Adding input offset on a non-inverting opamp.

[[email protected]]

Hello!
I need to make a small-gain amplifier with extremely
high impedance. Because of the latter requirement, I've
chosen the TL08x opamp in a non-inverting configuration:


+5V
|
|
---------------------- + |
TL08x --*-------------
.-- - | |
| | |
| -5V |
| |
*---/\/\/\/---'
| 10K
Input | Output
\
/
\ 20K
/
\
|
|
-------------------*---------------------------
Gnd


The gain of the above circuit will be x3.

I would really appreciate adding some offset, so
that e.g. Output=(Input+0.1)*3, but at the same
time I cannot give up on the high impedance and
non-inverting configuration. Also, I'd like to do
it using these TL08x parts, just because I have
plenty of them.

The following circuit seemed to work well (but all
the equipment I had to test it was a multimeter and
a voltage source):

+5V
|
|
---------------------- + |
TL08x --*-------------
.-- - | |
| | |
| -5V |
| |
+5V --/\/\/\/---*---/\/\/\/---'
20K | 10K
Input | Output
|
-5V --/\/\/\/---'
22K


-----------------------------------------------
Gnd


Now my problems are:

1) Does it *really* work well? It seems so at least!
But I don't have neither the equipment nor the
theoretical background to make sure it really is.
2) How do I calculate the gain now?
3) How do I calculate the offset now?

I.e. calculate the values of the 3 resistor for a
formula like:

Output=(Input+0.1)*3

(or any other values)

Thanks!
TPM

 

[Jim Thompson]

Hello!
I need to make a small-gain amplifier with extremely
high impedance. Because of the latter requirement, I've
chosen the TL08x opamp in a non-inverting configuration:


+5V
|
|
---------------------- + |
TL08x --*-------------
.-- - | |
| | |
| -5V |
| |
*---/\/\/\/---'
| 10K
Input | Output
\
/
\ 20K
/
\
|
|
-------------------*---------------------------
Gnd


The gain of the above circuit will be x3.

The gain is (10K + 20K)/20K = 1.5

I would really appreciate adding some offset, so
that e.g. Output=(Input+0.1)*3, but at the same
time I cannot give up on the high impedance and
non-inverting configuration. Also, I'd like to do
it using these TL08x parts, just because I have
plenty of them.

The following circuit seemed to work well (but all
the equipment I had to test it was a multimeter and
a voltage source):

+5V
|
|
---------------------- + |
TL08x --*-------------
.-- - | |
| | |
| -5V |
| |
+5V --/\/\/\/---*---/\/\/\/---'
20K | 10K
Input | Output
|
-5V --/\/\/\/---'
22K


-----------------------------------------------
Gnd


Now my problems are:

1) Does it *really* work well? It seems so at least!
But I don't have neither the equipment nor the
theoretical background to make sure it really is.
2) How do I calculate the gain now?

"Gain" is (10K + 20K||22K)/(20K||22K)

3) How do I calculate the offset now?

I.e. calculate the values of the 3 resistor for a
formula like:

Output=(Input+0.1)*3

(or any other values)

Thanks!
TPM

If 10K Resistor = R1
20K Resistor = R2
22K Resistor =R3
Input = Vin
Output = Vo
and Vs=5V

Then (summing currents at -Input of OpAmp):

(Vo-Vin)/R1 + (Vs-Vin)/R2 = (Vs-Vin)/R3

Solving is left as an exercise for the STUDENT ;-)

(Visually, it is obvious that the output is negative when Vin=0, so
your choice of resistors is incorrect.)

...Jim Thompson

 

[Chris]

Hello!
I need to make a small-gain amplifier with extremely
high impedance. Because of the latter requirement, I've
chosen the TL08x opamp in a non-inverting configuration:


+5V
|
|
---------------------- + |
TL08x --*-------------
.-- - | |
| | |
| -5V |
| |
*---/\/\/\/---'
| 10K
Input | Output
\
/
\ 20K
/
\
|
|
-------------------*---------------------------
Gnd


The gain of the above circuit will be x3.

I would really appreciate adding some offset, so
that e.g. Output=(Input+0.1)*3, but at the same
time I cannot give up on the high impedance and
non-inverting configuration. Also, I'd like to do
it using these TL08x parts, just because I have
plenty of them.

