op-amp nV input offset voltage

[Paul]

Hi,

As you know, the *input* offset voltage is the voltage required across
the op-amp's input terminals to drive the output voltage to zero.
Although it has been my experience that for most op-amps the input
offset voltage is due to the "-" input pin for the *most* part. For
example, according to Spice the input offset voltage on the "+" input
pin on a LMC660A op-amp for a non-inverting amp circuit is a few
nanovolts, disregarding thermoelectric effects mind you, but a few
millivolts on the "-" input pin. Although as you know the input signal
is not applied to the "-" input pin for a non-inverting amp circuit,
which means there's just a few nanovolts on the input of such a
circuit if we disregard thermoelectric effects.

I have a INA116PA Instrumentation op-amp where Ib typ = 3fA, Ib max =
25fA, and Vos typ = 0.5mV. Now it seems to me in order for there to be
0.5mV on the input of this Instrumentation op-amp circuit with 3fA
bias current that the DUT input impedance would have to be 0.50mV /
3.0fA = 170 Gohms. On the other hand, if the DUT input impedance is
say 200 Kohms then would the input offset voltage be 3.0fA * 200Kohms
= 0.6nV, disregarding thermoelectric effects?

INA116PA datasheet:
http://focus.ti.com/lit/ds/symlink/ina116.pdf

Regards,
Paul

 

[Paul]

Hi,

As you know, the *input* offset voltage is the voltage required across
the op-amp's input terminals to drive the output voltage to zero.
Although it has been my experience that for most op-amps the input
offset voltage is due to the "-" input pin for the *most* part. For
example, according to Spice the input offset voltage on the "+" input
pin on a LMC660A op-amp for a non-inverting amp circuit is a few
nanovolts, disregarding thermoelectric effects mind you, but a few
millivolts on the "-" input pin. Although as you know the input signal
is not applied to the "-" input pin for a non-inverting amp circuit,
which means there's just a few nanovolts on the input of such a
circuit if we disregard thermoelectric effects.

The offset voltage is *differential*. You can blame it on either pin,
or both pins... it doesn't matter who you blame, the result is the
same: offset voltage becomes measurement error.

I have a INA116PA Instrumentation op-amp where Ib typ = 3fA, Ib max =
25fA, and Vos typ = 0.5mV. Now it seems to me in order for there to be
0.5mV on the input of this Instrumentation op-amp circuit with 3fA
bias current that the DUT input impedance would have to be 0.50mV /
3.0fA = 170 Gohms. On the other hand, if the DUT input impedance is
say 200 Kohms then would the input offset voltage be 3.0fA * 200Kohms
= 0.6nV, disregarding thermoelectric effects?

The offset voltage error is a different thing from the input bias
current. They are unrelated [1]. You can of course generate a real,
external-to-the-opamp error voltage by dumping the bias current into
real external resistance, but that's a different matter entirely.

John

[1] Some opamps have low offsets and high bias currents, and some vice
versa. Chopper amps are low on both; cheap bipolars are high on both.

Here's my main concern. If I build the INA116PA for DC application,
which is an internal Instrumentation op-amp chip (3 op-amps), and the
impedance of my DUT is 200 Kohms, then what bias currents could a good
EE such as yourself expect? I mean, for a 200K ohm DUT input source we
cannot have both 0.5mV offset and 3fA bias on the DUT. I think V=I*R
applies, so if the bias current is 3fA then V = 3fA * 200Kohms = 0.6
nV.

Thanks,
Paul

 

[Paul]

Hi,

As you know, the *input* offset voltage is the voltage required across
the op-amp's input terminals to drive the output voltage to zero.
Although it has been my experience that for most op-amps the input
offset voltage is due to the "-" input pin for the *most* part. For
example, according to Spice the input offset voltage on the "+" input
pin on a LMC660A op-amp for a non-inverting amp circuit is a few
nanovolts, disregarding thermoelectric effects mind you, but a few
millivolts on the "-" input pin. Although as you know the input signal
is not applied to the "-" input pin for a non-inverting amp circuit,
which means there's just a few nanovolts on the input of such a
circuit if we disregard thermoelectric effects.

