5V output circuit

[Roger Hamlett]

The maxim chip seems great but maybe a little complex, I got my ideas
from here:
http://www.instructables.com/id/EGBQJPLCB2EP287KTZ/#CEVEIRM4776EP288H78

And would now like to expand on the idea, I dont want to buy the kit
because I want to marvel at my own design, I have a basic idea but now
need to put th eplan into action. How woul 4 AA's 6V with a regulator
compare with the 3V step up, the 6V would be much easier.

For efficiency, you would want to use a switching regulator for both
routes. The 'down' switcher, removed the need for the inductor, but
remember that you would only be able to discharge the batteries to just
over 1.25v, before you lose the ability to deliver 5v, which will give
less utilisation of the available power in the cells. Realistically, the
boost regulator, done with an IC, designed to do this, should quite easily
get over 90% efficiency, and use ost of the available power in the cells
as well. However remember that board design round a switching regulator,
needs to be carefully done...
The battery in the last Ipod Ilooked at, was only a 150mAHr cell, and this
was quoted to give 3 hours typical life (giving a consumption of perhaps
only perhaps 50mA to 60mA - given the extra losses at the '3 hour' rate,
versus the power normally quoted at the '10 hour' rate). The circuitry
drew about 400mA, _when charging_ the battery.

Best Wishes

 

[mc]

And would now like to expand on the idea, I dont want to buy the kit

because I want to marvel at my own design, I have a basic idea but now
need to put th eplan into action. How woul 4 AA's 6V with a regulator
compare with the 3V step up, the 6V would be much easier.

It's hard to downregulate 6 V to 5 V because there's not enough drop. If
you had five AA's, giving 7.5 V, a 7805 chip would do the whole job. (I
suggest adding a 0.1 uF capacitor on the input and output sides.) That
gives you a 3-component circuit.

 

[mc]

I should add that if you use a 7805, be sure to disconnect it from the
battery when it's not in use; the same goes for other voltage regulator
chips; they consume a small amount of power of their own.

 

[Byron A Jeff]

It's hard to downregulate 6 V to 5 V because there's not enough drop. If
you had five AA's, giving 7.5 V, a 7805 chip would do the whole job. (I
suggest adding a 0.1 uF capacitor on the input and output sides.) That
gives you a 3-component circuit.

But it's a pennywise and pound foolish solution because you burn up a
significant amount of power in heat.

A stepdown switcher will have a limited number of parts too. Less than
6. It will have way better efficiency of conversion which means that the
batteries will last longer.

Another suggestion is that unless size is of the utmost importance, that
you'd be better off using D cell batteries. They have much more capacity
than AAs.

BAJ

 

[Walter Harley]

It's hard to downregulate 6 V to 5 V because there's not enough drop.

20 years ago it might have been hard. Now, LDOs ("low dropout regulators")
are as common as dirt.

But the thing to hold on to here is that a nominal 1.5V battery has to
discharge to about 0.8V before it is considered "discharged", that is,
before its full energy capacity is used. That's almost a 2:1 ratio. A
linear-regulator based solution that can still provide 5V at the end of the
discharge cycle must therefore start out with around 10V, half of which it
must drop as heat.

So a linear regulator is *never* going to be an efficient approach to this
problem.

An aside to Roger H: respectfully, although there are simple switcher chips
capable of 90% efficiency, that's with good magnetics, good PCB design, and
the like, and it's with input voltage at optimum (that is, 90% is usually
the peak of the efficiency curve, not the mean). The OP appears to be a
beginner, and it seems unreasonable to expect him to get the magnetics and
PCB design just right, at least the first time. Thus my earlier estimate of
70%.

 

[Roger Hamlett]

20 years ago it might have been hard. Now, LDOs ("low dropout
regulators") are as common as dirt.

But the thing to hold on to here is that a nominal 1.5V battery has to
discharge to about 0.8V before it is considered "discharged", that is,
before its full energy capacity is used. That's almost a 2:1 ratio. A
linear-regulator based solution that can still provide 5V at the end of
the discharge cycle must therefore start out with around 10V, half of
which it must drop as heat.

So a linear regulator is *never* going to be an efficient approach to
this problem.

An aside to Roger H: respectfully, although there are simple switcher
chips capable of 90% efficiency, that's with good magnetics, good PCB
design, and the like, and it's with input voltage at optimum (that is,
90% is usually the peak of the efficiency curve, not the mean). The OP
appears to be a beginner, and it seems unreasonable to expect him to get
the magnetics and PCB design just right, at least the first time. Thus
my earlier estimate of 70%.

Yes.
I'd say that these days, if you follow the guidelines published, with many
of the chips, something in the order of 80-85%, is not 'unreasonable' to
expect for somebody with at least general electronic competence, but 90%,
would require care, and 70%, is a much 'safer' realistic figure,
especially for use as a guideline. With care, you can get a mean over 90%
though. I have a switcher used in a somewhat similar application, and this
manages just over 93% on test in the first production units, over the top
1/3rd of the discharge curve, and only drops below 90%, as the battery
voltage goes below 1v.

Best Wishes

 

[jasonbot]

So, how do I make this 70-90% efficent circuit?

 

[Walter Harley]

So, how do I make this 70-90% efficent circuit?

You already got probably as much of an answer as you'll get on this group.
The answer, from MC, was:

See this data sheet:
http://www.maxim-ic.com/quick_view2.cfm/qv_pk/1327
And this:
http://pdfserv.maxim-ic.com/en/ds/MAX641-MAX643.pdf

But you didn't like that, because it seemed too complex. Unfortunately,
switching regulator design *is* complex, as power supply circuitry goes.

Sorry, I doubt someone is going to hand this one to you on a silver plate,
for you to "marvel at your own design" as you put it. I think your choices
are either to plunge in, or to buy it from someone else. The former will
perhaps be more rewarding; the latter will be much less expensive, in the
long run, and will be done sooner and be more reliable.

 

[mc]

You already got probably as much of an answer as you'll get on this group.
The answer, from MC, was:


But you didn't like that, because it seemed too complex. Unfortunately,
switching regulator design *is* complex, as power supply circuitry goes.

Sorry, I doubt someone is going to hand this one to you on a silver plate,
for you to "marvel at your own design" as you put it. I think your
choices are either to plunge in, or to buy it from someone else. The
former will perhaps be more rewarding; the latter will be much less
expensive, in the long run, and will be done sooner and be more reliable.

To elaborate on this a little: In order to actually post a diagram, I'd
have to breadboard the circuit myself first, and to do that, I'd have to get
one of these Maxim chips and a suitable inductor. Or preferably several, to
try. Then I'd have to specify exactly what you should buy. Notoriously,
inductors have lots of properties besides inductance (they also have current
carrying capacity, Q, resistance, and a strange interaction of current with
inductance) so you generally want to do more than just get the same
inductance that someone else used; they're often specified by make and
model.

I could spend $5 on parts, wait for them to arrive, and do it. Eventually
I'll probably do so, since I need to get experience with switching power
supplies.

But the reason I was able to give you a 3-component, inefficient but
reliable, 7805-based circuit off the top of my head is that I've build
hundreds of similar circuits. I don't have that level of experience with
swtichers.

 

[jasonbot]

Well, thanx everyone for all the help. I'll get to it next week. I'll
post pics when i'm done!