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13th November 2006, 09:48 AM  #1 
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Join Date: Jan 2002
Location: somewhere in Australia

to mach1: tubecad regulator using tl431
you pointed me to tubecad article before. http://www.tubecad.com/july99/page13.html
can you tell me what triode did you use? what is the current output of the CCS and B+ voltage? thank you. 
14th November 2006, 12:11 AM  #2 
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Join Date: Sep 2004
Location: Brisvegas

Jarthel,
I used one half of a 5687. It is fed by a CCS set for 21mA. The regulated voltage out is 200V (Vpk = approx 191V). 16mA passes through the 5687, and the remaining 5mA is used to feed two CCS', each drawing 2.5mA. The current through the 5687 was set by ear: the reg sounds much better at 16mA than lower currents. The 5687 has a max power dissipation of 4.2W (although it is a very tough tube and would undoubtably take more over short periods), so I would restrict current draw on the device you are powering to no more than 1012mA to account for no load conditions. For higher current applications look at the 6SA4, which is very well suited to shunt reg duties. pm 
14th November 2006, 02:02 AM  #3  
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Location: somewhere in Australia

Quote:
I checked frank and tdsl and cannot find such a tube. typo? thanks again 

14th November 2006, 04:09 AM  #4 
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14th November 2006, 05:49 AM  #5 
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jarthel,
you appear to be having some difficulty in selecting shunt tubes and determining appropriate operating conditions. If you follow the following instructions you should be able to get a halfdecent shunt reg up and running. Once you have experimented a bit and blown up a few TL431s you will be able to determine your own approach. 1. Determine the voltage (Vdevice) and current requirements (Idevice) of the device you are attempting to power and calculate the power consumed in watts (Pdevice). 2. Peruse tube data sheets and select a triode (or triode strapped pentode) with the ability to dissipate around 2 x Pdevice (or more (more is preferable)) on its plate  let's call the maximum power dissipation for the tube Pmax. 3. Determine the maximum power you can safely dissipate through the shunt tube under a no load condition. This is equal to Pmax  Pdevice  let's call this Ptube 4. Calculate Ptube/Vdevice  this is the maximum safe current flow through the shunt tube under a no load condition  let's call this Ishunt. 5. Plot the point Vdevice and Ishunt on the shunt tube plate curves 6. Move this point to the left on the curves until the plate voltage shown on the xaxis plus the gridcathode voltage indicated by the curves (count this as a positive, even though it is shown as a negative) = Vdevice. 7. Read off Vplate and Vgridcathode at this point. 8. Multiply Vgridcathode by Ishunt to determine the power that will be dissipated by the TL431 9. If the power is greater than the maximum dissipation for the TL431 (this will depend on the type of package your TL431 comes in), insert a voltage dropping resistor from the shunt tube cathode to the TL431 of sufficient value to drop the power dissipated by the TL431 to within its recommended max. 10. Place a CCS between your psu and the shunt reg and set at Idevice+Ishunt. Let's call this Iccs. There are some assumptions and approximations built into this approach, but it will work and also help you select a suitable shunt tube. You can probably increase Iccs (and hence Ishunt) by at least 10% with no problems, as the calculations err on the safe side due to one of the approximations (not taking Vgridcathode for the shunt tube into account when calculating Ishunt). You can do another iteration and determine a more accurate figure than 10% if you want. Be warned  some shunt tubes can sound very ordinary when run at only half their maximum power dissipation. The best ones generally display high plate dissipation, high transconductance, and a low Vgridcathode for any given plate voltage. pm 
15th November 2006, 01:32 AM  #6 
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Jarthel,
I forgot to add that you need to check the TL431 dissipation under no load conditions. To determine this draw a horizontal line through the shunt tube plate curves where I = Iccs. Find the point on this line where Vp (on the xaxis) plus Vgridcathode (count this as a positive number) = Vdevice. Multiply Vgridcathode by Iccs to find the power dissipated by the TL431. If the power is greater than the max specified for the device insert a voltage dropping resistor from the shunt tube cathode to the TL431 as per previous instructions. pm 
15th November 2006, 02:09 AM  #7  
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Quote:
for example: http://www.mif.pg.gda.pl/homepages/f...030/e/EL34.pdf (page 7) an EL34 triode has a max power of 16.5W. say Pdevice is 150V * 68mA = 10.2W 16.5  10.2 = 6.3W. so 6.3W isn't the max but the allowance I have before I burn up the tube. also isn't the max power dissipated on the tube under a no load condition is equal to Vdevice * Iccs? because all the current would be passing through the shunt? Quote:
again under a no load condition, I thought all current from the CCS circuit will be passing through shunt tube? Quote:
============================== I apologise if my questions sounds/is silly. Thanks for the help 

