# Offset voltage on voltage follower

Discussion in 'Electronic Design' started by Tuurbo46, Aug 20, 2004.

1. ### Tuurbo46Guest

Hi
I need a bit of help on a circuit im trying to build. I will try to explain
my problem below.

Circuit requirements:

*input to op-amp = 0 to 5v
*output from op-amp = 0.5v to 3.5v

Im am currently using a CA3140E opamp with a 0.5v offset in it, which is
quoted in the data sheet.

Pin wirring on chip:

Pin 1: to left leg of 10k pot
pin 2: connected to pin 6 with no series resistor
pin 3: varible input voltage, 0 to 5v
pin 4: to middle leg of above 10k pot and connected to GND
pin 5: to right leg of above 10k pot
pin 6: connected to pin 2 with no series resistor, and then this voltage is
dropped across 2 resistors in series to produce the output voltage ratio

Im at the stage in the circuit, that when you increase the input voltage
from o to 5v, the output is alway 0.5v higher than the input which is
correct. At this point i drop the output voltage over a potential divider
and i get the correct output voltage ratio over this. The problem starts
when a 1k load resistor is taken from the potential divider. At this point
the voltage drops. When the load resistor is removed the circuit produces
the correct output voltage again.

I think the problem is the circuit cant provide enough corrent, but i cant
see how to change this.

Any ideas?

Anyone have any better ideas?

2. ### Tim WescottGuest

I would not be very trusting of a 1/2-volt offset produced by severely
yanking the "offset _null_" line of an Op-Amp.

Your circuit indeed cannot provide current with it's potential divider
without it's output voltage changing. Assuming that your input
impedance must be high, I would suggest that you use the first stage as
a voltage buffer, with a potential divider in the middle, followed by
another voltage buffer. Further, I would put the 1/2 volt offset in the
potential divider, like so:

Vref
+
|
|
.-.
| |
| |
input |\ '-'
-----------|+\ ___ |
| >--o---|___|---o |\
.--|-/ | o----------|+\ output
| |/ | | | >---o----------
| | .-. .---|-/ |
'--------' | | | |/ |
| | | |
'-' '----------'
|
|
===
GND
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

Use a dual op-amp and you won't increase your package size. You'll need
to use a reference voltage that's right for the level of precision that
you need; this may range anywhere from whatever you're using for your
raw supply voltage to a high-precision 2.5V reference. Setting the
values of the three resistors in the voltage drop/shift network are left
as an exercise to the reader, of course.

3. ### Jonathan KirwanGuest

I made a similar comment in .basics, where the OP also asked this question
separately.
I said as much there, too.

Without really considering the load on the input source, assuming a 5V only
supply, and trying to keep to just one rail-to-rail opamp, I got something like
this:

(0-3V)
30k 27k 90k
0V to 5V input >-----/\/\/----+----/\/\/----, ,-----/\/\/-----,
| | | |
| | | +5 |
\ | | |\| |
/ 45k | +-----|-\ |
+5V \ | | | >------+--> OUT
--- | | | ,--|+/
| | | \ | |/|
| --- | 45k/ | |
\ gnd | \ | ---
/ 180k | | | gnd
\ | | |
| | --- |
| (0.5V) 27k | gnd |
+------------/\/\/----------+ |
| +-------'
| |
\ \
/ 20k / 45k
\ \
| |
| |
--- ---
gnd gnd

Jon

4. ### Tim WescottGuest

I was going to send him to 'basics -- I should have looked, I suppose.

I think I win the "fewest resistor" contest -- you can remove my first
opamp and still use my little network, assuming low source impedance.

5. ### Spehro PefhanyGuest

Or leave out both op-amps if the load is high-Z.

+5.0V
o
|
.-.
| |
| |24.0K 1%
'-'
0-5V ___ | 0.5-3.5V
Vin o ----|___|----+-------> Vout (eg. to ADC)
|
4.02K 1% .-.
| |
| | 8.06K 1%
'-'
|
|
===
GND

Best regards,
Spehro Pefhany

6. ### Tim WescottGuest

The OP complained about his Z=1k load pulling the voltage down, so you
at least need the follower.

Nice of you to calculate the values, though.

7. ### John FieldsGuest

---
LOL!!!

I knew it was there somewhere, but I couldn't wrap my head around
it...

Very damned clever!

+5 +5 +5
| | |
[4.02K] [24.0K] [3.443K]
| | |
+---------+ = +----3.497V
| |
[8.06K] [8.06K]
| |
GND GND

and

+5 +5
| |
[24.0K] [24.0K]
| |
+---------+ = +----0.503V
| | |
[4.02K] [8.06K] [2.682K]
| | |
GND GND GND

8. ### John WoodgateGuest

I read in sci.electronics.design that Tim Wescott
Just make the resistors ohms instead of kohms.

9. ### Frank BemelmanGuest

How did you find the values? Brute force with three nested for 0->96 loops?

