Hoffman Amplifiers Tube Amplifier Forum
Amp Stuff => Tube Amp Building - Tweaks - Repairs => Topic started by: bruno on July 05, 2010, 02:18:20 pm
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I've got a marshall style 100w pt with a heater secondary that reads 3,50V per side, instead of the more usual 3.15V, what do you guys consider to be a safe error margin for this value is 7V usable or should I be looking for another pt?
Thanks
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If you are worried about it, just install 1N5408 nad it will drop voltage around 0.7V on each filament side. just be sure to check current rating on diodes in order to avoid burning them up.
Hope this helps
Best Regards
Rzenc
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forgot to add the schem. :rolleyes:
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7V no load is within normal. Maybe there's nothing to fix. 7 Volt under load is acceptable.
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JJ's right - 7 volts AC w/o load is fine. Heater supply of 7 volts under load may be OK, but it's my understanding that higher heater voltage can shorten tube life.
If you are worried about it, just install 1N5408 nad it will drop voltage around 0.7V on each filament side. just be sure to check current rating on diodes in order to avoid burning them up.
Hope this helps
Best Regards
Rzenc
The diode approach to reducing heater voltage throws me a bit. Wouldn't that act like a half-wave rectifier on the heater supply? Not saying it doesn't work, just curious.
As an alternative, could you use very small value power resistors? I know they'd have to have a high wattage rating (7 amp +/- current draw). If you put resistors in place of the diodes as shown above, would you calculate the net resistance of the two as being in parallel or in series? If in parallel, I think two 0.15 ohm/3 watt resistors would drop the voltage about 1/2 volt.
R = V/I = 0.5/7 = 0.071 ohms
P= (V^2) * I = 0.25 * 7 = 1.75 watts
PLEASE have someone verify my thinking before trying this! Also, I had trouble finding any power resistors in small enough values with high enough power ratings.
Chip
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I've got a marshall style 100w pt with a heater secondary that reads 3,50V per side, instead of the more usual 3.15V, what do you guys consider to be a safe error margin for this value is 7V usable or should I be looking for another pt?
Measure the winding DCR, add all heater current in your projected build, use Ohm's law: with 3A - just two EL34's ! - a .25R DCR will drop .75V, bringing heater voltage very close to 6V3.
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7V no load is within normal. Maybe there's nothing to fix. 7 Volt under load is acceptable.
yeah, we should have asked it... Bruno, did you measure it under load?
The diode approach to reducing heater voltage throws me a bit. Wouldn't that act like a half-wave rectifier on the heater supply? Not saying it doesn't work, just curious.
Chip
sand rectos have a constant voltage drop when operated above some microamps.
In order to act as a rectifier, it would be necessary that both diodes were oriented in the same way, so that each diode conduct positive cicles of current while the other would be off. When sine wave inverts, then the other diode starts to conduct and the very first diode would quit conducting.
On the drawing, the upper diode conducts positive cicles while the bottom diode conducts only negative cicles.
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rzenc, your top circuit cannot work. One of the diodes will always be reverse biased, and since they are in series with the load, no current can flow thru the load (filament circuit). The bottom circuit will work, but each diode will drop 0.7 volt and the resultant current will be unfiltered, halfwave rectified dc.
A better way would be to put the diodes parallel but back to back, then in series with the filament string. That way you can drop about 0.7v and still have AC voltage applied to the filaments.
However, IMO, 7VAC on a receiving tube filament designed to work at 6.3VAC is fine, whether that 7 VAC is loaded or unloaded.
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rzenc must be thinking of the diode method shown by Merlin: http://www.freewebs.com/valvewizard/heater.html
You get longer filament/tube life if yo run heates a tad low say, 5.9V. But that really applies to amps that are on a lot. A guitar amp may be on only a few hours a week. Maybe better for a guitar amp is a slow start heater circuit to limit voltage spikes or current surges at turn-on. See KOC, TUT1. I don't know of anyone actully implementing such a circuit. I guess I could see the point for vintage tubes.
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sluckey, does the attached show what you described?
FYL
Measure the winding DCR, add all heater current in your projected build, use Ohm's law: with 3A - just two EL34's ! - a .25R DCR will drop .75V, bringing heater voltage very close to 6V3.
I ASS-U-ME'd four EL-34s for 100 watts plus at least three 12AX7s, giving me 6.9 amps.
Can someone tell me whether two resistors placed in the second drawing are in series or in parallel? I think that they are effectively in series and that would be "good" in that you could cut the wattage rating in half. Of course, with resistors in series, we have to add the combined resistance. Is that on target or half-baked?
