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Old 25th April 2004, 07:21 PM   #1
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Default SRPP Help Needed

Ive decided to build an SRPP line stage as my first tube project. I have the power transfo as well as the tubes purchased. The amp will use 5687wb nos jan tubes with a b+ of about 330v. The problem I have run into is the heater biasing on the upper section. I was planning on using ac to heat the tubes, I have 2 6.3v windings at 2.5A and 1.25A. So I have enough current to supply both the heaters on seperate windings, 0.9A per tube @ 6.3v. My question is how to reference the supply for the upper tube without over coming the cathode heater potential limits, which is about 90v in my case. Should I create a voltage divider off of my b+ @ around 1/2 of the B+ voltage? Will A DC reference volage cause problems or should I create the reference before the rectifier tube? I havnt recieved the transfo yet so Im not 100% sure if the 6.3v windings are center tapped or not, Im a pretty sure they are not though. Any advice would be appretiated, or sketches even

Thanks Mark
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Old 25th April 2004, 09:46 PM   #2
SY is offline SY  United States
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You're pretty close. Set up a voltage divider string with the output at roughly 20-30V above where you think the cathode will be. Use that to float the heater voltage. It's best to use a CT winding with a hum adjust if you really want to use AC. But you might want to seriously consider DC.

If you're worried about complication, compromise by going from SRPP to a plain vanilla voltage amp. It will do fine for that magic First Volt.
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Old 25th April 2004, 10:02 PM   #3
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Hi,

Heater to cathode insulation is indeed 90 V either way (+/-) but with 900mA AC running through the preamp it's going to be tricky to make it inaudible.

Running DC is a lot quieter but regulating it will require a low drop regulator.


6.3*1.4 = 8.82 so the reg can only lose about 2.5V, not impossible.

To bias the upper triodes you can use something along these lines provided you use a twin triode envelope there and the other for the bottom part:

Cheers,
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Old 25th April 2004, 10:22 PM   #4
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I think I will go for DC just to try to reduce noise as best as possible. For example if I have a cathode value of 160v If I had my heater set up for roughly 180 to 186.3 that would be sufficient? Would I bennifit from running a solid sate rectifier before my rectifier tube then running the voltage regulator, or would running right of the rectified B+ be sufficient? Is it possible to float DC biased heaters if so what would that look like? Is a pot a good idea for adjusting DC bias as well or is a fixed value ok? Just trying to iron this power supply out at the moment, its starting to make more sence though Just for interest sake I have 420vct which I want to tube rectify with a choke input, the rectifier and choke havnt been picked out yet, any advice on that would be great too, I have 2A @ 5v available to me. The idea is to get a nice quite power supply with a B+ of 330vdc , haha a worthy goal.
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Old 26th April 2004, 01:18 AM   #5
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Markedward,

Use a 5AR4 as the B+ rectifier. Your trafo's 5 V./2 A. winding is quite adequate for the 5AR4's heater requirements. Another good thing about the 5AR4 is its relatively low forward voltage drop.

For the P/S filter, a 10 H. unit of appropriate current rating feeding a 47 muF. cap. paralleled by a 10 KOhm POWER bleeder resistor forms the 1st section. Follow the 1st section with individual LC sections for each channel. 10 H. and 100 muF. in the paired 2nd sections should do nicely.

If you go with a DC heater supply, use Schottky diodes for rectification. Schottky diodes have a LOW forward voltage drop and ZERO switching noise.

An "easy" way to construct a DC heater supply is to use a full wave voltage doubler made from Schottky diodes feeding a cap. stack of 2X 4700 muF. 15 WVDC 'lytics. The doubler feeds a 7812 3 terminal regulator. Follow the regulator with a 10 muF. 25 WVDC 'lytic. Snub each 5687 heater at the socket with a 4.7 muF. 25 WVDC 'lytic in parallel with a 10 nF. CERAMIC cap. Apply B+ bias to the heater's center tap. BTW, connect the 2X 6.3 VAC windings in parallel. You need 4X the DC draw in AC RMS.
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Old 26th April 2004, 01:32 AM   #6
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Hi,

Quote:
I think I will go for DC just to try to reduce noise as best as possible.
IMO that would be best.