The offset voltage is *differential*. You can blame it on either pin,
or both pins... it doesn't matter who you blame, the result is the
same: offset voltage becomes measurement error.

I have a INA116PA Instrumentation op-amp where Ib typ = 3fA, Ib max =
25fA, and Vos typ = 0.5mV. Now it seems to me in order for there to be
0.5mV on the input of this Instrumentation op-amp circuit with 3fA
bias current that the DUT input impedance would have to be 0.50mV /
3.0fA = 170 Gohms. On the other hand, if the DUT input impedance is
say 200 Kohms then would the input offset voltage be 3.0fA * 200Kohms
= 0.6nV, disregarding thermoelectric effects?

The offset voltage error is a different thing from the input bias
current. They are unrelated [1]. You can of course generate a real,
external-to-the-opamp error voltage by dumping the bias current into
real external resistance, but that's a different matter entirely.

John

[1] Some opamps have low offsets and high bias currents, and some vice
versa. Chopper amps are low on both; cheap bipolars are high on both.

The LMC660A has a typical voltage offset of 1mV and bias current of
2fA, but that depends what type of op-amp circuit. According to Spice
the input voltage offset for an inverting or differential circuit is
about what the Vos spec says, but for a non-inverting circuit it's a
few nanovolts on the "+" input pin. I'm wondering if the Vos in
datasheets is referring to a certain type of op-amp circuit such as
the inverting type (http://hyperphysics.phy-astr.gsu.edu/Hbase/
Electronic/opampvar.html#c2).

Regards,
Paul

 

[Jamie]

Hi,

As you know, the *input* offset voltage is the voltage required across
the op-amp's input terminals to drive the output voltage to zero.
Although it has been my experience that for most op-amps the input
offset voltage is due to the "-" input pin for the *most* part. For
example, according to Spice the input offset voltage on the "+" input
pin on a LMC660A op-amp for a non-inverting amp circuit is a few
nanovolts, disregarding thermoelectric effects mind you, but a few
millivolts on the "-" input pin. Although as you know the input signal
is not applied to the "-" input pin for a non-inverting amp circuit,
which means there's just a few nanovolts on the input of such a
circuit if we disregard thermoelectric effects.

I have a INA116PA Instrumentation op-amp where Ib typ = 3fA, Ib max =
25fA, and Vos typ = 0.5mV. Now it seems to me in order for there to be
0.5mV on the input of this Instrumentation op-amp circuit with 3fA
bias current that the DUT input impedance would have to be 0.50mV /
3.0fA = 170 Gohms. On the other hand, if the DUT input impedance is
say 200 Kohms then would the input offset voltage be 3.0fA * 200Kohms
= 0.6nV, disregarding thermoelectric effects?

INA116PA datasheet:
http://focus.ti.com/lit/ds/symlink/ina116.pdf

Regards,
Paul

The last time I checked, the offset voltage would be the difference
between the (-) and (+) input with the Op-amp in (-) loop back mode.

So, if you were to put an op-amp in (-) loop back and lets say 5 volts
on the (+) input, the (-) input should be offset no more than what the
spec's state.

Or, I guess if you were using a +/- to common supply, you can simply
tie the (+) to common with op-amp in (-) loop back. You should be seeing
that offset factor at the (-)/output..

Maybe things have changed but that is what I go by..

http://webpages.charter.net/jamie_5"

 

[Helmut Sennewald]

Hi,

As you know, the *input* offset voltage is the voltage required across
the op-amp's input terminals to drive the output voltage to zero.
Although it has been my experience that for most op-amps the input
offset voltage is due to the "-" input pin for the *most* part. For
example, according to Spice the input offset voltage on the "+" input
pin on a LMC660A op-amp for a non-inverting amp circuit is a few
nanovolts, disregarding thermoelectric effects mind you, but a few
millivolts on the "-" input pin. Although as you know the input signal
is not applied to the "-" input pin for a non-inverting amp circuit,
which means there's just a few nanovolts on the input of such a
circuit if we disregard thermoelectric effects.

