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### Network # Offset voltage on voltage follower

T

#### tuurbo46

Jan 1, 1970
0
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?

C

#### CFoley1064

Jan 1, 1970
0
Subject: Offset voltage on voltage follower
From: [email protected] (tuurbo46)
Date: 8/19/2004 11:27 AM Central Daylight Time
Message-id: <[email protected]>

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
Any ideas?

This type of thing is called level shifting. You might want to try something
like this (view in fixed font or M$Notepad): Summing Op Amp Level Shifter Vin(0-5V) ___ o-|___|-o---. 2K | | 3K.-. | .----------. +5V | | | | VCC | + | | | | + | | '-' | | |\| | .-. | | '---|-\ | 18K| | === | ___ | >---o | | GND '-|___|--o---|+/ Output (0.5V to 3.5V) '-' 33K | |/| | ___ | VEE o----------|___|--' | 33K .-. 2K| | | | '-' | === GND created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de (By the way, the shareware program AAC is a good way to do this rather than listing nodes) Looking at the circuit above, you can see that your 0-5V signal is attenuated to a 0-3V signal (the span of your final output is 3V) by the 2K/3K divider. Then you add 0.5V with the 18K/2k divider. The two identical 33K resistors cause a summing action to get your 0.5 to 3.5V. This is buffered by your CA3140 set as a voltage follower to give you a low impedance output. Use 1% resistors for better accuracy. Good question. The non-inverting summing amplifier is your friend. One of the best basic tutorials on how to use op amps is National Semiconductor's App Note 31. http://www.national.com/an/AN/AN-31.pdf Read it -- a lot of what you need to use op amps is right there. Good luck Chris J #### Jonathan Kirwan Jan 1, 1970 0 Summing Op Amp Level Shifter Vin(0-5V) ___ o-|___|-o---. 2K | | 3K.-. | .----------. +5V | | | | VCC | + | | | | + | | '-' | | |\| | .-. | | '---|-\ | 18K| | === | ___ | >---o | | GND '-|___|--o---|+/ Output (0.5V to 3.5V) '-' 33K | |/| | ___ | VEE o----------|___|--' | 33K .-. 2K| | | | '-' | === GND This looks wrong. It boils down to: 34.2k 0-3V V1 O-----/\/\/----, | 0.5V V2 O-----/\/\/----+-----> V3, to opamp + input 34.8k That should yield: V3 = (V1*34.8k + V2*34.2k) / (34.2k+34.8k) = 247.83mV + 0.50435*V1 at the + input. I'd tend to guess, then, that the output would vary from about 245mV to about 1.76V. Not quite the desired 0.5 to 3.5V. Jon C #### CFoley1064 Jan 1, 1970 0 Subject: Re: Offset voltage on voltage follower From: Jonathan Kirwan [email protected] Date: 8/19/2004 3:40 PM Central Daylight Time Message-id: <[email protected]> This looks wrong. It boils down to: 34.2k 0-3V V1 O-----/\/\/----, | 0.5V V2 O-----/\/\/----+-----> V3, to opamp + input 34.8k That should yield: V3 = (V1*34.8k + V2*34.2k) / (34.2k+34.8k) = 247.83mV + 0.50435*V1 at the + input. I'd tend to guess, then, that the output would vary from about 245mV to about 1.76V. Not quite the desired 0.5 to 3.5V. Jon It is wrong. I guess I was thinking (?) about an inverting summer with the virtual ground input, which isn't applicable. The OP should try something like this (view in fixed font or M$ Notepad):

100K
___
o-|___|----.
Vin (0-5V) | .---------.
| | |
| | |\| |
| '--|-\ |
| | >---o------o
Vo o-----|+/
+5V | |/|
+ |
| |
.-. |
| | |
| |3K |
'-' |
| ___ |
o--|___|--'
| 150K
.-.
| |
| |1K
'-'
|
===
GND

created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

The 1K and 3K resistors give a reference voltage of 1.25V. The 100K and 150K
resistors give a summing voltage of 0.5V when the voltage input is 0V, and 3.5V
when the voltage input is 5V. This assumes a small (about 1%) error for the
source impedance of the 1.25V reference from the power supply. If that's
unacceptable, a separate 1.25V reference can be devised which is not dependent
on a resistive voltage divider or power supply variations. The math for a
non-inverting summer is a little more complicated, but it's in the app note,
too.

Cheez. A senior moment, I suppose. Thanks for the spot, Jon.

Chris

R

#### Robert C Monsen

Jan 1, 1970
0
Jonathan Kirwan said:
This looks wrong. It boils down to:

34.2k
0-3V V1 O-----/\/\/----,
|
0.5V V2 O-----/\/\/----+-----> V3, to opamp + input
34.8k

That should yield:

V3 = (V1*34.8k + V2*34.2k) / (34.2k+34.8k)
= 247.83mV + 0.50435*V1

at the + input. I'd tend to guess, then, that the output would vary from about
245mV to about 1.76V.

Not quite the desired 0.5 to 3.5V.

Jon

First, scale the input to 1V using a voltage divider. Then, use a
non-inverting amplifier with a gain of 3 and a range of 0.5 to 3.5.

.-----------------.
| |
.-. |
Vin | | 2k |
0-5V | | | |
| I | '-' |
.-. V | |
4k | | | |
| | | |\ |
'-' o----|-\ |
| | | >---------o--- Vout
o---------|----|+/ 0.5 to 3.5V
| | |/
| |
.-. .-.
1k | | | | 1k
| | | |
'-' '-'
| |
GND Vx = -0.25V reference... :?

created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

Since Vin goes from 0 to 5, V+ goes from 0 to 1. Then, due to the
negative feedback, V- goes from 0 to 1.

At Vin = 0, we have V- = 0, so I = .25/1k = 250uA. Thus, Vout = 2k *
250u + 0V = 0.5.

At Vin = 5, V+ = 1, so I = 1.25/1k = 1.25mA. Thus, Vout = 1.25m * 2k +
1 = 3.5V

The problem, of course, is manufacturing the -1/4 V low impedance
reference.

Regards,
Bob Monsen

J

#### Jonathan Kirwan

Jan 1, 1970
0
First, scale the input to 1V using a voltage divider. Then, use a
non-inverting amplifier with a gain of 3 and a range of 0.5 to 3.5.

.-----------------.
| |
.-. |
Vin | | 2k |
0-5V | | | |
| I | '-' |
.-. V | |
4k | | | |
| | | |\ |
'-' o----|-\ |
| | | >---------o--- Vout
o---------|----|+/ 0.5 to 3.5V
| | |/
| |
.-. .-.
1k | | | | 1k
| | | |
'-' '-'
| |
GND Vx = -0.25V reference... :?

created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

Since Vin goes from 0 to 5, V+ goes from 0 to 1. Then, due to the
negative feedback, V- goes from 0 to 1.

At Vin = 0, we have V- = 0, so I = .25/1k = 250uA. Thus, Vout = 2k *
250u + 0V = 0.5.

At Vin = 5, V+ = 1, so I = 1.25/1k = 1.25mA. Thus, Vout = 1.25m * 2k +
1 = 3.5V

The problem, of course, is manufacturing the -1/4 V low impedance
reference.

I had been thinking more along the following lines, earlier, after reading
Chris's post which had the right 'idea' but the wrong implementation details. I
think I knew where Chris was coming from, though.

I'll post the schematic I was thinking under another response here that is
similar to Chris's, but I think arranged with the topology I believe he was
reaching for.

Jon

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