Guest
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 Posted:
Tue Feb 15, 2005 7:07 pm Post subject:
Adding input offset on a non-inverting opamp. |
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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
|
|
John Popelish
Guest
|
| Posted:
Tue Feb 15, 2005 10:30 pm Post subject:
Re: Adding input offset on a non-inverting opamp. |
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|
tpm@againstallspam.com wrote:
| Quote: |
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.
|
Not unless you switch the values of the two resistors.
| Quote: | 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?
|
The gain is 1+Rf/Rg, where Rf is the feedback resistance and Rg is the
grounded resistance. But Rg can be made up of a parallel combination
of two resistors to other voltages (I'll call them R1 and R2), as you
show above. In that case, Rg is 1/((1/R1)+(1/R2))
| Quote: | 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 you want to make the offset adjustable, tie a trim pot across the
two supply voltages (say, a 10k) and connect the wiper to a fixed
resistor to the - input.
The gain will vary only slightly, if the pot has a much lower
resistance than the resistor to -. For instance, if the feedback
resistor were 100k and the wiper to - resistor were 47k, and the pot
was 10k, the gain would vary from about 3.128 with the pot at either
extreme to 3.02 with the pot at the center position. That variation
is less than the tolerance of most resistors.
--
John Popelish |
|
Robert Monsen
Guest
|
| Posted:
Wed Feb 16, 2005 12:52 am Post subject:
Re: Adding input offset on a non-inverting opamp. |
|
|
tpm@againstallspam.com wrote:
| Quote: | 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
|
Opamps with negative feedback work by adjusting the output so the
negative input nearly equals the positive input. Thus, in your example,
the V- input is held at Vin by the opamp. There is (almost) no current
going into the V- input. Thus, any current going into the junction goes
out one of the other resistors. So, if R1 = 20k, R2 = 22k, and Rf = 10k,
we have
(5-Vin)/R1 + (-5 - Vin)/R2 + (Vout - Vin)/Rf = 0 (eqn 1)
which just means that all the current going into that node, summed up,
is zero.
Solve for whichever variables you don't know. In this case, you know R1,
R2, and R3, so
(5-Vin)/20k + (-5 - Vin)/22k + (Vout - Vin)/10k = 0
so the answer is
Vout = 43/22 * Vin - 5/22
or equivalently
Vin = 22 * Vout / 43 + 5/43
Thus, the output is zero when the input is 5/43, and the output gain
from input that differs from this voltage is 43/22, or about 2.
You can use that eqn 1 above to pick your resistor values for any input
offset you prefer, up to a point.
Another issue this has is that any noise or supply variation will be
coupled into the output. Opamps are generally designed to avoid coupling
the supply rails into the output. This is called their PSRR, and is
usually very high, like 70dB or more (meaning they only couple 316uV
into the output for a 1V excusrion!) Sadly, your circuit makes that
effort in vain, by coupling in any noise through those biasing resistors.
That trick about summing up the current into a node is called the
Kirchoff Current Law (or KCL for short) and is pretty handy. You can use
it to figure out the DC gain of most any opamp feedback connection.
--
Regards,
Robert Monsen
"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God.
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