On Sat, 21 Jun 2008 08:59:10 -0700 (PDT), the renowned Paul
<energymover@[EMAIL PROTECTED]
> wrote:
>On Jun 21, 8:41 am, John Larkin
><jjlar...@[EMAIL PROTECTED]
> wrote:
>> On Sat, 21 Jun 2008 08:25:59 -0700 (PDT), Paul <energymo...@[EMAIL PROTECTED]
>
>> wrote:
>>
>> >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
Think of the "offset voltage" as something like an internal battery
(ideal floating voltage source*) of voltage equal to the "input offset
voltage" connected in *series* with one of the inputs. End of story.
Now, bias current can be thought of as current sources (or sinks)
connected to each of the input pins. The value of the bias currents on
each input need not be similar to the other (on bipolar op-amps it
often is fairly well matched, on MOSFET input op-amps where it's just
leakage, it could be anything). They are unrelated to the "input
offset voltage", and voltage drop resulting from input source
resistance could add or subtract from Vos.
So, as you can see, the bias current certainly can be xx fA and offset
voltage can be xx mV, regardless of source resistance.
In your example, with source impedance of 200K, and if Ib is really
3fA (including PCB leakage) then the bias current effect is
negligible.
* of course it will vary from unit to unit, and is a function of
temperature, time and (sometimes) previous history of differential
voltage, but to keep it simple as a first approximation, it's constant
(different from unit-to-unit, and may have a 'hole' in the
distribution if you choose to buy a low grade unit), as a better
approximation, it is a function only of temperature.
Best regards,
Spehro Pefhany
--
"it's the network..." "The Journey is the reward"
speff@[EMAIL PROTECTED]
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