10. ### Jonathan KirwanGuest

Just so the OP knows what's going on here:

Vin * R1 * R2 + 5.0 * R2 * R3 + 0.0 * R1 * R3
Vout = ---------------------------------------------
R1 * R2 + R1 * R3 + R2 * R3

Vin * R1 * R2 + 5.0 * R2 * R3
= -----------------------------
R1 * R2 + R1 * R3 + R2 * R3

Vin * R1 * R2 5.0 * R2 * R3
= --------------------------- + ---------------------------
R1 * R2 + R1 * R3 + R2 * R3 R1 * R2 + R1 * R3 + R2 * R3

Since this also must be:

Vout = 0.5 + Vin * (3/5)

it's then clear that:

3 R1 * R2
- = ---------------------------
5 R1 * R2 + R1 * R3 + R2 * R3

R2
0.6 = R1 * ---------------------------
R1 * R2 + R1 * R3 + R2 * R3

and also,

5.0 * R2 * R3
0.5 = ---------------------------
R1 * R2 + R1 * R3 + R2 * R3

R2 * R3
0.1 = ---------------------------
R1 * R2 + R1 * R3 + R2 * R3

R2
0.1 = R3 * ---------------------------
R1 * R2 + R1 * R3 + R2 * R3

which means that,

R1 = 6 * R3
R2 = 2 * R3

Call R3 just R, for now, and recheck:

Vin * 6 * R * 2 * R + 5.0 * 2 * R * R
Vout = -------------------------------------
6 * R * 2 * R + 6 * R * R + 2 * R * R

Vin * 12 * R^2 + 10 * R^2
= ----------------------------
12 * R^2 + 6 * R^2 + 2 * R^2

Vin * 12 * R^2 + 10 * R^2
= -------------------------
20 * R^2

Vin * 12 + 10
= -------------
20

= (3/5) * Vin + 0.5

Just the desired answer!

So,

+5V
|
|
\
/ 6*R
\
|
0V to 5V R | 0.5V to 3.5V
Vin O---/\/\/----+-------> Vout
|
|
\
/ 2*R
\
|
|
===
GND

Jon

11. ### Frank BemelmanGuest

Beautiful ;-)

12. ### Spehro PefhanyGuest

No, I did it numerically with a steepest descent algorithm, and added
the constraint that the output impedance should be just lower than
2.5K (arbitrary, but the maximum recommended for many PIC ADCs, as you
know) and the resistor values standard 1% series. But Jonathan's
solution is FAR nicer. ;-)

Best regards,
Spehro Pefhany

13. ### Jonathan KirwanGuest

Output impedance is:

R*(2R)*(6R) 12*R^3 3
----------------------- = ------ = - * R
R*(2R)+R*(6R)+(2R)*(6R) 20*R^2 5

So, R = (5/3) * desired impedance, in this case.

Not much left for the OP to do, I suppose, except get rid of that opamp-twisting
pot on the offset null pins.

Jon

14. ### Fred BloggsGuest

Well if Vout=3/5*Vin + 0.5 then the sum of currents at the Vout node :

View in a fixed-width font such as Courier.

3 3 3
- Vin + 0.5 - Vin - Vin + 0.5 - 5 - Vin + 0.5
5 5 5
------------------- + ----------------- + ------------- =0

R R R
3 1 2

Since this must be true for 0<=Vin<=5, the Vin and constant

coefficients must both be zero:

3 1 1 2 1 1 1 4.5
- ( -- + --) - - -- =0 and 0.5 (-- + --) - --- =0
5 R1 R2 5 R3 R3 R2 R1

You call R3=R then call x=1/R1 and y= 1/R2 so that the above

equations become:

2 1 1
x + y = - - and 9x - y = -
3 R R

5 1 1 1
so that by inspection 10x= - - or x= - -
3 R 6 R

2 1 1 1 1 1
and then y= - - - - - = - -
3 R 6 R 2 R

which makes R1=6R and R2=2R confirming your solution.

15. ### Jonathan KirwanGuest

-----
I like that observation. Neatly tidies things up. And... of course!
-----

-----
Also, you can now go back and re-verify your observation -- the part you
excluded because it 'must be zero' given that Vin can vary over that range:

0.5 4.5 0.5
--- - --- + --- = 0
R3 R1 R2

Substituting:

0.5 4.5 0.5
--- - --- + --- = 0
R3 6R3 2R3

6 9 3
--- - --- + --- = 0
R3 R3 R3

Which, of course, it is for all values of R3.

Nice when you can flexibly change points of view with equal success.

Jon

16. ### John WoodgateGuest

I read in sci.electronics.design that Frank Bemelman <[email protected]
Three unknowns, so three equations are needed. For example, zero and
maximum input voltage, and any other value.

17. ### Rich GriseGuest

It's a straight line, so you can cancel out at least one unknown.

Y = m * x + b

Cheers!
Rich

18. ### John WoodgateGuest

I read in sci.electronics.design that Rich Grise <>
The straight line criterion is equivalent to a third equation. There is
NO way of solving two equations with three unknowns.

19. ### Frank BemelmanGuest

It sure is. I have to admit I'm surprised that it results in such
a nice 1-2-6 ratio. Too good to be true, sort of

20. ### Fred BloggsGuest

The most common pencil paper approach would probably be:

View in a fixed-width font such as Courier.

Thevenin Equivalent

R RTH
Vin >---/\/\--+--/\/\----< Vth
|
|
|
+---------> Vout

Vin x RTH VTH x R
Vout= --------- + ---------
R + RTH R + RTH

Equate coefficients to required Vout=3/5*Vin + 0.5 :

(1) (2)
RTH VTH x R
--------- =3/5 and --------- = 0.5
R + RTH R + RTH

By inspection:

(1)=> RTH=R/( 5/3-1)=3/2*R and

(2)=> VTH*(1-3/5)=0.5 or VTH=0.5*5/2

Then equate coefficients:

R1 R2
5V >---/\/\--+--/\/\----0V
|
|
|
+---------> VTH

R1*R2 3 5 *R2 5
(1)=> ------- = - R (2)=> ------- = 0.5 * -
R1 + R2 2 R1 + R2 2

R1 3 2 1
(1)/(2) => -- = - R - --- or R1= 6R
5 2 5 0.5

1 2 1 1
Then -- = --- - --- = --- or R2=2R
R2 3*R 6*R 2*R