24V Power supply

If i wanted to build a 24v power supply what voltage rating should the
torodal be rated at? is there any rule of thumb for this,

Thanks

 

Is this running off the mains directly our are you making a switcher.

You should assume that the mains are about 20% low, the transformer
has the highest leakage inductance and resistance, the diodes have the
most drop, the load current is at maximum, and the boss just entered
the room. You then work backwards to find out what turns ratio the
transformer must have.

 

Even better than a 'rule of thumb' is to calculate it ! It's not difficult.

Since you haven't specified if the supply is to be regulated or unregulated it's
not possible to do that yet.

Graham

 

Ok it is regulated using a LT1083CP, it is supplying 3.5Amps i was going to
use a GBPC2502A as i have one laying around and a 22000uf smoothing cap with
a 10uf on the o/p of the regulator. i see that torodals only come in 5, 6,
9, 12, 15, 18, 25v so i went for the 25v option but i think i am dropping
far to much voltage across the regulator? the 18v torodal doesn't seem to
deliver enough grunt?

Thanks

 

Dropout voltage is 1.2V. That means the input terminal must not drop below 25.2V
(assuming the 24V output is trimmed exactly).

Let's do the sums.

You need not less than 25.2V at the input taking into account ripple and low
line conditions.

The ripple is given by I_load . t_discharge / C_reservoir cap. At 50Hz
t_discharge (the length of time the reservoir cap is supplying the load current)
is ~ 8ms.

That gives 3.5 x 8.10^-3 / 22.10^-3 =~ 1.3V.

So, the reservoir cap needs to charge to 25.2 + 1.3 volts = 26.5V

Now check the forward voltage of the rectfier bridge. I tend to use 1V per diode
as a rule of thumb for around 1 Amp rectifiers btw. Let's see what the datasheet
says. 1.1V @ 12.5A. The peak current will actually be a little more than this so
let's say 1.2V.

There are 2 diodes conducting, so we need to add 2.4V to the previous 26.5V >>
28.9V

So the transformer needs to supply 28.9V peak == 20.4 Vrms (multiply peak by
0.7071) on load.

It'll need to do this @ low line conditions too or you'll lose regulation. Low
line is 230V -6% officially IIRC (216V) but I like to play safe and use 208V.

If we target 20.44 V @ 208V that gives us 22.6V rms @ 230V in.

What we don't know, since very few transfomers manufacturers tell you is a
detailed figure for regulation, but already it's clear that an 18V transformer
has no chance at all. A 25V secondary will give you some decent headroom in all
probability. You could even calculate the loading of the capacitor input filter
by extrapolating the effective winding resistance from the transfomer data but
that doesn't seem necessary here.

Graham

 

That was very well explained, ok so you are saying the 25v torodal? but im
dropping some serous volts across the regulator, i then have thermal
management issues? approx 11v?

 

Where did you get 11V from ?

Graham

 

25 x 1.414 = 35 .25v pk from torodal , maybe i over exgarareted that a bit
then i know the regulator gets fairly warm quickly not able to touch so i
was hoping to try and reduce the torodal voltage to a minimum avoiding
thermal management issues?

 

Did you not actually read my post ? Where's the voltage drop due to the
rectifier for example ?

Yes, you'll have to dissipate around 35W. You're going to need a decent
heatsink. That was obvious surely ?
http://uk.farnell.com/jsp/search/browse.jsp?N=411+1000162&Ntk=gensearch_001&Ntt=2.1+c/w&Ntx=

Graham

 

Sorry i did read your post and found it to be very well explained and
educational, im still getting my head round the maths bit though, i know
that there is an increase in voltage from my own practical experiments from
the cap and that there is a peak voltage drop across the bridge, i
calculated that an ideal torodal voltage of 21vrms 21x1.414 = 29.69vpk tying
in with you previous post and the calculation you made.

Have you ever used torodal in series i.e.: a 12v and 9v to give me the
required 21v?

 

I can't recall an occasion when I've needed to but it would certainly work and would help minimise
thermal overhead.

One thing you should be aware of is that those voltages quoted are likely to be a bit 'nominal'.
They're probably going to be the on-load voltages at the stated *resistive* load. From that you can
estimate the actual open-circuit secondary voltage and the equivalent source resistance. You see the
reservoir cap won't charge to the peak *open-circuit* secondary voltage when you're drawing full
load.

You'll also want to be be sure whether they've been specified @ 230 or 240V input.

Good luck with that. You *will* have to make some detailed measurements if you're to keep the
dissipation to a bare minimum. Also, if you know where it's going to be used, you could cut back on
the allowance for low line voltage if it's not going to be required.

Graham

 

I think I would still like to keep the allowance for the low line voltage
just in case, thanks very much for your help, I going to give it ago and see
how we get on.

Thanks Paul