If I have two different DC voltage sources, do they simply add up if I connect them?

Question

This question specifically relates to guitar amplifiers. I have one source of voltage coming in from the guitar input, and another from a wall socket. The purpose of the wall socket is to provide power to the vacuum tubes which amplify the signal from the guitar input, but several connections are made between the power supply and the guitar signal. A good example of this is seen in this schematic (pdf) of a Fender Princeton 5F2.

http://www.schematicheaven.com/fenderamps/princeton_5f2_schem.pdf

As you can see in the schematic, the guitar signal comes in at the left, and moves right, and the power supply comes in at the bottom, and moves up. There are two connections made that come directly from the power supply, via two 100k resistors (underneath and slightly to the left of the bold "12AX7", and the other long vertical line in the center-right of the diagram). According to this, they each supply 150 volts, but the voltage of the incoming guitar signal is unknown because it varies proportionally with the volume level of the guitar. My question is do they add volts to the signal? Or do they keep it at a constant 150 volts?

If anyone has audio DIY experience and would like to exchange ideas please email me ([e-mail])

Answer 1

I think you have the wrong concept of vacuum tubes as amplifiers, or any other circuit.

A vacuum tube requires a high voltage to be applied to the "plate", to cause the electrons generated at the "cathode" to be attracted to it.

In the amplifier circuit shown, the signal from the guitar is led to the "GRID", which is exactly what it is, a wire grid placed in the above electron flow.

Because the grid voltage varies with the guitar signal, the grid either allows more or less electrons to flow to the plate.

In the high- voltage line from the power supply to the plate, is a resistor. This resistor has a constant (150) volts applied to it, the other side connects to the plate.

Because the grid is varying the electron flow to the plate, an amplified reconstruction of the guitar signal causes the voltage at the plate and the top of the resistor to rise and fall from the approx. 150V level.

The amplified signal is picked from the plate, therefore. But it is "de- coupled" from the power supply voltage. The guitar signal has simply caused a greater voltage difference in respect of the 150 volts. There is NO direct connection with the signal from the guitar and the power supply. It cannot be "added" or "subtracted".

Answer 2

The large dc voltage is needed for the tubes to operate. The guitar signal is used to control the grids of the tubes so as to cause the voltage at the plate to vary as an amplified version of the guitar signal. But the plate signal also has a large DC offset so that although it varies, it is always positive. The capacitors that couple the plate of one stage to the grid of the next block the high dc voltage so the next stage will operate properly. The output transformer also blocks the DC offset and transforms the high voltage low current signal at the tube plate to the low voltage high current needed by the loudspeaker.

Answer 3

in words of idealised circuit elements it is an identical as divide by using 0. The nodes both component of the batteries are at 20V and 15V jointly, which isn't accessible. authentic batteries are modelled by using an appropriate voltage source plus a chain resistance (spoke of as the inner resistance). for that reason even as 2 battery fashions are correct jointly the voltage drop between the appropriate sources is shared throughout both inner resistances as a accessible divider. ordinarily this is a foul theory to connect any batteries in parallel except for ones that are nominally same.

Answer 4

the voltage input is constant the variable is the resistors some of which are control knobs the diagram shows voltage transformer is first on both circuits

Answer 5

The signal from the guitar pick-up is not d.c. It is .a.c., and it modulates the anode voltage of the first tube.