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5th April 2015, 11:39 PM  #1 
diyAudio Member
Join Date: Aug 2009

Drawing loadline using interstage / choosing interstage
So, I've been playing around with different ideas for driver stage with an 845 amp I'm still in the prototyping stages with.
I came across this design: 845 Amplifier with only two stages that I like the idea of.. Using the 6HV5A beam triode tube, allowing me to have just a single stage driver. Now I get the concept of using the interstage transformer, but I'm having trouble figuring out the operating points of using one. In my power stage I'll require about 120V of gain (240Vpp). The 6HV5A seems to be able to easily provide this, if I were to use a resistive load with a supply voltage of 950V, I drew a loadline which would give me about 210V of gain.. More than I need.. And I do like the idea of using a nice interstage transformer as well.. So I got thinking about using a 2:1 interstage transformer, or 1.5:1. I drew up this loadline in the same way I'd draw a loadline for an output transformer.. Loadline from 1600V / 0mA  0V / 50mA. Quiescent voltage of 950V (My existing power supply), and biased at around 2025mA.. So it would be biased at 2V using a ~100ohm resistor. Swinging between roughly 400V  1400V with 2V (4Vpp) at the grid. So with 1V (2Vpp) of signal I'd have 250V (500Vpp) of swing.. Using a 2:1 interstage that would be 125V (250Vpp).. Which seems right on target.. Am I understanding that correctly? Now the part I really don't understand is the Load impedance.. If I were using a resistive load I'd calculate the resistor value at (1600 / 50mA) 32K ohms. With this inductive load I'm just not sure how that plays into everything.. If I try to compare the operation of the interstage transformer to the operation of an output transformer, my load impedance is the reflected impedance of the load based on the turns ratio of the output transformer.. So, say with an output transformer it's got a turns ratio of 25:1, I've got an impedance ratio of 625:1 (25 ^2).. So with the interstage I've got a turns ratio of 2:1, and so I've got an impedance ratio of 4:1?.. And so in the case of the 845 with a plate resistance of 1700ohm, I've got a reflected impedance load of 6800 ohm? Is that correct? If so, that's much too low for that loadline.. For a 6800 ohm load, my loadline would have to run from 0mA / 1600V  236mA / 0V... I'm hoping someone can help me understand this better.. I've been reading quite a bit over the past few days, but still struggling with this. 
6th April 2015, 02:25 AM  #2 
diyAudio Member
Join Date: Oct 2006
Location: Charlotte, NC

Your load impedance is that looking into the grid of the 845, which is very high (megohms) until you reach 0V and positive, where the grid begins conducting current. At that point the grid "resistance" gets very low, stressing the driver. You do not consider the plate resistance of the 845 at this stage.
As long as you don't plan on running A2, you could assume the transformer reflects a nearly infinite load as seen at the grid, with one caveat: the grid capacitance should be considered, which IIR is Cgk + Cga(1+mu). This can be very significant, so run the calculation. In contrast to a resistor load, a capacitance load will change with frequency, so you would effectively see an elliptical pattern on your load line which will vary with frequency. And yes, it will be reflected as the square of the turns ratio, or 1/4 with a 2:1 stepdown. 
6th April 2015, 02:27 AM  #3 
diyAudio Member
Join Date: May 2005
Location: USA

If you stick with Class A_{1} operation, then the secondary of the IST basically disappears. If R_{sec} >> (wM)^{2}, this holds regardless of whether the coupling is very loose (small M) or tight, but with a very HiZ load.
Under these conditions, the only thing the triode sees is the primary inductance, and it will behave as though loaded with a CCS, and a very shallow loadline. That's what inductance does: opposes changes in current. That'll give the triode nearly ufactor voltage gain X whatever the SEC : PRI turns ratio is. The one potential bugaboo with this design is the capacitor bypassing of the cathode resistor. Capacitive currents in the cathode mean capacitive currents in the plate load, and these can resonate with the primary inductance to cause nasty distortion. Leaving the cathode resistor unbypassed isn't really a solution as this drives up the r_{p}, and you don't want that. You may have to go for fixed bias on the driver, and a DC bias servo would be a good idea, especially if your triode tends to be drifty. Either that, or play with possible cathode bypass capacitors to find one that doesn't make for nasty resonances. 
6th April 2015, 04:36 AM  #4 
diyAudio Member
Join Date: Aug 2009

Hm.. Okay, that clears things up a bit. I suppose I was planning to run class A1, although the thought of some grid current being available for the 845 was kind of a nice thought.
I guess part of the problem here is that I'm not familiar with using a choke load.. You say that it'll give the triode nearly ufactor gain X the ratio.. How do I determine what inductance I need on the primary of the insterstage? And how about bias? Is the current it's biased for important at all... Like am I just looking to get it high enough to get into the straight part of the curves? Lastly, I'm having a hell of a time finding a 2:1 interstage transformer.. I've looked through the list of Lundahl's products Other Tube Transformers  Product Categories  Lundahl Transformers And not finding a configuration on any of them for a 2:1 SE application.. 
6th April 2015, 08:21 AM  #5  
diyAudio Member
Join Date: Nov 2013

Joining in the discussion for the sake of learning something myself. Somebody correct me if i am wrong.
Quote:
fc = rp / 2*pi * L fc = cutoff frequency.. should be 20Hz or less for good bass response. Bartola went with 7.7Hz rp = tube's internal resistance which is approx 35004500R. Bartola went with 4600R. Using the values above, Bartola end up with 100H of inductance. Assuming we are all homo sapiens, we can only hear as low as 20Hz. At 20Hz, 100H equals to 12.5k of reactance. I think this would be your loadline, but only at 20Hz and only if you use 100H. Higher frequency equals higher reactance and your loadline would be closer to horizontal, i.e closer to that of CCS loadline. Again, somebody correct me if i am wrong. Quote:


6th April 2015, 03:20 PM  #6  
diyAudio Member
Join Date: Aug 2009

Quote:
In any case, I'm definitely not finding a suitable interstage transformer anywhere.. 100H, 25mA current handling, 2:1 ratio, 1000V+ voltage rating.. Does anyone have a suggestion on where to find a suitable one? The author seems to have had a custom one made from Bartolucci, which I can only imagine was quite pricy. 

6th April 2015, 06:01 PM  #7  
diyAudio Member
Join Date: May 2005
Location: USA

Quote:


6th April 2015, 06:34 PM  #8  
diyAudio Member
Join Date: Aug 2009

Quote:
I got looking at the Lundahl LL1692a http://www.lundahl.se/wpcontent/upl...3/05/1692A.pdf One configuration shows a 4:3.5 winding ratio.. I don't quite understand if this is different than say 2:1.75.. Is there some significance to it being listed as 4:3.5? I suppose this ratio would be fine, Would still be way more voltage swing than I'd need, but would reduce it somewhat.. Would the specs be suitable? 18mA max (I could probably work with that) 125H Max output voltage = 175V RMS (Not exactly sure how this translates to the voltage swing that I'd see with the 6HV5A.. I guess if the 6HV5A with inductive load has an amplification factor of 300, or swinging 600Vpp the interstage would be unsuitable for full swing.. But if I'll be getting full power from the 845 @ 120V or 240Vpp would this RMS rating be okay?) It also shows voltage isolation rating of 4 kV / 2 kV. 

6th April 2015, 07:06 PM  #9  
diyAudio Member
Join Date: May 2005
Location: USA

Quote:
175_{RMS} * sqrt(2)= 247V_{p} The 845 needs a grid voltage of 155, so you're good to go there if you stay within Class A_{1} conditions, as you said you would. 

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