The offset voltage is *differential*. You can blame it on either pin,
or both pins... it doesn't matter who you blame, the result is the
same: offset voltage becomes measurement error.

I have a INA116PA Instrumentation op-amp where Ib typ = 3fA, Ib max =
25fA, and Vos typ = 0.5mV. Now it seems to me in order for there to be
0.5mV on the input of this Instrumentation op-amp circuit with 3fA
bias current that the DUT input impedance would have to be 0.50mV /
3.0fA = 170 Gohms. On the other hand, if the DUT input impedance is
say 200 Kohms then would the input offset voltage be 3.0fA * 200Kohms
= 0.6nV, disregarding thermoelectric effects?

The offset voltage error is a different thing from the input bias
current. They are unrelated [1]. You can of course generate a real,
external-to-the-opamp error voltage by dumping the bias current into
real external resistance, but that's a different matter entirely.

John

[1] Some opamps have low offsets and high bias currents, and some vice
versa. Chopper amps are low on both; cheap bipolars are high on both.

The LMC660A has a typical voltage offset of 1mV and bias current of
2fA, but that depends what type of op-amp circuit. According to Spice
the input voltage offset for an inverting or differential circuit is
about what the Vos spec says, but for a non-inverting circuit it's a
few nanovolts on the "+" input pin. I'm wondering if the Vos in
datasheets is referring to a certain type of op-amp circuit such as
the inverting type (http://hyperphysics.phy-astr.gsu.edu/Hbase/
Electronic/opampvar.html#c2).

Regards,
Paul

Hello Paul,
Maybe it helps if you think about the transistor circuit
of an opamp.

The first stage of an opamp consists of a differential
amplifier made by a pair of two well matched transistors.
The difference of the Vgs(Mosfet opamp) or Vbe(bipolar opamp)
of these two transistors in the input stage is the main
contributor for the offset voltage.

Offset voltage is always measured between the + and - input.
What you have measured at the +input is the bias(leakage)
current multiplied by the value of the resistor connected
to the +pin.

Best regards,
Helmut

 

[Paul]

Hi,
As you know, the *input* offset voltage is the voltage required across
the op-amp's input terminals to drive the output voltage to zero.
Although it has been my experience that for most op-amps the input
offset voltage is due to the "-" input pin for the *most* part. For
example, according to Spice the input offset voltage on the "+" input
pin on a LMC660A op-amp for a non-inverting amp circuit is a few
nanovolts, disregarding thermoelectric effects mind you, but a few
millivolts on the "-" input pin. Although as you know the input signal
is not applied to the "-" input pin for a non-inverting amp circuit,
which means there's just a few nanovolts on the input of such a
circuit if we disregard thermoelectric effects.
The offset voltage is *differential*. You can blame it on either pin,
or both pins... it doesn't matter who you blame, the result is the
same: offset voltage becomes measurement error.
I have a INA116PA Instrumentation op-amp where Ib typ = 3fA, Ib max =
25fA, and Vos typ = 0.5mV. Now it seems to me in order for there to be
0.5mV on the input of this Instrumentation op-amp circuit with 3fA
bias current that the DUT input impedance would have to be 0.50mV /
3.0fA = 170 Gohms. On the other hand, if the DUT input impedance is
say 200 Kohms then would the input offset voltage be 3.0fA * 200Kohms
= 0.6nV, disregarding thermoelectric effects?
The offset voltage error is a different thing from the input bias
current. They are unrelated [1]. You can of course generate a real,
external-to-the-opamp error voltage by dumping the bias current into
real external resistance, but that's a different matter entirely.
John
[1] Some opamps have low offsets and high bias currents, and some vice
versa. Chopper amps are low on both; cheap bipolars are high on both.

