5v MCU powered from 5-24v

[Zero999]

However from the look of some data sheets I should be able to get around 4.8v out from a 5v source

Is 4.8V your minimum voltage or will it work at lower voltages?

Are you using the 5V as a reference? Could it be reduced to 3V?

If it'll work down to 3V You could just about get away with the L317L which comes in a small TO-92 of SOIC8 package but don't expect great regulation when the input voltage is only 5V.

The LM317 would be better but takes up more room so use the H or MDT version which comes in the smaller TO-39 or TO-252 packages.

I still think the LM431 is the best solution suggested so far.

[mobbarley]

Either way 200mW needs to dissipated, it either going to be the series resistor or the series voltage regulator.

So here is where I gave up on the shunt regulator, but perhaps I am doing it wrong:
Series resistor:
5V at least 10mA required = ~500R
@25V 500R = 50mA
50mA through 500R = 1.25 W.

This makes for a pretty large resistor. I was hoping to do this all SMD.

[mobbarley]

Is 4.8V your minimum voltage or will it work at lower voltages?

Are you using the 5V as a reference? Could it be reduced to 3V?

The MCU could run at 3V but the outputted data signals must be pretty close to 5V. The device that this controller is plugged into could supply 5, 12 or 24V but requires 5V data signals. If it took the input voltage for data I would settle for a 3.3v reg and use a logic level mosfet for the data line.

[Zero999]

So here is where I gave up on the shunt regulator, but perhaps I am doing it wrong:
Series resistor:
5V at least 10mA required = ~500R
@25V 500R = 50mA
50mA through 500R = 1.25 W.

Where did you get the 50mA from?

A shunt regulator such as the LM431 only requires 1mA to regulate properly and the PIC requires 10mA so that's 11mA in total.

The problem you now have is that you need a constant current source to power the shunt regulator - I didn't think of this before. If you select a resistor to give 11mA at 25V 20/11 = 1.8k the current will fall below that when the voltage drops below 25V. If you select a resistor which will only drop 0.5V at 11mA then 0.5/0.011 = 45, 43 Ohh being the nearest preferred value, you have 20/43 = 465mA, when the supply voltage is 25V. Maybe the LM431 wasn't such a bright idea after all.

The constant current source will need to be low drop out too so you might as well go with a low drop out voltage regulator - it's the only sensible option.

[RayJones]

Well it is certainly looking like the conventional linear regulators are all limiting in one aspect or another for the original reason I stated - heat dissipation.
You stated you are not after efficiency, but you don't want the heat....
You could just whack a few R's is parallel and share the heat love....

It really does start to make a transformer based switcher like Scott's design, actually National's, look more and more appealing.
That should EASILY fit in 2" square, especially if you wield the SMD stick.

It crossed my mind last night on the train you need something like a fully enhanced mosfet as a pre regulator for 5V and increase the Rds with increasing input volts.
Ala a heroic constant current source, at least the power is then being dissipated in a more heatsink friendly package, and the current rise does not get stupid giving such high power dissipation.
A fully enhanced mosfet at 5V is however back to front in term of Vgs, but hey it is only 6AM right now and the brain is still half asleep!

[Zero999]

Well it is certainly looking like the conventional linear regulators are all limiting in one aspect or another for the original reason I stated - heat dissipation.
You stated you are not after efficiency, but you don't want the heat....
You could just whack a few R's is parallel and share the heat love....

What are you talking about? The maximum current is only 10mA and with a 25V supply is only a miserable 200mW which should be easy to sink with a SMT component on a PCB copper plane.

The resistors also want to go in series with the regulator but they're not required in this case and will only increase the drop-out voltage so would be a bad idea.

[RayJones]

Oh FFS, the context I'm talking about was a shunt regulator.
The OP has already pointed out the current would be much higher with a 25V input, and he cannot tolerate the drop over a linear regulator at 5V input.

BTW, the current in the OP's current calcs for a 25V input to a 5V shunt regulator would not be 50mA, but 40mA with a 500R resistor:  => 20V drop over 500ohms.

20V x 20V / 500 = 800mW, certainly not as bad as the OP's original 1.25W, but still requires a meaty R to dissipate the heat.

So put 3 or four smaller R's in parallel to spread the heat.

