Understanding How a Tube Works
by Tom Taylor

In an effort to help our new members understand the action of a tube we would like to offer a mechanical perspective of its action.   If we think of the VOLTAGE across a resistor increasing and decreasing in accordance with the CURRENT changes through it (using the equation V = I x R where I is the current and R is a fixed resistor value) we see that V goes up and down proportionally as  I  goes up and down.   In a similar fashion the length of a spring increases and decreases in accordance with forces placed on it.   Now let us connect together three springs placing a longer spring at the top in series with a second but controllable center spring in series with a third shorter spring at the bottom.   Can you see how length changes in the center spring inversely affect both the top and bottom springs.   As the center spring collapses the top and bottom springs expand and visa versa.   You squeeze the balloon in the center and the other two areas get larger.   Now let us look at that center spring as a triode tube where the voltage across it, cathode to plate, is controlled by its control grid voltage AND each of the other two springs long and short are instead large and small value resistors differing in value by a factor of say 100, ie.  50,000 and 500 ohms.    Let’s call the large value resistor a plate load placed between the plate and the most positive supply voltage and the small value one a cathode resistor tied to ground.   In the tube perspective, a change of control grid voltage affects its ability to flow current in an inverse manner.   An increase in grid voltage increases tube current conduction thereby increasing the voltage across each of the other two resistors (V= I x R again) WHICH in turn decreases the voltage across the tube.   The mechanical perspective of that is our manipulation of the center spring resulting in opposite length changes on the other two springs.   A compression of the center spring stretches the other two springs and visa versa.   Knowing there is equal current through a series string of electrical components there will most certainly be a larger voltage swing across the 50,000 ohm plate load resistor than there will be across the 500 ohm cathode resistor, V = I x R again.  If R is bigger then the voltage V across it is bigger.   Going just a step further, IF we take steps to prevent voltage change across the bottom resistor (length changes across the bottom spring) we can get a small increase in change of voltage across the top resistor for the same grid voltage change (same as a small increase in length change of the top spring).   Watch one end of the balloon get still bigger if the size of the other end is maintained while the center is squeezed as before.  THAT is accomplished in the electrical perspective by putting a capacitor across the cathode resistor.   Capacitors hold a charge and therefore resist changes in voltage across themselves.   For a change in grid voltage provided to the tube we will find the largest change in voltage at the tube output or right at its plate and its value is influenced by the tube’s ability to provide current, the value of the plate load resistor and the control grid voltage.   In our radio circuits that plate load may be a transformer of some kind the impedance (similar to but not the same as resistance) of which is largely determined by the frequency in use according to Z = 2 x Pi x F x L.