Using 3A5 Tubes?

[Dr. Toobz]

Every once in a while, I see mention of 3A5 dual-triode battery tubes in a DIY project (or occasionally, an all-DHT amp by a Chinese manufacturer). I'm sort of curious to see what the Quickie would sound like with these tubes used in place of the triode-wired 3S4 beam tubes used in the stock circuit. Being an electrical novice, I was wondering if a more hardcore DIY'er might chime in re: circuit changes?

Here's the tube's numbers: max. plate dissipation = .5W, plate resistance = 8.3k, transconductance = 1800 micromhos, amplification factor is 15. There is a common cathode, so it looks like I would either use one half per channel, or parallel the halves.

As somebody with only a bare minimum of electrical knowledge, I'm not sure I understand how to read plate curves, or draw a load line. The curves for this tube show an intersection of 0V on the grid with 36V plate voltage at 2mA plate current. 36V and 2mA are the same characteristics as the Quickie. Does this mean that 36V on the plate, with the grid held at 0V (yet still negative in relation to the cathode), yields a plate current of 2mA? Where does the whole "drawing a load line" come into the picture?

Also, how to calculate the cathode bias and resistor value (esp. in this case, since the cathode is the filament, which already has 1.5VDC on it), how to hold the grid negative in relation to the cathode, and how to calculate the proper anode load is where I start to get confused. For example, 36V/2mA would give an 18k plate resistor value. Yet, the stock Quickie uses something closer to 4k. Also, the Quickie uses a cathode resistor of 1k, which must develop a voltage across it, since it's referenced to ground. Yet, there's the battery voltage already on the cathode...so does the bias equal the battery voltage plus whatever is developed across the cathode resistor?

DHT's appear to be more confusing to think about when one is trying to understand the basic operation of a triode! Any help in understanding some of this basic theory would be appreciated.

[Paul Joppa]

I don't think I can squeeze a whole lesson on load lines into a short post, but as long as you have the curves in front of you ...

Let's assume you have 35 volts available, just to make the arithmetic easier because we'll need to divide by 7. 3/7 of 35v is 15v, so identify the zero-bias point at 15v - it's just about 0.4mA. Now the zero-bias line is approximately a 3/2 power curve, so the plate resistance at that point, which is the slope of the zero-bias line, is 2/3 of 15v/0.4mA, or 25K ohms. A good load line is twice the plate resistance, which would be 50K ohms. The 50K load line goes from the point you just found, down and to the left arriving at zero current and 35 volts. The symmetrical operating point would be halfway along that line, 0.2mA at 25v. As it happens, this load line gives the maximum power output that you can get from this tube with 35v available. Notice the bias is -1v, so the bias resistor is 1v/0.4mA or 2500 ohms. (Also notice that the bias plus the plate to cathode voltage equals exactly 36 volts - pretty clever, isn't it?)

This tube will run at 1/10 the current of the stock Quickie, with just about 10 times the output impedance, so it won't drive much cable. But it should work.

[Dr. Toobz]

Thank you for making this a bit more comprehensible to us non-engineering folk!

That said, I can see why you guys didn't bother with this tube to begin with. It looks like it would be much happier with a higher plate voltage than the arrangement in the Quickie.