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Old 17th November 2010, 08:49 PM   #1
tdmur is offline tdmur  United States
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Default 6.3 filament from 5vac winding

Hi All,
This is my first post to a forum so please bear with me.I found this idea in the book Tonnes of Tone by Kevin O'Conner and thought I would try running my 3 preamp tube filaments on 6vdc to see if it made a difference in the hum of the amp.The article stated it could run up to 5 preamp tubes on a 5vac 2amp winding.I am putting it into a tweed chassis and with it being a little tight I changed the bridge rectifier from a KBP602 to a BR104.I also changed the transistor from a 2n3005 to a TIP3005G.the circuit runs at 6.28 vdc with no load but drops to 4.2vdc when I connect a .45amp load.Any ideas on the problem would be greatly appreciated.Please see the attachment for the schematic.Thanks,Tom
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Old 17th November 2010, 09:48 PM   #2
tomchr is offline tomchr  United States
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Theory says that you should get sqrt(2)*VAC on the reservoir cap. However, this assumes that the rectifier conducts infinite current for only an instant. In reality, you end up with 1.2~1.3*VAC. So on your reservoir cap, C4, you should see about 5*1.25 = 6.25 V.

At no load, you'll get very close to that voltage on the output. However, as the load current increases, the transformer output voltage will start to droop slightly due to losses in the transformer. In addition, Q1 starts drawing base current. Now, I don't recall the hfe of a 2N3055 off the top of my head, but I bet it's lower than 50 - especially at low collector currents. The increase in base current will cause the voltage across R6 to increase, so the base voltage on the 2N3055 drops. As the base voltage drops, the emitter voltage will drop as well. In addition, as output current is drawn, ripple voltage develops on the reservoir cap. As the regulator doesn't have any headroom (input voltage is the same as output voltage), this ripple will go straight through to the output which begs the question, why have a regulator there in the first place.

I'm sorry to say, but that design is fundamentally flawed if it's designed to operate off of a 5.0 V winding.

If you want the circuit to work with minimum changes, I suggest adding a 6.8 V zener diode in parallel with C5 (cathode of the diode to the positive terminal of C5). That'll give you a fairly stable 6.1~6.2 V output voltage. The drawback will be that the output voltage will depend on the temperature of the 2N3055, but as the transistor heats up, the output voltage will rise (about 2 mV per degree C) so you may get closer to 6.3 V out as the supply heats up. You still need to provide a higher input voltage, though. You might be able to get by running off of a 6.3 V winding, but I'd suggest using an 8~9 VAC transformer.

If you want to be more exact about it and get 6.3 V across temperature, I suggest looking at the LM317. It should be available at just about any electronic component shop near you. You'll need to provide a few external components (two resistors, one capacitor) and put it on a heat sink. You still need to provide a higher transformer voltage, though... The LM317 needs about 2.5 V across it to regulate properly.

Depending on what kind of transformer you're operating with, you could also add a winding to it... Like I did here.

~Tom

Last edited by tomchr; 17th November 2010 at 09:54 PM.
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Old 11th December 2010, 06:35 AM   #3
zega is offline zega  United States
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I guess the idea was to use transistor as smoothing ripple (active cap), but considering low starting voltage proposed design is not a good one.
Even using 317 s bad solution because of sharp cutoff when voltage sags below adjusted value (in this case) maybe some low drop regulator like 1084 (less then 0.8V drop @1A)or something even better, mayybe...
I would use better active cell with PNP serial transistor (so you don't loose 0.7V Vbe at least), just pure huge cap with some inductor in series before to smooth high diode/winding peak current spikes, or little step up (boost) smps (i know i know nooise, but actually it is not that bad)...
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Old 11th December 2010, 09:37 PM   #4
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You could use a full-wave voltage doubling rectifier before the voltage regulator.
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Old 11th December 2010, 10:09 PM   #5
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You may be making this too hard. 5vac through a bridge rectifier will give you 7v -
the voltage drop across the bridge which will probably be ~1v. So you probably
will end up with about 6vdc. Usualy we forgit about this voltage drop when dealing
with higher voltages but in this case it's an inportant factor.
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Old 12th December 2010, 03:11 AM   #6
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Hello,
6 Volts DC from a full wave bridge rectifier from a 5 V AC winding is not going to happen. You may not get 5V DC. Somehow I cannot remember that I cannot even get 6 V DC from a 6.3 volt AC transformer. Remember the typical diode drops over 1 volt and a full wave bridge rectifier drops voltage across two diodes or nearly a 3 volt drop (real world). You might get a voltage doubler to work but that is brute force and will require not only twice the number of capacitors, big ones because of the stacked current peaks with the big valley in between.
I use 8 Volt AC transformers for 6.3 Volt heaters, that leaves a little room for drop out on the regulator.
DT
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Old 12th December 2010, 05:09 AM   #7
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Make a full wave doubler using shottkey diodes and feed the regulator with that. At such low voltage large value capacitors are quite small and it does not matter it there is one or two volts of ripple, the regulator has enough headroom to remove it.
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Old 14th January 2011, 11:28 PM   #8
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Please realize that if a transformer is rated for 6.3VAC, then it will likely be higher than that with no load. Proper manufacturers compensate for typical loading. But, remember this is RMS. 6.3VAC is really around 8.9Vp or 17.8Vpp. When passed through a bridge rectificer; there are some voltage drops in the diodes (silicon is 0.7V). This is a greater percentage of your voltage than it would be for a high voltage supply. The output of the rectifier should be 8.9V ripple DC at 120Hz with no losses in the bridge rectifier and no load if the transformer puts out 6.3VAC.

If you are coming up short on your supply voltage, you can use a voltage doubler into an adjustable voltage regulator or just a voltage divider network using resistors. Please take wattage and heat dissipation into consideration. Good luck!
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Old 15th January 2011, 11:09 AM   #9
AndrewT is offline AndrewT  Scotland
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Be careful you don't overheat the secondary winding and the transformer.

Heater current is continuous duty. The transformer secondary must be rated to supply this continuous duty.

A 6.3Vac 500mAac secondary can supply a 500mAac heater continuously.

A 5Vac 500mA secondary when rectified and smoothed can only supply ~250mAdc continuously.

You must choose a secondary AC current rating at least double the total heater current requirement. Preferably choose 4times heater current to maintain cool running of the secondary winding.

At these low voltages the Vf of the rectifying diodes is significant.
Choose your voltages and diodes to give the output power you really need.
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Old 15th January 2011, 11:31 AM   #10
AJT is offline AJT  Philippines
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yor best bet is to forgo the regulators, use big caps, use schottky bridge for lower losses and see what happens....

depending on transformer construction, a heater winding rated at 2amps can usualy be drained 3amps without serious problem....
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