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AMP

 
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amp
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AMP
PostPosted: Mon Mar 26, 2007 7:55 pm     Reply with quote

Hi everybody
I have a little bit question about my amplifier
I am using non inverting amplifier for my sensor.
The thing is it cannot amplify the mV before entering my PIC and display

Or should i program in the PIC to detect the negative vaue
I Can amplify the positive input.
For example:
pH 1 = 70mV
pH 2 = 60mV
until pH 7=0

Then when it starts with pH = 8 whereby the output from the sensor is -10mV it cannot amplify

Can someone answerme.
Thanx
PCM programmer



Joined: 06 Sep 2003
Posts: 21708

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PostPosted: Mon Mar 26, 2007 8:18 pm     Reply with quote

What is the range of the input signal ? Give the minimum and
maximum expected input values, in volts.
kender



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Posts: 768
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Re: AMP
PostPosted: Mon Mar 26, 2007 8:42 pm     Reply with quote

amp wrote:
Then when it starts with pH = 8 whereby the output from the sensor is -10mV it cannot amplify

I guess, your amp has only one positive power supply. The negative power supply is at the ground potential. As a result, your amp "rails". You can solve the problem by making an artificial ground in the moddle of your output voltage range. Then negative input will result in a positive output. It's hard to give you a more detailed suggestion without knowing the schematic (or at least type) of your amp.
SherpaDoug



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PostPosted: Tue Mar 27, 2007 5:51 am     Reply with quote

Do you have an RS232 driver on the board? If so you may be able to steal just enough negative voltage from that to power a simple op amp. You still need to offset the voltages so the voltage going into the PIC A/D is always positive.
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PostPosted: Wed Mar 28, 2007 12:47 am     Reply with quote

My maximum input is 70mV which is pH 0
meanwhile min input is -70mV which is pH 14

I am using LMC6001 for my amplifier and LMC6041 for my offset
My gain is about 35.7 and my gain is 2.5 so that when pH 7 = OV then it will up to 2.5V

Yup. i connect only one power supply to +5V
But for pin no 4, i connect to ground.
Is it i have to connect to -5V?
Then, i must make a circuit for -5V but i dont know. if using dc power supply then i can use -5V..

"As a result, your amp "rails". You can solve the problem by making an artificial ground in the moddle of your output voltage range"

I dont understand the meaning of above sentence. Can u explain to me???
amp
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PostPosted: Wed Mar 28, 2007 12:50 am     Reply with quote

typo mistake. Just now the 2.5 is not the gain. It is my offset. Wink
pijang
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PostPosted: Fri Mar 30, 2007 10:02 pm     Reply with quote

Hi.
I think you should supply -5V to your amplifier.
you can use inverter to get -5V
For example your dc supply is 5V then connect to inverter circuit.
U can find the circuit via internet.
Then, for your offset, i think u can use differential amplifier.
you must use -2.5V for your Vfixed. and your output from your amplifier is set as Vin.
I hope this can help Laughing
Ttelmah
Guest







PostPosted: Sat Mar 31, 2007 3:54 am     Reply with quote

Lets look at the actual voltages involved:

Ph 14 -70mV to Ph0 +70mV

Now the output required for these extremes, is:

-0.07*35.7 + 2.5 = 0.001v

and

0.07*35.7 + 2.5 = 4.999v

Now look at the output swing spec for the op-amp. Typical output swing, is 0.2 to 4.75v. So straight away,there is a problem at the end of the range.
Now, it is not clear from your description, how you are combining the functions of offset, and amplification. You talk about two op-amps, suggesting that you are amplifying, then offsetting (or vice versa). If the former, then you have a really large problem with the first amplifier, since after amplification, the signal before offsetting, will be -2.499v to +2.499v. Not even close to possible without a -ve supply...
If instead you are offsetting first, then errors in the offset, will be amplified as well, making this potentially a very inaccurate solution.
Though it is 'difficult' without a negative rail,it is actually possible, provided you accept that the system will not handle the very limits of the Ph range (because of the output swing limitiations). The 'answer', is to have the probe referenced to half of your supply rail voltage.
If you generate 2.5v, at a low impedance from one op-amp, and feed this as the 'ground' for the probe, and to the ip- on the other amplifier, then have the second op-amp give the required gain, the system will function. Even better, don't use an op-amp at all to generate the hlf supply rail, but a chip designed to do this (both Phillips, and Maxim, do 'half rail' generator chips, and I'd expect others exist as well). This is what is meant by the line you don't enderstand from the earlier poster about this solution. The output involved, suggests you are intending to use the supply rail as your reference, and obviously, the overall accuracy, will be limited by how good this is.

Best Wishes
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