However, the original calculation of 500 ohms is seriously flawed. 
For sure you'll get 10mA with 5V across 500R, leaving NO VOLTS INTO THE MCU!

So make it 50 ohms = 0.5V drop @ 10mA, but 400mA at 25V input
That's now 8W in the series R.

 - SHUNT IS NOT GOOD FOR WIDE INPUT RANGE -

The LDO is a possible answer, but not the only one.
http://www.national.com/ds/LP/LP2954.pdf shows a reasonable dropout voltage, best you could probably hope for, but pay due caution to the stability requirements in the DS.

And yes I'll grant you 200mW for a LDO voltage regulator.

At the end of the day it is the OP's choice which way he goes.
If he truly desires 5V at 5V in, a SMPS is the only way out.

[mobbarley]

BTW, the current in the OP's current calcs for a 25V input to a 5V shunt regulator would not be 50mA, but 40mA with a 500R resistor:  => 20V drop over 500ohms.

I tried to make a worst-case generalisation, i'm not sure if that is correct or not. In reality the shunt regulator would also burn off excess power into heat spreading some of the heat over its transistor (might need to be external) - but with about a watt constant i'm sure my little board would get quite warm.

[Zero999]

In a shunt regulator, most of the power will be burnt off in the series resistor when the input voltage is high and in the MCU when the input voltage is low.

I also explained the problem of the shunt regulator current not being constant a few posts ago.
https://www.eevblog.com/forum/index.php?topic=1548.msg20805#msg20805

[slburris]

So if no one likes the National switcher idea, how about this?

Since the current requirements are so low, build a switched capacitor
voltage doubler (maybe with a CMOS 555?) that's only enabled when
the input voltage is below, say 8 volts.  Feed the output of the voltage
doubler through a diode to a conventional linear regulator. 

Also, connect the original input voltage through a diode to the regulator
as well.  So the higher of the two voltages powers the regulator.  If the
input voltage is above 8 volts, then the regulator cuts that down to 5 volts.
If the input voltage is below 8 volts, the capacitor doubler makes that
V x 2 and that powers the regulator which regulates it back down to 5 volts.

Or something like that.

Scott

[RayJones]

So if no one likes the National switcher idea, how about this?

Scott

I like the switcher Scott and still think it is the most suitable answer and would not take up that much more space.

Heck it will even probably deliver 5V out with an input voltage below 5V.

[Zero999]

I disagree, I think a low drop out linear regulator is the best solution: one IC, a few capacitors and no inductors, why add extra complexity when it's not required?

[Strube09]

I too think it is major over kill for a 10ma source. What kind of efficiancy will that produce at such a low load. I bet the dive current of the controller and the gate of the switching FET will kill your efficency at this low of power. @ 5V we are only looking at a 500mW load.

[slburris]

The original problem was produce 5 volts out with an input voltage as low as 5 volts.
If that's true, then I don't see how you get away from some sort of switcher/capacitor
switchover design.

If the problem really is produce 4.75v out with an input voltage as low as 5 volts,
then some LDO solution will work fine.

I'm just not seeing shunt designs working well with such a wide input voltage range,
although I suppose you could drop a several watt zener after a 0.1ohm resistor and
get things to work.

Re: the question about the switcher efficiency. Attached is the efficiency graph
of the webench design.  It's at least 70% efficient or better.

Scott

[Zero999]

The original problem was produce 5 volts out with an input voltage as low as 5 volts.
If that's true, then I don't see how you get away from some sort of switcher/capacitor
switchover design.

If the problem really is produce 4.75v out with an input voltage as low as 5 volts,
then some LDO solution will work fine.

The regulator I suggested at the start of the thread, LM2936Z-5, will give 4.85V out when the current is 10mA. The LM2940 is overkill but should output over 4.9V when the output current is just 10mA but the quiescent current is also 10mA which is probably a bit higher than we'd want.

I've found a better IC, the LP2980 which will only drop 90mV at 10mA (worst case) and the quiescent current is negligible.
http://www.national.com/ds/LP/LP2980.pdf

The original poster explained that the requirement is due to the circuit interfacing with a 5V circuit which should be able to accept 4.9V inputs with no problem.

