Amplifier transistor matching?

[Spehro Pefhany]

On Tue, 14 Jun 2005 08:44:01 -0400, the renowned "Arny Krueger"
<arnyk@hotpop.com> wrote:

I don't know what they use for power internally. Based on
the limited headphone drive they provide, they may in fact
be running off the 1.5 volt directly.

No, *definitely* not. The last one I bought (Creative) has a blue LED
display backlight. That requires 3.5V. ;-) Similarly with the
Nike/Philips.

Trouble with the Li-ion cells is that every device seems to
want a different one.

Which may not trouble the OEM so much. 8-(


Best regards,
Spehro Pefhany
--
"it's the network..." "The Journey is the reward"
speff@interlog.com Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog Info for designers: http://www.speff.com

 

[Spehro Pefhany]

On Tue, 14 Jun 2005 08:56:56 -0400, Spehro Pefhany
<speffSNIP@interlogDOTyou.knowwhat> wrote:

On Tue, 14 Jun 2005 08:44:01 -0400, the renowned "Arny Krueger"
arnyk@hotpop.com> wrote:

I don't know what they use for power internally. Based on
the limited headphone drive they provide, they may in fact
be running off the 1.5 volt directly.

No, *definitely* not. The last one I bought (Creative) has a blue LED
display backlight. That requires 3.5V. ;-) Similarly with the
Nike/Philips.

P.S.

I just looked, the Nike/Philips unit uses a Sigmatel 3410L SOC which
has the DC-DC converter logic on chip (along with the 65MHz DSP, USB
interface and lots of other stuff). It uses a (relatively) large
external 4.7uH inductor and what appears to be quite a few other
discrete (cheap) components. The only ICs are the SOC and two Toshiba
TC58DVM92A1FT00 512Mb flash memory chips, the rest of the 150 or so
parts are discretes, not even an IC audio amplifier.

(photos about 200K each on 0.2" quadrille paper)
http://server2.hostingplex.com/~zstoretr/mp3top.jpg

(the inductor is the large black square to the lower right of the SOC)

http://server2.hostingplex.com/~zstoretr/mp3bot.jpg


http://www.toshiba.com/taec/components/Datasheet/TC58DVM92A1FT_030110.pdf
http://www.sigmatel.com/documents/App-Brief1-Flash-MP3-7-1.pdf


Best regards,
Spehro Pefhany
--
"it's the network..." "The Journey is the reward"
speff@interlog.com Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog Info for designers: http://www.speff.com

 

[Ken Smith]

In article <83ita11u0nspc331atu0ljndp8rjsaj167@4ax.com>,
Spehro Pefhany <speffSNIP@interlogDOTyou.knowwhat> wrote:
[...]

Chances are you're going to use a switching regulator anyway if you
want good battery life. Otherwise you're probably throwing away more
than half the energy.

If you are doing modest quality audio, you may not put the regulator in
the circuit but instead use a class-D audio output.


--
--
kensmith@rahul.net forging knowledge

 

[Ken Smith]

In article <42ae674a$0$16492$afc38c87@news.optusnet.com.au>,
Mr.T <MrT@home> wrote:

"Ken Smith" <kensmith@green.rahul.net> wrote in message
news:d8k3nn$cra$1@blue.rahul.net...
[....]
I suspect that the cells in the 9V are actually a little better than the
AA in energy per volume but I don't think it is by enough to make up for
the about 1/3rd of the space that is not actually a cell.

I note you have included NO data to back up your assertion.
Your claim of more energy per unit volume is looking dubious then.

The numbers are in another part of this thread, kindly posted by
someone else and they support my suggestion.




--
--
kensmith@rahul.net forging knowledge

 

[Mr.T]

"Spehro Pefhany" <speffSNIP@interlogDOTyou.knowwhat> wrote in message
news:hheta1lfgh57ggjorg38i9ld9omv5d898k@4ax.com...

This manufacturer has a wide line with three types of 9V cell alone.
You think they're going to improve the 9V cell just for you?

It's not ME bitching about them. I only use them when appropriate.

If you now use D cells instead, the
figures are MUCH greater in favour of D cells than AA.
Obviously by your reasoning nobody should ever use AA cells since D cells
are far better!

Only in energy storage per battery.

No.

*If* you read and understood my entire post, you would have seen the
other two comparisons-- of energy storage per gram and energy storage
per unit volume. In *all* cases, the best available AA cell of that
brand/series beats the best available 9V cell of that brand/series by
a considerable margin.