The following circuit seemed to work well (but all
the equipment I had to test it was a multimeter and
a voltage source):

+5V
|
|
---------------------- + |
TL08x --*-------------
.-- - | |
| | |
| -5V |
| |
+5V --/\/\/\/---*---/\/\/\/---'
20K | 10K
Input | Output
|
-5V --/\/\/\/---'
22K


-----------------------------------------------
Gnd


Now my problems are:

1) Does it *really* work well? It seems so at least!
But I don't have neither the equipment nor the
theoretical background to make sure it really is.
2) How do I calculate the gain now?
3) How do I calculate the offset now?

I.e. calculate the values of the 3 resistor for a
formula like:

Output=(Input+0.1)*3

(or any other values)

Thanks!
TPM

Wow. First, questions of this type usually get a better reception at
sci.electronics.basics.

Second, you've got a couple of resistors bass-ackwards on your first
diagram. A non-inverting amplifier with a gain of 3 might look like
this (view in fixed font or M$ Notepad):

|
| Vcc
|
| V(in) |\|
| o--------|+\ V(out) = 3*V(in)
| | >------o---o
| .---|-/ |
| | |/| |
| | |
| | Vee |
| | ___ |
| V(in) o-----|___|---'
| | 20K
| .-.
| | |
| 10K | |
| '-'
| |
| GND
|
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

Now, for your problem. You're not talking about an offset voltage
(usually used to describe cancelling out the op amp V(os), but adding
or summing two voltages in a non-inverting amplifier.

As is frequently the case, National Semiconductor AN-31 calls out with
an answer. On page two, it describes a non-inverting summing amplifier
which will do the job for you (assuming the output impedance of your
V(in) is fairly low):

| ___
| .---|___|---.
| | 100K |
| | Vcc |
| ___ | |\| |
| .--|___|--o---|-\ |
| | 20K | >----o----o
| === .--|+/
| GND | |/|
| | Vee
| |
| |
| V(in) ___ |
| o---|___|-o
| 100K |
| |
| ___ |
| o---|___|-'
| 33.3mV 100K
|
|
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

Here's the way it works: The op amp non-inverting input sees the
average of V(in) and 33.3mV, or [V(in) + 33.3mV]/2. The two resistors
at the inverting side are set up for a gain of 6. That results in a
total gain of [V(in) * 3] + 100mV, which is what you were looking for
to begin with.

With a 5V supply, you might get close enough for government work by
using a 5.1K/33 ohm resistive voltage divider to get the 33mV. The
important thing is to keep both impedances low.

http://www.national.com/an/AN/AN-31.pdf

Download this app-note and keep it with you. It's a really good
cookbook for the most basic op amp circuits.

Have fun, and good luck
Chris

 

[Ken Smith]

Hello!
I need to make a small-gain amplifier with extremely
high impedance. Because of the latter requirement, I've
chosen the TL08x opamp in a non-inverting configuration:

This is the simple version
Modified schematic:

+5V
|
|
---------------------- + |
TL08x --*-------------
.-- - | |
| | |
| -5V |
R1 | |
+5V --/\/\/\/---*---/\/\/\/---'
| 10K
Input | R3 Output
|

R2 !
GND --/\/\/\----

Now my problems are:

1) Does it *really* work well? It seems so at least!
But I don't have neither the equipment nor the
theoretical background to make sure it really is.

The way you drew it the difference between the +5V and -5V voltages
effects the output a lot. The change I've done. Reduces the effect of
the supply voltage.

2) How do I calculate the gain now?
3) How do I calculate the offset now?

Basically the answer to both is apply Ohm's law. You know that the inputs
of an op-amp are at equal voltage when the op-amp is doing its thing.

Assume that the R3 is really 10K.

You know the output voltage when the input is zero.

Assume the input voltage is 0V and figure out the currents through R3

Once you have the current apply the fact that none of this flows into the
op-amp so it must go through the R1 resistor. R2 must have 0V on it and
thus no current in it. Apply Ohms law to find the resistor.

Now you know the output voltage when the input is 1V (Ignore any clipping)

Use Ohm's law to find the current in R1 and R3. You now know how much
current flows in R2 with 1V on it so apply Ohm's law.

 

[Chris]

Have fun, and good luck
Chris

By the way, if the 100K input impedance of the above circuit isn't good
enough, take another amplifier and use it as a voltage follower to give
you the high input impedance.

Good luck
Chris

 

[Jim Thompson]

By the way, if the 100K input impedance of the above circuit isn't good
enough, take another amplifier and use it as a voltage follower to give
you the high input impedance.

