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Old 15th May 2007, 08:55 PM   #1
Klimon is offline Klimon  Belgium
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Default Tie cathode to heater??

Hi,

I'm building a double dc darling and a while back a forummember told me that I should tie the heaters of the 1626's to their cathodes so as not to exceed max. cathode-heater voltage.

Before proceeding I'ld like to double-check; the four 1626 are run from one heater supply (all 4 parallel); if I connect cathodes to heaters wouldn't that mean that the cathodes are de facto connected with each other (parallelled) through the heater lines?

Which would mean that:

- the cathode bypass cap value would double (parallel) thus being much larger than necessary

- I could just us well use one cathode resistor shared by all four tubes instead of one for each channel (and take half of that value)

Does this make sense?

Cheers -- Simon
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Old 16th May 2007, 12:13 AM   #2
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Err....

Why would you Want to attach the cathodes to the heaters....? (When the Maker has gone to all that trouble of insulating them..

This could cause you to experience Hum problems if not done with care, and also, as you say-All your cathodes would be then connected together--I'm assuming the valves you quote are the O/P ones...? IF the cathodes were all connected together, and being a less than 'perfect-world' your L-R channel 'crosstalk' will suffer terribly....

A little trick Ive found that works very well to prevent heater 'induced' hum is to 'bias' the Heaters around 30-40V more Positive than the Cathodes normally run at You then keep all cathodes separate, or at least isolated to their own channel....Works a treat, A simple voltage divider from +B, and a de-coupling cap to the heaters is good....
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Old 16th May 2007, 01:23 AM   #3
Klimon is offline Klimon  Belgium
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Quote:
Why would you Want to attach the cathodes to the heaters....? (When the Maker has gone to all that trouble of insulating them..
Maximum allowable heater-cathode voltage for 1626 is 100V; in this DC-coupled circuit there is 154V between both. See the (two identical) last schematics in this link:

http://www.geocities.com/TimesSquare/1965/jeremy.html

Simon
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Old 16th May 2007, 03:05 AM   #4
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Simon,

Biasing the heaters off B+ to a value in the 85-90 V. range should make things fine. Both cathodes will be < 100 V. away from the heater's potential. The idea is to put the heater at a potential approx. 1/2 way between the 2 cathodes, as that gives you a "200" V. window. The 154 V. difference between the 2 cathodes fits easily into the window.
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Old 16th May 2007, 09:33 AM   #5
thomsva is offline thomsva  Finland
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Quote:
a while back a forummember told me that I should tie the heaters of the 1626's to their cathodes so as not to exceed max. cathode-heater voltage
The forummember probably meant that you should "tie" the heaters to the cathodes with 100-200 ohm resistors. The aim is just to elevate the heater voltage without disturbing the cathodes.

As you have four tubes you would have to decide if you connect the resistors only to one cathode or to all four. I would not like to connect only to one cathode and connecting to all four would need lots of resistors and wires. A simpler solution would be to leave cathodes alone and connect the heaters to a voltage divider made up of two resistor between b+ and ground. By choosing the right ratio of resistors you can have the heaters referenced to any voltage you want.
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Old 16th May 2007, 03:49 PM   #6
Klimon is offline Klimon  Belgium
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Thanks for the elucidation gentlemen, I get the principle and after some thought SHOULD be able to implement it... I'll get back to you if I prove to be more helpless than I thought

Simon
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Old 16th May 2007, 11:34 PM   #7
Klimon is offline Klimon  Belgium
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Okay, I overestimated myself..

Quote:
The forummember probably meant that you should "tie" the heaters to the cathodes with 100-200 ohm resistors. The aim is just to elevate the heater voltage without disturbing the cathodes.
This sounds like the easiest solution to me, now step by step:

- should I just connect for each tube as follows "heater pin (one of both) -- resistor -- cathode" ?

- What power rating should the resistor have?

- Given that the amp uses two parallel tubes/ channel that share one cathode resistor I presume that I only need to 'tie up' one tube (heater to cathode) per channel?

