307a DHT in triode and Schade feedback

IMG_0339Vegard Winge kindly sent me some great DHTs for tracing including the 307a directly heated pentode. The sample traced is not an original Western Electric but a lovely Raytheon RK 75 307a NOS. There is limited information of this valve in triode mode and the folks at DIYaudio are looking at potentially using it for a DHT headphone amp.

This valve has a filament of 5.5V and 1A and an anode dissipation of 21W in class A (including screen dissipation) when triode-connected.

Let’s see how this valve performs in triode-mode:

307a triodeSMALL

 

 

How well can we match a triode model for this valve?

307a triode small

Here is the 307a triode composite, would be great for someone to test it as haven’t had time to do so yet:

**** 307A TRIODE DHT Composite ******************************************
* Created on 03/16/2013 10:33 using paint_kit.jar 
* www.bartola.co.uk/valves
* Curves image file: 307a triodeSMALL.jpg
* Data source link: 307a triodeSMALL.jpg
* Created by Ale Moglia [email protected]
*----------------------------------------------------------------------------------
.SUBCKT TRIODE_307A-Composite 1 2 3 4 ; Plate Grid K1 K2
+ PARAMS: CCG=15P CGP=0.5P CCP=12P 
+ MU=5.08 KG1=2610 KP=29 
+ KVB=186 VCT=-2.26 EX=1.54 RGI=2000
* Vp_MAX=200 Ip_MAX=0.08 
* Vg_step=4 Vg_start=0 Vg_count=10
* END PARAMS -----------------------------------------------------------------------
* cathode resistor is 5.5 ohm, the pins K1 and K2 are 1.375 ohms from the ends of it
RFIL_LEFT 3 31 1.375
RFIL_RIGHT 4 41 1.375
RFIL_MIDDLE 31 41 2.75
E11 32 0 VALUE={V(1,31)/KP*LOG(1+EXP(KP*(1/MU+V(2,31)/SQRT(KVB+V(1,31)*V(1,31)))))}
E12 42 0 VALUE={V(1,41)/KP*LOG(1+EXP(KP*(1/MU+V(2,41)/SQRT(KVB+V(1,41)*V(1,41)))))}
RE11 32 0 1G
RE12 42 0 1G
G11 1 31 VALUE={(PWR(V(32),EX)+PWRS(V(32),EX))/(2*KG1)}
G12 1 41 VALUE={(PWR(V(42),EX)+PWRS(V(42),EX))/(2*KG1)}
RCP1 1 3 1G
RCP2 1 4 1G
C1 2 3 {CCG} ; CATHODE-GRID
C2 2 1 {CGP} ; GRID=PLATE
C3 1 3 {CCP} ; CATHODE-PLATE
D3 5 3 DX ; FOR GRID CURRENT
D4 6 4 DX ; FOR GRID CURRENT
RG1 2 5 {RGI} ; FOR GRID CURRENT
RG2 2 6 {RGI} ; FOR GRID CURRENT
.MODEL DX D(IS=1N RS=1 CJO=10PF TT=1N)
.ENDS
*$

Anode resistance is higher than 300B but better than many other DHP triode-strapped which typically are around 1.2KΩ and 2KΩ Let’s see how this valve is expected to perform with a load of about 5KΩ:

307a triode loadline Zaa=5K

Biased at about 260V/80mA can achieve 1W@THD=2.1% (mainly H2 as expected) with a driving signal of about 50Vpp. Linearity of this valve in triode is not that impressive. Sure that overall performance with driver H2 cancellation will do better than this.

So, what about anode to grid feedback a la Schade? Well, tried it with various screen voltages (150V, 200V and 250V) to see what was the impact of a greater pentode performance. Let’s see the first example at screen biased at 250V. All tests were made with 10% feedback:

307a schade2 SMALL

 

We can see above that anode resistance doubles, mu increases and transconductance decreases with a more pentode-like curves. Using Dmitry’s tool is difficult to match the curves with the triode model as expected:

307a Schade2 model

If we decrease the screen voltage to 200V:

307a schade1SMALL The curves are easier to match now with the triode-model with the exception of the knee at low currents which is not critical as we will not operate the valve in this region:
307a schade1SMALL model

Finally if we look at the screen at 150V:

307a schade3 SMALL

The anode resistance is lowered whilst mu remains the same and transconductance increases.  The model matches quite well as we can see below:

307a Schade3 model

 

So with a load now of 6K, biasing the Schade-connected 307a at 300V/70mA we can get 1W @ THD=0.3% (with similar H3 and H2 levels as expected due to the pentode contribution) driven with a 42Vpp grid signal:307a Schade FB0.1 Loadline Zaa=6K Vg2=150V

Further tests are yet to be done with FB=13% and other operating points and screen voltages, but initially it looks like a better candidate to me with Schade feedback…

Hope this helps and looking for your comments…

Ale

Author: Ale Moglia

"A mistake is always forgivable, rarely excusable and always unacceptable. " (Robert Fripp)

14 thoughts on “307a DHT in triode and Schade feedback”

  1. When time allows :
    Use the lowest possible vg2
    And check for higher values of feedback
    Load 0.9 x va/ua

  2. The curves in the diagram with a loadline for 5K show that I(a+g2) would be around 90 mA at Va = Vg2 = 200 V, and Vg1 = -15 V.