Yes I read it, and still say a D cell has more energy per unit volume.
Possibly per unit weight, but I'll let you do the calculation for that if
you like

Personally I would NEVER consider six AA cells as a good replacement for
a
9V battery when a **FAR** better option exists.

Who said anything about six? It takes seven to equal the voltage of a
typical 9V battery

Six alkaline cells = 9V, so I assume you are now talking NiMH. Of course
that only makes the situation worse size wise anyway.
However comparing apples with oranges does confuse the argument somewhat.

And a D cell much further still.

Got any numbers to back up that (quite dubious) assertion?

Yes, they are on the major battery manufacturers web sites.

Now where are the figures for six AAA cells and battery holder?

Waiting for you to calculate them. I have no need of the numbers at
present.

Nor I.

Shame you cannot use that wasted space between cells.

In some cases some of it can be used by a clever designer.

I can't say I've seen any of those clever designs then.

The AA cell is >3 times better than a 9V battery in *energy per cubic
millimeter*. "Size difference" is already taken into account in this
number. You want to take it into account again?

If you weren't stuck in the same groove, you might realise putting something
THREE times the size into a device which is supposed to be SMALL, is not
necessarily an option.
In this instance I am talking about SIZE period, not volumetric efficiency.

Run it off a Car battery if size is NOT an issue!

MrT.

 

[Mr.T]

"Ken Smith" <kensmith@green.rahul.net> wrote in message
news:d8mp0d$bev$4@blue.rahul.net...

In article <42ae674a$0$16492$afc38c87@news.optusnet.com.au>,
Mr.T <MrT@home> wrote:

"Ken Smith" <kensmith@green.rahul.net> wrote in message
news:d8k3nn$cra$1@blue.rahul.net...
[....]
I suspect that the cells in the 9V are actually a little better than
the
AA in energy per volume but I don't think it is by enough to make up
for
the about 1/3rd of the space that is not actually a cell.

I note you have included NO data to back up your assertion.
Your claim of more energy per unit volume is looking dubious then.

The numbers are in another part of this thread, kindly posted by
someone else and they support my suggestion.

The figures are rather misleading, but you are probably correct this time.
Your "I suspect that the cells in the 9V are actually a little better than
the AA in energy per volume" is still doubtful though.
I still think a DC-DC inverter is the best solution, as you can have small
and/or higher capacity as well as lower battery costs.
However this thread is getting too far into comparing apples with oranges,
IMO.

MrT.

 

[Jonathan Westhues]

"Mr.T" <MrT@home> wrote in message
news:42b114b3$0$16493$afc38c87@news.optusnet.com.au...

*If* you read and understood my entire post, you would have seen the
other two comparisons-- of energy storage per gram and energy storage
per unit volume. In *all* cases, the best available AA cell of that
brand/series beats the best available 9V cell of that brand/series by
a considerable margin.

Yes I read it, and still say a D cell has more energy per unit volume.
Possibly per unit weight, but I'll let you do the calculation for that if
you like

I used the alkaline Energizer Industrial series, because Digikey sells
them. All capacities are to 0.8 V per cell, at 25 mA. That puts 9V
batteries at a slight disadvantage, because that is more watts with a 9V
than a 1.5V, but it looks like the error is small enough that I wouldn't
do better trying to read a capacity off the graph.

The prices are in ones. They provide a volume, which appears to be more
or less the volume of the cylinder, but I also calculated the `bounding
box' volume.

AA cell:
m=23 g, Q=2779 mA*h, E=4168 mW*h, $=0.75 CAD, V=8.1 or 9.8 cm^3
E/m = 181, E/V = 515 or 425, E/$ = 5557

D cell:
m=142 g, Q=20500 mA*h, E=30750 mW*h, $=1.69 CAD, V=55.9 or 69.4 cm^3
E/m = 216, E/V = 550 or 443, E/$ = 18195

9V battery:
m=45.6 g, Q=625 mA*h, E=5625 mW*h, $=2.86 CAD, V=21.1 cm^3
E/m = 123, E/V = 267, E/$ = 1966

6V lantern battery:
m=665 g, Q=18000 mA*H, E=108000 mW*h, $=13.02 CAD, V=440.4 cm^3
E/m = 162, E/V = 245, E/$ = 8295

The energy figures are wrong because the voltage declines during the
life of the cell, but they will be equally wrong for all types so I
think that the numbers can be compared.

