Need to drop 0.2 V with 2-ternimal series device

Winfield Hill wrote:
Tim Hubberstey wrote...

Winfield Hill wrote:


I'd use 1% resistors. If you move the 1.2M and change its value,
you can make the 1.25V reference current independent of the battery
voltage by running it from the 1.35V output. Then if you select a
different opamp, you can run it from your 1.55V silver oxide cell.

. +---+-----+-------+--------+----
. | | | | |
. === | |LT1389 | 1.00M
. Camera | \_|_ | |
. Body | /_\ ,--- |--------+
. _|_ | | | |
. - | | _| 80.2k
. 1.55V | | '--|- \ | to
. | | | >---+--+---> Meter
. | +-----|+_/ | -1.35V
. | | | |
. '---- |-------' |
. '-- 68k ------'
. 1.5uA --

Looks interesting... but won't it need a large-value resistor to the
anode of the LT1389 to get it to start up? Looks to me like it would
come up with the output = 0 and just stay there. Or could we rely on
noise to give it enough of a "push" to get it started?


I posted a note concerning startup. This type of bootstrap circuit
start up fine if the opamp's output has more BJT saturation voltage
than the input's offset (divided by the feedback-resistor ratio), so
when the power is applied, the + input sees a bit more voltage than
the - input and drives the output further toward the turned-on state.

There are two problematic issues. First, the feedback resistors to
common tend to reduce the BJT saturation voltage. Using high-value
resistors helps; e.g., here they draw a current of 30nA for say 30mV
of output-transistor saturation. We note that 30nA is probably much
less than the base-drive current for the PNP output transistor.

Second, when the output is only modestly-higher than the reference,
the "offset voltage divided by the feedback-resistor ratio" parameter
becomes painful. Here we would divide our estimated 30mV by 12.5 to
get a 2.5mV max offset-voltage spec, which might be hard to meet.
Click to expand...
Another consideration for these modern LV amps is that the amplifier
gain approaches 0 at rail saturation voltages- so that Vin,diff=Vout/Aol
is untenable, the differential cannot exist with enough magnitude to
support that locked up state.
 
On 16 Mar 2005 05:42:21 -0800, Winfield Hill
<hill_a@t_rowland-dotties-harvard-dot.s-edu> wrote:

Tim Hubberstey wrote...

Winfield Hill wrote:

I'd use 1% resistors. If you move the 1.2M and change its value,
you can make the 1.25V reference current independent of the
battery
voltage by running it from the 1.35V output. Then if you select a
different opamp, you can run it from your 1.55V silver oxide cell.

. +---+-----+-------+--------+----
. | | | | |
. === | |LT1389 | 1.00M
. Camera | \_|_ | |
. Body | /_\ ,--- |--------+
. _|_ | | | |
. - | | _| 80.2k
. 1.55V | | '--|- \ | to
. | | | >---+--+---> Meter
. | +-----|+_/ | -1.35V
. | | | |
. '---- |-------' |
. '-- 68k ------'
. 1.5uA --

Looks interesting... but won't it need a large-value resistor to the

anode of the LT1389 to get it to start up? Looks to me like it would

come up with the output = 0 and just stay there. Or could we rely on

noise to give it enough of a "push" to get it started?

I posted a note concerning startup. This type of bootstrap circuit
start up fine if the opamp's output has more BJT saturation voltage
than the input's offset (divided by the feedback-resistor ratio), so
when the power is applied, the + input sees a bit more voltage than
the - input and drives the output further toward the turned-on state.

There are two problematic issues. First, the feedback resistors to
common tend to reduce the BJT saturation voltage. Using high-value
resistors helps; e.g., here they draw a current of 30nA for say 30mV
of output-transistor saturation. We note that 30nA is probably much
less than the base-drive current for the PNP output transistor.

Second, when the output is only modestly-higher than the reference,
the "offset voltage divided by the feedback-resistor ratio" parameter
becomes painful. Here we would divide our estimated 30mV by 12.5 to
get a 2.5mV max offset-voltage spec, which might be hard to meet.
Click to expand...
Assuming the anode of the zener and the 1Meg output resistor are
earthed V(+) = V(-) at the opamp inputs would be = Vee. If we had a
opamp configuration like a 741 the input transistors would be cut off
as would the darlington stages. That would force the opamp output into
a positive direction . With other Opamps any other simple solution
would do.
 
Winfield Hill wrote...
Tim Hubberstey wrote...

Winfield Hill wrote:

I'd use 1% resistors. If you move the 1.2M and change its value,
you can make the 1.25V reference current independent of the battery
voltage by running it from the 1.35V output. Then if you select a
different opamp, you can run it from your 1.55V silver oxide cell.

