On 11/7/2014 7:18 PM, rickman wrote:
On 11/7/2014 5:07 PM, Jerry Stuckle wrote:
On 11/7/2014 4:40 PM, rickman wrote:
On 11/7/2014 4:23 PM, Jerry Stuckle wrote:
On 11/7/2014 3:07 PM, rickman wrote:
On 11/7/2014 1:53 PM, Jerry Stuckle wrote:
On 11/7/2014 1:26 PM, rickman wrote:
On 11/7/2014 1:17 PM, Jerry Stuckle wrote:
On 11/7/2014 1:02 PM, rickman wrote:
On 11/7/2014 10:49 AM, Jerry Stuckle wrote:
On 11/6/2014 11:45 AM, rickman wrote:
On 11/6/2014 10:04 AM, Jerry Stuckle wrote:
On 11/5/2014 1:29 PM, rickman wrote:
On 11/4/2014 9:42 PM, Jerry Stuckle wrote:
On 11/4/2014 6:29 PM, rickman wrote:
I am working on a project for receiving a very narrow
bandwidth
signal
at 60 kHz. One of the design goals is to keep the power
consumption to
an absolute minimum. I'm trying to figure out how to run a
pre-amplifier on less than 100 uW. So far I have found
nothing.
Any
suggestions?


I agree with Jim. We need many more specifics to provide a
meaningful
answer. There are a lot of micropower opamps out there now,
but
the
devil is in the details.

I've only found one detail that is giving me the devil. That
is the
bandwidth. The signal is 60 kHz. I can't think of any other
issues I
would have with any amp capable of amplifying this signal with
a low
power level. What more info do you feel is needed? Can
you ask
questions? Better yet, just point me to any amp that will
meet
my two
stated requirements!


The other posts you made had the info - things like
impedance and
gain
are important, as is frequency of operation (but we already
know
that).

A couple of things to consider, however. The higher the
impedance, the
more susceptible it will be to ambient noise pickup. You're
starting
with a very small signal and may need to add shielding to limit
external
noise.

The other problem is you're asking for low impedance
output. Low
impedance limits noise pickup, but increases current drain. So
how low
of an impedance do you want?

I don't follow on this. How does a low output impedance
drive the
current drain?


There are op amps with very high (in the gigaohm range) input
impedance
and pretty low quiescent current drain. How much it draws
during use
will be greatly dependent on the output current required, which
obviously depends on output voltage and impedance.

Consider the current used only by the amp, not the load.


I don't have time right now, but later today I'll look through
some of
my data sheets on op amps to see what I can find.

Thanks.


Total current is not just dependent on output current; it also is
affected by the design of the chip. Op amps are not just single
transistor devices; a lower output impedance also means more
current to
drive the output stage, which affects other components. So even
if you
have a high impedance load, the lower the output impedance of
the op
amp
(i.e. the more current it can source/sink at a specific supply
voltage),
the more overall current the op amp will draw.

With that said, I did some looking around (sorry for not getting
back to
you quicker - yesterday was pretty busy). Depending on your
needs,
there are hundreds you can choose from. I might recommend you
check
out
http://www.mouser.com/Semiconductors...mps/_/N-6j73m/





. You can pick and choose the parameters you want. Another one
I've
used is http://www.newark.com/operational-amplifiers.

Between the two I found several hundred possibilities, but you
know the
details of what you want better than I do, so rather than
guess at
what
you might want, I think this would be better. It should give
you a
start.

I have done this before and found nothing. But I did it again at
both
Mouser and Digikey and found several. One listed by Mouser looked
especially good only to find rather than 0.75 uA of supply
current, it
had 0.75 mA of supply current. lol

But then the next part, same thing... another one... and
another... one
part I'm not sure what to make of it. The selection table shows
supply
current of 0.034 mA and the data sheet shows 25 A! Yes, that's
right,
the data sheet shows between 25 and 300 Amps for typical supply
current!!! I would contact TI about this obvious typo, but this
part is
not suitable because of the GBW which is also incorrect in the
selection
table.

Same thing at Digikey, everything in the selection table that
meets
these two requirements is a mistake.


A couple of things.

First of all, I've found minor errors in the listings at Mouser (I
don't
use Digikey much), but never real glaring errors. And this is th
first
time I've seen a TI datasheet that far off. Looks like someone
dropped
a decimal point . However, I've found Mouser is interested in
correcting errors; they are input by humans, after all, at some
point in
time, and errors do creep in.

Yes, when you list millions of parts there will be errors. I have
written digikey many times about listing errors and they always
thank
me. I'm sure Mouser is no different.


Secondly, the current shown is going to be max current, which will
depend on the output impedance (and the amount that has to be
sourced/sunk). It's not going to pull this all the time; I would
expect
your actual current draw to be much less since you're 1) going
into a
high impedance load and 2) not going from rail to rail.

I find the opposite. The current listed is under specified
conditions
which usually *do not* include output drive. In fact, it usually
listed
as a quiescent current.


Well, yes and no. Op amps typically sink more than they source, and
the
sink current does not come from the chip. Source current at the
output
is supplied by the chip, of course.

