[sdiy] weird voltage regulator problem / question

[harrybissell]

Hi Karl...

s-diy is the BEST school :^P

Regulator problems are just applied ohms law.

1)  If your meter is reading DC without much ripple... that
effectively IS RMS. RMS really applies only to AC voltages...
and it is the "effective heating value" expressed in DC volts.

So 12V AC (RMS) makes the same amount of heat in a resistive 
load as 12VDC.

2) Your condition:

14V (meter) is 14V.  14 - 9 = 5V (drop across the regulator)

Take your current (whatever you draw, in amps) and multiply by the drop.
(70mA =)  .07A * 5V = .35W

Multiply this value by the thermal resistance of the regulator.
(assuming the 7809 in a TO-220 package, from the data sheet). 
This would be Tj-a  (resistance junction to ambient).  This is 
54 degrees C per watt (using National Semiconductor data sheet)...

54 + .35 = 18.9 degrees C above ambient. If the ambient is 25C the
regualtor
will be 44C.

If this is TOO hot, add a heatsink. Usually the bigger, the better.
The heatsink will have its own thermal resistance.

In this case... you would use the Tj-c value (4C / w) and the heatsink
value.
Add them to get the total. (there will be some added rise because of
contact
resistance, if you use thermal grease you can usually ignore this loss).

The Heatsink value might be 10C / W (for a small heatsink) or as low as
a fraction of 1C / W for a large heatsink, in a hurricane (moving air
reduces
temperature a LOT.

Such a heatsink with 10C / W + 4C /W (Tj-c) = 14C / W * .35W = 4.9C rise
over ambient. 

WAY cooler, yes ???   :^P

3)  "Internally regulated" means that if you get the junction TOO hot,
the
regulator will shut down, and protect itself from overheating. This is a
'dead synth' condition. It does not apply to you. It means that if you
get the
junction to the limiting temperature... it will stop functioning until
it cools
again. BAD regulator.  (really, BAD designer :^)

4) Wobbling in pitch.

The 9V should be steady. Measure the input to the regulator on AC
volts...
it should be a small number.  Usually these regulator need 3V more than
the output
voltage... minimum. You have only 5V... any AC ripple might make the
regulator
drop out.

Solutions : Bigger filter cap, or higher input voltage (which makes
everything
HOTTER again :^). Recalculate the new power and temperature.

If you have a scope, you can see the ripple voltage directly. If not,
you have
to rely on the meter and it is tough to know exactly how it will report
the
ripple (AC) voltage.

This variation would usually be 60Hz or 120Hz... more an FM growl than
a warble, IMHO

Are you with me ???

Check this out, then reply

H^) harry


Karl Ekdahl wrote:
> 
> Okay, so now i have a 12v line to my regulator(s), 14v
> says my multimeter so i guess 12v would be RMS? Anyhow
> a lot less, but maybe still too much since the
> heatsink i added which is very small. From what i
> gather this would result in (14-9)*0.07=350mW (70mA
> app. usage) that needs to be dissipated. Unfortunley
> the datasheet only says "internally regulated" on
> power dissipation.
> 
> If i understand the calculation below regarding the
> voltage drop correctly it's 24v - 9v = 15v and then
> what? 15v * 10v = 150ohm, yes, but why?? Anyhow, i
> guess then that what i need for 12v would be something
> in the vicinity of (14-9) * 10 = 50 ohm (@ 1W) ...
> right?
> 
> sorry for asking such fundamental questions, i really,
> really need to go to school again....
> 
> Anyhow. Why i'm asking is also because i noticed the
> oscillator (powered by said psu) wobbling in pitch,
> and especially if touching any powercable. With my
> limited
> experience this tells me there's a powersupply problem
> and i can't really think of anything being wrong with
> it except maybe above issue. (Yes, i have tried
> putting filter caps everywhere; no luck so far)
> 
> thanks people
> 
> Karl
> 
> ----------------------------
> 
> The heat given out by the regulator is solely
> dependant on the voltage
> across the regulator and the current through it.
> 
> If you have 24V going in, and 9V out, that's 15V your
> regulator needs to
> drop.
> 
> At 60mA, you have a power dissipation of 60mA x 15V.
> Which is 900mW.
> 
> Power dissipation = [Vin - Vout] x Iout
> 
> It is no surprise that your 79L09 is getting warm, in
> fact, I'm
> surprised it hasn't burned up yet. Check the maximum
> heat dissipation
> for the device on the datasheet and you will see that
> you need to be a
> lot lower than that, even if you can encourage some
> air flow.
> 
> I would add a dropper resistor in series with the
> input to help with
> heat dissipation. If you try to drop around 10V across
> the resistor,
> then this should be fine. R will be roughly 150R or
> so.
> 
> Don't forget that the resistor also needs be able to
> dissipate that heat
> too, so a 1W resistor is probably best.
> 
> Your other option is to use a 7809 and heatsink it.
> 
> Tony