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25th May 2009, 02:53 AM  #1 
diyAudio Member
Join Date: Apr 2007

cathode biasing question
I have a Hammond 1650R, 5000k plate to plate to integrate into a pair of push pull KT88 as shown. R18, R19, R20 and R21 have been chosen to be 1k ohm which effectively gives a 500 ohms cathode bias.
Planning to run 500V B+ so (pls verify if) Ia = B+/(Rc + Rpp). this becomes Ia=500v/(500ohms + 5000ohms) = 0.09A. not sure should I take 5000ohm or 2500ohm on the Hammond and the first 500ohm or both on the cathode as this is push pull cct. Can someone help here, thks. 
25th May 2009, 04:21 AM  #2  
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Join Date: Apr 2009

Re: cathode biasing question
Quote:
Looking at the datasheet curves we see that at 500v and the grid biased to 30v the anode current is about 80mA. Your tube will be biased somewhere in the middle of the graph, roughly, so you would expect an anode current of anywhere from 50mA to 250mA. When there is no signal, the transformer has very little resistance, but more than zero. A few volts will drop across it. But the resistance is nowhere near 5K. The 5K is the reflected AC impedance under AC signal conditions. So if your V++ is 500 Volts, almost all of that is on the anode. Each KT88 will see half of the transformer, and so under AC conditions will see 1/4 of the 5K (because the impedance is related to the square of the turns ratio). You draw your AC load line from the bias point and at a slope of 1/4 the stated impedance, which in your case would be 1/4 of 5K, which is 1250 ohms. This is partially true for your pushpull, since both tubes are in the circuit on their own half of the transformer. The load line isn't straight, but the 1/4 factor gets you in the ballpark. Along that load line is a defined voltage and current for the anode during AC signals. The datasheet can be found at http://tubedata.tigahost.com/tubedat...163/k/KT88.pdf Read these two web pages. First, read the Single Ended article for important background info, then read the pushpull article. http://www.freewebs.com/valvewizard1/se.html http://www.freewebs.com/valvewizard1/pp.html 

25th May 2009, 05:00 AM  #3 
diyAudio Member

I love the TC push pull calculator.
I ran an equivalent circuit through it. Results: 90 mA per tube. 40 Watts plate dissipation at idle 100% of plate dissipation. Dissipation in the 4 1k resistors 8 watts total. Max class AB1 output 10.5 watts. Max input voltage 46v peak Predicted distortion 3% 3rd, .1% fifth.
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25th May 2009, 05:04 AM  #4 
diyAudio Member

Here is the output Wave.
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Scienta sine ars nihil est  Science without Art is nothing. (Implies the converse as well) Mater tua criceta fuit, et pater tuo redoluit bacarum sambucus 
25th May 2009, 03:03 PM  #5  
diyAudio Member
Join Date: Apr 2007

Re: Re: cathode biasing question
thks for the valuable info. the links were very useful.
Quote:


25th May 2009, 04:49 PM  #6  
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Join Date: Apr 2009

Re: Re: Re: cathode biasing question
Quote:
Your grid is tied to ground, and the cathode voltage is raised due to the current through the cathode resistors. So it is a net negative bias on the grid. Just two different ways of looking at the numbers relative to each other. Let's assume that the designer originally wanted 500V on the plate/anode from the Vpp supply. When the current flows through the tube and out the cathode into ground, the cathode will rise above ground by, for example, 30 volts. The effective voltage from anode to cathode is now 470V, not 500V. The grid is tied to ground, so it is 30V below the cathode. This would be the same as putting 470V on the anode, grounding the cathode, and putting a fixed voltage bias on the grid of 30V. In any event, the current through the cathode will be anode current plus screen current. Screen current will be a small % of anode current, maybe about 5% for a very rough estimate. You could ignore screen current if you want, but there will be some error in resistor value. Anode current is dependent on the anode voltage and the grid bias. You look on the datasheet curves to find this point if it is known and then read the anode current. If you don't know the original design spec, you can fiddle around with numbers until you zero in on it. Once you know anode current you can then figure out the resistive drop in your transformer if the manufacturer publishes a DC resistance number. If not, take your DC multimeter and read the resistance from center tap to one end of the primary. That is the DC resistance and you can then calculate the DC voltage drop during zero signal conditions. 

3rd June 2009, 09:49 AM  #7 
diyAudio Member
Join Date: Apr 2007

while i wait for the power transfo to be ready, i'm simulating the duncan's ps software. i do not have the power supply cct for this so i'm guessing here that 460V for the power stage, 350V for the invertor stage and 300V for the voltage amplifier stage.
i'm not too sure about the invertor stage as most designs use 400 plus volts but the 12AX7 datasheet says Va max=330V. i plan to run the power stage on one winding (400Vac) and both the invertor and voltage ampli stage on one winding (250Vac) on the power trans. Is this OK? 
3rd June 2009, 04:24 PM  #8 
diyAudio Member
Join Date: Apr 2007

im trying to figure this out but do not seem to zero in. i have a 250Vac winding which gives 350Vdc to be fed into the +B2 pin (12AX7 invertor cct) below. if you see the cct, the 350Vdc path goes thru 220k, then the tube Va, 33k and finally 51k before GND.
so the 350Vdc must be dropped in that path but by just taking 5mA, the calculations do not make sense. How do i calculate the voltage across the tube, Va? thks for any help. 
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