And I've found a wide difference between how op amp specs are listed;
some show quiescent current, some show average current under typical
operating conditions. Some even show maximum current which can be
drawn. So I'll retract that statement above. Wasn't thinking
clearly.


Also, if you use a bipolar supply, then current drain should be
less
because you'll be operating near ground, instead of the midpoint
of a
single supply voltage (where the output would be at 1/2 Vcc).
Some of
these are quite low voltage, and I would think a couple of the
larger
lithium coin batteries should last quite a while.

Not sure how the ground level would affect the bias currents. When
the
supply voltage is lowered the GBW lowers as well.


If the output is at ground level, no current will be pulled from
either
rail (at the output). Shifting above or below that will draw a
little
current, reference zero. However, if you're running a single ended
supply, your output will be at 1/2 Vcc, and will always be pulling
some
current to maintain that level. The signal will change that
slightly,
increasing or decreasing. But unless you have a square wave with
a 50%
duty cycle, you'll end up needing more current from the single ended
supply.

What you are saying is only true if your load is ground connected.
The
load for this circuit will be a voltage source through a high
impedance.
The input is differential and to make it as sensitive as possible a
bias will be applied to one input sufficient to offset the input bias
voltage. So in reality the load will be biased to approx 1/2 Vcc.


True, but with a bipolar supply, the input is referenced to ground and
no current flows with no input. The output is also referenced to
ground, so no current flows their, either. And with both input and
output at ground potential, there is less current flowing internally.

I just explained a scenario where the load will draw current from the
amp regardless of power supply arrangement. You are making an
assumption that the input and output are ground referenced. That is
independent of the supply arrangement.


But if they aren't ground referenced, then they must be referenced to an
artificial ground, i.e. 1/2 Vcc. And creating that artificial ground
will require a certain amount of current.

For instance - it's common to bias the input of an op amp running from a
single ended supply at 1/2 Vcc. This is generally done with a couple of
resistors, in various configurations. But you will always have a small
current through those resistors. The lower the impedance of the input,
the lower the resistors must be.

Output in this case will also be referenced at 1/2 Vcc, which means the
op amp output is conducting some current all of the time. Even if the
output is capacitive coupled to the load, internally the op amp must
draw some current to maintain that 1/2 Vcc. Again, the amount of
current is dependent on the output impedance, but it is still there.

With a bipolar supply, the op amp doesn't draw input current with no
input signal, and doesn't have to source or sink any current when you
have 0V output.

I hope this is a bit clearer.

It is not a question of clear. It isn't relevant to the power
consumption of the opamp. No matter what the reference, somebody,
somewhere even if it is in the power supply, is using power sometime
unless there are no voltages on any of the resistors in the design. But
none of that is relevant to the power consumed by the opamp when in the
quiescent state.


Yes and no. Op amps are by design bipolar devices; they go plus and
minus from some value. It can be zero volts (ground), or it can be some
value between Vcc and ground. In the latter case, an artificial ground
must be established; by definition this takes current to establish a
voltage between Vcc and ground.

I'm trying to pick an opamp. I have no need to evaluate the rest of the
design when the problem is trying to find an amp with sufficient GBW and
low quiescent current.


But when you're talking very low power like you are, low quiescent
current is very dependent on the power supply, as well as input and
output impedance and the chip used.

The real point is that this is *LOAD* current, not amplifier
current and
is independent of the amplifier and so considered separately since
selection of the amplifier has no impact on it.


That is true - to an extent. When the op amp is sinking current, the
current comes from the load, not the op amp. However, internally there
must still be current flowing to provide that sink.

You aren't grasping the issue. It is not about which direction the
current is flowing, it is about what is responsible for setting the
amount of current. The load determines the current that flows in or out
of the load and is independent of the opamp characteristics or the power
supply arrangement. What I can control by picking the opamp is the
current that flows through the opamp that is independent of the load or
input.


But the direction is important, also. You wanted to know how much the
op amp itself will draw; when sourcing the load, you will find more
current on Vcc then when the op amp is sinking the load.4

No, I am not asking what the opamp will "draw". I'm asking about low
power amplifiers. I'd be happy knowing how it is done in the chips in
the radio controlled clocks since I'm pretty sure I'm not going to find
a standard opamp that will do this.


You can; there are a number with that match your requirements. But you
need to design the entire circuit around your requirements, not just the
op amp.


But when the op amp is sourcing current, the amplifier has to provide
the source current plus the drive current. Now you can bias the output
so that the op amp is always sinking, but then you have a steady drain
from the standby current.

When you're using a bipolar supply and input is at ground, the output
will also be at ground, and very little current will be flowing.

You can create an artificial ground at 1/2 Vcc, but even creating that
artificial ground draws current.

The difference in current drain is not important in the vast
majority of
cases. But they become important when you're talking the low drain you
wish.

None of this is relevant to the issue of picking an amplifier.


It is if you're trying to minimize power usage to the maximum extent. I
admit my op amp theory is around 40 years old, but I don't think the
laws of physics have changed in that time

No, and the laws of relevancy haven't changed either. The power
consumed by the load doesn't impact my selection of amplifier.


In many cases, it really doesn't matter much. But it does, when you're
down in the microamp range. That's what you don't understand.

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