How does an OPA work?

[Zed]

I am a bit confused about how OPAs work. My questions are as follows:

1. Let's say I use an OPA with dual, simmetric supply. I do not put
anything to its inputs. If there is no voltage difference between the
two input pins, how will output work? Will it be 0V or something else?

2. What if I use the OPA with single supply? The same case ( no
voltage difference between inputs ). What voltage will I measure on
the output? Near 0V? ( I know, the voltage on the output cannot reach
0, except the OPA is rail-to-rail. Let's say it is a non rail-to-rail
kind. ) Half of the supply voltage?

3. What if I use the OPA with +10V for positive and -3V for negativ
supply. Would it work at all?


Thanks for your replies, Zsolt ( from Hungary )

 

[Eeyore]

I am a bit confused about how OPAs work. My questions are as follows:

1. Let's say I use an OPA with dual, simmetric supply. I do not put
anything to its inputs. If there is no voltage difference between the
two input pins, how will output work? Will it be 0V or something else?

If there is NO connection to the inputs, the output will be indeterminate.
Both inputs MUST be connected for any valid function.

If both inputs were *grounded* the output would be at Vios x Avol (input
offset voltage times DC open loop gain ) And will typically be 'stuck' at
either V+ or V-.

Op-amps require negative feedback for any vaild function (other than as a
comparator but you're best to use a comparator to do that).

2. What if I use the OPA with single supply? The same case ( no
voltage difference between inputs ). What voltage will I measure on
the output? Near 0V? ( I know, the voltage on the output cannot reach
0, except the OPA is rail-to-rail. Let's say it is a non rail-to-rail
kind. ) Half of the supply voltage?

Op-amps will run fine on a single supply with the restrictions imposed by
their output voltage swing and input common-mode range values.

The term 'single supply' op-amp is used to mean an op-amp that has values
for the above that include ground ( V-) or very close to ground.

3. What if I use the OPA with +10V for positive and -3V for negativ
supply. Would it work at all?

If it can work from a supply of 13V total it'll be fine but again subject
to the above constraints.

You need to read a book on op-amps. This takes some beating !
http://focus.ti.com/lit/an/slod006b/slod006b.pdf

Graham

 

[MooseFET]

I am a bit confused about how OPAs work. My questions are as follows:

1. Let's say I use an OPA with dual, simmetric supply. I do not put
anything to its inputs. If there is no voltage difference between the
two input pins, how will output work? Will it be 0V or something else?

Ideally, the output is at zero volts when the difference between the
two inputs is zero.
In real life, this isn't what happens.


For many op-amps the output really follows a rule like:


Vout = Vneg + K1 * (IN1 - IN2 + Voff)

where:
Vneg is the minus supply of the op-amp
K1 is a huge number like a million
IN1 is the non-invering input
IN2 is the inverting input
Voff is the offset voltage of this op-amp

Because K1 is so huge, even a tiny voltage difference between IN1 and
IN2 matters more than it does, so we can drop it from the equation
when we don't need a perfectly accurate description. This makes the
equation:


Vout = K1 * (IN1 - IN2 + Voff)

Voff is usually such a small number that in most cases, we don't need
to worry about it so we can drop it too.

Vout = K1 * (IN1 - IN2)

Now lets use this to figure the gain of a simple inverting op-amp.

First we need to say that the inputs of an op-amp draw no current.
This isn't really true but the current they do draw is usually so
small that it can be ignored.

ASCII ART

R1 R2
Vin-----/\/\------+-------\/\/-------+-------Vout
! !
! !
--!-\ !
! >------------
GND-------!+/


Lets make R1 and R2 equal and 1K and K1 a million.

We will work from the output back to the input. Assume Vout = 1V

Remember:
Vout = K1 * (IN1 - IN2)

Rearrange:

(IN1 - IN2) = Vout/K1

IN1 = 0 so IN2 = -1V/K1 = -0.000001V

R2 has a voltage of 1V - (-0.000001V) = 1.000001V on it

R2 has 1.000001V and is 1K so I= 1.000001mA

R1 has 1.000001mA and is 1K so it drops 1.000001V

IN2 is at -0.000001V so the Vin must be at:

(-0.000001) - 1.000001 = 1.000002V

The gain of this circuit is Vout/Vin so it is

1V / 1.000002V = 0.999998

This is so close to R2/R1 that there is no reason not to just say it
is R2/R1.