. +---+-----+-------+--------+----
. | | | | |
. === | |LT1389 | 1.00M
. Camera | \_|_ | |
. Body | /_\ ,--- |--------+
. _|_ | | | |
. - | | _| 80.2k
. 1.55V | | '--|- \ | to
. | | | >---+--+---> Meter
. | +-----|+_/ | -1.35V
. | | | |
. '---- |-------' |
. '-- 68k ------'
. 1.5uA --

Looks interesting... but won't it need a large-value resistor to the
anode of the LT1389 to get it to start up? Looks to me like it would
come up with the output = 0 and just stay there. Or could we rely on
noise to give it enough of a "push" to get it started?

I posted a note concerning startup. This type of bootstrap circuit
start up fine if the opamp's output has more BJT saturation voltage
than the input's offset (divided by the feedback-resistor ratio), so
when the power is applied, the + input sees a bit more voltage than
the - input and drives the output further toward the turned-on state.

There are two problematic issues. First, the feedback resistors to
common tend to reduce the BJT saturation voltage. Using high-value
resistors helps; e.g., here they draw a current of 30nA for say 30mV
of output-transistor saturation. We note that 30nA is probably much
less than the base-drive current for the PNP output transistor.

Second, when the output is only modestly-higher than the reference,
the "offset voltage divided by the feedback-resistor ratio" parameter
becomes painful. Here we would divide our estimated 30mV by 12.5 to
get a 2.5mV max offset-voltage spec, which might be hard to meet.
Click to expand...
At which point Tim's suggested resistor is the way to go.

.. +---+-----+-------+---------+----
.. | | | | |
.. === | |LT1389 | 1.00M
.. Camera | \_|_ | |
.. Body | /_\ ,--- |---------+
.. _|_ | | | |
.. - | | _| 80.2k
.. 1.55V | | '--|- \ | to
.. | | | >-----+-+---> Meter
.. | +--+--|+_/ | -1.35V
.. | | | | |
.. | 10M '--- |- 68k -'
.. | | |
.. '-----+-------'

Assuming a good-quality reference, this two-resistor bias scheme has
little advantage over one resistor to the battery, unless a variety of
battery voltages is expected and power conservation is very important.


--
Thanks,
- Win
 
Tim Hubberstey wrote...
Winfield Hill wrote:

I'd use 1% resistors. If you move the 1.2M and change its value,
you can make the 1.25V reference current independent of the battery
voltage by running it from the 1.35V output. Then if you select a
different opamp, you can run it from your 1.55V silver oxide cell.

. +---+-----+-------+--------+----
. | | | | |
. === | |LT1389 | 1.00M
. Camera | \_|_ | |
. Body | /_\ ,--- |--------+
. _|_ | | | |
. - | | _| 80.2k
. 1.55V | | '--|- \ | to
. | | | >---+--+---> Meter
. | +-----|+_/ | -1.35V
. | | | |
. '---- |-------' |
. '-- 68k ------'
. 1.5uA --

Looks interesting... but won't it need a large-value resistor to the
anode of the LT1389 to get it to start up? Looks to me like it would
come up with the output = 0 and just stay there. Or could we rely on
noise to give it enough of a "push" to get it started?
Click to expand...
I posted a note concerning startup. This type of bootstrap circuit
start up fine if the opamp's output has more BJT saturation voltage
than the input's offset (divided by the feedback-resistor ratio), so
when the power is applied, the + input sees a bit more voltage than
the - input and drives the output further toward the turned-on state.

There are two problematic issues. First, the feedback resistors to
common tend to reduce the BJT saturation voltage. Using high-value
resistors helps; e.g., here they draw a current of 30nA for say 30mV
of output-transistor saturation. We note that 30nA is probably much
less than the base-drive current for the PNP output transistor.

Second, when the output is only modestly-higher than the reference,
the "offset voltage divided by the feedback-resistor ratio" parameter
becomes painful. Here we would divide our estimated 30mV by 12.5 to
get a 2.5mV max offset-voltage spec, which might be hard to meet.


--
Thanks,
- Win
 
Winfield Hill wrote:

I'd use 1% resistors. If you move the 1.2M and change its value,
you can make the 1.25V reference current independent of the battery
voltage by running it from the 1.35V output. Then if you select a
different opamp, you can run it from your 1.55V silver oxide cell.

