I think the tube is not actually biased nearly off. What happens instead is that the cathode is trying to faithfully follow the grid but the load R meter is quite high and slugged by a large C. This means that the tube can easily lift the cathode voltage but is incapable of making it go lower, it just slows decays due to large RC. This asymmetrical follower action is what makes it a rectifier. Of course there is a standing bias offset which is what the meter zero adjustment removes.
Clearly I don't have the understanding to know if it is biased off, but that 100M ohm on the grid is a pretty good size resistance. If you're wrong, I'm sure someone will tell you!
That circuit does "grid leak bias", where the hot electrons from the cathode make a small negative grid voltage. That pulls the grid a bit negative through the 100M resistor, which partially turns the tube off. You can easily measure the quiescent cathode voltage, check the tube curves, and estimate the grid-cathode voltage. Or briefly short the 100M resistor abn see how the cathode voltage changes.
With high AC voltage input, the tube will act like a peak rectifier. At low voltages, it will act like a square-law detector. The meter scale is nonlinear to adjust for the detector behavior.
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It isn't even marked below Q=40, 10 on the low scale.
Old-timey wound L's had high Qs. That Boonton wouldn't measure the Q of many modern surface-mount parts.
--
John Larkin Highland Technology, Inc
lunatic fringe electronics
Yeah, the 100Meg R minimises loading on the tuned circuit, the tube is not afraid of such high impedances! With no AC signal the grid sits at close to 0V and the cathode will bias itself up just enough. That positive offset with zero signal is what the zero meter adjustment then cancels. When a whiff of AC comes along the cathode is able to follow the positive peaks but cannot fall back in time to reproduce the troughs
- the follower is "crippled" if you prefer.
I read somewhere that infinite impedence detectors used as AM demodulators were prized for their low audio distortion compared to more basic diode envelope detectors. Was that due to their lower loading on the tuned circuit or better linearity at low signal levels?
Everyone loves telling me when I'm wrong but hardly any when I not wrong ;)
Looking at the schematic I see no DC blocking capacitor so I guess sometimes the grid is grounded through the test inductor to the LO or GND terminals. Does that change anything ?
Your point of non-linearity may be pertinent to my quest. You may have read this previously, but it may be answerable now.
My plan was to double the FS voltage of the meter, in my quest to measure Q's over 625. the limit of the Boonton 260A. I'm hoping to measure up to 1250.
Using a series resistor and a adjustable power supply I set the needle at 250 on the Q meter scale. I found adding a parallel 1,210 ohm resistor the needle dropped to 125. or 1/2. But, now you have pointed out the meter scale is not linear, so 1,210 ohms may not have been the real internal resistance of the meter. 125 is not in the middle of the scale.
But, to continue,
The meter has an internal resistance of 1,210 ohms. (now questionable) I paralleled it with a 1,210 ohm resistor and then added a series 605 ohm resistor to keep the same resistance in the circuit hopefully keeping the circuit aligned. This messed up the linearity when adjusting the injection voltage. An injection voltage of 13.33mv gives a Q of 325. An injection voltage of 8mv gives a Q of 388. Without the meter modification, the Q's read the same. Seems odd. Removing the series resistor didn't solve the problem. I did this with 4-9" wires run out to a switch, so I could have a normal and a times two setting. I don't see how the wires could affect linearity with a dc circuit. Maybe the filtering is not as good as I thought.
I don't know, using the wrong internal resistance would have changed my multiplier above or below 2X, but I don't see how it could affect linearity.
More thoughts, I'm not sure about the square law region, Because, you can set the injection voltage to 20mv, 10mv, and 8mv. At any of these millivolt settings you use part or all of the Q meter scale. So, I don't see how you could use a single scale to read correctly with multiple injection voltages with square law linearity in the mix.
FWIW, tubes really do /produce/ power. Normally, an open circuit grid will float around -2 to -3V, held there by a balance of leakage currents on the order of low ~uA (giving a DC input resistance in the gigs).
Here's a model of the 6AL5:
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The BSRC models the Child-Langmuir law. The exponent is closer to 1.3 than the theoretical 1.5, probably because theory is for plane-parallel geometry, not cylindrical.
The diode and resistor model the statistical (exponential) cutoff region, and leakage between electrodes.
The voltage source corresponds to the high-energy tail of cathode emission: emission physically "boils" off electrons, with some excess kinetic energy on the order of 1eV. (By coincidence, the high-energy tail of the optical emission spectrum is ~1.5eV. That is, the cathode glows red. ;-) )
The peak power point of this model is a few microwatts, not a bad yield for a 6.3V 0.3A heater. :-p
(Okay, so this is for a diode, but the grid leakage of any tube will be extremely similar -- it's the closest electrode to the cathode, and is itself a damn fine plate, until it melts!)
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