EL83 OT impedance question

[Chuck H]

Not at all.

Ok, to be more clear, I though the ideal load impedance was proportionate to tube plate resistance. I do know that they can be considered separately.

[Helmholtz]

I though the ideal load impedance was proportionate to tube plate resistance

No, the Rp is the inverse of the slope of one of the output characteristics (Rp=dVp/dIp) at a given operating point. It easily varies by a factor of 10 or more along the same curve. The flatter the characteristic the higher the Rp.
When doing a load line construction the Rp is not considered. No such thing as impedance matching involved.

[tubeswell]

The EL84 is rated at 12W plate dissipation, 2W screen dissipation in my datasheets.



This simple formula ignores the actual shape of the characteristic and gives somewhat high Zout values.

I’m not discussing an EL84. I was talking about the two tubes linked in the OP’s enquiry.

The ‘simple formula’ for load impedance is close enough for working out the ballpark load for centre-bias Class A operation in an SE guitar amp. In the real world, no two tubes of the ‘same type’ are going to have exactly the same rp characteristics. The chart is an approximation. There seems little point in getting too choosy about the shape of the grid curves.

[Old Tele man]

I’m not discussing an EL84. I was talking about the two tubes linked in the OP’s enquiry.

The ‘simple formula’ for load impedance is close enough for working out the ballpark load for centre-bias Class A operation in an SE guitar amp. In the real world, no two tubes of the ‘same type’ are going to have exactly the same rp characteristics. The chart is an approximation. There seems little point in getting too choosy about the shape of the grid curves.

Exactly! Both the datasheets and charts only 'represent' a BOGEY tube, which is a 'theoretical' tube that matches 100% of all specifications. Most tubes, however, only come 'close' on most parameters, but true "bogey/calibration" tubes do exist, albeit culled (1 in a million?) from huge production lots...and they are COSTLY!

[Helmholtz]

I was talking about the two tubes linked in the OP’s enquiry.

So did I. His is comparing the EL84 to the EL83/6CK6.

[J M Fahey]

FWIW this is the load line I chose based on datasheet posted above.



Others are possible and in fact you´ll notice a previous one I erased, based on smaller idle current of some 27mA but to be closer to the original question and first datasheet suggestion, I chose 250V plate (duh!) , and accepted 35mA idle current (point A), only because I saw 6CK6 can reach 70mA with an acceptable (for me) saturation voltage drop of only 50V (point B).

This is "half the load line" , going from idle to saturated, we must then draw the other half, reaching almost cutoff current and double +V, but since slope is the same, half linee is enough to find that.

So we have:
* delta I= 70mA-35mA=35mA
* delta V=250V-50V=200V

So on first approach, optimum load impedance is: 200V/0.035A=5714 ohm so your 5K winding will be *perfect*.

EDIT: peak power will be 200*.035=7W so 3.5W RMS .

[Helmholtz]

I just realized that I overlooked the 200V Vg2 chart and used the Vg2= 170V one instead. Higher Vg2 reduces optimum load impedance. But the simple formula of tubeswell above ignores the Vg2 effect as well as the finite plate saturation voltage and thus gives somewhat high Zout results. Zout higher than optimum in turn results in asymmetric clipping, produces high peak plate voltages and increases screen dissipation.

[nickb]

Deleted - I thought we were talking about a push pull (since AC30 was mentioned)

[tubeswell]

I just realized that I overlooked the 200V Vg2 chart and used the Vg2= 170V one instead. Higher Vg2 reduces optimum load impedance. But the simple formula of tubeswell above ignores the Vg2 effect as well as the finite plate saturation voltage and thus gives somewhat high Zout results. Zout higher than optimum in turn results in asymmetric clipping, produces high peak plate voltages and increases screen dissipation.

Which is why I calculated my load line using Vg2= 250 being 50V lower than the plate voltage I calculated my centre-bias Class A load line for. This assumes the g2 supply node will be fed by a dropping resistor that will inhibit g2 dissipation. Also, I calculated for the plate to be idling at 100% which is why the bias voltage came out a smidgeon over 6V. This produces a nice Class A load line that doesn’t cross Pmax. By all means, run a lower load, but you will want to increase Vg1 to compensate.