. +---+-----+-------+--------+----
. | | | | |
. === | |LT1389 | 1.00M
. Camera | \_|_ | |
. Body | /_\ ,--- |--------+
. _|_ | | | |
. - | | _| 80.2k
. 1.55V | | '--|- \ | to
. | | | >---+--+---> Meter
. | +-----|+_/ | -1.35V
. | | | |
. '---- |-------' |
. '-- 68k ------'
. 1.5uA --
Click to expand...
Looks interesting... but won't it need a large-value resistor to the
anode of the LT1389 to get it to start up? Looks to me like it would
come up with the output = 0 and just stay there. Or could we rely on
noise to give it enough of a "push" to get it started?
--
Tim Hubberstey, P.Eng. . . . . . Hardware/Software Consulting Engineer
Marmot Engineering . . . . . . . VHDL, ASICs, FPGAs, embedded systems
Vancouver, BC, Canada . . . . . . . . . . . http://www.marmot-eng.com
 
T

Tim Hubberstey

Guest
Hi all,

I've got a problem that I've been wracking my brain over for a while
now. I need a small (under 5 parts) 2-terminal circuit to drop 0.2 V at
up to 100 uA.

The situation is that I have an old but high quality camera that used a
1.35 V mercury cell for the light meter. These cells are no longer
available due to a world-wide ban on mercury cells. I had a few extras
stashed away but those are now gone. :(

The closest thing I can find with similar discharge characteristics is a
silver oxide cell at 1.55 V. Compounding the problem is the fact that
the camera uses the metal body as the positive ground and getting access
to the on-off switch would require far more disassembly than I'm
comfortable with. Hence the need for a 2-terminal solution. The battery
check circuit in the camera considers a cell to be "good" when the
terminal voltage is between 1.27 and 1.35 V at 90 uA load.

A 3 V manganese dioxide lithium cell is also an option if paired with a
3-terminal positive-ground regulator with very low (<10 uA) quiescent
current.

The standard "solution" involves putting a low-current Schottky in
series with the cell, but reports are that this still causes significant
errors in the meter readings on this particular camera model.

The 5 parts limitation comes from the small amount of space available in
the base of the camera, in between various gears.

I don't usually do this type of design so I'm looking for some help.
Anyone got some ideas?

Thanks
--
Tim Hubberstey, P.Eng. . . . . . Hardware/Software Consulting Engineer
Marmot Engineering . . . . . . . VHDL, ASICs, FPGAs, embedded systems
Vancouver, BC, Canada . . . . . . . . . . . http://www.marmot-eng.com
 
On Mon, 14 Mar 2005 07:07:28 GMT, Tim Hubberstey <bogus@bogusname.com>
wrote:


Thanks, Robert!

I didn't use your circuit but you did give me an idea and the lead on
the LT1389. This is what I ended up with:

+---+-----+-------+--------+
| | | | .|.
=== | | | | |
Camera | | | | |1M
Body | | | '-'
| |LT1389 | |
| _|_/ +---|--------+
| / A | | .|.
--- | | | | |
3V - | | | | |68k
LiMnO2 | | | |\| '-'
| | +-|-\ | to
| | | >-------+---> Meter
| +-----|+/
| | |/| LT1494
| .-. |
| | | |
| | |1M2 |
| '-' |
| | |
+-----+-------+
Click to expand...

---
Shouldn't it be wired like this?


+-------+----------+-- Camera body
| | |
| [1M] |
|+ | | LT1494
[BAT] +---------|+\
| |K | >--+-- To meter
| [LT1389] +--|-/ |
| | | | [68K]
| | | | |
+-------+-[1M]-+---|----+
| |
+------------------+

--
John Fields
 
John Fields wrote:
Shouldn't it be wired like this?

+-------+----------+-- Camera body
| | |
| [1M] |
|+ | | LT1494
[BAT] +---------|+\
| |K | >--+-- To meter
| [LT1389] +--|-/ |
| | | | [68K]
| | | | |
+-------+-[1M]-+---|----+
| |
+------------------+
Click to expand...
No, this circuit sets the "to meter" point at 1.35 V above the negative
rail. The output needs to be referenced to the positive rail (body).
Thanks for checking it over, though.
--
Tim Hubberstey, P.Eng. . . . . . Hardware/Software Consulting Engineer
Marmot Engineering . . . . . . . VHDL, ASICs, FPGAs, embedded systems
Vancouver, BC, Canada . . . . . . . . . . . http://www.marmot-eng.com
 
On Mon, 14 Mar 2005 07:07:28 +0000, Tim Hubberstey wrote:

Robert Baer wrote:
Tim Hubberstey wrote:

Hi all,

I've got a problem that I've been wracking my brain over for a while
now. I need a small (under 5 parts) 2-terminal circuit to drop 0.2 V
at up to 100 uA.

The situation is that I have an old but high quality camera that used
a 1.35 V mercury cell for the light meter. These cells are no longer
available due to a world-wide ban on mercury cells. I had a few extras
stashed away but those are now gone. :(

The closest thing I can find with similar discharge characteristics is
a silver oxide cell at 1.55 V. Compounding the problem is the fact
that the camera uses the metal body as the positive ground and getting
access to the on-off switch would require far more disassembly than
I'm comfortable with. Hence the need for a 2-terminal solution. The
battery check circuit in the camera considers a cell to be "good" when
the terminal voltage is between 1.27 and 1.35 V at 90 uA load.