The LMC660A has a typical voltage offset of 1mV and bias current of
2fA, but that depends what type of op-amp circuit. According to Spice
the input voltage offset for an inverting or differential circuit is
about what the Vos spec says, but for a non-inverting circuit it's a
few nanovolts on the "+" input pin. I'm wondering if the Vos in
datasheets is referring to a certain type of op-amp circuit such as
the inverting type (http://hyperphysics.phy-astr.gsu.edu/Hbase/
Electronic/opampvar.html#c2).

Regards,
Paul

Hello Paul,
Maybe it helps if you think about the transistor circuit
of an opamp.

The first stage of an opamp consists of a differential
amplifier made by a pair of two well matched transistors.
The difference of the Vgs(Mosfet opamp) or Vbe(bipolar opamp)
of these two transistors in the input stage is the main
contributor for the offset voltage.

Offset voltage is always measured between the + and - input.
What you have measured at the +input is the bias(leakage)
current multiplied by the value of the resistor connected
to the +pin.

Best regards,
Helmut- Hide quoted text -

I appreciate all of the replies! All of these years I've had this
false idea about the datasheets Vos burnt into my head. I've always
assumed that if the datasheet said the op-amps Vos was say 50uV then
that's the lowest input voltage (by my def: the voltage applied on the
input device due to the op-amp) one can expect with a typical op-amp
circuit such as an inverter or non-inverter.

So it's true that one could achieve input voltages in the nanovolt
region on a 200K ohm DUT from an Instrumentation op-amp chip such as
INA116PA even though the datasheet Vos spec is 2mV?

Thanks,
Paul

INA116PA datasheet:
http://focus.ti.com/lit/ds/symlink/ina116.pdf

 

[Helmut Sennewald]

Newsbeitrag

On Jun 21, 8:41 am, John Larkin


As you know, the *input* offset voltage is the voltage required
across
the op-amp's input terminals to drive the output voltage to zero.
Although it has been my experience that for most op-amps the input
offset voltage is due to the "-" input pin for the *most* part. For
example, according to Spice the input offset voltage on the "+" input
pin on a LMC660A op-amp for a non-inverting amp circuit is a few
nanovolts, disregarding thermoelectric effects mind you, but a few
millivolts on the "-" input pin. Although as you know the input
signal
is not applied to the "-" input pin for a non-inverting amp circuit,
which means there's just a few nanovolts on the input of such a
circuit if we disregard thermoelectric effects.

The offset voltage is *differential*. You can blame it on either pin,
or both pins... it doesn't matter who you blame, the result is the
same: offset voltage becomes measurement error.

I have a INA116PA Instrumentation op-amp where Ib typ = 3fA, Ib max =
25fA, and Vos typ = 0.5mV. Now it seems to me in order for there to
be
0.5mV on the input of this Instrumentation op-amp circuit with 3fA
bias current that the DUT input impedance would have to be 0.50mV /
3.0fA = 170 Gohms. On the other hand, if the DUT input impedance is
say 200 Kohms then would the input offset voltage be 3.0fA * 200Kohms
= 0.6nV, disregarding thermoelectric effects?

The offset voltage error is a different thing from the input bias
current. They are unrelated [1]. You can of course generate a real,
external-to-the-opamp error voltage by dumping the bias current into
real external resistance, but that's a different matter entirely.

John

[1] Some opamps have low offsets and high bias currents, and some vice
versa. Chopper amps are low on both; cheap bipolars are high on both.

The LMC660A has a typical voltage offset of 1mV and bias current of
2fA, but that depends what type of op-amp circuit. According to Spice
the input voltage offset for an inverting or differential circuit is
about what the Vos spec says, but for a non-inverting circuit it's a
few nanovolts on the "+" input pin. I'm wondering if the Vos in
datasheets is referring to a certain type of op-amp circuit such as
the inverting type (http://hyperphysics.phy-astr.gsu.edu/Hbase/
Electronic/opampvar.html#c2).

Regards,
Paul

Hello Paul,
Maybe it helps if you think about the transistor circuit
of an opamp.

