Simple Power Supply Question - Redeaux

[mydementia]

Hi Guys.
I'm starting to really get into playing with space charge tube circuits and realize that when I add another 12AL8, I'm going to exceed the current capability of my wall wart... so I bought a RS transformer and rectifier and am trying to calculate my required filter cap per RG's article here:
http://www.geofex.com/Article_Folders/Power-supplies/plus-9v3.gif

The equation is confusing me a bit...
Here's what I understand:

My power transfomer has 12.6VAC secondaries at 3A (37.8VA).
My rectified DC power should be 12.6*1.414-(4*0.7)~15VDC (assuming 4 0.7V diode drops in the rectifier).
This is sufficient voltage to feed my LM7812 to get a nice, steady 12VDC.

The equation:
Vout = (Vsecondary+1.414)-(2*Vdiode)-(Iload+(8ms)/(2+C)
When I tried to solve for 'C' I noticed that the parentheses don't work out so I don't know if the last term is:
(Iload+(8ms)/(2+C))
or
Iload+(8ms)/(2+C)
or something else

I also read several other sources that rectified DC will be approximately 1.414*Vsecondary (what's with the Vsecondary + 1.414 term?).

I'm trying to do the work to solve this problem myself (small as it may be) - but I want to solidify my education here so I'll know what to do with the next one (something about giving a man a fish vs. teaching him to fish?  )

Here's the tranny I'm working with:
http://www.radioshack.com/product/index.jsp?productId=2102702&cp=&sr=1&origkw=12.6v+transformer&kw=12.6v+transformer&parentPage=search
And the rectifier for reference:
http://www.radioshack.com/product/index.jsp?productId=2062580&cp=&sr=1&origkw=rectifier&kw=rectifier&parentPage=search

Thanks for looking.
Mike

EDIT:
I was just looking at the LM7812 data sheet again and noticed that it outputs 1A... I saw the 2.4A peak and interpreted it as 'max' - I'm assuming this is wrong!  I need about 2A for my next space-charge amp... any other ideas on how to get this current with at least 12V out of the parts I have?  I suppose I could try to filter the PS and live with the line variations... but I'm sure there's a better way...

[Mandrake754]

with regard to Vs * 1.414:

1.414 is the ratio between an RMS voltage and peak voltage (when dealing with ac). so when you multiply rms*1.414 you get peak, and when you multiply peak*0.707 you get RMS

I'm pretty sure this only applies to a sine wave ac signal, like that of the wall, not to any audio signal.

[R.G.]

The equation:
Vout = (Vsecondary+1.414)-(2*Vdiode)-(Iload+(8ms)/(2+C)

Well, first, you have the equation wrong.

that's  Vout = (Vsecondary*1.414) - (2*Vdiode) - ((Iload*8ms)/(2*C))

Let's take it a step at a time.

If you have 12.0000Vac, then full wave rectifying it with perfect diodes will produce a peak voltage of 12*1.414 = 16.968V.

However, no real diodes are perfect. Real 1A diodes will have a voltage drop of about 0.7V each in this kind of duty, so the real diodes will subtract twice the diode drop from that, or V = 16.968 - 1.4V = 15.568V.

That's the peak of the rectified DC, and if you put that into ANY cap, the cap will charge up to that and stay as long as there is no load removing charge between charging pulses.

But there is no point in charging a cap that can't discharge. We want to power a load. This will make some load current Iload.

The caps ONLY charge when the AC waveform is near its peak. In between, Iload comes out of the cap. A cap with a constant current load has a voltage which decreases linearly with time. So for approximately 8 of the 8.3333mS in a half-wave of 60 cycle AC, the load simply runs the cap down. The current to charge a cap is I = C*dv/dt. Solving that for dv, we get dv = I*dt/C. So a cap with a constant current load will change its voltage downwards by a voltage of delta-V = I * delta-time/C.

That gets us the Iload*8mS/C. Where did that 2 come from?

Well, the cap runs down by Iload*8mS/C and then is charged up to the peak. The AVERAGE voltage is half the difference between Vpeak and Vpeak-(Iload*8mS/C), or half of the run-down voltage.

[mydementia]

OK - now we're cooking!
RG - did I just read the equation wrong on your diagram? (linked in my first post).

So...
Solving for C...
C=  (Iload x 0.008sec)/2((Vsec x 1.414)-(2 x Vdiode) - Vout)

Iload=2A (shouldn't need more than this)
Vsec=12.6VAC
Vdiode=0.7V
Vout=12.0VDC (this is about what I want)

So... C= 1811uF (or if I want 12.6VDC, C=2096uF)

Look good?  Seems about right for PS filter caps...

Thanks for the clarification.
Mike

[R.G.]

Not quite.

The "solving for C" you're doing is technically correct, but the equation is obscuring something from you - the ripple voltage.

If Vout is 12.0V average, and you're getting 15.568V peak, the DC voltage is going up to 15.568V and then down to 8.432V to average to 12.0. That's a problem. You'll get massive hum from the 7V of ripple voltage.

Worse yet, you've not provided a minimal DC voltage for your regulators to work from.

What you need is for your regulators, presumably some variant of the 7812, to have 12.0v out and at least 2V across the regulator. That must come out of the raw rectified DC as well. So after losses for diodes, you get 15.568V peak. Your DC must never dip below 14.0V or your regulators will come out of regulation. That leaves you with a ripple voltage of AT MOST 15.568-14 = 1.568V peak to peak. That means that the downwards run of the capacitor voltage must never get below there.

So C = Iload *8ms/1.568V = 2A*8mS/1.568V = 0.0102F       um... is that -farads-?

Yes.  You need 10,200uF to keep ripple that low on a 2A supply.

For regulators the minimum voltage is what matters.

[mydementia]

Thanks for the detailed response!!
Now it's becoming clearer...
I had abandoned the idea of using a 7812 regulator since the spec sheet indicates a max current output of 1A (with max peak at 2.4A).  Are you saying these regulators will handle 2A just fine or is there more electrical trickery afoot?

I'm now starting to understand why the chipamp power board (from brian at chipamp.com) has two 10,000uF filter caps... same situation but no regulators...

Thanks again.
Mike

[Minion]

you can easilly increase the current capability of a regulator by useing a Pass Transistor, It should only require adding a resistor or 2 and a NPN Power transistor...the 7812 datasheet should have a curcuit example.....

You could also allways use two 7812 regulators on a good heatsink to get your 2A of current......

[mydementia]

Interesting stuff Minion...
From the LMXX data sheet, I see this schematic for high current applications:


So I'm passing current through the regulator AND through the transistor... does the transistor have an effect on output voltage?  What's up with R1?
Looks interesting...

I might go with the 'easy' way out and use two regulators... I suppose I could run about 1A worth of heaters from one and the other 0.5A heaters and the circuit off the other... at least I'd be moving forward...

Thanks for your input.
Mike

[Minion]

When useing a Pass transistor to increase the current of a regulator the Regulator regulates the Voltage and the Transistor is configured for Current gain so the transistor increases the current, most of the heat will be disapated in the transistor, The extra resistor is for biasing the transistor......The transistor will have no effect on the Voltage.....