A 3 V manganese dioxide lithium cell is also an option if paired with
a 3-terminal positive-ground regulator with very low (<10 uA)
quiescent current.

The standard "solution" involves putting a low-current Schottky in
series with the cell, but reports are that this still causes
significant errors in the meter readings on this particular camera model.

The 5 parts limitation comes from the small amount of space available
in the base of the camera, in between various gears.

Are you saying that the 1.55 V silver oxide cell messes up the reading
or functionality of the camera?

It throws the meter off by about 1.5 stops. The camera is totally
mechanical, except for the meter.

Other than a resistor (if current load is relatively constant), theer
is nothing that would give a 0.2V drop.
A schottky or germanium diode will give about 300-350mV drop,
depending on current and diode rating.
Now using a 3-terminal regulator with a 3V source seems to be a
solution, but you need one with two attributes: Low Drop-Out (or LDO in
the trades) *and* low drain.
Maxim advertises a shunt regulaor that supposedly works down to 1uA,
but they can be completely discounted, as Maxim/Dallas parts are vaporware.
However, Linear technology does have the LT1389 shunt regulator,
available at 1.25V, 2.5V and 4.096V, speced to operate from 0.8uA to
2mA; curves show operation down to around 0.4uA.
Then their LT1634 shunt regulator, available at 1.25V, 2.5V, 4.096V
and 5V, speced to operate from 10uA to 2mA.
Their LT1004 seems to have similar specs, but available only at 1.23V
and 2.50V.
The LM285-2.5 works from 20uA to 20mA, and there are adjustable
versions made by National Semi.
So, you could make your own 3-term regulator, as most of what is cited
above is available in SOT-23 or SOIC-8.
Idea: NPN pass transistor, base to cathode of adjustable, FB of
adjustable to tap of voltage divider: 1.25 meg from FB to ground (anode)
and 100K from FB to NPN emitter.
Now all one needs is base drive from NPN collector that will run the
regulator and the NPN.
Say one allows a LM285 regulator current of 10uA and a base drive of
1uA (yes, that is high - so the extra also goes thru the LM285).
The base will be 1.35V (ouput or emitter voltage) plus a Vbe of about
600mV, or about 1.95V; making the Vcb about 1.05V.
Obviously, no constant current device (JFET or DMOS) is available, so
a resistor "pullup" will have to do.
E = I * R or 1.05(V) = 11(uA) * R(megohms); or roughly 100K.
The resistors can be SMD 0805; easily available in 1% values to 10
megs; the NPN is available in SOT-23 so it can be done in a small
profile PCB which could be 20mils thick if desired.
That totals to the 5 parts max.

Thanks, Robert!

I didn't use your circuit but you did give me an idea and the lead on
the LT1389. This is what I ended up with:

+---+-----+-------+--------+
| | | | .|.
=== | | | | |
Camera | | | | |1M
Body | | | '-'
| |LT1389 | |
| _|_/ +---|--------+
| / A | | .|.
--- | | | | |
3V - | | | | |68k
LiMnO2 | | | |\| '-'
| | +-|-\ | to
| | | >-------+---> Meter
| +-----|+/
| | |/| LT1494
| .-. |
| | | |
| | |1M2 |
| '-' |
| | |
+-----+-------+

(created by AACircuit v1.28.5 beta 02/06/05 www.tech-chat.de)

LTspice says it draws 3.8 uA quiescent, so my battery should last a
couple of years. I may do a bit more tweaking on the resistor values
after further analysis but I think that this is a pretty good starting
point.

Any obvious problems?

Thanks.
Click to expand...
I don't have any nano-power design experience either, but it looks to me
like your circuit has a shot at working. I think better values for the
resistors are 1 meg and 80.6k, though.

Of course you will use 1% tolerance resistors, which will give you a
variance of around 1% in the output voltage. If you want to, you could use
a 100k trimpot instead of the 80.6k resistor. Then you can trim to the
desired voltage. Or you could use your 68k resistor, then add a 20k
trimpot, but then the parts count will be up to 6 parts.

I also haven't checked the deviations you will get due to leakage currents
at the inputs, but the 1494 seems to have pretty good specs in that
regard. If you use a trimpot, this probably won't matter anyway.

I was going to recommend the lm185 (or lm285), but this circuit seems to
have that one beat in terms of current consumption. But if you did use the
lm185, the part count might be as low as 3 parts. The current consumption
would be right around 10 uA.

regards,
Mac
 
Tim Hubberstey wrote:

Robert Baer wrote:

Tim Hubberstey wrote:

Hi all,

I've got a problem that I've been wracking my brain over for a while
now. I need a small (under 5 parts) 2-terminal circuit to drop 0.2 V
at up to 100 uA.

