Grid current in pentagrid mixer - Page 1
Forum section: electronics / projects

author: TimFox
author_rank: Super Contributor
timestamp: February 10, 2023, 09:48:48 pm
content: Those not familiar with vacuum tubes need not reply.For those old enough to have dealt with pentagrid mixer tubes, specifically 6SA7/6BE6 and 6SB7/6BA7:In general, frequency mixers have a strong AC voltage applied from the local oscillator, and a weak AC voltage applied to the "RF" port, so that the LO voltage produces "large-signal" or switching behavior, and the RF voltage produces a response proportional to that lower voltage.In the pentagrid mixer/converter tubes mentioned, the LO voltage is applied to grid 1 (either from an external oscillator or from a Hartley oscillator using G1 and the cathode).The signal voltage goes to G3, which does not affect the cathode current strongly, but modulates the plate current (the rest of the cathode current flows to G2 and G4.In the recommended operating conditions of the data sheets, G1 is driven strongly enough to draw grid current, and the bias voltage on that grid results from the grid current through a grid-leak resistor (after a coupling capacitor).For both types, the recommended grid-leak resistor is 20 k ohms. The recommended G1 current is 0.5 mA for the 6SA7 and 0.35 mA for the 6SB7.The data sheets do not specify AC or DC for that current, so I am asking for clarification.I assume they mean the DC component of the G1 current over the LO cycle, and that the DC component of the voltage across the grid leak is therefore -10 V (6SA7) or -7 V (6SB7), and that voltage can be measured by placing a high-value (> 200 k) resistor from the grid to a test point.Assuming the G1 positive excursion is small (forward bias of G-K diode), the -10 V DC component for the 6SA7 therefore implies approximately 20 V pk-pk for the actual voltage at G1 (-7 V implies 14 V pk-pk for the 6SB7).For separately-excited mixers, the grid leak goes from G1 to ground, and the cathode either goes to ground or has a small bias voltage from a cathode resistor.Question: is my understanding of the data-sheet parameters correct?

author: trobbins
author_rank: Frequent Contributor
timestamp: February 10, 2023, 10:59:55 pm
content: The GE datasheet for 6BE6 does say typical test set with 10Vrms for g1, which implies 28Vpp but the drive would likely be clipped as g1 current increases.I've only had some cursory use of 6BE6 in audio circuitry - the below article may add some insight into static conditions over a range where g3 signal is modulated to keep distortion to reasonable levels of audio application.https://dalmura.com.au/static/6BE6audi.pdf

author: TimFox
author_rank: Super Contributor
timestamp: February 11, 2023, 05:34:30 am
content: I'll check that datasheet. Mine is a frequency mixer application for a theremin, rather than an audio gain-control application.

author: floobydust
author_rank: Super Contributor
timestamp: February 11, 2023, 07:30:26 am
content: Snippet of all I could find that made sense:(2) The pentagrid converter tube can be considered as made up of two parts, a triode and a tetrode. In the triode section, K is the cathode, G1 is the control grid, and G2 operates as the plate. Physically, G2 consists only of a pair of vertical rods across which no grid wires are strung. For the tetrode section, G4 is the control grid, G3 and G5 are the screen grids, and P is the plate. Grids G3 and G5 are connected internally as shown. G3 prevents interaction between the oscillator circuits and the rf input signal circuits. G4 prevents interaction between the i-f (intermediate-frequency) output on the plate and eg. The cathode for the tetrode section exists virtually between G3 and G4.(3) Electrons emitted from the surface of the cathode, K, are influenced by the various elements within the tube. In practice, G1 receives 3 to 6 percent of the electrons emitted, G2 receives about 40 percent, G3 and G5 receive about 25 to 30 percent, and the plate receives the remaining electrons, about 30 percent. The tube is biased so that G4 does not draw current.(4) Oscillation is produced because of feedback from G2 to G1 in the oscillator circuit. This causes G1 to become alternately positive and negative. Consequently, the current flow through the tube is alternately increased and decreased. Many of the electrons which approach G2 flow beyond it, because of its open structure, even though it is highly positive. Electrons passing G2 are further accelerated by the high positive potential of G3. Grid G4 is always negative, and therefore tends to retard the electrons approaching it. This action produces a virtual cathode between G3 and G4. The electron flow from this virtual cathode is varied or modulated by the rf signal, e g, that is injected on G4. The electrons then are drawn to the plate, which is highly positive.(5) If G1 is slightly negative, or even somewhat positive, the virtual cathode is able to supply the plate of the tetrode section with sufficient electrons. On the other hand, if G1 is made very negative, a deficiency of electrons exists at the virtual cathode which may even cut the tube off. As a result of this variation in the virtual cathode, a current is produced at plate P which varies at the oscillator frequency. However, the electron stream is further modulated by the signal injected on G4. In this way, both the oscillator signal and the rf input signal on G4 modulate the electron stream, with the result that i-f signals are produced in the plate circuit. The L-C tuned circuit selects the desired intermediate frequency."sourceVacuum Tube Manual: Basic Theory of Electron Tubes (Army 1952)