The first stage of an opamp consists of a differential
amplifier made by a pair of two well matched transistors.
The difference of the Vgs(Mosfet opamp) or Vbe(bipolar opamp)
of these two transistors in the input stage is the main
contributor for the offset voltage.

Offset voltage is always measured between the + and - input.
What you have measured at the +input is the bias(leakage)
current multiplied by the value of the resistor connected
to the +pin.

Best regards,
Helmut- Hide quoted text -

I appreciate all of the replies! All of these years I've had this
false idea about the datasheets Vos burnt into my head. I've always
assumed that if the datasheet said the op-amps Vos was say 50uV then
that's the lowest input voltage (by my def: the voltage applied on the
input device due to the op-amp) one can expect with a typical op-amp
circuit such as an inverter or non-inverter.

So it's true that one could achieve input voltages in the nanovolt
region on a 200K ohm DUT from an Instrumentation op-amp chip such as
INA116PA even though the datasheet Vos spec is 2mV?

Thanks,
Paul

INA116PA datasheet:
http://focus.ti.com/lit/ds/symlink/ina116.pdf

Hello Paul,

Yes you can apply voltages as small as you like.
they will be still amplified by the gain G, set with
the feedback resistors. The drawback of any Vos
is that you will have an output voltage of (Vos+Vin)*G .
This menas you have to either adjust the offset voltage
already at the input or you have to subtract Vos*G at
the output.

Best regards,
Helmut

 

[Paul]

Newsbeitrag
On Jun 21, 8:41 am, John Larkin
Hi,
As you know, the *input* offset voltage is the voltage required
across
the op-amp's input terminals to drive the output voltage to zero.
Although it has been my experience that for most op-amps the input
offset voltage is due to the "-" input pin for the *most* part. For
example, according to Spice the input offset voltage on the "+" input
pin on a LMC660A op-amp for a non-inverting amp circuit is a few
nanovolts, disregarding thermoelectric effects mind you, but a few
millivolts on the "-" input pin. Although as you know the input
signal
is not applied to the "-" input pin for a non-inverting amp circuit,
which means there's just a few nanovolts on the input of such a
circuit if we disregard thermoelectric effects.
The offset voltage is *differential*. You can blame it on either pin,
or both pins... it doesn't matter who you blame, the result is the
same: offset voltage becomes measurement error.
I have a INA116PA Instrumentation op-amp where Ib typ = 3fA, Ib max =
25fA, and Vos typ = 0.5mV. Now it seems to me in order for there to
be
0.5mV on the input of this Instrumentation op-amp circuit with 3fA
bias current that the DUT input impedance would have to be 0.50mV /
3.0fA = 170 Gohms. On the other hand, if the DUT input impedance is
say 200 Kohms then would the input offset voltage be 3.0fA * 200Kohms
= 0.6nV, disregarding thermoelectric effects?
The offset voltage error is a different thing from the input bias
current. They are unrelated [1]. You can of course generate a real,
external-to-the-opamp error voltage by dumping the bias current into
real external resistance, but that's a different matter entirely.
John
[1] Some opamps have low offsets and high bias currents, and some vice
versa. Chopper amps are low on both; cheap bipolars are high on both.
The LMC660A has a typical voltage offset of 1mV and bias current of
2fA, but that depends what type of op-amp circuit. According to Spice
the input voltage offset for an inverting or differential circuit is
about what the Vos spec says, but for a non-inverting circuit it's a
few nanovolts on the "+" input pin. I'm wondering if the Vos in
datasheets is referring to a certain type of op-amp circuit such as
the inverting type (http://hyperphysics.phy-astr.gsu.edu/Hbase/
Electronic/opampvar.html#c2).
Regards,
Paul
Hello Paul,
Maybe it helps if you think about the transistor circuit
of an opamp.
The first stage of an opamp consists of a differential
amplifier made by a pair of two well matched transistors.
The difference of the Vgs(Mosfet opamp) or Vbe(bipolar opamp)
of these two transistors in the input stage is the main
contributor for the offset voltage.
Offset voltage is always measured between the + and - input.
What you have measured at the +input is the bias(leakage)
current multiplied by the value of the resistor connected
to the +pin.
Best regards,
Helmut- Hide quoted text -

I appreciate all of the replies! All of these years I've had this
false idea about the datasheets Vos burnt into my head. I've always
assumed that if the datasheet said the op-amps Vos was say 50uV then
that's the lowest input voltage (by my def: the voltage applied on the
input device due to the op-amp) one can expect with a typical op-amp
circuit such as an inverter or non-inverter.

