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VALVE Magazine
VALVE Magazine Online!
Triode Amplifier Operating Points
Article By Paul Joppa

From VALVE Volume 2 Number 5 May 1995

 

  How do you choose a good operating point (supply voltage, current, and load impedance) for a vacuum tube? For ordinary tubes, used in an ordinary way, it's simple -- you just look it up in your RCA handbook. But suppose you want to use an unusual tube (e.g. the 6CK4 television vertical amplifier)? Or suppose you just want to try breaking the rules?

I struggled with this problem recently, and I've concluded that then really is a single "reference" operating point for any given (triode) tube, and that  variations around that point have similar effects for all triodes. I've only studied single-ended Class A triode circuits with zero grid current (actually Class A1), but that covers a lot of audio circuits of interest. Mostly, I've looked for conditions that provide the maximum power output. Even for preamps, I think that's a good tradeoff operating condition. Yes, you can get a few more volts out of a tube, but only at the price of a much higher output impedance. Most of the numerical results are based on a "perfect" triode, following the 312 power law.

Figure 1 shows a typical power triode load line, drawn on the plate curves -- in this case, a 2A3 operating at the handbook recommended conditions of 250voIts, 60 mA, and 2500 ohms. Here's what happens if you vary these conditions:

If you raise the impedance, the load line becomes more horizontal, so the output voltage increases while the current decreases. The power will go down, though slowly at first, and the distortion will also go down. The distortion reduction is caused by the load line end point moving up out of the non-linear region (low current region). If you reduce the load impedance, the available input voltage swing will be limited by cutoff so the power will go down again, this time with increasing distortion. Figure 2 shows a typical result. Notice especially the relatively wide range of load impedances over which substantial power is available - a good thing, considering how much speaker impedance can vary with frequency! Powertubes are usually run at about twice the maximum-power impedance to reduce distortion without losing too much power.

If you increase the supply voltage, you will have to decrease the current to keep within the plate dissipation limit. The load impedance for maximum power will then increase rapidly, and also the range of impedances for full power will get narrower. If the tube remains linear at high voltages and low currents, the distortion in this case will be less and the efficiency and power output will increase.

Sounds like a winner, right? Unfortunately, two things limit how far you can go with this. You can't safely exceed the maximum rated plate voltage. And most tubes are not linear at high voltages and low currents; the plate curves start to bunch up and flatten out, causing more distortion. (Incidentally, this distortion is mostly second-harmonic, which is cancelled in push-pull operation. This is why P-P amps are often run at higher voltages than SE, and why they often have higher efficiency. Eventually this will lead to Class AB and Class B operation.)

If you reduce the supply voltage, you must also lower the current to keep the operating point away from the start of grid current. The power output drops quickly, although the efficiency remains the same. The load impedance will increase, but not by a lot. This often happens with voltage amplifiers, where a low supply voltage requires less current - and higher circuit impedances -- than the tube is otherwise capable of.

Figure 3 shows how load impedance, current and power output vary with supply voltage, relative to the reference operating point. I have shown a range of impedances, from maximum power to 1% distortion (for perfect triodes, of course). So, how do you find the reference operating point? I have defined it as the minimum supply voltage at which the maximum power operating point uses the maximum plate dissipation. Based on my computer models of a perfect triode following a 3/2 power law, this is how it works out: If you know rp (plate resistance) at current I, and choose a dissipation power
P, then:

At this operating point, the maximum power output Pref is about 23% of the plate power P.

As you can see, the 2A3 handbook conditions are about right. You can also see why the 211 really wants to see 1200 volts Notice the high plate voltage on typical voltage amplifier triodes. Remember, this is the plate voltage -- for RC coupled amplifiers, the plate resistor will drop nearly as much voltage as the tube, so the supply voltage must be nearly twice this much. Few of us want to run our 12AX7's on 1000 volts You can either find a low-voltage tube like the 6DJ8/6922, or operate at lower power (a 12AX7 triode at 200 Volts and 0.5 mA, for example), or give up some maximum output capability by running a smaller plate resistor.

I hope you have as much fun with these formulas and curves as I have!

 

 

 

 

 

 

 

 

 

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