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Old 13th April 2008, 08:00 PM   #1
slor is offline slor  United States
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Default Novice PS Question: Diode Dilemma

I'm building an active tube-powered DI box for my studio, and I'm looking for a little clarification on the PS, specifically what happens to voltages during rectification and why I should use one arrangement or another. (There is a LOT of information on the forum about this, I know, but I'm trying to winnow this down to easy bites!)

For starters, here's a schematic a tech provided me; it's slightly modded from the original (based on the Altec 1566) to accomodate a new power transformer:

Click the image to open in full size.

The diodes are 1N4007; I don't know what the bridge rectifier for the filament supply is supposed to be. Here are my questions:

-If the filament winding produces 6.3VAC, what will the DC output be? Another poster writes elsewhere:
Quote:
"When using a full-wave bridge rectifier arrangement, the peak DC voltage output will be 1.414 times the AC voltage input to the rectifiers, but this does need to be reduced by 2x the rectifier 'losses' to give the final voltage level for the circuit. This will depend upon the rectifiers used, but as another generalisation, 0.6V would be average here for each diode, so the final output from such an arrangement would be approx. 1.2 volts less than the calculated 1.4-ish times the AC from the transformer's secondaries."
In addition, I know from my last PS adventure that there will be losses across the filtering caps. A helpful poster wrote:

Quote:
You are discharging those capacitors and the change of voltage is: V = It/C = 0.3 x 0.00833/1000 x 10-6 = nearly 3V each
-Now to the B+. This is less confusing to me; if I spend enough time scratching my head, I think I can figure out voltage gains/losses across the diodes, caps and resistors. Question is: Do the voltage gains called out for a bridge rectifier above also hold true for the arrangement used for the B+? Again, I think I have all the tools I need, perhaps minus a few brain cells; what I'm asking for is an experienced pair of eyes to tell me: "You're on the right track, just do the math and adjust values accordingly," or, "This is the worst idea since the Edsel."

As always, replies and wisdom are most appreciated.

-Seth
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Old 13th April 2008, 08:22 PM   #2
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Hi Seth

Download the PSUD software from here.
http://www.duncanamps.com/psud2/index.html

You can than simulate the PS and most (all) relevant values will be calculated!

Erik
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Old 14th April 2008, 12:59 AM   #3
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Default Re: Novice PS Question: Diode Dilemma

Quote:
Originally posted by slor
The diodes are 1N4007; I don't know what the bridge rectifier for the filament supply is supposed to be. Here are my questions:

-If the filament winding produces 6.3VAC, what will the DC output be?
With a C input, the unloaded Vdc will be nearly the peak voltage, less the diode forward drop. In this case:

Vp= 6.3 * sqrt(2)= (6.3)(1.414)= 8.91Vdc

8.91 - (2 * 0.7)= 7.51Vdc (unloaded)

This is getting pretty thin here. The voltage under load will be less due to copper and leakage reactance losses in the xfmr, and voltage averageing as the filter capacitor discharges under load. If the load isn't too severe, it will probably give enough DC for the heaters. Still, I would eliminate it since a poorly filtered DC heater supply is worse than a relatively clean AC one. 1000uF with a small RC filter doesn't look like it'll get rid of enough AC harmonics for clean sonics.

Quote:
Do the voltage gains called out for a bridge rectifier above also hold true for the arrangement used for the B+? Again, I think I have all the tools I need, perhaps minus a few brain cells; what I'm asking for is an experienced pair of eyes to tell me: "You're on the right track, just do the math and adjust values accordingly," or, "This is the worst idea since the Edsel."
That arrangement is a full wave voltage doubler circuit. Like all voltage multipliers, it charges the capacitors in parallel and discharges them in series to give approximately twice the peak AC value. Since it discharges in series, the capacitors require twice the value for the required amount of ripple filtration.

This looks like a really old schemo. You can use bigger filter capacitors with a SS power supply these days.

Download a copy of this: Rectifier Applications Handbook.
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Old 14th April 2008, 03:32 AM   #4
slor is offline slor  United States
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Erik and Miles:

Thanks much for the info and links; I've downloaded the Handbook and will study up! A question, Miles, about the filament circuit. You write:
Quote:
This is getting pretty thin here. The voltage under load will be less due to copper and leakage reactance losses in the xfmr, and voltage averageing as the filter capacitor discharges under load. If the load isn't too severe, it will probably give enough DC for the heaters. Still, I would eliminate it since a poorly filtered DC heater supply is worse than a relatively clean AC one. 1000uF with a small RC filter doesn't look like it'll get rid of enough AC harmonics for clean sonics.
What do you mean by "eliminate it"? You mean, stick with AC? The draw, so you know, is only 2 x 12AX7.

Best,
Seth
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Old 14th April 2008, 04:03 AM   #5
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^^^^

Stick with AC for heating. With good balance, AC heater noise shouldn't be a problem unless it's a very low level stage.
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Old 14th April 2008, 06:33 AM   #6
slor is offline slor  United States
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Default Hmmm.....

Interesting, and thanks! I have the parts for two of these DI boxes, with the idea that a friend will buy the second (and, ideally, better) one. Maybe I'll try AC heaters in the first and DC in the second, if there are undue noise issues....

Also, if it makes any difference, I believe the filament circuit is oversupplied in terms of current: It's rated at 1A, and will only be supplying 2 x 12AX7, which I believe is around 600mA filament current draw. I'm told that because of the I/V relationship, higher current capacity coupled with lower current draw can manifest in the form of higher filament voltage. Any truth to this?
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Old 14th April 2008, 07:29 AM   #7
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Default Re: Hmmm.....

Quote:
Originally posted by slor
Also, if it makes any difference, I believe the filament circuit is oversupplied in terms of current: It's rated at 1A, and will only be supplying 2 x 12AX7, which I believe is around 600mA filament current draw. I'm told that because of the I/V relationship, higher current capacity coupled with lower current draw can manifest in the form of higher filament voltage. Any truth to this?
Yup. With capacitor input filters, the Vdc will increase with decreasing current demand. Ripple will also decrease as well.
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Old 14th April 2008, 02:19 PM   #8
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That looks to be a full wave voltage doubler in the B+. Since a doubler's caps only see half of the doubled output voltage, you can use the peak of your secondaries voltage. This is 120 in your case, so I would say 200v-250v for C9 and C7 should be fine. All other capacitors should be standard 350-450v since your output will be around 250ish volts from the doubler. You could go larger if you wanted. For most power supplies for tube stuff you should use the 1n4007 because it has a PIV of 1000v @ 1 amp. This gives you plenty of protection from back EMF. You could use another device, but the 4007 is widley available and very cheap. UF4007 is even better as it has a faster recovery time.
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Old 14th April 2008, 05:05 PM   #9
slor is offline slor  United States
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Gents:

Again, thanks for the info! The schemo I have is indeed old, and not knowing a great deal about diodes I had wondered if using different types would make a great deal of difference.

For instance, would using a lower-rated bridge, say 50V (the filament winding only puts out 6.3VAC) lessen the voltage drop and thereby increase the margin for error in the filament circuit?

Thanks much,
Seth
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