neo Posted January 11, 2004 Report Posted January 11, 2004 Guys, for my project work I need to make an EEG amplifier. Hope you all know what an EEG signal is (Electro Encephalogram), which for my greatest sorrow, is an extrmely small signal, something around 10 to 50 microvolts, which is very small compared to the noise signals that I will encounter. By amplifier I mean any differential amplifier which can amplify such a small signal at the same time reject the common noise signal. I expect an output of about 3 - 4.5 V.Can you guys suggest something.Thankyou. ??? Quote
MP Posted January 11, 2004 Report Posted January 11, 2004 I know nothing about EEG, but in instrumentation it is common to convert the voltage to current and read the current or even convert it back to a voltage in order to keep the noise level to a minimum.Hope it helps you a little. Another thought is that fiber optics are pretty much immune to noise.MP Quote
neo Posted January 12, 2004 Author Report Posted January 12, 2004 thanx buddy, let me try that way Quote
MP Posted January 13, 2004 Report Posted January 13, 2004 The following link might be an excellent building block to get you started. It is a low voltage (0-1V) voltage to current converter with single supply rail. See Fig. 1 for the schematic.http://www.elecdesign.com/Articles/Index.cfm?ArticleID=2985Happy soldering!MP Quote
Kevin Weddle Posted February 24, 2004 Report Posted February 24, 2004 Watch for the input power. If the input power does not matter, then use any amplifier. If it does matter, use an opamp because the low current will help to establish voltage. Quote
Ldanielrosa Posted February 26, 2004 Report Posted February 26, 2004 Probably best to use an instrumentation amplifier. These can be a bit spendy, but one can be cobbled from three opamps (and they'd best be from a quad for better device matching). From that a current loop driver would be good, and an optoisolator too. I'll try to find an illustration I have laying about this afternoon, but I'm off to work in a few minutes.Okay, I'm a day later than I promised. The image is from page 1-19 of Designer's Handbook of Integrated Circuits by Arthur B. Williams. The relevant text indicates that the gain will be [2(R2/R1) + 1](R3/R4). "Resistor values should be kept low to minimize DC offset." Another illustration shows 10k in positions R2, R3, and R4. The gain can be varied with R1 alone. Pretty neat, I think I'll use it on something. I hope this helps. Quote
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