Updated: Mar 29, 2020
Part of a blog series Tube Amplifier Circuits Explained
From the last few posts, you have the basics. Now it’s time to broaden our circuit a bit more and deal with a few more issues. The triode tube we have seen so far is used as a driver stage in our amplifier with still a relatively low level of current and output voltage. We will need to put it through a power amplification stage to really get it to the level of current and amplification needed to drive our speakers and move some air on the journey of sound back to our ears!
We will take the amplified voltage from the anode, which you already know operates around 175V DC, fluctuating up and down with the audio signal. We can consider this an AC signal riding on top of the operating point DC voltage. We will want to feed this signal into another tube—in our case an EL34 power tube—for further amplification, but we can’t put 175V onto the grid. This high voltage will cause all sorts of problems on the grid which is intended to be negative, likely destroying it. We want only the AC portion.
The coupling capacitor is our solution. As you know, capacitors block DC and allow AC. The capacitance value can be small. We simply want to block the DC and allow all audio frequencies to pass. In our case, we will use 0.22uF as a commonly used and available value.
The coupling capacitor is directly in the signal path and so it is a very important component, and we want to use something high quality that will pass the signal without noise or distortion. Some type of film capacitor is best, and there are very high-end audiophile grade coupling capacitors produced, some at outrageous prices. I often use a Solen brand that I believe is high quality, but reasonable in cost. Note that the voltage rating of this capacitor needs to be high enough to handle the entire B+ because the tube takes time to warm up before any current flows, so the full voltage will appear on the anode, at least for a few seconds at startup.
We will get into the output stage later, but we need to go back now and consider a few things with our driver stage. Our load line was fine for defining an operating point when there is DC voltage without audio (AC) on the grid, but since we are allowing AC voltage to pass on to the next stage through the coupling capacitor, we need another load line to understand the load under operating conditions: an AC load line.
The AC load will be the combination of two things: the anode load resistance we have already been dealing with, but also the impedance of the next stage, in this case a path to ground through the next stage’s grid leak resistor. Do you see how Rg2 is in parallel with our driver tube (with regard to AC voltage, not DC)? Let’s pretend that Rg2 is 220k Ohms. The formula to solve for parallel resistance is:
R = (Ra X Rg2) / (Ra + Rg2)
Solving this with 50,000 and 220,000 as our Ra and Rg2 values tells us that the AC resistance would be about 40.7k Ohms. Now, let’s find our new AC load line. The operating point will stay the same, and we can choose to plot some other point to get our line. Let’s pretend under AC voltage the anode drops by 100 volts. Ohms law will tell us that if our AC resistance is 40.7k Ohms, then the current difference for this voltage drop is 2.46mA. So we can plot a point that is 100V lower than our operating point, and 2.46mA higher in current. Now this is what our true working load line will be.
Our circuit is coming together! We will move on to the power stage soon, but first, one more problem to deal with. Remember how we wanted our cathode to be +2V relative to the grid. But think for a moment what is happening as the grid rises and falls with an audio signal—use a sine wave as an example and think about the voltage rising and falling over time. When the input signal is +1V, then the grid becomes less negative with respect to the cathode, and more current is allowed to flow. When more current flows, what will happen across our cathode resistor? Dang it. Voltage on the cathode will rise, right? Ohm’s law again. On our AC load line, if the grid rises to -1V, then current is somewhere around 5.75mA. Remember how we solved to find our cathode resistor value and came to 444 ohms so our cathode would be at 2V. Well, 5.75mA across 444 ohms using Ohm’s law is about 2.5V, instead of our intended 2V. The opposite is true when the signal goes the other direction: move down the AC load line, less current flows, which reduces the voltage across the cathode resistor. So the cathode voltage is fluctuating along with the input signal to some degree. Not what we want, and the net effect is that it counterbalances the amplification of the tube, reducing gain.
In some circuit designs, this can be used to an intended advantage (referred to as cathode degeneration), but in our design, we don’t want this. Our solution is a cathode bypass capacitor. By putting a capacitor alongside the cathode resistor, we can hold the cathode at a desired DC voltage, but allow AC to pass through the capacitor. As the grid changes in voltage at audio frequencies, and current rises or falls accordingly, the AC currents will no longer impact the cathode voltage, it will remain at the intended 2V DC while the AC current can bypass it.
We will use an electrolytic capacitor because capacitance needs to be relatively high to allow current from low audio frequencies to pass. I am using 100uF in this particular circuit giving a cutoff somewhere around 3-4Hz. Lower cutoff might be better in some cases, so you could increase this to some higher value like 220uF or more, but our final frequency response measures only 1dB down at 20Hz, which is still very good. The voltage rating needed is low given the expected voltage on the cathode. In this kit, I oversized the voltages a bit to use components that are a little bit larger and easier to work with physically (25V for input stage, and 100V for the power stage).
We will talk next about the output stage and pentode tubes.