author: TimFox
author_rank: Super Contributor
timestamp: February 11, 2023, 02:40:48 pm
content: I think your reference is to an earlier pentagrid tube, the 6A8.My topic is just a question of datasheet reading. I'm preparing to measure the excitation available in my circuit.My thinking, summarized in my original post, is that the only tools readily available before 1940, when the inner-grid injection pentagrid was introduced, were DC voltmeters, especially the VTVM with a 1 megohm resistor in the probe and total resistance of 11 megohm. Therefore, the specification was written for DC voltage measurement to establish a DC grid current value.Applying a large LO voltage to G1 modulates the current from the cathode to the rest of the tube, a small RF voltage at G3 directs that current between the plate and the other grids (G2 and G4).Looking at detailed curves shows a peak in the "conversion transconductance" (G3 to one sideband at the plate) as a function of the G1 current.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: February 11, 2023, 06:39:41 pm
content: It's DC.Consider how they would've measured it back in the day. Clip the grid leak resistor to your VOM or VTVM if you're fancy, and measure the current. Or with a high impedance probe, measure grid voltage and calculate the drop through the grid leak, same thing.The peak current could be measured, with some difficulty even with today's instruments, perhaps using a sensitive current transformer and preamplifier. The bandwidth isn't very much at least, but it would be difficult to measure on a VHF let alone UHF converter for example.More likely they would specify peak indirectly, such as a minimum grid circuit (AC) impedance, which is easily enough measured with AC bridge techniques and an AC/Vpk probe. That, combined with minimum grid leak resistance or DC bias (in whatever combination), defines peak current. Which they might never measure directly anyway, just determined based on performance and lifetime.Tim

author: TimFox
author_rank: Super Contributor
timestamp: February 12, 2023, 03:21:38 am
content: DC makes sense to me.It was probably patently obvious to interested readers in 1940, therefore RCA did not elaborate.The basic question is related to Class C amplifiers driven heavily into grid current, but my experience is mainly in negative-grid tube amplifiers.

author: TimFox
author_rank: Super Contributor
timestamp: February 14, 2023, 10:17:31 pm
content: Finally got to making measurements on the 6SB7Y frequency mixer, with +125 V plate supply.It took more G1 drive than I expected to obtain 0.35 mA DC grid current.At 2 MHz, 21.5 V pk-pk into the 100 pF grid-leak capacitor gave 7 V DC voltage at G1, for the specified 0.35 mA current in the 20 k grid-leak resistor.Increasing the G1 current from 0.08 mA to 0.40 mA decreased the DC cathode current from 16 mA to 12.5 mA.(A small cathode resistor, heavily bypassed, gave roughly +1 V on the cathode over that range of current.)Scaling up to the 0.5 mA G1 current spec for a 6SA7/6BE6 gives approximately the 28 V pk-pk mentioned above by trobbins.At that G1 drive, I measured 0.85 mA/V conversion transconductance from G3 to the plate (LSB output at approximately 2 kHz), close to the datasheet value of 0.90 mA/V.