Cascading linear regulators

[rosborne]

I'm building a board that has a 5 VDC input. I need 3.3, 2.5, and 1.2 and am thinking of using linear regulators. Last night it occured to me that it might be advantageous to cascade the voltages down from 5 to 3.3 then from 3.3 to 2.5 etc. The kicker is that I think this is a better way to do it, but can't think of why it's better. Input?<p>-Rick

[cato]

Better than what? <p>On the plus side, the low voltage regulators will run cooler.<p>On the minus side, the 5 volt regulator will have to deal with the current flow through all the regulators. Therefore, it will run hotter. <p>Furthermore, any noise generated on, say the 3 volt circuit, will be couple through regulator to the 5 volt circuit. <p>For example, if there are switching currents in the 3 volt circuit (like a processor clock for instance) those currents will have to be supplied by the 5 volt regulator, therefore, the 5 volt supply will wiggle up and down at the rate of the switching.<p>If the 3 volt regulator is powered directly from a transformer secondary circuit,inparallel with the 5 volt regulator, those switching currents will be supplied by a lower impedance source and cause fluctuations only on the input side of the 5 volt regulator, where they are less significant and more easily filtered.

[rosborne]

Better than having all of the regulators operating off of the 5 V supply.

[cato]

How much headroom do the low voltage regulators need? I'm guessing that the 2.5 volt regulator will need an input voltage greater than 3.3 volts. The 1.2 volt regulator will probably work with a 2.5 volt regulator..... check the spec sheets.<p>[ October 25, 2003: Message edited by: cato ]</p>

[rosborne]

cato said,
>If the 3 volt regulator is powered directly from >a transformer secondary circuit,inparallel with >the 5 volt regulator, those switching currents >will be supplied by a lower impedance source and >cause fluctuations only on the input side of the >5 volt regulator, where they are less >significant and more easily filtered. <p>The daughter board with the 3.3, 2.5, and 1.2 voltage requirements will have a 5VDC supply tapped from the mother board. So, regulating the 3V in parallel with the 5V is not an option.<p>-Rick

[MrAl]

Hello there,<p>Rather then think about this, i wrote a quick
program to simply calculate all the powers
in three regulators, first hooking them up
in cascade, then hooking them up directly across
the input power supply.
The voltages used were rounded to 5, 3, and 1
volts, instead of 5, 3.3, and 1.2 .
The input supply voltage was a constant 9v.
The current supplied to each regulators load
was 0.1 amps.<p>The numbers below are in this order:
{p1, p3, p5, pT}
where
p1 is power dissipated in the 1v regulator,
p3 is power dissipated in the 3v regulator,
p5 is power dissipated in the 5v regulator,
pT is the total power dissipated by all
three regulators.
Power is in watts.<p>Here are the actual numbers with the cascade connection:
{0.2, 0.4, 1.2, 1.8}<p>Here are the numbers for all regulators connected
directly across the 9v input power supply:
{0.8, 0.6, 0.4, 1.8}<p>From this we can conclude that the total power
stays the same, although the distribution of
that power changes with the way we connect the
regulators, and that that distribution tells us
that the 5v regulator has to dissipate a lot
of power when they are all connected in cascade.<p>Of course this assumes adequate bypass is used
also.<p>
Good luck with your circuits,
Al

[dyarker]

Cascading is the way to go. Use LDO regulators.
Start with like Fairchild KA278R33TU, a TO-220 3.3V with 0.5V dropout. Then ILC7082AIM525X for 2.5V. ILC7801AIM5ADJX adjustable for 1.2V. Check the Mouser catalog if you need different current ratings and Very Low Drop-Out regulators.<p>Each following stage will not have more noise out because ... well ... they're regulated. Just be sure to use the caps recommended by the regulator's manufacturer. For an EWAG; 20uF tantalum and 0.01uF ceramic on 3.3V reg input, 10uF tantalums and 0.01uF ceramics between 3.3V 2.5V and 1.2V and on 1.2V output. 0.01uF or 0.0047uF ceramics near each logic IC.

[cato]

<blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr><p>Each following stage will not have more noise out because ... well ... they're regulated. [/QB]<hr></blockquote><p>My point is that each preceding stage will have more noise than it otherwise would. Regulators are relatively slow devices and non-ideal ones at that.<p>[ October 26, 2003: Message edited by: cato ]</p>

[rosborne]

The trouble is that I'd like at least 2 amps from the 1.2 V regulator.
-Rick