# Ground

Discussion in 'General Electronics Chat' started by Voltboy, Aug 25, 2007.

1. ### Voltboy Thread Starter Active Member

Jan 10, 2007
197
0
Hello..
I'm trying to read a circuit and build it.. There is a +5v,a -5v a +6v and -6v and one ground. The +5v is connected to the cathode of a 5v battery and the -5v is connected to the anode of the battery, right? So the +6v and -6v are connected same but on a 6v battery. But now the ground, where is it connected?

2. ### hgmjr Moderator

Jan 28, 2005
9,030
214
It would be helpful if you could post the schematic.

hgmjr

3. ### beenthere Retired Moderator

Apr 20, 2004
15,815
283
There have to be more batteries. One 6 volt battery has one output voltage. With a pair of them, connected so one's anode is to the other's cathode and measuring from that common point, there is a + 6 volt and a - 6 volts. The cathode is the negative voltage, by the way. One might be able to run a voltage regulator off the +/- 6 volts to produce +/- 5 volts.

4. ### Voltboy Thread Starter Active Member

Jan 10, 2007
197
0
The schematic is in a book so i cant post it but well the +/-6v isn't really part of the +/-5v because the 5v circuit is from the digital circuit and then it activates the 6v circuit to power up a motor but the ground (my bad on the first question).
To activate the circuit there is a IR phototransistor that when there is IR light it let current pass to the pin 4 of an LM339 (the - input) and to the pin 5 is connected a potentiometer that comes from a +5vdc to the ground so in this circuit to the output be the Vcc the voltage from the IR phototransitor must be bigger than the one from the potentiomenter, right?
Sorry I cant get a schematic.

5. ### hgmjr Moderator

Jan 28, 2005
9,030
214
If you have access to a scanner, you could sketch the circuitry that you are trying to figure out and post it here.

hgmjr

6. ### Voltboy Thread Starter Active Member

Jan 10, 2007
197
0
I made it in Eagle and some paint.. the only thing is the that the pins 4 and 5 go
the other way, the 4 goes on the 5 and the 5 on the 4, i didnt know how to rotate it, srry.
And R2 is a potentiometer

Yoda

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7. ### Distort10n Active Member

Dec 25, 2006
429
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Are the anodes of the battery the common of the circuit?

8. ### Voltboy Thread Starter Active Member

Jan 10, 2007
197
0
I dont know, this is how the circuit is in the book, i just copy it using eagle.

9. ### techroomt Senior Member

May 19, 2004
198
1
with the four power sources you mention, a battery will be required for each. a 5v battery which has it's cathode (-) conected to the ground will have +5v potential at it's anode (+) with respect to ground. a 5v battery with it's anode (+) connected to ground will have -5v potential at it's cathode (-) with respect to ground, etc..

10. ### mOOse Member

Aug 22, 2007
20
0
Couldn't you use a 12V battery and a voltage divider with two
equal resistors? The point between the resistors is the common.
From the anode (+) to common is +6V and
from the cathode (-) to common is -6V.
I suppose exactly how the divider resistors are loaded would be
rather important here, or you won't get the voltage you expect.
But if the loads were constant it might work.

11. ### RiJoRI Well-Known Member

Aug 15, 2007
536
26
The schematic is that of a comparator.
So the pot will set the sensitivity of the circuit, and when it is exceeded, the OpAmp will change state. Note that in the circuit you show there is only +5V and ground, no -5V (in reference to ground).

--Rich