[Pedro Vecino]

Thank you very much to all!

Deleted - I thought we were talking about a push pull (since AC30 was mentioned)

I would have liked to read it. Both examples maintain a relationship.

[Helmholtz]

By all means, run a lower load, but you will want to increase Vg1 to compensate.

Some thoughts about the "simple formula". To begin with, it isn't bad at all and easily gets you in the ballpark while the results are somewhat on the high side. (The Radiotron Designer's Handbook (Langford-Smith) recommends to multiply the result by 0.9.)

The formula actually states that Zout=Vq/Iq, where Vq und Iq are the quiescent point values. This implies ideal tube characteristics with zero saturation voltage and results in a somewhat asymmetrical loadline for real tubes.
For Vp=250V and a plate dissipation Pp of 9W we get Iq=Pp/Vq=36mA and Zout= 6944 Ohm (resp. 6250 Ohm acc. to the RDH).

For better symmetry I suggest a simple modification to the simple formula taking account of the fact that plate voltage excursion for high plate currents is limited by the saturation voltage Vps, leading to
Zout= (Vq-Vps)/Iq, where the saturation voltage has to be taken/estimated from the tube characteristics. Using Juan's value of Vps=50V, I thus get Zout= 5555 Ohm. The effect of different screen voltages would be included in the choice of Vps.

Generally speaking there is always some freedom to optimize Zout for different key aspects like symmetry, distortions characteristics, max. power, peak plate voltage or screen dissipation. It is up to the designer to check and make his choice.

[Chuck H]

Some thoughts about the "simple formula". To begin with, it isn't bad at all and easily gets you in the ballpark while the results are somewhat on the high side. (The Radiotron Designer's Handbook (Langford-Smith) recommends to multiply the result by 0.9.)

The formula actually states that Zout=Vq/Iq, where Vq und Iq are the quiescent point values. This implies ideal tube characteristics with zero saturation voltage and results in a somewhat asymmetrical loadline for real tubes.
For Vp=250V and a plate dissipation Pp of 9W we get Iq=Pp/Vq=36mA and Zout= 6944 Ohm (resp. 6250 Ohm acc. to the RDH).

For better symmetry I suggest a simple modification to the simple formula taking account of the fact that plate voltage excursion for high plate currents is limited by the saturation voltage Vps, leading to
Zout= (Vq-Vps)/Iq, where the saturation voltage has to be taken/estimated from the tube characteristics. Using Juan's value of Vps=50V, I thus get Zout= 5555 Ohm. The effect of different screen voltages would be included in the choice of Vps.

Generally speaking there is always some freedom to optimize Zout for different key aspects like symmetry, distortions characteristics, max. power, peak plate voltage or screen dissipation. It is up to the designer to check and make his choice.

Does NOTHING change when applied to an AB circuit? Keeping in mind that either tube is off for some portion of the waveform and the OT must swing at more than twice the power +/- ??? Does nothing change WRT ideal primary impedance?

[tubeswell]

Does NOTHING change when applied to an AB circuit? Keeping in mind that either tube is off for some portion of the waveform and the OT must swing at more than twice the power +/- ??? Does nothing change WRT ideal primary impedance?

ABsolutely ;-)

[Helmholtz]

Does NOTHING change when applied to an AB circuit? Keeping in mind that either tube is off for some portion of the waveform and the OT must swing at more than twice the power +/- ??? Does nothing change WRT ideal primary impedance?

This is all and only about single ended (see OP) transformer output, (almost) symmetrical operation at max. plate dissipation. Never heard of class AB in single ended audio amps. PP is very different.

[Chuck H]

I know. But I've noticed that it's common practice among guitar amp builders and even manufacturers to use the same primary load for both (one tube se and two tubes push/pull). I was openly wondering why and thought the higher minds might some insight