I'm just not seeing shunt designs working well with such a wide input voltage range,
although I suppose you could drop a several watt zener after a 0.1ohm resistor and
get things to work.

Yes, I think we ruled out shunt regulators a few posts ago.

Re: the question about the switcher efficiency. Attached is the efficiency graph
of the webench design.  It's at least 70% efficient or better.

I don't doubt the efficiency but it's run from a wall so it doesn't matter - the extra complexity and expense of an SMPS simply don't justify the benefits.

[cybergibbons]

If I'm ever trying to solve a problem, and I find that no-one else has already solved the problem, then I probably need to restate the problem, maybe break it down into several distinct issues.

What are you trying to achieve? I can't think of anything that has such a voltage range.

If it is trying to work with a range of wall warts, that probably isn't enough of a range. I've got 3V up to 28V here, and then you need to deal with polarity and AC.

[NiHaoMike]

You could use a PTC with a shunt regulator, so it self adjusts, but a LDO is a better choice.

[RayJones]

The original problem was to produce 5 volts out with an input voltage as low as 5 volts.
If that's true, then I don't see how you get away with not using from some sort of switcher/capacitor
switchover design.

If the problem really is to produce 4.75v out with an input voltage as low as 5 volts,
then some LDO solution will work fine.

Scott

I've taken the liberty to edit in Italics some words to better convey what I presume was Scott's original intent....

Yes I agree entirely, and that has been the basis of my assertions.

If the OP wants 5.0V out with 5.0V in then it must be a switcher.
If the OP is happy with 4.75V (5% low) then a suitable LDO will do.
If the OP needs a regulated 5V from an unregulated 5V, then once again the switcher is the only one that could possibly cope with Vin dipping below 5V.

There has been plenty advise already espoused in this thread.
We have explored the shunt regulator and determined that due to the widely spec'd input range it is 100% unsuitable despite initial appearances.
The LDO will work if the design is OK with a "5V" rail likely to dip by 5% to 10%.

By the sound of it, the design all really hinges on the device required to be driven.
Is it logic? If so, what's The minimum acceptable level for a logic "1"?
It would be all academic if it external input is actually the ubiquitous TTL compatible, nominally 5V but Vin(min) = 2V!
A 5V coming back can be easily handled either by level translators eg 5V to 3V3 stuff, or micros that will cope with input levels higher than Vcc.

Only the OP knows these answers, we've supplied heaps of answers, tell us more please

[mobbarley]

Here we go.

Yes I agree entirely, and that has been the basis of my assertions.

If the OP wants 5.0V out with 5.0V in then it must be a switcher.
..
if the OP needs a regulated 5V from an unregulated 5V, then once again the switcher is the only one that could possibly cope with Vin dipping below 5V.

No arguments there - but I was trying to keep the parts / size / cost to a minimum as it needs to be small - its a 'circuit on a plug' of sorts. The power is supplied by the plug and will be either a 5, 12 or 24v regulated source depending on the brand / configuration of the equipment it is plugged into. The data lines input back to the device and must only be 5v.

If the OP is happy with 4.75V (5% low) then a suitable LDO will do.

I am coming to the realisation that 5v in and 5v out is not really possible over this input range without a switcher

There has been plenty advise already espoused in this thread.
We have explored the shunt regulator and determined that due to the widely spec'd input range it is 100% unsuitable despite initial appearances.
The LDO will work if the design is OK with a "5V" rail likely to dip by 5% to 10%.

I've done testing and found that anywhere around the 3v mark is recognised by my equipment, so this is sounding like the simplest option

By the sound of it, the design all really hinges on the device required to be driven.
Is it logic? If so, what's The minimum acceptable level for a logic "1"?
It would be all academic if it external input is actually the ubiquitous TTL compatible, nominally 5V but Vin(min) = 2V!
A 5V coming back can be easily handled either by level translators eg 5V to 3V3 stuff, or micros that will cope with input levels higher than Vcc.

If only it was that simple. The 'specification' (if I can call it that) is 5V high 0V low - as there is a lot equipment this could be plugged into and I only have one brand to test with the aim was to get as close as possible - of course the difference between 4.5 and 5v is nitpicking so an LDO looks like the best option.

Only the OP knows these answers, we've supplied heaps of answers, tell us more please