Re: gEDA-user: HPS announce

[John Sheahan <jrsheahan@optushome.com.au>]

I didn't look to see the expected load current guessed at 1A.

using a LDO reg and applying a PC-cpu style heatsink and/or heatsink fan
would also be avenues to consider for this style of power supply -if its 
getting warm
that is. The back of an aluminium case is not half bad as a heatsink.

(although I must admit its pretty hard to beat using an old PC supply
as an el-cheapo source of plenty of 5v and 12v)

For a variable bench supply - a switching power supply driving a linear
cleaner-upper stage  is hard to beat when the current goes up.

actually the most used supply on my bench is a venerable uA723 with an 
emitter follower,
as it has got range switched  current limit and monitoring down to mA.

digital meter modules are nice too on thepower supply.

john


Robert Riemer wrote:

> True, more transformer voltage = more Pd.  And worst case the 12V regulator is 
> dissipating 3.4 watts already.  With out doing the heatsink calculations, the 
> spec sheet suggests  a 10C/watt heatsink with an ambient of 35 C and Pd of 
> 7.5W.
>
>
> On October 29, 2004 07:24 pm, John Sheahan wrote:
>  
> > 47,000uf  ? thats huge.
> > lets think. 10ms period (50hz full wave)
> > 1A (for this reg)
> > i = c dv/dt => 
> > c = i * dt / dv
> >   = ,01/2
> >   = 5000uf
> >
> > I suspect you have a kind of a lot of significant figures 
> > for the accuracy of supply voltage too :)
> >
> > an extra couple of volts  from the transformer would help the cap size
> > and hurt power.  Thats linear supplies  for you.
> >
> >
> > john
> >
> >
> >
> >
> > Robert Riemer wrote:
> >
> >    
> > > After a few quick calculations regarding the power supply filter 
> > >      
> capacitance, 
>  
> > > I think you should spec a different transformer.  With the transformer you 
> > > have specified:  Vfl = 16.968Vp, I used this for my calculations even 
> > >      
> though 
>  
> > > you are not using full rated current draw for the transformer.  Vout = 
> > >      
> 12.0V.  
>  
> > > According to National Semiconductor the voltage accros the regulator could 
> > >      
> be 
>  
> > > 2.5V worse case.  I used Vf = 0.7V  for the diode.  Subtract all the drops 
> > >      
> > > from the transformer voltage, this leaves you with the maxinum ripple 
> >    
> voltage 
>  
> > > accros the capacitor.  dV = 1.768V.  Assuming a full load current the 
> > >      
> filter 
>  
> > > capacitor for the 12 volt supply should be >47000 mfd.  The 5V supply will 
> > > require an even larger capacitor.
> > >
> > > I may be under estimating the transformer ratings, and over estimating the 
> > > dropout voltage, but is that what a good designer should do?
> > >
> > >
> > >
> > >
> > >
> > > On October 29, 2004 06:16 pm, Robert Riemer wrote:
> > >
> > >
> > >      
> > > > Some technical tips:
> > > >
> > > > 1)  Use fewer 500 mfd capacitors.  There may be too much capacitance.  If 
> > > >   
> > > >        
> > > not, 
> > >
> > >      
> > > > use fewer larger value capacitors.  Probabaly the same price and size 
> > > > approx..  The value of capacitance depends on the transformer voltage, 
> > > >   
> > > >        
> > > amount 
> > >
> > >      
> > > > of allowable output ripple, and current drawn from the supply.  If the 
> > > > capacitance is too large, the regulator could dissapate too much power.  
> > > >        
> If 
>  
> > > > it is too small the output ripple may be large, our the regulator can drop 
> > > > out of regulation.
> > > >
> > > > 2)  The transformer is potentially unevenly loaded.  There is a split 12v 
> > > > supply but not 5v.
> > > >
> > > > 3) The terminal labled +Ureg (unregulated) should be as variable, not 
> > > > unregulated.
> > > >
> > > >
> > > >   
> > > >        
> >