This is a nice thread Ed has got going here. It helps clarify many things for me. However, one question I had (probably many other newbies like me also), is how
does a triode amplify a signal? So, I read up a little bit and came up with two ways of explaining the phenomenon – a graphical method using the triode characteristics, and an analytical method based on a equivalent circuit representation of the triode.
In this post, I would like to try and explain the graphical method, which I think can be understood more easily than the equivalent circuit method.
For many of our resident gurus, this is, I am sure old hat, and something they could talk about in their sleep. I hope you will bear with me, and correct any errors or omissions. OK, here goes.
Tube Components
A triode has an anode (positively-charged), a cathode (negatively-charged) and a control grid. When the cathode is heated (either directly or indirectly), electrons emanate from the cathode and travel towards the anode or plate.
The control grid’s function is to somehow regulate this flow of electrons. By charging the control grid to be more negative than the cathode, some of the electrons emanating from the cathode are repelled back to the cathode. Others make it through to the plate (anode).
For the electrons to pass through to the plate, the grid must be more negative than the cathode. If the grid was positive with respect to the cathode, it will attract all the electrons from the cathode, and none will get through to the plate (my interpretation).
The signal to be amplified is impressed on the grid. This signal is a time-varying signal with components at various frequencies. To keep the tube in conduction, we have to ensure that the grid is more negative than the cathode at all times. That is, at the positive-most excursion of the signal to be amplified, the grid must still be more negative than the cathode, for the tube to conduct.
Tube Parameters
The tube amplifies the signal imposed on its grid due to the following properties of the tube:
- Plate Resistance: Denoted by the symbol “Rp”, the plate resistance is defined as the change in plate voltage divided by the corresponding change in plate current, at a given grid voltage. It is important to note that if you change the grid voltage from one value to another, the same change in the voltage at the plate need not produce the same change in plate current. That is, let us say that at a grid voltage of -3V, a change in plate voltage of 10V produces a change in plate current of 1mA. If the grid voltage is changed to -6V, the same 10V change in plate voltage need not produce the same 1mA change in plate current. The tube’s non-linear nature is the reason for this. Thus, the plate resistance is dynamic – keeps changing. To maintain a constant plate resistance at all operating points, the tube must be operated in its linear region, where the tube characteristics are denoted by straight lines parallel to each other.
- Transconductance: Denoted by the symbol “gm”, the transconductance is defined as the change in plate current divided by the corresponding change in the grid voltage, at a given plate voltage.
- Amplification factor: This is the product of Rp and gm. It is usually denoted by the symbol for mu, and is also defined as the change in plate voltage divided by the corresponding change in grid voltage, at a given (constant) plate current.
It is the amplification factor mu that multiplies (amplifies) the voltage (analog signal) impressed on the grid, and makes it available as a change in the voltage at the plate.
Here is a triode, connected to the high-voltage B+ supply through the load resistor RL. The cathode of the triode is grounded. Vout denotes the point where the amplified output is obtained.
Triode Plate Characteristics
Now consider the plate characteristics of a popular triode.
The y-axis shows the current through the plate in mA. The x-axis shows the plate voltage in V. The characteristic curves show the relationship between the plate voltage and the plate current at different values of the grid voltage.
We want the grid to be more negative than the cathode for the tube to conduct. Also, when a signal is imposed on the grid, we want the grid to continue to be more negative than the cathode even at the positive-most excursion of the signal to be amplified.
So, we have to choose a base-line for operation of the triode. That is, when there is no signal present on the grid, what should be the plate voltage and therefore the current through the tube? Any signal now imposed on the grid will cause the operating point of the tube to shift from its base-line value.
So, if we pick the following base-line points:
- Plate voltage = 300V when plate current = 0
- Plate current = 10mA when the plate voltage = 0.
The two points above determine points on the x-axis and y-axis respectively. We can join these two points by a straight-line to get:
The line is also called the load line. The operating point with no-signal on the grid is determined by the choice of the grid voltage. So, if we pick a grid voltage of -6V, the plate voltage is around 160V, and the plate current is around 5mA, and this is the base-line or quiescent operating point of the tube.
It is important to remember that the grid must be negative with respect to the cathode. Let us say that the audio signal imposed on the grid is a 2V (peak-to-peak) sine wave. Thus, the grid voltage will rise to -5V when the input signal is at its maximum, and fall to -7V when the input signal is at its minimum.
Now, we impress on the grid the 2V (peak-to-peak) sine wave.
With no signal, the grid is at -6V. With the passage of time (shown by the arrow), the voltage on the grid rises to -5V, and then drops down to -7V. (The dashed lines enclosing the sine-wave on the grid are perpendicular to the load line.)
The change in plate voltage is obtained by projecting the grid voltage points to the load line, and then further projecting those points to the plate voltage axis (x-axis).
You can see that the voltage at the plate changes between 170V and 145V (approximately). That is, a 2V (peak-peak) input signal produces a 25V (peak-peak) signal on the plate. This gives an amplification factor of 25/2 = 12.5.
As long as the plate is operated in the linear portion of its characteristics, the amplification factor will remain constant.
Further, notice that as the grid signal undergoes a positive excursion (rises to -5V), the plate voltage reduces to its low of 145V. As the grid signal undergoes a negative excursion, the plate voltage rises to its maximum of 170V. That is, the signal has undergone a phase inversion as it gets amplified.
The next step is to analyze the triode using an equivalent circuit, but I will leave that for later, if there is interest.
References
- Basic Electronics and Linear Circuits, N. N. Bhargava, D. C. Kulshreshtha, S. C. Gupta
- Beginner's Guide to Tube Audio Design, Bruce Rozenblit
. It hasn't left the bathroom since I bought it.