    However, the Western Electric datasheet for the 307A shows I(a=g2) = 39.9 mA at Va = Vg2 = 200 V, and Vg1 = -15 V.

    How can this big difference be explained? I would think that the Western Electric datasheet is correct.

    1. Hi Robert
      The curves I measured are for triode and “Schade-like” plate-to-grid feedback, which have nothing to do with the pentode curves from the data sheet. You can’t expect Ik=Ia+Ig2 to be the same for the same Vgk as the valve is connected and behaves differently. Unfortunately the WE data sheet are missing a triode-strapped connection plot.
      Cheers, Ale

      1. Hello Ale,

        Unless I’m mistaken, the set of curves with the 5K loadline show the curves of a triode connected 307A, so without Schade feedback applied yet.

        I know that the dynamic behaviour of a pentode and a pentode connected as triode differ a lot. But if you only look at static dc-conditions, I think a triode connected 307A should still pass 39.9 mA (being Ia + Ig2) at Va = Vg2 = 200 V, and Vg1 = -15 V like the Western Electric datasheet indicates. The tube under these static dc-conditions ‘doesn’t know’ wether it’s connected like a pentode or like a triode. It just has 200 V on both its anode and its screen grid.

        1. I think you are mixing things here. You can’t compare the behaviour of a triode with a pentode with just a DC static point arbitrarily with same Vgk and Vak conditions. The triode has a dynamic resistance rp below 2.2K and the pentode around 40K. Of course the triode Rp will vary widely depending on which operating point is and behaves like a voltage source whereas the pentode has more consistent rp and behaves more like a current source.

          Just applying ohms law doesn’t work here as you have active devices which can be modelled as voltage/current sources instead.

          1. Again: I’m not saying anything about the dynamic behaviour of the tube because it is not relevant for the point I try to make. My point is that according to the Western Electric datasheet, a 307A will pass 39.9 mA at Va = Vg2 = 200 V, and Vg1 = -15 V. For this static condition it doesn’t matter if the the anode and the screen grid are connected to eachother or not. They just both need 200 V. This static condition has to be true for both situations, so wired as a pentode and wired a triode. But your curves show around 90 mA in this static condition. So for me your curves must be wrong (maybe the scaling is wrong?), unless the Western Electric datasheet would be wrong. I had a set of curves of a triode connected 307A on my PC that I found on the internet some time ago. Those curves show 36.5 mA under this condition, so much closer to the 39.9 mA in the Western Electric datasheet. I posted those curves in post #12 of the thread “DHT headphone amp” on diyAudio.

          2. Additional: Take a look in a Philips datasheet for the EL84 and compare the curves for pentode mode with Vg2 = 250 V and the curves for triode mode. In both you will find that at Va = Vg2 = 250 V, and Vg1 = -7.5 V I(a+Ig2) is about 53 mA. The same goes for Vg2 = 300 V. Iin both modes you will find that at Va = Vg2 = 300 V, and Vg1 = -12.5 V I(a + g2) = about 30 mA.

          3. After you wrote that I couldn’t just apply Ohm’s law here (while I didn’t mention Ohm’s law at all) and after you wrote that you thought I was mixing things up, it has been silent for more than a month now since my last replies.

            Did you take a look in the datasheet for the EL84 by now? Or in the datasheet(s) of any pentode that also contain(s) curves for triode connection? If so, than you must have seen that I’m right.

            Could we have closure on this topic?

          4. Hi Robert
            Apologies for late reply. COVID knocked me out and recovery has been long for several weeks and everything piled up between work and family. Sadly not much time spent on Audio am afraid.

            You’re right. Having thought about I did miss the point of same Va=Vg2 which is the static point of the pentode behaving as a triode. Curves should coincide at this point only.

            Here is the proof with a model of 6GE5 (which posted long ago) and was the first one I randomly picked up to validate this. See attached.

            Regarding the 307a curves I guess something was wrong (either the valve or measurement). Unfortunately don’t have the time to re-test now but will do at some time in future when I get the 307a again out of the attic storage.

            Thanks for pushing my COVID brain to think again 🙂

            Cheers
            Ale

          5. Hello Ale,

            Good to hear you recovered. Thank you for your reply.

            Greetings,
            Robert

    1. Hello Felipe,

      I also think that the chance they are equivalents is very big. That is why I also think that there is something wrong with the triode curves of the Raytheon 307A/RK75 on this site. They just don’t match with the Western Electric 307A datasheet.

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