So that means that a AA cell is twice as good as (E/V) or 50% better
than (E/m) a 9V battery. A D cell is about as good as (E/V) or 20%
better than (E/m) a AA cell. That is sort of like what I expected.

What's wrong with the lantern battery though? Is there a good reason why
it's worse or did I make a silly mistake somewhere? Different capacity vs.
internal resistance tradeoff?

Jonathan

 

[Mr.T]

"Jonathan Westhues" <google-for-my-name@nospam.com> wrote in message
news:Vj9se.7292$yU.572117@news20.bellglobal.com...

"Mr.T" <MrT@home> wrote in message
news:42b114b3$0$16493$afc38c87@news.optusnet.com.au...
Yes I read it, and still say a D cell has more energy per unit volume.
Possibly per unit weight, but I'll let you do the calculation for that
if
you like

I used the alkaline Energizer Industrial series, because Digikey sells
them. All capacities are to 0.8 V per cell, at 25 mA. That puts 9V
batteries at a slight disadvantage, because that is more watts with a 9V
than a 1.5V, but it looks like the error is small enough that I wouldn't
do better trying to read a capacity off the graph.

The prices are in ones. They provide a volume, which appears to be more
or less the volume of the cylinder, but I also calculated the `bounding
box' volume.

AA cell:
m=23 g, Q=2779 mA*h, E=4168 mW*h, $=0.75 CAD, V=8.1 or 9.8 cm^3
E/m = 181, E/V = 515 or 425, E/$ = 5557

D cell:
m=142 g, Q=20500 mA*h, E=30750 mW*h, $=1.69 CAD, V=55.9 or 69.4 cm^3
E/m = 216, E/V = 550 or 443, E/$ = 18195

Thanks for showing that I was right and a D cell does in fact have more
energy per unit volume, and is FAR better in energy per dollar.
Frankly I couldn't be bothered going to the trouble you have, so I'm
impressed you did even though it proves you were wrong.

9V battery:
m=45.6 g, Q=625 mA*h, E=5625 mW*h, $=2.86 CAD, V=21.1 cm^3
E/m = 123, E/V = 267, E/$ = 1966

6V lantern battery:
m=665 g, Q=18000 mA*H, E=108000 mW*h, $=13.02 CAD, V=440.4 cm^3
E/m = 162, E/V = 245, E/$ = 8295

The energy figures are wrong because the voltage declines during the
life of the cell, but they will be equally wrong for all types so I
think that the numbers can be compared.

So that means that a AA cell is twice as good as (E/V) or 50% better
than (E/m) a 9V battery. A D cell is about as good as (E/V) or 20%
better than (E/m) a AA cell. That is sort of like what I expected.

Which puzzles me why you claimed otherwise then?

What's wrong with the lantern battery though? Is there a good reason why
it's worse or did I make a silly mistake somewhere? Different capacity vs.
internal resistance tradeoff?

They have always been a rip off such that adapters were even made to fit 4 D
cells instead. The E/V is pretty similar to the other rip off, the 9V
battery though.

MrT.

 

[Ken Smith]

In article <42b1187d$0$16705$afc38c87@news.optusnet.com.au>,
Mr.T <MrT@home> wrote:
[....]

The figures are rather misleading, but you are probably correct this time.

I think the figures are no more misleading than many.

Your "I suspect that the cells in the 9V are actually a little better than
the AA in energy per volume" is still doubtful though.

Pry apart a 9V battery and look at how the cell is constructed. There is
very little there that is not the active part of the cell.

--
--
kensmith@rahul.net forging knowledge

 

[Jonathan Westhues]

"Mr.T" <MrT@home> wrote in message
news:42b13882$0$13943$afc38c87@news.optusnet.com.au...

AA cell:
m=23 g, Q=2779 mA*h, E=4168 mW*h, $=0.75 CAD, V=8.1 or 9.8 cm^3
E/m = 181, E/V = 515 or 425, E/$ = 5557

D cell:
m=142 g, Q=20500 mA*h, E=30750 mW*h, $=1.69 CAD, V=55.9 or 69.4 cm^3
E/m = 216, E/V = 550 or 443, E/$ = 18195


Thanks for showing that I was right and a D cell does in fact have more
energy per unit volume, and is FAR better in energy per dollar.
Frankly I couldn't be bothered going to the trouble you have, so I'm
impressed you did even though it proves you were wrong.

[...]

So that means that a AA cell is twice as good as (E/V) or 50% better
than (E/m) a 9V battery. A D cell is about as good as (E/V) or 20%
better than (E/m) a AA cell. That is sort of like what I expected.

