Negative voltage

Thread Starter

killerfish

Joined Feb 27, 2009
29
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
Example:

Take two 12-volt batteries, connect in series - negative terminal of one battery to the positive terminal of the other battery. Now, for purposes of reference and this explaination, drive a ground rod into the ground and connect the "tied end" of the batteries (negative-to-positive connection you made earlier) to the ground rod. The ground rod is the reference or common connection for our 12-volt battery arrangement.

So, if you measure between the ground rod and negative terminal of one battery (black lead to ground rod & red lead to negative terminal) you will read "- 12 volts". The voltage at this point is more negative with respect to ground. Conversely, you could say that ground is more positive with respect to that batteries negative terminal.

Electrons always flow from negative to positive. The positive-most point in any circuit represents a depletion of electrons and the negative-most point represents a surplus of electrons. The surplus (-) condition is always trying to satisfy or equilize the depleted (+) condition. Thus, electron current flow.

"Conventional current flow" depicts current flowing from positive to negative as a matter of conveinience. All is good and well as long as you realize that it is actually the movement of electrons doing the work.

"when and why you must use a negative voltage?."

The reasons are as varied as the circuits themselves. One common application of a negative voltage is for biasing circuits. A negative voltage may be used to turn a device on whereas a positive voltage may be used to turn it off and visc-versa.
Thanks.
 
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Thread Starter

killerfish

Joined Feb 27, 2009
29
fial.JPG

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
 

Jony130

Joined Feb 17, 2009
5,487
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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Thread Starter

killerfish

Joined Feb 27, 2009
29
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?

fial.JPG
 

retched

Joined Dec 5, 2009
5,207
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]
 
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