So I went to full voltage tests with the 2 tubes in place. With zero bias I measure

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·???????? CW voltmeter 1700 V measures 236*8=1888 V intensity 175 mA; the tubes do not redden 298 W (2 tubes)

·???????? SSB voltmeter 2500 V measures 329*8=2632 V intensity 220 mA; the tubes redden 550 W (2 tubes)

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In low voltage tests 231 W (plate current 390 mA under 592 V) for one tube redden. The result therefore seems consistent.

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In my country (France) the power is limited to 500 W.

·???????? I could operate the L-7 on the CW position and by limiting the excitation 500 W RF output.

·???????? Periodically, run the amp in SSB position, without RF, with just the 220 mA quiescent current to make the tubes glow and thus maintain optimal degassing.

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I am concerned that by placing diodes in the cathode circuit to bias the grid slightly negatively, the quiescent current in SSB mode will not be sufficient

Hello

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Please do not think that I am not interested in your relevant information. I have to integrate them and, often, I redo the experiments not to verify their validity but to adapt them to my environment.

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In a previous theard, I showed how I was able to make the plate of a single tube red under a relatively low voltage with a very positive grid voltage. One of the next steps is to move to high operational voltages (CW and SSB) and again to be able to make the plate red in this new condition.

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The removal of the 50k + 50k +5k bleeder resistors worries me.

1.???? It quickly makes the high voltage disappear

2.??? It regulates the high voltage a little

3.??? It avoids stressing the amplifier with a no-load HT which will be equal to the peak voltage

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Here are the measured data:

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With the bleeder resistors and a zero anode current I note:

·???????? CW: 1500 V DC

·???????? SSB: 2500 V DC

At the secondary of the transformer, depending on the transformation ratio and at no load:

·???????? CW: 230 x 2.95 = 678.5 V (960 V peak)

·???????? SSB: 230 x 4.14 = 952.2 V (1347 V peak)

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The high voltage at no load, without bleeder could be (estimated):

·???????? CW: 1920 V

·???????? SSB: 2694 V DC (we are therefore 200 V above the voltage without bleeder - Is this serious ?

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Concerning the polarization, I am a little in the dark. What you are proposing to me is a little like what I did a long time ago, in AM, with a "clamp" modulation on the screen of a tetrode. The idea was then to send the carrier only when the operator spoke. When he spoke weakly the carrier was not at full power. I did this by rectifying the BF to obtain a continuous voltage which was superimposed on the modulation on the screen. This artifice allowed the tube to be used at high power in the modulation peaks only.

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For these 3-500Z, I am trying to apply this principle. When the operator does not speak, the quiescent current allowing to switch from class B to class AB is not useful.

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·???????? When the BF is weak, is it useful? the distortions in % will be higher but since the output power is low, is it serious.

·???????? When the BF is strong (PEP peak), it seems important to be clearly in class AB. Please confirm.

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Do you have a diagram showing where you place the diodes on the L-7?

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73

What should happen if I add the red components in the attached schematic. I have the impression that it would do the opposite of the desired effect, we would block the tube on a strong RF input.

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I added the 100 k? for the cut-off of the tube in reception

Hi Francois.

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In my experience, the 7.5V Zener provides a little bit of bias to limit the quiescent current of a pair of 3-500Zs running at around 3kV to 150mA. This is the requirement for Class AB2 linear operation.

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You could run them with Zero Bias (as was Eimac’s recommendation) to reduce circuit complexity, but the quiescent current would be much higher – for no improvement in linearity – hence the Zener diode.

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Most of us shy away from Zener diodes these days as a few decent size silicon diodes (1N5408 or 6A10) can be placed in series and provide the same bias voltage as the expensive Zener.

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Good luck with it. I do love all of my amps – 2 x 3-500Z on 6m, FU728F tetrode in Emtron DX-2SP for HF, GS35B triode in homebrew 2m amp for EME and YC156 triode for 160m/80m. All the triodes have silicon diodes providing the quiescent current bias voltage.

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Best 73,

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John VK6JX

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Hello

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Now that both of my 3-500Zs have been de-gassed, I have a new question.

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·???????? In the L7 (Drake) the bias is at 0 volts

·???????? In the Kenwood TL-522 it is -7.5 V as shown in the following diagram

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We read a bit of everything

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The objective is to put the L-7 into operation in SSB at medium power.

1-2 hrs max...per tube.

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Alright. Then I'll do the second 3-500Z valve.

Then I'll do the full voltage tests with the Drake's HT transformer

--

F1AMM

Fran?ois

Hello

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Cooking the tubes.

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It was necessary to modify the tube polarization system. As shown previously, the HT power supply is floating relative to ground. This trick allows the cathode voltage (the filaments) to be modified relative to ground and as the grids are galvanically connected to ground, it allows the grid - cathode voltage to be adjusted. This is what happens in reception where the tubes are blocked with a grid - cathode voltage of -60V.

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To be able to positively polarize the grid relative to the cathode, in operational operation, the original diagram does not allow it. This diagram must therefore be modified. The positive polarization voltage of the grid can only come from an external source; it is shown as a green battery on the diagram

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Obviously this involves some work including drilling the back panel to get the two wires out of the green source. I put in two female banana plugs and a bridge to place a short circuit in place of the auxiliary power supply to find the original diagram (0 V polarization).

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I remind you of the objective: to cook the anodes to red under a low anode voltage and a high current. All this to degas the tubes. It will certainly also be useful to improve the linearity by increasing the quiescent current which is possible since in France we are limited to 500 W HF (against the almost 2 kW that the PA can output).

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The first tests are done with a 230/230 transformer which unfortunately only makes 400 VA (it must not exceed 1.7 A on the secondary). The polarization is done from a 48 V transformer and a Graetz bridge rectifier.

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·???????? With a polarization at 0 the plate voltage is at 688 volts and the quiescent current (since there is no HF in excitation) is 30 mA

·???????? With a polarization at 18 V DC (but it is rectified bi-alternation) the HT is 611 V, the plate current 300 mA (therefore 183 W) and the grid current (DC) 85 mA

·???????? With a polarization at 23 V DC the HT is at 593 V the plate current at 400 mA (237 W) the grid current 260 mA and I am at the limit of the transformer 1.6 A RMS

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That is not enough to make the plates red.

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I switched to a single 3-500Z tube. As a precaution, there is a load at the input and output but the system seems quite stable.

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Now it works

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Oddly enough, it only takes 20 Ω in series in the primary to bring the filament voltage back to 5.0 V. I plugged the hole caused by the missing lamp so as not to short-circuit the ventilation that I forced to the max. I placed a glass plate above the lamp box in order to leave the ventilation as when the cover is in place but being able to keep an eye on the anode of the tube (otherwise, you can't see the tubes).

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Two tests that give almost the same result: the anode in its lower part is dark red.

·???????? Bi-alternation rectified voltage on the grid Vg DC = 38 V Ig DC 110 mA plate current 390 mA under 592 V (231 W) secondary consumption: 359 VA

·???????? Bi-alternation rectified voltage + 6300 ?F capacitor so, basically DC on the grid Vg DC = 41.9 V Ig DC 120 mA plate current 400 mA under 584 V (234 W) secondary consumption: 389 VA

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I am limited by the 400 VA max of the 230/230 transformer supplying the voltage doubler.

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The measured values ??are questionable; I expected more difference when I added the capacitor. I am surprised that with 234 W on the anode, we are already in the dark red; the tube is given for 500 W of anode dissipation. I don't dare remove the anode radiator which also serves as a connector; I suppose it is this which prevents the entire anode from being red, from top to bottom.

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How long should I cook it for?