Need to drop 0.2 V with 2-ternimal series device

[lemonjuice]

Tim Hubberstey wrote...

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.

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.

 

[Fred Bloggs]

Tim Hubberstey wrote...



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.

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.

 

[lemonjuice]

Winfield Hill wrote...

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.

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.

 

[Fred Bloggs]

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.

 

[lemonjuice]

No it's not- look at the battery polarity again.

Yes. But with a .15V reverse bias I don't see it working.

 

[lemonjuice]

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.

 

[Winfield Hill]

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.

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.

 

[lemonjuice]

lemonjuice wrote...

+---+-----+-------+--------+----
. | | | | |
. === | |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?
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.

 

[Winfield Hill]

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.

 

[Winfield Hill]

Winfield Hill wrote...

lemonjuice wrote...

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.

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.

 

[lemonjuice]

lemonjuice wrote...

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...

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.

 

[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.

 

[Robert Baer]

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.

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.