Offset voltage on voltage follower

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

(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

2. formatting link

Read it -- a lot of what you need to use op amps is right there.

Good luck Chris

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

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

from about

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

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

What is your power supply?

And why the choice of the CA3140E?

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?

Or does anyone have any better circuit ideas?

Harris/Intersil military temperature range version of the 741, perhaps. But it may also just be that it's a part in hand. Also, note he's using the "offset null" to set the offset to 0.5V with that 10k pot!

Jon

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

Well, you are using a divider on the output, after all. Adding a load resistor changes the divider, so you get a different output.

You could, if you want, still bother cranking that poor opamp's offset voltage to 0.5V with that 10k pot you are using and also then add another opamp at that divider output, I suppose. But there is a more traditional way, I think, and it doesn't depend on you using a pot nor the use of offset nulling to push your offset way the heck to one side.

30k 27k 90k 0V to 5V input >-----/\/\/----+----/\/\/----, ,-----/\/\/-----, | | | | | | | +5 | \ | | |\| | / 45k | '-----|-\ | +5V \ | | >------+--> OUT --- | +----------|+/ | | | |/| LT1783 | --- | | \ gnd | --- / 30k | gnd \ | | | | 27k | +----/\/\/------------------+ | | | | \ \ / 45k / 45k \ \ | | | | --- --- gnd gnd

I'm assuming a 5V power supply, here, and you can use a variety of rail-to-rail in/out opamps if that's the case. But you don't say what's on pin 7, so hard to know for sure.

Jon

On Fri, 20 Aug 2004 16:06:57 GMT, Jonathan Kirwan wrote:

I made an error in copying down the resistor values, forgetting to change some of them. I also forgot to add a resistor! The schematic *should be*:

30k 27k 90k 0V to 5V input >-----/\/\/----+----/\/\/----, ,-----/\/\/-----, | | | | | | | +5 | \ | | |\| | / 45k | +-----|-\ | +5V \ | | | >------+--> OUT --- | | | ,--|+/ | | | \ | |/| LT1783 | --- | 45k/ | | \ gnd | \ | --- / 180k | | | gnd \ | | | | | --- | | 27k | gnd | +----/\/\/------------------+ | | +-------' | | \ \ / 20k / 45k \ \ | | | | --- --- gnd gnd

Oh, well.

Jon

Thanks for your help guys!