Which puzzles me why you claimed otherwise then?

You may wish to check the From: header. That was my first post to the
thread.

Jonathan

 

[Joerg]

Hello Rich,

Hasn't someone already mentioned the 6X AAA pack? Only a few mm bigger
than a 9V, but lasts considerably longer? ...

Yes, but that would be a kludge. It isn't rocket science to design stuff
so it can work off a couple AA NiMH. What's the big deal?

Regards, Joerg

http://www.analogconsultants.com

 

[Joerg]

Hello Arny,

Readily available DC-DC converters seem to want at least 2.7
volts input voltage, to produce 5 volts. So, you're stuck
with using 2 AA cells, which are somewhat longer and wider
than one 9 volt cell.

You can build your own DC-DC with cheap logic chips and that works just
fine down to 2V.

Also, nearly any function can be achieved with circuitry that works at
2V without a converter. I am glad that Sennheiser has finally done it as
well. Their G2 mics can take two AA NiMH. Way to go. Thing is, if they
hadn't done it we wouldn't have bought their stuff anymore.

It's probably the cost that dominates most design decisions
for consumer products. A DC-DC converter could add $20-30 to
the final cost.

Nah, the last one I designed was well under a Dollar. Ok, mass
quantities but even with low qties it wouldn't have been a lot more.

It seems like a single-AA cell solution could be practical
for higher end wireless mics and earphones for pro audio,
for example.

Yes. But it requires that engineers learn to design down to transistor
level again. With opamps it usually isn't going to work without the cost
going through the roof.

Regards, Joerg

http://www.analogconsultants.com

 

[Jim Thompson]

On Thu, 16 Jun 2005 21:25:03 GMT, Joerg
<notthisjoergsch@removethispacbell.net> wrote:

[snip]

I am glad that Sennheiser has finally done it as
well. Their G2 mics can take two AA NiMH. Way to go. Thing is, if they
hadn't done it we wouldn't have bought their stuff anymore.

[snip]

Joerg, Which G2 model (lapel mike, with receiver) do you recommend?
Thanks!

...Jim Thompson
--
| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona Voice480)460-2350 | |
| E-mail Address at Website Fax480)460-2142 | Brass Rat |
| http://www.analog-innovations.com | 1962 |

I love to cook with wine. Sometimes I even put it in the food.

 

[Joerg]

Hello Jim,

Joerg, Which G2 model (lapel mike, with receiver) do you recommend?
Thanks!

Don't know yet but I will have to figure that out anyway for our church.
Probably this weekend. When I found out I'll let you know. First I'll
have to find a reasonable dealer (good prices) but that's the easy part.

Regards, Joerg

http://www.analogconsultants.com

 

[Mr.T]

"Jonathan Westhues" <google-for-my-name@nospam.com> wrote in message
news:Vulse.7613$yU.616660@news20.bellglobal.com...

You may wish to check the From: header. That was my first post to the
thread.

My sincere apologies.

MrT.

 

[Tim Martin]

"Joerg" <notthisjoergsch@removethispacbell.net> wrote in message
news6mse.2844$NU5.601@newssvr13.news.prodigy.com...

Yes. But it requires that engineers learn to design down to transistor
level again. With opamps it usually isn't going to work without the cost
going through the roof.

By the way, while many op-amp circuits published specify a supply coltage of
+- 15 volts, or +- 9 volts, many circuit designs will work with +- 3 volts,
which is easily achieved with two pairs of AA cells.

Also, many circuits will take more power from one side of the power supply
than the other; so if you're using batteries to provide a +- power supply,
it's worth testing them to see if only half need replacing.

Tim

 

[Watson A.Name - \"Watt Su]

"Tim Martin" <tim2718281@ntlworld.com> wrote in message
news:0GCCe.1147$yH4.833@newsfe2-win.ntli.net...

"Joerg" <notthisjoergsch@removethispacbell.net> wrote in message
news6mse.2844$NU5.601@newssvr13.news.prodigy.com...

Yes. But it requires that engineers learn to design down to
transistor
level again. With opamps it usually isn't going to work without the
cost
going through the roof.

By the way, while many op-amp circuits published specify a supply
coltage of
+- 15 volts, or +- 9 volts, many circuit designs will work with +- 3
volts,
which is easily achieved with two pairs of AA cells.

Also, many circuits will take more power from one side of the power
supply
than the other; so if you're using batteries to provide a +- power
supply,
it's worth testing them to see if only half need replacing.

