage on VA without diode D1.
I = ( E1 - VD2 -E2 )/ ( R1 + R2) = (10V- 0.75V-(-15V) )/30K=24.25V/30K=808uA
So VA=-15V + 0.75V + I*R2=-15V+0.75V+12.12V=-2.13V
So D1 can not conducts.

If we move the GND the VA=12.87V

We can do similarly to example 2.

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5. ### killerfish Thread Starter Member

Feb 27, 2009
17
4
Thanks Jony130 it very clear to me now... especially with illustrations

I recap again. If both diode assumed to be conducting(refer to image), i could find the currents, and realise that the extra current from the ground to make up the 1mA is impossible because D1 is reversed bias. So the assumption is wrong. am i right?

6. ### retched AAC Fanatic!

Dec 5, 2009
5,201
313
As long as the current coming from ground (like in a short circuit or ground fault) is lower than the breakdown voltage of the diode, it will not pass.

[ed]
D1 is not reversed biased in the diagram.

If the (+) and ground pin were switched, it would be reverse-biased. In the case you have it, Any back-flowing current will stop at D1. At least until the avalanche or breakdown voltage is reached.
[/ed]

Last edited: Feb 14, 2010