The situation is that I have an old but high quality camera that
used a 1.35 V mercury cell for the light meter. These cells are no
longer available due to a world-wide ban on mercury cells. I had a
few extras stashed away but those are now gone. :(

The closest thing I can find with similar discharge characteristics
is a silver oxide cell at 1.55 V. Compounding the problem is the
fact that the camera uses the metal body as the positive ground and
getting access to the on-off switch would require far more
disassembly than I'm comfortable with. Hence the need for a
2-terminal solution. The battery check circuit in the camera
considers a cell to be "good" when the terminal voltage is between
1.27 and 1.35 V at 90 uA load.

A 3 V manganese dioxide lithium cell is also an option if paired
with a 3-terminal positive-ground regulator with very low (<10 uA)
quiescent current.

The standard "solution" involves putting a low-current Schottky in
series with the cell, but reports are that this still causes
significant errors in the meter readings on this particular camera
model.

The 5 parts limitation comes from the small amount of space
available in the base of the camera, in between various gears.


Are you saying that the 1.55 V silver oxide cell messes up the
reading or functionality of the camera?


It throws the meter off by about 1.5 stops. The camera is totally
mechanical, except for the meter.

Other than a resistor (if current load is relatively constant),
theer is nothing that would give a 0.2V drop.
A schottky or germanium diode will give about 300-350mV drop,
depending on current and diode rating.
Now using a 3-terminal regulator with a 3V source seems to be a
solution, but you need one with two attributes: Low Drop-Out (or LDO
in the trades) *and* low drain.
Maxim advertises a shunt regulaor that supposedly works down to
1uA, but they can be completely discounted, as Maxim/Dallas parts are
vaporware.
However, Linear technology does have the LT1389 shunt regulator,
available at 1.25V, 2.5V and 4.096V, speced to operate from 0.8uA to
2mA; curves show operation down to around 0.4uA.
Then their LT1634 shunt regulator, available at 1.25V, 2.5V, 4.096V
and 5V, speced to operate from 10uA to 2mA.
Their LT1004 seems to have similar specs, but available only at
1.23V and 2.50V.
The LM285-2.5 works from 20uA to 20mA, and there are adjustable
versions made by National Semi.
So, you could make your own 3-term regulator, as most of what is
cited above is available in SOT-23 or SOIC-8.
Idea: NPN pass transistor, base to cathode of adjustable, FB of
adjustable to tap of voltage divider: 1.25 meg from FB to ground
(anode) and 100K from FB to NPN emitter.
Now all one needs is base drive from NPN collector that will run
the regulator and the NPN.
Say one allows a LM285 regulator current of 10uA and a base drive
of 1uA (yes, that is high - so the extra also goes thru the LM285).
The base will be 1.35V (ouput or emitter voltage) plus a Vbe of
about 600mV, or about 1.95V; making the Vcb about 1.05V.
Obviously, no constant current device (JFET or DMOS) is available,
so a resistor "pullup" will have to do.
E = I * R or 1.05(V) = 11(uA) * R(megohms); or roughly 100K.
The resistors can be SMD 0805; easily available in 1% values to 10
megs; the NPN is available in SOT-23 so it can be done in a small
profile PCB which could be 20mils thick if desired.
That totals to the 5 parts max.


Thanks, Robert!

I didn't use your circuit but you did give me an idea and the lead on
the LT1389. This is what I ended up with:

+---+-----+-------+--------+
| | | | .|.
=== | | | | |
Camera | | | | |1M
Body | | | '-'
| |LT1389 | |
| _|_/ +---|--------+
| / A | | .|.
--- | | | | |
3V - | | | | |68k
LiMnO2 | | | |\| '-'
| | +-|-\ | to
| | | >-------+---> Meter
| +-----|+/
| | |/| LT1494
| .-. |
| | | |
| | |1M2 |
| '-' |
| | |
+-----+-------+

(created by AACircuit v1.28.5 beta 02/06/05 www.tech-chat.de)

LTspice says it draws 3.8 uA quiescent, so my battery should last a
couple of years. I may do a bit more tweaking on the resistor values
after further analysis but I think that this is a pretty good starting
point.

Any obvious problems?

Thanks.
Click to expand...