Tim

I just had to bash the hell out of a mini Maglite to get the innermost
AA cell out of it. It had started to leak and the corrosion on the end
was just enough to prevent it from coming out. I had to smakc it
against a concrete floor, with a few layers of paper towel on it to give
just a small amount of give.

Reason for saying this is that it'e really poor advice to give to reuse
half spent batteries. Works really great if the equipment takes a
single cell. But if you put a half spent battery in series with a new
one, it's likely that the half spent one will be discharged to zero and
beyond, leaking in the process. I think if I had two of four cells run
down faster than the other two, I would swap the sets halfway thru their
life. Better yet, get a piece of equipment that's not so poorly
designed that it runs down one set faster than the other.

 

[jakdedert]

"Watson A.Name - "Watt Sun, the Dark Remover"" <NOSPAM@dslextreme.com> wrote
in message news:11dnmgl96bnou12@corp.supernews.com...

"Tim Martin" <tim2718281@ntlworld.com> wrote in message
news:0GCCe.1147$yH4.833@newsfe2-win.ntli.net...

"Joerg" <notthisjoergsch@removethispacbell.net> wrote in message
news6mse.2844$NU5.601@newssvr13.news.prodigy.com...

Yes. But it requires that engineers learn to design down to
transistor
level again. With opamps it usually isn't going to work without the
cost
going through the roof.

By the way, while many op-amp circuits published specify a supply
coltage of
+- 15 volts, or +- 9 volts, many circuit designs will work with +- 3
volts,
which is easily achieved with two pairs of AA cells.

Also, many circuits will take more power from one side of the power
supply
than the other; so if you're using batteries to provide a +- power
supply,
it's worth testing them to see if only half need replacing.

Tim

I just had to bash the hell out of a mini Maglite to get the innermost
AA cell out of it. It had started to leak and the corrosion on the end
was just enough to prevent it from coming out. I had to smakc it
against a concrete floor, with a few layers of paper towel on it to give
just a small amount of give.

Reason for saying this is that it'e really poor advice to give to reuse
half spent batteries. Works really great if the equipment takes a
single cell. But if you put a half spent battery in series with a new
one, it's likely that the half spent one will be discharged to zero and
beyond, leaking in the process. I think if I had two of four cells run
down faster than the other two, I would swap the sets halfway thru their
life. Better yet, get a piece of equipment that's not so poorly
designed that it runs down one set faster than the other.

Measuring and matching used batteries is not that difficult, if you need to
use them in pairs. Also, some sources for used batteries are applications
where they were originally used in pairs. I collect and cull AA's from RF
mics all the time with little ill-effect. Sometimes I come in from a gig
with my pockets literally bulging with AA's. I usually just match them by
open-circuit voltage and put them in different piles matched within a 0.1 or
..05 volts of one another. Anything below 1.4 volts, I usually either
relegate to the kids' CD/Mp3 players or discard.

It saves me hundred$ a year, at least.

jak

 

[Tim Martin]

"Watson A.Name - "Watt Sun, the Dark Remover""

Better yet, get a piece of equipment that's not so poorly
designed that it runs down one set faster than the other.

OK. Suppose we have a circuit that draws 20 milliamps at -3v, and 220
milliamps at +3 volts. And suppose our power supply is four 1.5v 2200mAh
batteries, wired so one pair provides +3v, and the other pair provides -3v.

Is it possible to redesign the power supply to do better than replace the
+3v pair of batteries every 10 hours, and the -3v pair of batteries every
110 hours? (So for 110 hours of operation, we'd need to replace 12 pairs of
batteries.)

Tim

 

[Kristian Ukkonen]

Tim Martin wrote:

"Watson A.Name - "Watt Sun, the Dark Remover""
Better yet, get a piece of equipment that's not so poorly
designed that it runs down one set faster than the other.

OK. Suppose we have a circuit that draws 20 milliamps at -3v, and 220
milliamps at +3 volts. And suppose our power supply is four 1.5v 2200mAh
batteries, wired so one pair provides +3v, and the other pair provides -3v.

Is it possible to redesign the power supply to do better than replace the
+3v pair of batteries every 10 hours, and the -3v pair of batteries every
110 hours? (So for 110 hours of operation, we'd need to replace 12 pairs of
batteries.)

Make a switching powersupply using 6V input and outputing
two 3V voltages?

Use a latching relay to switch the batteries (3V <> -3V)
every once and a while?