Need to drop 0.2 V with 2-ternimal series device

[lemonjuice]

On 17 Mar 2005 06:15:52 -0800, Winfield Hill
<hill_a@t_rowland-dotties-harvard-dot.s-edu> 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.
For example

According to the zener specs it should drop 1.25V. That would mean .3
and 1.35 between the output and the V(+) node , across the 68K
resistor. Meaning current through resistor is 1.05/60.8K which doesn't
add up to whats written there.

I tried a Microcap simulation and i got V(-) almost equal to the
supply voltage, which is more or less what I'd expect from Positive
feedback increasing to infinite the output impedance which somewhat
agrees with my earlier calculation.


Why not ground both the zener anode and the 1M node connected to the
supply. Less fidgeting and circuit is more predictable. Vout is then
easily determined and controlled by the expression Vzener*(1 + R2/R1)?
R2, R1 now are the 1M and 80.2K resistors

 

[Winfield Hill]

lemonjuice wrote...

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.
For example
According to the zener specs it should drop 1.25V. That would mean
.3 and 1.35 between the output and the V(+) node , across the 68K
resistor. Meaning current through resistor is 1.05/60.8K which
doesn't add up to whats written there.

You seem rather badly confused, there's 100mV across the resistor
when the circuit is at equilibrium. The circuit is an output-voltage
regulator, not a fixed-voltage-dropper, as the OP originally proposed,
perhaps that explains your confusion?

I tried a Microcap simulation and i got V(-) almost equal to the
supply voltage, which is more or less what I'd expect from Positive
feedback increasing to infinite the output impedance which somewhat
agrees with my earlier calculation.

The overall feedback is negative, not positive at equilibrium. That's
because the zener dynamic impedance is much much lower than 68k. The
output impedance is certainly NOT infinite. I'm going to sign off.


--
Thanks,
- Win

 

[lemonjuice]

You seem rather badly confused, there's 100mV across the resistor
when the circuit is at equilibrium. The circuit is an output-voltage
regulator, not a fixed-voltage-dropper, as the OP originally
proposed,
perhaps that explains your confusion?
The circuit I proposed is a fixed output voltage regulator so its all

clear on my side.
I see confusion somewhere else. Unless we are looking at 2 different
circuits.
Stating that there's 100mV across the resistance doesn't help the
argument much.
Writing the loop equation you get -1.55 +1.25+ IR*VR + Vout =0;
Now you say Vout is 1.35Volt. Show me how that gets you 100mV for
IR*VR?

 

[Mark Weber]

Check with Russ @ Kominek camera service 416 416 977 2132. They used
to supply the originals when you couldn't get them thru the regular
channels.
Maybe they still do.

On Sun, 13 Mar 2005 02:21:32 GMT, Tim Hubberstey <bogus@bogusname.com>
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.

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

Thanks

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[Robert Baer]

Mark Weber wrote:

Check with Russ @ Kominek camera service 416 416 977 2132. They used
to supply the originals when you couldn't get them thru the regular
channels.
Maybe they still do.

On Sun, 13 Mar 2005 02:21:32 GMT, Tim Hubberstey <bogus@bogusname.com
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.

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

Thanks



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This problem with the camera battery has been beaten to death.

A number of solutions within the parts count and size limitations
have been proposed.