There is no formula for calculating the required primary Z of the OPT from the operating point. The way it usually is done is by drawing this point on the anode curves, and choosing a load line (the slope of which is the primary impedance) such that a reasonable tradeoff is made between distortion and power output.
Having a look at the "Of loadlines, power output and distortion" series on Steve Bench's website should make things clearer.
Having a look at the "Of loadlines, power output and distortion" series on Steve Bench's website should make things clearer.
Additionally, this is also excellent reading as the example directly relates to the 211.
Also go the the WE site and get the 300B datasheet and look at the loadlines and distortion (2H and 3H) for differing loads.
Tubes, especially triodes are really tolerant of primary impedance variations, otherwise how would they deal with the impeadance variations presented by most speakers?
Rule of thumb = 3x Rp.
Also go the the WE site and get the 300B datasheet and look at the loadlines and distortion (2H and 3H) for differing loads.
Tubes, especially triodes are really tolerant of primary impedance variations, otherwise how would they deal with the impeadance variations presented by most speakers?
Rule of thumb = 3x Rp.
For a single-ended pentode, assuming the bias is good, V/I (plate voltage over plate current) will get you going. This will be near maximum power, but may be far from minimum distortion.
This also works in triodes IF the plate resistance is VERY much less than load resistance.
But triodes never have plate resistance as low as we would like.
And many of the transmitter triodes are optimized for best overall efficiency, which points to low current (small heater power) with HIGH voltages. Some of these just won't make much power unless you drive the grids positive. And in audio (unlike radio), grid current is often more trouble than it is worth.
valve is 211 GE
plate volts = 950
grid = -45
bias = 75ma
Always look at the tube-maker's data-chart. Some bow-tied guy slaved over a slide-rule, French-curve, and manual typewriter for hours, all for your benefit. Do not ignore the wisdom of the ancients.
http://www.triodeel.com/images/211.pdf
Your operating point is similar to RCA's 1000V 53mA condition, except 0.95 times the voltage and 1.4 times the current. RCA liked 7,600 ohms for 1,000V 53mA; multiply 7,600 times 0.95 then divide by 1.4 to get a better match for your condition. 5,157 ohms, and triode loading is all +/-50% work, so use 5K to 8K.
I suspect that the generous 75mA is too much current for this tube at this supply voltage, for best power. However, it errs in the direction of low distortion, which may be more important than raw power.
Why do people use the (IMHO) wretched 211? It is like so dead below 500V (unless driven into grid current), and can't be worked above 1,250V. Lots of heat, not much noise.
> what is the formula please.
If tubes had no maximum rating, and distortion did not matter: triode Rp*2 gives best power.
When you have to stay within ratings, and meet a distortion goal, there is no formula. Several books work through a whole chapter of derivations, and finally prove that you must work with plate curves to make the most of a tube.
This also works in triodes IF the plate resistance is VERY much less than load resistance.
But triodes never have plate resistance as low as we would like.
And many of the transmitter triodes are optimized for best overall efficiency, which points to low current (small heater power) with HIGH voltages. Some of these just won't make much power unless you drive the grids positive. And in audio (unlike radio), grid current is often more trouble than it is worth.
valve is 211 GE
plate volts = 950
grid = -45
bias = 75ma
Always look at the tube-maker's data-chart. Some bow-tied guy slaved over a slide-rule, French-curve, and manual typewriter for hours, all for your benefit. Do not ignore the wisdom of the ancients.
http://www.triodeel.com/images/211.pdf
Your operating point is similar to RCA's 1000V 53mA condition, except 0.95 times the voltage and 1.4 times the current. RCA liked 7,600 ohms for 1,000V 53mA; multiply 7,600 times 0.95 then divide by 1.4 to get a better match for your condition. 5,157 ohms, and triode loading is all +/-50% work, so use 5K to 8K.
I suspect that the generous 75mA is too much current for this tube at this supply voltage, for best power. However, it errs in the direction of low distortion, which may be more important than raw power.
Why do people use the (IMHO) wretched 211? It is like so dead below 500V (unless driven into grid current), and can't be worked above 1,250V. Lots of heat, not much noise.
> what is the formula please.
If tubes had no maximum rating, and distortion did not matter: triode Rp*2 gives best power.
When you have to stay within ratings, and meet a distortion goal, there is no formula. Several books work through a whole chapter of derivations, and finally prove that you must work with plate curves to make the most of a tube.
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