So it's true that one could achieve input voltages in the nanovolt
region on a 200K ohm DUT from an Instrumentation op-amp chip such as
INA116PA even though the datasheet Vos spec is 2mV?

INA116PA datasheet:
http://focus.ti.com/lit/ds/symlink/ina116.pdf

Hello Paul,

Yes you can apply voltages as small as you like.
they will be still amplified by the gain G, set with
the feedback resistors. The drawback of any Vos
is that you will have an output voltage of (Vos+Vin)*G .
This menas you have to either adjust the offset voltage
already at the input or you have to subtract Vos*G at
the output.

Best regards,
Helmut- Hide quoted text -

Thanks! As you said the output offset can always be corrected, but
it's great to know that a 2mV op-amp chip such as the INA116PA can
apply DC voltages as low as a few nanovolts on the input device
without adding shunt resistors. Of course one can always add a shunt
resistor to lower the input voltage across the DUT, something I knew
about, but of course that has obvious effects of decreasing the DUT's
effective input voltage to the op-amp.

I'm wondering if there are any op-amps or perhaps a BiFET amp circuit
that could achieve a few nanovolts across say a 200K ohm device while
consuming no more than a few microwatts. The idea is that such a
microwatt amp would have considerably less input thermoelectric
effects. Thermoelectric effects can generate a half dozen or more
microvolts on the DUT unless carefully balanced with dummy resistors.
I believe Linear Tech has some microwatt op-amps, but nothing near
25fA bias current.

Thanks,
Paul

 

[Paul]

Paul,
a thermoelectric effect means you get a voltage
from a temperature difference in case different
metal combinations are involved. They act as
input offset voltage, independent on the bias
current.
These thermoelectric effects are in the microvolt
per Kelvin region. and thus are only to be
considered in high DC-gain applications.

While FET Input opamps have far lower bias currents,
they don't achieve the low input offset voltage
common to bipolar input OpAmps.

There are Fet input opAmps that get rid of the
input offset voltage by  trading bandwidth against
the chopper feature.

Rene

Hi,

I'll try to clarify:

I am referring to the input voltage on the *DUT* caused by the op-amp,
and therefore if the bias current through the DUT is decreased then
the offset voltage on the DUT will be less-- ohms law.

The op-amps I am working with have offsets around 0.5uV to a few uV.
Therefore thermoelectric effects should be considered. As far as I
know instrumentation op-amp appear to have to least thermoelectric
effects since both input pins go to the same polarity on both op-amps,
the + pin, but there are still thermoelectric effects since both op-
amps are not 100% identical. Other circuits such as the inverter
require dummy resistors and such to help reduce the thermoelectric
voltages on the DUT.

My interest in BiFET's is to design a low power amp circuit with low
bias current.

Thanks,
Paul

 

[neon]

Hi,

As you know, the *input* offset voltage is the voltage required across
the op-amp's input terminals to drive the output voltage to zero.
Although it has been my experience that for most op-amps the input
offset voltage is due to the "-" input pin for the *most* part. For
example, according to Spice the input offset voltage on the "+" input
pin on a LMC660A op-amp for a non-inverting amp circuit is a few
nanovolts, disregarding thermoelectric effects mind you, but a few
millivolts on the "-" input pin. Although as you know the input signal
is not applied to the "-" input pin for a non-inverting amp circuit,
which means there's just a few nanovolts on the input of such a
circuit if we disregard thermoelectric effects.