No fiddling needed.
Instead of 1 meg, use an 0805 1.24 meg 1%; Digikey cut tape price is
an outrageous $0.80; and thus instead of 68K use an 0805 100 K 1%; same
price.
If the reference was exactly 1.24V, then the op-amp would be at 1.35V
within 1% (the resistors in practice are inside the limits of the spec,
so the average error of the output voltage works out to within 1%.
Since the reference is 1.25V "exactly", the derived voltage will be
slightly higher than 1.35V, and will be well within 1.4% (worst case)
and probably better than 1%.
Then, might as well change that 1.2 meg and use 1.24 meg 1% from the
same cut tape to save money.
Lay out a PCB and have a bunch made (Express PCB or similar), assemble
them yourself and sell the solution!
 
Winfield Hill wrote...
I'd use 1% resistors. If you move the 1.2M and change its value,
you can make the 1.25V reference current independent of the battery
voltage by running it from the 1.35V output. Then if you select a
different opamp, you can run it from your 1.55V silver oxide cell.

. +---+-----+-------+--------+----
. | | | | |
. === | |LT1389 | 1.00M
. Camera | \_|_ | |
. Body | /_\ ,--- |--------+
. _|_ | | | |
. - | | _| 80.2k
. 1.55V | | '--|- \ | to
. | | | >---+--+---> Meter
. | +-----|+_/ | -1.35V
. | | | |
. '---- |-------' |
. '-- 68k ------'
. 1.5uA --
Click to expand...
BTW, in this circuit the opamp shouldn't necessarily be a RRIO type.
In operation the input and outputs are both near the negative rail,
an easy opamp spec (even tho low-power 1.5V operation is not). But
to insure startup, you want an opamp whose input-stage works properly
with CM voltages closer to the + rail than the output is at startup,
or that pushes the output negative until the input CM is working.
If you turn the circuit upside-down, and are working at higher power
levels, many classic single-supply opamps meet these unusual specs.
I'm not sure if any offered opamps will work in my 1.55V circuit...


--
Thanks,
- Win
 
On 16 Mar 2005 06:33:40 -0800, Winfield Hill
<hill_a@t_rowland-dotties-harvard-dot.s-edu> wrote:

Winfield Hill wrote...

Tim Hubberstey wrote...

Winfield Hill wrote:

I'd use 1% resistors. If you move the 1.2M and change its value,
you can make the 1.25V reference current independent of the
battery
voltage by running it from the 1.35V output. Then if you select a
different opamp, you can run it from your 1.55V silver oxide cell.

. +---+-----+-------+--------+----
. | | | | |
. === | |LT1389 | 1.00M
. Camera | \_|_ | |
. Body | /_\ ,--- |--------+
. _|_ | | | |
. - | | _| 80.2k
. 1.55V | | '--|- \ | to
. | | | >---+--+---> Meter
. | +-----|+_/ | -1.35V
. | | | |
. '---- |-------' |
. '-- 68k ------'
. 1.5uA --

Looks interesting... but won't it need a large-value resistor to
the
anode of the LT1389 to get it to start up? Looks to me like it
would
come up with the output = 0 and just stay there. Or could we rely
on
noise to give it enough of a "push" to get it started?

I posted a note concerning startup. This type of bootstrap circuit
start up fine if the opamp's output has more BJT saturation voltage
than the input's offset (divided by the feedback-resistor ratio), so

when the power is applied, the + input sees a bit more voltage than
the - input and drives the output further toward the turned-on
state.

There are two problematic issues. First, the feedback resistors to
common tend to reduce the BJT saturation voltage. Using high-value
resistors helps; e.g., here they draw a current of 30nA for say 30mV
of output-transistor saturation. We note that 30nA is probably much
less than the base-drive current for the PNP output transistor.

Second, when the output is only modestly-higher than the reference,
the "offset voltage divided by the feedback-resistor ratio"
parameter
becomes painful. Here we would divide our estimated 30mV by 12.5 to
get a 2.5mV max offset-voltage spec, which might be hard to meet.

At which point Tim's suggested resistor is the way to go.

. +---+-----+-------+---------+----
. | | | | |
. === | |LT1389 | 1.00M
. Camera | \_|_ | |
. Body | /_\ ,--- |---------+
. _|_ | | | |
. - | | _| 80.2k
. 1.55V | | '--|- \ | to
. | | | >-----+-+---> Meter
. | +--+--|+_/ | -1.35V
. | | | | |
. | 10M '--- |- 68k -'
. | | |
. '-----+-------'

Assuming a good-quality reference, this two-resistor bias scheme has
little advantage over one resistor to the battery, unless a variety
of
battery voltages is expected and power conservation is very
important.


--
Thanks,
- Win
Click to expand...
The Zener is forward biased in the circuit above . You have to invert
it and I'd think a better solution is connecting it up the way I
suggested in my previous post.
 
lemonjuice wrote:
On 16 Mar 2005 06:33:40 -0800, Winfield Hill
hill_a@t_rowland-dotties-harvard-dot.s-edu> wrote:


Winfield Hill wrote...