15th November 2006, 05:35 AM  #8 
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Join Date: Sep 2004
Location: Brisvegas

Jarthel,
you wrote: 'somehow I feel that Pmax  Pdevice is wrong. or maybe the term you use "maximum power you can safely dissipate through the shunt tube under a no load condition"?' You are entirely correct, my brain does not operate well with a low caffeine level. Points 4 & 5 should have read: 4. Determine the maximum power you can safely dissipate through the shunt tube under normal operating conditions This is equal to Pmax  Pdevice, because under a noload condition the additional Pdevice will be shunted through the tube  let's call this Ptube 5. Calculate Ptube/Vdevice  this is the maximum safe current flow through the shunt tube under normal operating conditions  let's call this Ishunt. Page 4 of the EL34 datasheet gives the maximum power dissipation of the EL34 as 30W when strapped as a triode. Sorry  I don't know where you got 16.5W ? I assume that your Idevice is 68mA and Vdevice = 150V If Pdevice is 10.2 W then you have 19.8 W left which you can safely pass through the tube under normal operating conditions. With a Vdevice of 160V, this gives you a maximum Ishunt of 123mA. However, page 4 of the datasheet gives the maximum current (Ik max) for the EL34 as 150mA, so you will have to restrict Ishunt to Ik max  Idevice (150mA  68mA = 82mA) instead. Go to page 9 of the EL34 datasheet for the triode mode plate curves and draw a horizontal line at Ia = 82mA to represent Ishunt. Plot the point on this line where Vp (shown as Va on the chart) plus Vgrid cathode (it is shown on the plate curves as Vg1) = 150V (Vdevice). I make it approximately where Vp = 145V, and Vg1 = 5V. So with a Vdevice of 150V your EL34 cathode will be sitting at 5V. This means the TL431 will be dissipating 5 x 0.082 = 0.41W under normal conditions, which is fine, so you won't need a voltage dropping resistor. Under a no load condition Ishunt = 150mA. If you look at the plate curves again you will see that if we again assume a Vdevice of 150V, Vg1 goes to zero volts as this current is approached, which suggests the TL431 will be dissipating nothing under a no load condition. So in this case, as the power dissipated by the EL34 goes up, that dissipated by the TL431 goes down ! The only issue I can see is that the TL431 will stop regulating once its supply voltage goes below Vref for the device (2.5V). I don't know what will happen to the voltage supplied by the shunt reg under these conditions  it could rise to a voltage marginally below the CCS supply voltage. This would mean that the shunt tube is dissipating more than its recommended max. I suggest you build it and find out by measuring the voltage. Exceeding Pmax for a short period should not be an issue as the EL34 is a pretty rugged tube. This unknown aside, the EL34 seems to be an ideal shunt tube for the application you have chosen 
15th November 2006, 10:55 PM  #9  
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Join Date: Jan 2002
Location: somewhere in Australia

Quote:
http://www.mif.pg.gda.pl/homepages/f...129/e/EL34.pdf (Mullar's datasheet) on page 4 of that datasheet, it does say 30W max dissipation for an el34 triode but the page is titled "operating conditions of 2 tubes in pushpull with continuous sive drive". does this make a difference? ================ using the phillips datasheet: http://www.mif.pg.gda.pl/homepages/f...030/e/EL34.pdf (page 7). page is titled "operating conditions in triode connection (and no mention of pushpull) the Wo (which I assume is plate dissipisation is listed as 16.5W. 

16th November 2006, 12:26 AM  #10  
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Join Date: Sep 2004
Location: Brisvegas

Jarthel,
you wrote Quote:
The Wo you refer to in the Philips datasheet is the power output of a pushpull pair operated under the conditions described. pm 

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