I have a INA116PA Instrumentation op-amp where Ib typ = 3fA, Ib max =
25fA, and Vos typ = 0.5mV. Now it seems to me in order for there to be
0.5mV on the input of this Instrumentation op-amp circuit with 3fA
bias current that the DUT input impedance would have to be 0.50mV /
3.0fA = 170 Gohms. On the other hand, if the DUT input impedance is
say 200 Kohms then would the input offset voltage be 3.0fA * 200Kohms
= 0.6nV, disregarding thermoelectric effects?

INA116PA datasheet:
http://focus.ti.com/lit/ds/symlink/ina116.pdf

Regards,
Paul

the input offset voltage is never required actualy is an un-wanted situation. and either pin input just the [-] input

 

[Pieter]

Hi,

I'll try to clarify:

I am referring to the input voltage on the *DUT* caused by the op-amp,
and therefore if the bias current through the DUT is decreased then
the offset voltage on the DUT will be less-- ohms law.

The op-amps I am working with have offsets around 0.5uV to a few uV.
Therefore thermoelectric effects should be considered. As far as I
know instrumentation op-amp appear to have to least thermoelectric
effects since both input pins go to the same polarity on both op-amps,
the + pin, but there are still thermoelectric effects since both op-
amps are not 100% identical. Other circuits such as the inverter
require dummy resistors and such to help reduce the thermoelectric
voltages on the DUT.

My interest in BiFET's is to design a low power amp circuit with low
bias current.

Thanks,
Paul

To prevent thermoelecric voltages, keep all pins at the same
temparature. But also all surrounding resistors etc. A cooling airflow
gives temperature differences. And resistors and opamps that get warm
may give some effects.

Pieter

 

[Tom Bruhns]

The offset voltage is *differential*. You can blame it on either pin,
or both pins... it doesn't matter who you blame, the result is the
same: offset voltage becomes measurement error.

The offset voltage error is a different thing from the input bias
current. They are unrelated [1]. You can of course generate a real,
external-to-the-opamp error voltage by dumping the bias current into
real external resistance, but that's a different matter entirely.

John

[1] Some opamps have low offsets and high bias currents, and some vice
versa. Chopper amps are low on both; cheap bipolars are high on both.

Here's my main concern. If I build the INA116PA for DC application,
which is an internal Instrumentation op-amp chip (3 op-amps), and the
impedance of my DUT is 200 Kohms, then what bias currents could a good
EE such as yourself expect? I mean, for a 200K ohm DUT input source we
cannot have both 0.5mV offset and 3fA bias on the DUT. I think V=I*R
applies, so if the bias current is 3fA then V = 3fA * 200Kohms = 0.6
nV.

Thanks,
Paul

My apologies if this has been covered in the other branch of this
thread, which I don't have time to fully read...

As others have pointed out, the input bias current and input offset
voltage of the part are characteristics of the part. HOWEVER, the
external circuit strongly influences how well you can take advantage
of those characteristics. That is, the external circuit can
completely wipe out the potential benefits of either a low bias
current or a low offset voltage or both, even. For example, I used a
chopper-stabilized op amp to amplify the output of a diode RF
detector. The op amp has typically a pA of input bias current and
about a uV of input offset voltage. First, I had to be very careful
to guard the detector traces against currents leaking in from outside,
and then I had to be careful that the resistance between the guard
trace and the detector output trace (feeding the op amp input) was
high enough that a 1uV offset would not result in a current as large
or larger than the pA op amp bias current. For this part, that's only
a megohm or so, fairly easy to do, but in an earlier design using a
non-chopper amp where the offset voltage was up to a millivolt or so,
it was a killer. The RF detector diode is shunt between the guard and
the amplifier input, with RF fed in through a capacitor, and the zero
bias RF detector diode shows considerable current at a millivolt.

Seems like there should be some good references on applying low
offset, low bias amplifiers. I know that Bob Pease has had some good
articles on the trials and tribulations of testing amplifiers down in
the fA region--not trivial! His articles can be found with a search
on the web...

Cheers,
Tom