opamp with 0-5vinput and +/-5v output

Since the output voltage range is twice as big as the input range, you need a voltage gain of 2. You will also need an amplifier ssupply of at least +- 5 volts. Assuming you find a rail to rail opamp so that you can get by with just +-5, you connect your input signal to the + input, and use two equal resistors in a divider between the output and a +5 reference voltage (the positive supply if it is accurate enough). The center of the divider goes to the - input.

When the input is at 0 volts, the output must go to - 5 volts in order for the divider to produce a matching 0 volts in the - input.

When the input is at +5 volts, the output goes to +5 volts and there is no voltage across the divider, so the - input is also at +5.

Reply to
John Popelish
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We normally use a fixed width font, like Courier to make these sketches, since proportionally spaced fonts have such spacing variety. But I think I understand what you drew, and if so, it is correct.

Do you understand the principle that negative feedback must produce the desired output by matching the inputs?

Reply to
John Popelish

The point is that any amplifier programmed with feedback must arrive at the desired output by forcing the two inputs to the same voltage.

The ration of the resistors sets the gain. The formula for this non inverting configuration is gain=1+Routput/Rreference.

You change the offset by varying the voltage the reference resistor connects to.

The voltage gain magnitude and sign is set by the feedback network and the connections. Any opamp that is unity gain stable will have gain at DC of at least 100,000. you set the amplifier gain to something less than this with the feedback network.

Here is a basic opamp tutorial:

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If you pick one with a gain bandwidth product of 1 MHz, its gain will fall to 1 by the time the frequency rises to 1 MHz, since the gain times the frequency can;t be more than 1 million. You didn't mention what frequency this amplifier has to reproduce.

Here is a low cost dual amplifier that has a 1.5 MHz gain bandwidth product that works with 3 to 15 volts total supply:

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Reply to
John Popelish

Hello This is my first post to sci.electronics. I have no formal electronics training but have completed quite a few DIY projects ( Digital and Analogue) from the net and elsewhere.

At the moment I am working on a Printerport (lpt) dac using the Quad 8bit DAC Max506. This chip has a rail-to-rail output voltage (0-5V) for the application I need +/-5v. I know (think :) an opamp is the solution. However I am still bamboozled by some of the tech jargon on the datasheets. If anyone knows of a solution off the top of their heads I would really appreciate it. if not I will report back with a couple of chips data sheets and ask for some guidance.

If you need more info I can email my project so far :)

Kurt

Reply to
dimention11

Anything between those two values should work. However, 1k resistors will load the output quite a bit when it is trying to pull down to -5 (10 volts across 2 1k resistors in series takes 5 mA) so staying at the high end will get you closer to full rail voltage out. The opamp I suggested draws very tiny bias current, so even higher values could work, but stray capacitance starts to interfere with the frequency response. I think a pair of 100k 1% resistors are a good choice.

Reply to
John Popelish

thanks John I thought I would need a split power supply of at least +/-5v using a rail to rail opamp. (does not have to be exactly +/-5v just above +/-3.5 and not biased)

here is a rough ascii drawing of your description I hope.

------------------------------------------------+5v | | | R | (+5v) +--------| | R | | | (out) | ------------------------ | (+In) | | |

---------------| opamp | | | | | (-In) +-----| | | | | | | | ----------------------- | | | (-5v) | | | +--------------------------------------+

Reply to
dimention11

If you are asking if I understand why using the -ve input for feedback from the divider will change my signal from a 0-5v to a +/-5v I am a little sketchy and will do a little more reading. I do understand the voltage divider and hazzard a guess that changing one of the resistors value will change my offset or bias. I think I will need this for zeroing/calibrating if true.

I have a questions on the specs I will need(opamp)

Q1. what is the sign for voltage gain? (the symbol on datasheets)

Reply to
dimention11

Thanks for the links especialy the tutorial I had found a similar one and was slowly chewing my way through, yours is better though! :)

This project is to ultimatly control some high speed galvo's which have their own amp that req a +/-5v input. galvos are capable of 60k points per second (30khz i think) so anything in the mhz range should be sufficient. LOL.

Thank you so much on all your help you have realy set me in the right direction. One last question the two resistors have to be equal I understand this, as for a value anything in the range of 1K to 100K should not be to extreme? or does this need to be calculated?

Thanks again for all your help.

Reply to
dimention11

You need to *offset* the DAC's 0-5V output to give you a -5V to +5V span ?

An op-amp will indeed help you. :-) You won't need anything too exotic.

Sorry but I don't see all the posts in this thread ( news server playing up ).

You basically need a stage with a gain of 2 ( to increase the 5V span of the DAC to the 10V total span required ) with a -2.5V offset ( which will offset the DAC's output with an input code of 128 to 0V at your system output ).

Graham

Reply to
Pooh Bear

Reply to
dimention11

--
Geez, I posted this over on alt.electronics this morning and got
nothing back over there.  I guess he took my advice about posting!^)
Reply to
John Fields

--
This is a swamp.  I was referring to alt.electronics.
Reply to
John Fields

Which group are you referring to being a wasteland ?

I'm already reading this in s.e.d

Graham

Reply to
Pooh Bear

Reply to
dimention11

You can post a binary file to alt.binaries.schematics.electronic

Most of us also watch that group.

Otherwise, you will have to post it to some web host and give us a link for it. If the schematic is in a standard graphic format you may also email it to me, but that won't let anyone else correct my mistakes when I comment on it.

Reply to
John Popelish

No. if the input range or offset changes, you need to change the gain via the resistor ratio and the offset voltage the divider is connected to.

Reply to
John Popelish

The inverting version has the input going to the resistor that connects to the voltage reference in the non inverting version. And the offset reference goes to the + input. However, the gain for the inverting version is one less than the non inverting version. equal resistors gives a gain of 1 instead of 2.

This is covered in the opamp tutorial I pointed you toward, earlier.

Reply to
John Popelish

Ok I have finished my design and I am wondering if anyone would like to check it for me. I used eagle to make the schematic. I will send it via email just leave a post here or on my email. Thanks.

Reply to
dimention11

I wouldn't have thunk you could get 0.4% unadjusted initial accuracy off the printer port...What exactly is a multi-MHz "galvo" anyway?

Reply to
Fred Bloggs

Oh and also The amp I will be driving from this has either signal and gnd or signal and -ve signal as inputs.

If I wished use the -ve signal I would hook the feedback resistors from the output and then to the +ve input of the opamp. Is this correct?

Reply to
dimention11

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