Tim Hubberstey wrote...

Winfield Hill wrote:


I'd use 1% resistors. If you move the 1.2M and change its value,
you can make the 1.25V reference current independent of the

battery

voltage by running it from the 1.35V output. Then if you select a
different opamp, you can run it from your 1.55V silver oxide cell.

. +---+-----+-------+--------+----
. | | | | |
. === | |LT1389 | 1.00M
. Camera | \_|_ | |
. Body | /_\ ,--- |--------+
. _|_ | | | |
. - | | _| 80.2k
. 1.55V | | '--|- \ | to
. | | | >---+--+---> Meter
. | +-----|+_/ | -1.35V
. | | | |
. '---- |-------' |
. '-- 68k ------'
. 1.5uA --

Looks interesting... but won't it need a large-value resistor to

the

anode of the LT1389 to get it to start up? Looks to me like it

would

come up with the output = 0 and just stay there. Or could we rely

on

noise to give it enough of a "push" to get it started?

I posted a note concerning startup. This type of bootstrap circuit
start up fine if the opamp's output has more BJT saturation voltage
than the input's offset (divided by the feedback-resistor ratio), so


when the power is applied, the + input sees a bit more voltage than
the - input and drives the output further toward the turned-on

state.

There are two problematic issues. First, the feedback resistors to
common tend to reduce the BJT saturation voltage. Using high-value
resistors helps; e.g., here they draw a current of 30nA for say 30mV
of output-transistor saturation. We note that 30nA is probably much
less than the base-drive current for the PNP output transistor.

Second, when the output is only modestly-higher than the reference,
the "offset voltage divided by the feedback-resistor ratio"

parameter

becomes painful. Here we would divide our estimated 30mV by 12.5 to
get a 2.5mV max offset-voltage spec, which might be hard to meet.

At which point Tim's suggested resistor is the way to go.

. +---+-----+-------+---------+----
. | | | | |
. === | |LT1389 | 1.00M
. Camera | \_|_ | |
. Body | /_\ ,--- |---------+
. _|_ | | | |
. - | | _| 80.2k
. 1.55V | | '--|- \ | to
. | | | >-----+-+---> Meter
. | +--+--|+_/ | -1.35V
. | | | | |
. | 10M '--- |- 68k -'
. | | |
. '-----+-------'

Assuming a good-quality reference, this two-resistor bias scheme has
little advantage over one resistor to the battery, unless a variety

of

battery voltages is expected and power conservation is very

important.


--
Thanks,
- Win


The Zener is forward biased in the circuit above . You have to invert
it and I'd think a better solution is connecting it up the way I
suggested in my previous post.
Click to expand...
No it's not- look at the battery polarity again.
 
On Wed, 16 Mar 2005 15:41:56 GMT, Fred Bloggs <nospam@nospam.com>
wrote:

lemonjuice wrote:
On 16 Mar 2005 06:33:40 -0800, Winfield Hill
hill_a@t_rowland-dotties-harvard-dot.s-edu> wrote:


Winfield Hill wrote...

Tim Hubberstey wrote...

Winfield Hill wrote:


I'd use 1% resistors. If you move the 1.2M and change its value,
you can make the 1.25V reference current independent of the

battery

voltage by running it from the 1.35V output. Then if you select
a
different opamp, you can run it from your 1.55V silver oxide
cell.

. +---+-----+-------+--------+----
. | | | | |
. === | |LT1389 | 1.00M
. Camera | \_|_ | |
. Body | /_\ ,--- |--------+
. _|_ | | | |
. - | | _| 80.2k
. 1.55V | | '--|- \ | to
. | | | >---+--+---> Meter
. | +-----|+_/ | -1.35V
. | | | |
. '---- |-------' |
. '-- 68k ------'
. 1.5uA --

Looks interesting... but won't it need a large-value resistor to

the

anode of the LT1389 to get it to start up? Looks to me like it

would

come up with the output = 0 and just stay there. Or could we rely

on

noise to give it enough of a "push" to get it started?

I posted a note concerning startup. This type of bootstrap circuit
start up fine if the opamp's output has more BJT saturation voltage
than the input's offset (divided by the feedback-resistor ratio),
so


when the power is applied, the + input sees a bit more voltage than
the - input and drives the output further toward the turned-on

state.

There are two problematic issues. First, the feedback resistors to
common tend to reduce the BJT saturation voltage. Using high-value
resistors helps; e.g., here they draw a current of 30nA for say
30mV
of output-transistor saturation. We note that 30nA is probably
much
less than the base-drive current for the PNP output transistor.