Quote:
For example if I have a cathode value of 160v If I had my heater set up for roughly 180 to 186.3 that would be sufficient
If you measure 160V on the cathode with respect to the heater and your cathode to heater insulation maximum is 90V it means you're 160V-90V = 70V over the allowed maximum.
In order to counterbalance that you'll need to inject at least 70V from the B+ into the heater to bring it within limits.
A little more won't hurt at all, quite to the contrary.
In your example you'd be pushing the heater positive with respect to the cathode by 26.3 V.
You don't want that either, the heater can theoretically emit electrons although it would exhaust quite rapidly.

Quote:
Would I bennifit from running a solid sate rectifier before my rectifier tube then running the voltage regulator, or would running right of the rectified B+ be sufficient?
Not sure what you have in mind here....

Quote:
Is it possible to float DC biased heaters if so what would that look like? Is a pot a good idea for adjusting DC bias as well or is a fixed value ok?
If you mean to ask whether biasing the DC heater in the same way as shown for AC heater then, yes.
A pot is only useful if you want to toy with different tube types, other than that a fixed voltage divider works just fine.

Quote:
Just for interest sake I have 420vct which I want to tube rectify with a choke input, the rectifier and choke havnt been picked out yet, any advice on that would be great too, I have 2A @ 5v available to me.
A 5Y3 would be more than ample for the 5687.
Keep in mind that a choke input type filter will lose quite some more volts than a capacitor input.
Predicting how much is tricky but apparently PSUD can do it quite well in software.

Cheers,
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Old 26th April 2004, 11:28 AM   #7
Mike C is offline Mike C  United Kingdom
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Default Low Voltage Drop Regulator

Quote:
Originally posted by fdegrove
Hi,

Running DC is a lot quieter but regulating it will require a low drop regulator.


6.3*1.4 = 8.82 so the reg can only lose about 2.5V, not impossible.

Yes, as Frank says, a low drop regulator is necessary and quite possible.

When you allow for the drop across a diode bridge, the unregulated
output looks like about 7.7 volts, not 8.8?

With 1.5V drop from 7.7v to 6.3v, the normal regulators will not work;
L78S series, LM317 etc etc. These require about 2.5V.

However, there are some that will including MIC2940A-5 which I use successfully with a low drop.
If I read the data correctly, it will run with only 0.6V drop across
it. I have about 0.9V drop and it has run fine.
Note that this is a fixed 5V regulator, so I use it with two diodes,
forward biased, from the ground pin to earth. This raises the earth
reference by 1.2V givng 6.2V output; this is the Steve Bench circuit.

The use of a fairly low supply voltage and a low voltage drop across
the regulator has a big advantage; lower heat dissipation from the
regulator chip. I find this really useful as I've never managed to get
full current output from 3amp rated L78S series devices even with lots
of heatsinking; they simply overheat at well below the rated current.
The MIC2940A may be rated less, but in fact I can get a lot more out
of it.
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Old 26th April 2004, 12:18 PM   #8
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Hi,

Quote:
When you allow for the drop across a diode bridge, the unregulated output looks like about 7.7 volts, not 8.8?
Yeah...those mathematical models never take real life components into account...and they still complain about Murphy, huh?

About 8V with a big fat cap after the bridge should be possible....sun is shining, I can be optimistic, right?

The diode trick works fine with other three pinners as well so it's a real nice cudo.

Cheers,

Why is it I always end up with too many volts when I don't need them and too little when....oh well...
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Old 26th April 2004, 08:26 PM   #9
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>>When you allow for the drop across a diode bridge, the unregulated output looks like about 7.7 V.<<

That's why I favor a voltage doubler. 7.7 X2 = 15.4. 15.4 V. is more than enough to make a 7812 "strike". In addition, current requirements at 12 V. are 1/2 those at 6 V. Therefore, the regulator IC is stressed less.
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Old 27th April 2004, 12:05 AM   #10
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Eli

Your design sounds very good, I am planning on simulating it this week in circuit maker if i can manage to get the tubes simulation working properly. Just one question about your design what do you think the current output would be given a 100ma ac rms value from the transfo. Im a little conerned it might be a bit low to drive 2 double triods.

Thanks Mark
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