 

[D from BC]

If there is NO connection to the inputs, the output will be indeterminate.
Both inputs MUST be connected for any valid function.

If both inputs were *grounded* the output would be at Vios x Avol (input
offset voltage times DC open loop gain ) And will typically be 'stuck' at
either V+ or V-.

Op-amps require negative feedback for any vaild function (other than as a
comparator but you're best to use a comparator to do that).



Op-amps will run fine on a single supply with the restrictions imposed by
their output voltage swing and input common-mode range values.

The term 'single supply' op-amp is used to mean an op-amp that has values
for the above that include ground ( V-) or very close to ground.



If it can work from a supply of 13V total it'll be fine but again subject
to the above constraints.

You need to read a book on op-amps. This takes some beating !
http://focus.ti.com/lit/an/slod006b/slod006b.pdf

Graham

I have a related question:

As an example, the TL084 op amp datasheet on
http://focus.ti.com/lit/ds/symlink/tl081a.pdf
lists a typical Vos=3mV.

Why the positive polarity?
The Vos is with respect to what?

Does the model look like?

|<---internal------->|
-------+ 3mV (-)----|+\
| \______
| /
--------------------|-/

Or this?

-------(-) 3mv + ---|+\
| \______
| /
--------------------|-/

Or both?
Perhaps it's within +/- 3mV due to process variation..

Or is it a range?
An offset of 3mV means to null <+1.5mV or <-1.5mV has to be applied?


D from BC
British Columbia
Canada.

 

[Jim Thompson]

On Wed, 16 Jan 2008 09:59:02 -0800, D from BC

[snip]

The Vos is with respect to what?

Does the model look like?

|<---internal------->|
-------+ 3mV (-)----|+\
| \______
| /
--------------------|-/

Or this?

-------(-) 3mv + ---|+\
| \______
| /
--------------------|-/

Or both?
Perhaps it's within +/- 3mV due to process variation..

Or is it a range?
An offset of 3mV means to null <+1.5mV or <-1.5mV has to be applied?


D from BC
British Columbia
Canada.

+/- 3mV, if you're lucky ;-)

...Jim Thompson

 

[Eeyore]

I have a related question:

As an example, the TL084 op amp datasheet on
http://focus.ti.com/lit/ds/symlink/tl081a.pdf
lists a typical Vos=3mV.

Why the positive polarity?

It isn't. It's *magnitude* is 3mV.

The Vos is with respect to what?

The other input. Vios appears as a 'phantom signal' effectively.

Does the model look like?

|<---internal------->|
-------+ 3mV (-)----|+\
| \______
| /
--------------------|-/

Or this?

-------(-) 3mv + ---|+\
| \______
| /
--------------------|-/

I'd model it as a voltage between the + and - inputs. Your idea effectively
does the same.

Or both?
Perhaps it's within +/- 3mV due to process variation..

Or is it a range?
An offset of 3mV means to null <+1.5mV or <-1.5mV has to be applied?

Up to +/- 3mV.

Graham

 

[gearhead]

(snip)

As an example, the TL084 op amp datasheet onhttp://focus.ti.com/lit/ds/symlink/tl081a.pdf
lists a typical Vos=3mV.

Why the positive polarity?
The Vos is with respect to what?

Does the model look like?

     |<---internal------->|        
-------+ 3mV (-)----|+\
                    |  \______
                    |  /
--------------------|-/

Or this?

-------(-) 3mv + ---|+\
                    |  \______
                    |  /
--------------------|-/

Or both?
Perhaps it's within +/- 3mV due to process variation..

Or is it a range?
An offset of 3mV means to null <+1.5mV or <-1.5mV has to be applied?

D from BC
British Columbia
Canada.- Hide quoted text -

- Show quoted text -

Others answered your question; I'd just add that you can measure
offset pretty easily.