Good luck
Chris

The OP DID stress input-Z was critical ;-)

...Jim Thompson

 

[John Larkin]

Hello!
I need to make a small-gain amplifier with extremely
high impedance. Because of the latter requirement, I've
chosen the TL08x opamp in a non-inverting configuration:


+5V
|
|
---------------------- + |
TL08x --*-------------
.-- - | |
| | |
| -5V |
| |
*---/\/\/\/---'
| 10K
Input | Output
\
/
\ 20K
/
\
|
|
|
|

+5 ---r1---------+
|
|
|
r2
|
|
|
gnd

is better, because it's much less sensitive to power supply variation.
The original circuit adds noise and drift from both supplies into the
output. If r2 << 20K, the gain doesn't change much.

As a generalism, in the analog world, never try to make a small number
by subtracting two large numbers.

John

 

[John Larkin]

As a generalism, in the analog world, never try to make a small number
by subtracting two large numbers.

Oh, the classic bad example of this homily is...

+15
|
|
|
10k
|
|
|
/
/ 100r pot
/ <----------- trim voltage
/
/
|
|
|
10k
|
|
|
-15


Which I've seen far too many times, from people who should know
better.

John

 

[Chris]

The OP DID stress input-Z was critical ;-)

...Jim Thompson

Thanks. I caught that a little late, I guess. Possibly the OP might
prefer this...


| +5V
| Vin |\| Vout = 3*Vin +100mV
| o---------|+\
| | >--o---------o
| .-----|-/ |
| | |/| .-.
| | -5V | |200K
| | | |
| | '-'
| | |
| '-----------o
| |
| .-.
| | |100K
| | |
| '-'
| |
| -33mV o-----.
| | |
| .-. .-.
| | |33 | |5.1K
| | | | |
| '-' '-'
| | |
| === -5V
| GND
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

This is a variation on a circuit on page 6 of AN-31.

Thanks again for the spot, Mr. Thompson.

Chris

 

[Rich Grise]

Thanks. I caught that a little late, I guess. Possibly the OP might
prefer this...

I noticed he said he's got a lot of opamps. What if he keeps the
working circuit, for the X3 and high-impedance, and then resolve
Output=(Input+0.1)*3 to Output = (Input * 3) + (0.1 * .3), and
add it in another stage:

+5V
|
|
------------- + |
TL08x --*---/\/\/\---*----- +
.-- - | | | TL08x --*-----
| | | | .-- - |
| -5V | | | |
| | | `-------------'
*---/\/\/\/---' |
| 10K |
Input | | Output
\ +.3V --/\/\/---'
/
\ 20K
/
\
|
|
----------*---------------------------
Gnd

Good Luck!
Rich

 

[Rich Grise]

Oh, the classic bad example of this homily is...

+15
|
10k
|
/
/ 100r pot
/ <----------- trim voltage
/
|
10k
|
-15


Which I've seen far too many times, from people who should know
better.

Which is the right way to do this, then? I've seen this in military stuff.
Not that military stuff is designed better than other stuff, but it is
ostensibly reliable. Actually, in general, military stuff probably isn't
all that reliable electronically, when you consider there's about a
couple of hundred guys working full time to keep a couple of dozen planes
in the air. ;-)

Thanks,
Rich

 

[John Larkin]

I actually took some business away from some guys who did that. Their
constant-fraction discriminator just wouldn't stay stable!

Which is the right way to do this, then? I've seen this in military stuff.
Not that military stuff is designed better than other stuff, but it is
ostensibly reliable. Actually, in general, military stuff probably isn't
all that reliable electronically, when you consider there's about a
couple of hundred guys working full time to keep a couple of dozen planes
in the air. ;-)

Better:

+15
|
|
|
/
/ 20k pot
/ <--------10k-----+-------- trim voltage
/ |
/ |
| 100r
| |
| |
| gnd
-15



I actually once saw a guy use an inverting opamp with a gain of 0.001
to generate a +-15 mV offset for another amp. So he wound up adding
the offset and noise and drift of another opamp to the one he was
trying to trim.

John

 

[Ian]

"Gain" is (10K + 20K||22K)/(20K||22K)

Then (summing currents at -Input of OpAmp):

(Vo-Vin)/R1 + (Vs-Vin)/R2 = (Vs-Vin)/R3

Solving is left as an exercise for the STUDENT ;-)


...Jim Thompson

How mean of you to apply the Student's t-Test...

Regards
Ian
;-)