
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!