# Ground

Thread Starter

#### Voltboy

Joined Jan 10, 2007
197
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?

#### hgmjr

Joined Jan 28, 2005
9,029
It would be helpful if you could post the schematic.

hgmjr

#### beenthere

Joined Apr 20, 2004
15,819
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.

Thread Starter

#### Voltboy

Joined Jan 10, 2007
197
It would be helpful if you could post the schematic.

hgmjr
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.

#### hgmjr

Joined Jan 28, 2005
9,029
If you have access to a scanner, you could sketch the circuitry that you are trying to figure out and post it here.

hgmjr

Thread Starter

#### Voltboy

Joined Jan 10, 2007
197
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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#### Distort10n

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

Thread Starter

#### Voltboy

Joined Jan 10, 2007
197
Are the anodes of the battery the common of the circuit?
I dont know, this is how the circuit is in the book, i just copy it using eagle.

#### techroomt

Joined May 19, 2004
198
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..

#### mOOse

Joined Aug 22, 2007
20
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.

#### RiJoRI

Joined Aug 15, 2007
536
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
The schematic is that of a comparator.
One application is called the comparator. For all practical purposes, we can say that the output of an op-amp will be saturated fully positive if the (+) input is more positive than the (-) input, and saturated fully negative if the (+) input is less positive than the (-) input. In other words, an op-amp's extremely high voltage gain makes it useful as a device to compare two voltages and change output voltage states when one input exceeds the other in magnitude.
from http://www.allaboutcircuits.com/vol_3/chpt_8/3.html
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

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