How can you have a negative voltage? I don't really understand the concept of negative voltage, how can it exist?
Voltage is a difference of potential energy for electric charges, and potentials are defined from forces, so that $F=-\nabla V$, where $V$ is the potential and $F$ is the force. When you have determined the potential $V$ you can now add any constant you want, or any function $f$ that doesn't depend on coordinates, because $-\nabla V=-\nabla (V+f)$, as $f$ doesn't depend on coordinates. This last thing is what's known as setting an origin of potential energy, something you sure have heard about. Well, in that function $f$ where you set your origin for potentials, you can have negative voltages between two points.
Another way to see the voltage is the work you must do per unit charge to move that charge from one point to another, even here, when we're dealing with differences, you can also have negative voltages.
If you're talking about circuits, all of the above applies, you can set a potential difference $V$ by, let's say, a battery, then you can use some device so that the potential energy is even lower than the one set by the - sign of the battery, as you set the potential origin 0 for that battery, then that point will have a negative voltage.
The simplest thing to do here is remember that only diferences in potential matter, and that means that we can add or subtract a constant from every voltage in a system without changing it's behavior, so we can render any negative number positive to make you feel better or just as easily render all the positive ones negative.
In short, it's just a number and you shouldn't fret about it having a sign.
Let's say you want to know the electric potential difference r meters away from an electron sitting in a void of space.
The electric potential difference describes the difference in potential energy of a unit positive charge from one point in space to another and the work done on a unit positive charge to carry it from the same point to the other.
This gives the equation $V_a-V_b=(k Q/r_a)-(k Q/r_b)$, where Va is the final potential and Vb is the initial. In this scenario, it is conventional to make $V_b$ the zero reference point at $r$ equals infinity, which makes $r_b$ equal infinity and $V_b$ equal zero. Therefore, the equation becomes $V_a=kQ/r_a$.
Because we are looking at the potential difference of an electron, Q equals negative e, where e equals the elementary charge. So, the equation, in this case, is $V=k(-e/r)$, which gives a negative value.
We see here that no matter how far away the positive test charge is from the electron, the electric potential difference will always be negative. This makes sense because a positive point charge at $r = \infty$ loses potential energy as it is brought toward the electron (downstream), and negative work is done.