Finally, JJ's comments about the "soft start" for filaments makes a lot of sense. Merlin shows a heater standby switch at the bottom of This Page (http://www.freewebs.com/valvewizard1/standby.html) which looks a heck of a lot simpler than KOC's soft start approach. Actually, you could combine the heater standby with the opposing diodes and get the soft start plus a voltage drop as shown below. (only 1 pole of 2 shown - other would cut B+ from rectifier)
Thanks,
Chip
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sluckey, does the attached show what you described?
Yes. The link jjasilli posted shows it as well.
Can someone tell me whether two resistors placed in the second drawing are in series or in parallel?
The circuit is incomplete as drawn. But, if you connect each resistor to the filament pins of a tube, then the top resistor, tube filament, and bottom resistor are all in series. You could really just replace the two resistors with a single resistor.
Filament 'soft start' or standby is really wasted on a guitar amplifier. The standby was useful in the '60s when tv was still in the tube phase. Running a string of filaments thru a diode/switch would keep the filaments warm, resulting in very quick startup. Later, when tv receivers were mostly solid state, except for CRT, the diode/switch would allow the CRT filament to stay warm, giving a near 'instant on' feature. That may have been the beginning of the "we want it right now" mentality. :grin:
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you're right guys I measured it without load!
Thanks
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> 3,50V per side
Typical PT regulation is 10%. Under load, output is 90%. 90% of 7.0V is 6.3V. I'd say this is right-on.
> is 7V usable
Yes, on a new build, 7.0V even under load would not bother me. If I decided to keep the amp forever, I "might" someday do something. Or I might not.
> Measure the winding DCR
The full drop includes primary I and R. Primary I includes plate-power consumption. It is possible to measure this, but all those turns:ratios to factor!! You could assume that primary losses are similar to secondary losses: if you compute 5% sag from secondary I and R alone, the total sag may be near 10%. Or you could just not sweat the small stuff.
+1 for the series-diode plan NOT working.
It also gives a screwy waveform which may buzz more and which will not read correctly on many meters.
> "soft start" for filaments makes a lot of sense
Only if you see a lot of failed heaters, are not running looong strings of big-and-small tubes in series strings (TV sets), AND are not mechanically abusing the tubes.
The only failed heater I have seen in recent decades was a 200AB's 6SJ7 pentode. It had high hours (though few starts) followed by a 6-foot drop into a dumpster.
IMHO, adding "soft start" will not significantly increase heater life, and adds NEW parts to fail (at the worst time).
_IF_ the heater voltage under LOAD is actually over six and a half most days and nights I might add a couple resistors.
> the beginning of the "we want it right now" mentality.
Why did the automobile explosion-engine drive-out the steam engine? Despite the dangers of hand-cranking?
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Can someone tell me whether two resistors placed in the second drawing are in series or in parallel? I think that they are effectively in series and that would be "good" in that you could cut the wattage rating in half. Of course, with resistors in series, we have to add the combined resistance. Is that on target or half-baked?
The resistor drawing is correct for AC. The two resistors are in series and their resistance adds. Use Ohms' Law to find the resistance Value for the voltage drop you want. (Ohms' Law for AC is more complex, but the simple DC formula will work well enough.) Then divide that resistance Value by 2 to find the value for each resistor. Two resistors of equal value are needed to balance ea leg of the AC heater supply for proper hum cancellation. Use the Power formula for wattage rating. Remember there's a lot of current here. For a DC heater supply voltage drop, use 1 series resistor at the beginning of the +leg of the supply.
I guess the Merlin heater standby circuit also provides a good enough "soft start". Given, say, a 15V start-up voltage spike, in-phase with the single diode, then 7.5 volts would reach the filaments. Though as PRR says, there's no need for protection here.
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> Two resistors of equal value are needed to balance ea leg of the AC heater supply for proper hum cancellation.
Correct, if you use the hard CT on the PT winding.
If you use the 2-resistors CT and put it on the actual heater circuit (downstream of the dropping resistor), and the dropper is near the PT (before the heater system gets near any sensitive circuits), then you only need one dropping resistor. The PT winding floats off-center, but so what?
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rzenc, your top circuit cannot work. One of the diodes will always be reverse biased, and since they are in series with the load, no current can flow thru the load (filament circuit). The bottom circuit will work, but each diode will drop 0.7 volt and the resultant current will be unfiltered, halfwave rectified dc.
A better way would be to put the diodes parallel but back to back, then in series with the filament string. That way you can drop about 0.7v and still have AC voltage applied to the filaments.
However, IMO, 7VAC on a receiving tube filament designed to work at 6.3VAC is fine, whether that 7 VAC is loaded or unloaded.
Top circuit was only intended to show how a full-wave recto is set up, although it is not complete, my bad, I must have done it properly....sorry I was in a rush..and, indeed, as jjasilli mentioned, I was refering to Merlin arrangment, but draw it wrong.. :huh:
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Ah, the missing CT! That changes everything. :wink: