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    Building, troubleshooting and testing of these amplifiers should only be
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Novice Voltage Regulator Question

slor

Member
2007-07-01 9:56 pm
I'm hard at work rehabbing an Ampex 601 preamp module. Most of the work is done, but now I want to build a regulated DC power supply for the filament circuit.

A fellow modder was kind enough to give me a schematic for the circuit:

[IMGDEAD]http://thegoldenbears.net/images/filamentsupply.jpg[/IMGDEAD]

Makes sense to me, more or less. My real question is about heatsinking the LM350 regulator: The tab on the regulator is punched to allow it to be connected to a heatsink, but is also paralleled to the "Voltage out" pin. Would connecting this tab to a heatsink, and connecting the heatsink to the chassis, effectively short the output to ground? A former poster here:

Aikido Power Supply

cautions to isolate LM350 from heatsink and chassis, so clearly there's something here I'm misunderstanding.... Excuse my inexperience, this is my first stab at such a project.

Thanks,
Seth
 
You need to use an insulating pad.. either silicone or mica. If you don't want to order them, you can get them at radio shack if you have one close.
 

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woody

Member
Paid Member
2002-01-15 12:57 am
Tyrone Ga. U.S.A.
A couple of questions come up. What will the voltage drop be
across the regulator and what is the current draw. In other words
what will the power dissipation be for the regulator? If your
power disipation is only 2 or 3 watts you might mount the device
to the chassis with the proper hardware. You would need a insulating sholder washer that fits through the whole in the device and an insulator. If you only need one of these let me
know and I will send you a pair of them. I am trying to change
my holiday sperit. Just send me your adress.

Regards

Ba Humbug! (this is a joke)

Woody
 

slor

Member
2007-07-01 9:56 pm
New Information, New Questions

Gentlemen:

Thanks much for your replies. Embarassingly, soon after posting this I found an online source for purpose-built insulators and heat sinks. (Though next time i will certainly visit the Shack, should have tried there first.)

As for the specs of the circuit, bear with me, as I'm new and inexperienced. My understanding is that the filament circuit should be putting out roughly 6VAC. However, because most of the tubes in this module get pulled, when I fired it up I was measuring nearly 8VAC. So I need to drop 2VAC (though I don't know if I'll need to lose the same amount once it's rectified).

I haven't looked up the current draw of the tubes I'm heating, though I will once I get a spare moment. I'm running a 5879, a 12AY7 and two 12AU7.

Another thing I have to educate myself on is the routing of the filament circuit. Currently, it's set up in the "usual" (that is, parallel, right?) fashion for AC. I've been advised that I can leave the negative side of this new regulated filament supply floating (not tied to chassis ground, right?) or tie it to chassis ground for potentially lower noise. In any event, aside from removing the now-unused tubes, is there any rewiring of the filament supply needed?

Thanks much,
Seth
 

slor

Member
2007-07-01 9:56 pm
LM350 or LM317?

Thanks much for the advice folks; I hooked everything up today and found that there's too much of a voltage drop, the highest I can get is around 5.7. Can anyone recommend a way to boost this? Substituting an LM317 for instance?

Thanks much,
Seth
 

poobah

Member
2005-11-15 7:24 pm
It would help to know the voltage of the INPUT to the regulator as well.

BTW... 5.7 is not the end of the world... within 10%. The tube experts 'round here may know something good about being at the minimum of the voltage (cathode temperature) spec...

:)

OH BTW... you should tie one output leg, or the other, of your DC heater supply to something low impedance... what exactly to tie it to depends on the cathode voltages and the limits imposed on heater bias relative to them. The heater circuit should at a AC ground... with a correct DC bias. Keeps one tube from talking to another... or other weird things.
 

slor

Member
2007-07-01 9:56 pm
Hmm....

Thanks much for the replies; after I posted I found this handy calculator:

http://diyaudioprojects.com/Technical/Voltage-Regulator/

Perhaps by varying the value of the resistor between the adjustment pot and the output voltage, I can get closer to where I need to be. (By the way, I tested the preamp with the filament voltage at 5.7v; it sounded weak and terrible!)

Unfortunately, there's a lot I don't yet understand about regulator circuits (this is my first). The voltage coming off the transformer is 7.1, and I assume the diodes induce a voltage drop. But before I built the regulator, I was reading nearly 8v on the tubes!