,---------------|+\
| | \
| | >---+---Vout
| | / |
+---Ri--+-------|-/ |
| | |
| | |
| '----Rf---------'
gnd

Ground is somewhere about halfway between positive and negative
supply. Nothing critical. You could just use a single supply and a
resistive voltage divider to establish ground.
The offset will be amplified by the usual gain equation:
Vout = Vos x (1 + (Rf/Ri))
then
Vos = Vout / (1 + (Rf/Ri))
Rf should exceed Ri by maybe three orders of magnitude, enough to get
several volts on the output without railing the amplifier. fet op
amps can have pretty high offset for example, you might have to go
with gain of only a few hundred.

 

[John Popelish]

I have a related question:

As an example, the TL084 op amp datasheet on
http://focus.ti.com/lit/ds/symlink/tl081a.pdf
lists a typical Vos=3mV.

Why the positive polarity?
The Vos is with respect to what?

Does the model look like?

|<---internal------->|
-------+ 3mV (-)----|+\
| \______
| /
--------------------|-/

Or this?

-------(-) 3mv + ---|+\
| \______
| /
--------------------|-/

Or both?

Either. The polarity of the DC error is unknown until you
measure it. Wait a while and maybe change the temperature a
little, and it will change.

Perhaps it's within +/- 3mV due to process variation..
Bingo!

Or is it a range?
An offset of 3mV means to null <+1.5mV or <-1.5mV has to be applied?

Plus or minus, so a 6mV possible range.

 

[D from BC]

Either. The polarity of the DC error is unknown until you
measure it. Wait a while and maybe change the temperature a
little, and it will change.


Plus or minus, so a 6mV possible range.

Well...that clears that up..
Thanks to all for answers.

I was going to Google for it but decided to pop it in this post.

Not sure if this topic falls into the .basics group.
I don't know how basic that basic group is?
Ex: "Where can I buy electrons?"


D from BC
British Columbia
Canada.

 

[Fred Bartoli]

gearhead a écrit :

(snip)


Others answered your question; I'd just add that you can measure
offset pretty easily.


,---------------|+\
| | \
| | >---+---Vout
| | / |
+---Ri--+-------|-/ |
| | |
| | |
| '----Rf---------'
gnd

Ground is somewhere about halfway between positive and negative
supply. Nothing critical. You could just use a single supply and a
resistive voltage divider to establish ground.
The offset will be amplified by the usual gain equation:
Vout = Vos x (1 + (Rf/Ri))
then
Vos = Vout / (1 + (Rf/Ri))
Rf should exceed Ri by maybe three orders of magnitude, enough to get
several volts on the output without railing the amplifier. fet op
amps can have pretty high offset for example, you might have to go
with gain of only a few hundred.

Don't overlook the Ibias influence. With a 10nA bias current (OK that
mostly concerns bipolar opamps, but your answer was a bit general so...)
and a 1M feedback resistor, that's a 10mV additional offset.

This bias current can easily be measured and then factored out by
switching in or out some additional series resistors with each input.

 

[gearhead]

gearhead a écrit :









Don't overlook the Ibias influence. With a 10nA bias current (OK that
mostly concerns bipolar opamps, but your answer was a bit general so...)
and a 1M feedback resistor, that's a 10mV additional offset.

This bias current can easily be measured and then factored out by
switching in or out some additional series resistors with each input.

--
Thanks,
Fred.- Hide quoted text -

- Show quoted text -

Thanks for pointing that out.
Of course adding a resistor to the noninverting input with a value
equal to the parallel combination of Rf and Ri nulls a lot of the bias
current error (all of it, if the inputs have equal bias currents). I
should have put that resistor in the diagram.

 

[MooseFET]

Thanks for pointing that out.
Of course adding a resistor to the noninverting input with a value
equal to the parallel combination of Rf and Ri nulls a lot of the bias
current error (all of it, if the inputs have equal bias currents). I
should have put that resistor in the diagram.

... or perform the offset voltage measurement including the bias
current and then measure the bias current and back that out of your
numbers. Inserting an extra resistor in series with the inverting
input lets you measure the bias current on that input. This lets you
get the offset current out of the offset voltage measurement.

 

[Eeyore]

Of course adding a resistor to the noninverting input with a value
equal to the parallel combination of Rf and Ri nulls a lot of the bias
current error (all of it, if the inputs have equal bias currents).

But they don't do they ?

That why the bias current offset is specified too.

Graham