Many thanks,

Simon
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Old 17th May 2007, 12:21 AM   #8
drj759 is offline drj759  United States
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That forummember was me.

My comment is at the end of this post located here:
http://www.diyaudio.com/forums/showt...57#post1045657

The comment was provided based on the schematic here:
http://home.hetnet.nl/~a.van.waarde/id9.htm
which I belive is based on Jeremy Epstein's version.

In that schematic, the cathode, which is at elevated potential (154 volts) due to the 3200 ohm resistor, is connected to the centertap of the heater transformer, thus elevating the heaters to cathode potential. Thus my comment: "tie the heater to the cathode as shown in the DC Darling schematic "

Bob Danielak's DC version on his site (that site seems to be down at the moment) shows two 100k resistors in a voltage divider off the B+ and half the B+ applied to the heater.

For right or for wrong, I used the Jeremy Epstein method in my implementation. Seemed to work.

Dave
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Old 17th May 2007, 01:38 AM   #9
Klimon is offline Klimon  Belgium
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Hello again Dave,

Unfortunately the center tap of my heater winding is already taken: it's a 12.5-0-12.5V trafo. I'm not sure if I can tie the unused tap to cathode, instead of the center tap like on the schematic. I don't understand Epsteins solution either: the circuit says only to tie one channel to heater, what about exceeding cathode-heater voltage on the other channel?

I found another example of voltage divider to heater here: http://www2u.biglobe.ne.jp/~tossie/
See 71A SE amp; this one uses an additional cap between second resistor and ground.

What value (and power rating) of resistors would one recommend; with extra cap or not? B+ is 390V..

Feeling a bit silly...

Simon
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Old 18th May 2007, 12:38 AM   #10
drj759 is offline drj759  United States
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Hello Simon.

Quote:
Originally posted by Klimon


I don't understand Epsteins solution either: the circuit says only to tie one channel to heater, what about exceeding cathode-heater voltage on the other channel?

Actually, Epstein's schematic says to tie the heaters to the left channel cathode via the center tap of the heater transformer secondary. This has the effect of "lifiting" or "floating" the entire heater transformer secondary to the voltage of left channel cathode (154 VDC). All of the heaters for the four 1626's (both channels) will be offset to the 154 VDC because they all have a common connection to the same elevated heater secondary winding. Both 154 VDC and 12.6 VAC are simultaneously present on the heater secondary winding. Power for the heaters is provided by 12.6 V AC. No power is supplied by the 154 VDC, only a DC offset.

Connecting to the left channel cathode is arbitrary. You could achieve the same result by swaping the connection to the right channel.

Quote:
Originally posted by Klimon


I found another example of voltage divider to heater here: http://www2u.biglobe.ne.jp/~tossie/
See 71A SE amp; this one uses an additional cap between second resistor and ground.

What value (and power rating) of resistors would one recommend; with extra cap or not? B+ is 390V..

For the voltage divider application, the standard practice that I've seen elsewhere on this forum is to place a large value electrolytic capacitor across the lower resistor to provide a low impedance AC path to ground. I've seen recommended values in the range of 100 to 150 uF. Voltage rating of the cap needs to be compatible with the voltage across the lower resistor. As an example, with a B+ of approx 400 V and using two 100k resistors, the heater offset would be 200 VDC. In that case, you would need a 250 V rated capacitor.

For sizing of the resistors, there's not supposed to be any current drawn by the heaters from the resistors (in reality there's probably a tiny bit of leakage current draw that can be ignored). The resitors do need to be sized for the current draw through the resistors from B+ to ground. Voltage rating is important too. For the example of B+ 400V, two 100k resistors: current through both resistors is 400V/200k = 2 mA. Power dissipated by each resistor: 100k * (2 ma)^2 = 0.4 Watts. Use at least a 1 Watt resistor.

Connecting to your transformer:

I think you could still use the ungrounded 0V center tap to connect the heater offset on your 12.5-0-12.5 transformer.

Dave
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