Also, I'm afraid I don't understand your adivce poobah:

"OH BTW... you should tie one output leg, or the other, of your DC heater supply to something low impedance... " Are you talking about a bypass capacitor? As it is one leg of the filament circuit is tied to the voltage output of the LM350, the other to chassis ground....

Thanks much,
Seth
 

poobah

Member
2005-11-15 7:24 pm
First things first...

What is the voltage at the INPUT to the LM350? Across the 100-uF cap?

The resistors are easy to calculate. We need to find out if this will work in the forst place.

Can you post the complet e schematic... OR, a link. then we can figure out how to deal with where to "tie" the heater circuit.
 

EC8010

diyAudio Moderator Emeritus
2003-01-18 7:57 am
Near London. UK
That circuit isn't going to work with any reasonable valve heater across it. Even a back-of-envelope calculation says that 300mA (a reasonable heater current) will cause 3V drop across each 1000uF capacitor, so that's 6V in total. Make a few more deductions for diode drop, etc, and it won't work. Modelling it in PSUD confirmed the back-of-envelope calculation. Increase the first two 1000uF capacitors to 3,300uF and throw the third one away. Then, it'll work for a 300mA heater.

Edit: I'm assuming you started with a 6.3V heater winding.
 

slor

Member
2007-07-01 9:56 pm
Hmm...I'll try it

Gentlemen:

Thanks much for the advice!

EC8010: Yes, I am using the unit's 6.3v transformer, thanks for asking! I will try the mod you suggest first as it's closest to what I've already built. But can you explain what changing the cap value is actually doing? Am I correct in thinking that there's not typically voltage drop across capacitors?

Pikkujöpö, I thought of using a resistor but worried about excessive heat (plus the fact that the circuit would be non-adjustable). But I'll read the article (thanks for the link!) and reconsider.

Thanks much,
Seth
 

EC8010

diyAudio Moderator Emeritus
2003-01-18 7:57 am
Near London. UK
Re: Hmm...I'll try it

slor said:
Am I correct in thinking that there's not typically voltage drop across capacitors?

Unfortunately, no. There's a big drop. 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. Worse, they're in series, so that makes nearly 6V, which is why your circuit didn't work.
 

poobah

Member
2005-11-15 7:24 pm
Increasing the cap value allows them to store more energy with much less voltage change.

These cap only charge, or gain energy, during the peak of the sine wave... between those peaks the caps have to supply energy to your heater circuit.

What the cat has found out is that the voltage on these caps is dipping too low in between the recharge pulses. Your regulator can't give you 6.3 VDC if the input is dipping down to 5 volts... or even 8.

You can almost look at this like putting bigger batteries in a flashlight... so it will burn longer.

ENERGYcap = [VOLTAGEcap * VOLTAGEcap * CAPACITANCE] / 2

:)
 

slor

Member
2007-07-01 9:56 pm
I see!

Again, thanks for the explanations! I do understand enough about caps to get around a power supply, but not enough to understand the voltage drop phenomenon (insert some homily about the dangers of a partial education).

EC8010: Thanks much for the formula, that's exactly the kind of information I need to get my head around. But can you explain what "It" (value= .00833) is in the equation?

This speaks to a larger issue than this post can address, namely the difference between knowing what to do (in other words, getting this preamp to sound pretty good prior to the voltage regulator mod) and knowing WHY you're doing it (the formulas and theories I need to know so I'm not guessing in the dark)!

So, in summation: If I can get the input voltage to the regulator high enough (by fiddling with the filter cap values), the regulator will still be dropping 3V or so, but I'll end up with enough (6.3V) at the other side?

Thanks again,
Seth
 

EC8010

diyAudio Moderator Emeritus
2003-01-18 7:57 am
Near London. UK
Re: I see!

slor said:
So, in summation: If I can get the input voltage to the regulator high enough (by fiddling with the filter cap values), the regulator will still be dropping 3V or so, but I'll end up with enough (6.3V) at the other side?

Yes. As the poobah said, I = load current, and t = the time between half cycles of your mains (60Hz), so one cycle takes 1/60 seconds, and the time between half cycles is half that = 0.00833s.