Second, when the output is only modestly-higher than the reference,
the "offset voltage divided by the feedback-resistor ratio"

parameter

becomes painful. Here we would divide our estimated 30mV by 12.5
to
get a 2.5mV max offset-voltage spec, which might be hard to meet.

At which point Tim's suggested resistor is the way to go.

. +---+-----+-------+---------+----
. | | | | |
. === | |LT1389 | 1.00M
. Camera | \_|_ | |
. Body | /_\ ,--- |---------+
. _|_ | | | |
. - | | _| 80.2k
. 1.55V | | '--|- \ | to
. | | | >-----+-+---> Meter
. | +--+--|+_/ | -1.35V
. | | | | |
. | 10M '--- |- 68k -'
. | | |
. '-----+-------'

Assuming a good-quality reference, this two-resistor bias scheme has
little advantage over one resistor to the battery, unless a variety

of

battery voltages is expected and power conservation is very

important.


--
Thanks,
- Win


The Zener is forward biased in the circuit above . You have to
invert
it and I'd think a better solution is connecting it up the way I
suggested in my previous post.


No it's not- look at the battery polarity again.
Click to expand...
Yes. But with a .15V reverse bias I don't see it working.
 
oops that was 0.2V and not .15. Hard trying to work and play. It'd
seem the original purpose of the diode isn't there ... because you're
forcing a 0.2 voltage onto it. I'd opt for a non inverting amplifier
circuit.
 
lemonjuice wrote...
Assuming the anode of the zener and the 1Meg output resistor are
earthed V(+) = V(-) at the opamp inputs would be = Vee. If we had a
opamp configuration like a 741 the input transistors would be cut off
as would the darlington stages. That would force the opamp output
into a positive direction. With other Opamps any other simple
solution would do.
Click to expand...
Excuse me, you're considering an opamp "like a 741" in a circuit
with a 1.5V total supply voltage? Darlington stages? * Cough *
Have you studied low-voltage RRIO opamps? Ahem, internally these
are about as far from a 741 configuration, etc., as you can get.


--
Thanks,
- Win
 
On 16 Mar 2005 10:17:42 -0800, Winfield Hill
<hill_a@t_rowland-dotties-harvard-dot.s-edu> wrote:

lemonjuice wrote...

Assuming the anode of the zener and the 1Meg output resistor are
earthed V(+) = V(-) at the opamp inputs would be = Vee. If we had a
opamp configuration like a 741 the input transistors would be cut
off
as would the darlington stages. That would force the opamp output
into a positive direction. With other Opamps any other simple
solution would do.

+---+-----+-------+--------+----
. | | | | |
. === | |LT1389 | 1.00M
. Camera | \_|_ | |
. Body | /_\ ,--- |--------+
. _|_ | | | |
. - | | _| 80.2k
. 1.55V | | '--|- \ | to
. | | | >---+--+---> Meter
. | +-----|+_/ | -1.35V
. | | | |
. '---- |-------' |
. '-- 68k ------'
. 1.5uA --


Excuse me, you're considering an opamp "like a 741" in a circuit
with a 1.5V total supply voltage? Darlington stages? * Cough *
Have you studied low-voltage RRIO opamps? Ahem, internally these
are about as far from a 741 configuration, etc., as you can get.
Who said I was considering a 741 in that circuit?
Click to expand...
darlington?
Didn't you say that the opamp shouldn't necessarily be RRIO type?
My post was to show that an opamp with the same internal circuitry
like a 741 in the circuit above with other slight modifications
wouldn't have start up problems and even if it did they could easily be
solved.

BTW The circuit would have a worse problem. you've got a saturation
conditions at the opamp inputs.
V(-)= 1.434V and V(+)= 0.115V.
 
lemonjuice wrote...
BTW The circuit would have a worse problem. you've got a saturation
conditions at the opamp inputs. V(-)= 1.434V and V(+)= 0.115V.
Click to expand...
I'm not sure where you get those numbers, or what point you're
trying to make, but I'll say this, sub-100mV collector-saturation
voltages for sub-100uA currents (i.e., Rsat under 1k) is entirely
reasonable for some low-power opamp output stages.


--
Thanks,
- Win
 
Winfield Hill wrote...
lemonjuice wrote...

BTW The circuit would have a worse problem. you've got a saturation
conditions at the opamp inputs. V(-)= 1.434V and V(+)= 0.115V.

I'm not sure where you get those numbers, or what point you're
trying to make, but I'll say this, sub-100mV collector-saturation
voltages for sub-100uA currents (i.e., Rsat under 1k) is entirely
reasonable for some low-power opamp output stages.
Click to expand...
Notice I haven't suggested a specific low-voltage opamp - that's too
much like work (and it's the OPs task). There is a useful industry
term, sub-1V, which yields 1170 hits on Google. Good reading.


--
Thanks,
- Win
 
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