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I have an EEG brain signal measurement device. I need to calibrate it -- i.e., I need to be able to apply known voltages to it and see its output. After doing this many times, I want to be able to determine the relationship between an applied voltage at the terminals and the output it gives. Any ideas on how I could do this? Is there a particular device that I could use to apply the deterministic voltages? I understand the voltages read by such EEG devices are so low and hence susceptible to noise. Would this affect me so badly. I am from CS, no much knowledge on EE

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  • \$\begingroup\$ Calibrated function generator? Custom signal generation circuit? You're going to have a fair bit of electrical test equipment to get familiar with if you also have to do the input stimulus portion of this system calibration. \$\endgroup\$ – user2943160 Jun 25 '16 at 2:30
  • \$\begingroup\$ You need a 80 to 100 HZ sine wave of about 100 nanovolts as a differential voltage for a signal source just for alpha-wave simulation. You have picked an expensive project that requires great electronic assembly skills and/or expensive equipment. \$\endgroup\$ – Sparky256 Jun 25 '16 at 4:53
  • \$\begingroup\$ Since we are talking very low frequencies, will I be able to isolate the signal from the noise? I have read that the contacts between the terminals of the brain device and the signal generator would have way, too much noise. All I would register as my signal would be noise. Whats your take on this? \$\endgroup\$ – Minaj Jun 25 '16 at 6:51
  • \$\begingroup\$ we actually have an expert on these systems here @scott-seidman. \$\endgroup\$ – placeholder Jun 25 '16 at 16:15
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A device to apply deterministic voltages is fairly easy to achieve.

As an EEG needs to be very sensitive, you really need battery powered, which will isolate you from ground loops.

I suggest a 3v source, say 2xAA batteries or a CR2032 if you want it smaller, to drive an HC14 relaxation oscillator. This form of oscillator has very low power consumption, at least at low frequencies.

A hex chip contains 6 inverters, from which you can make 6 independent oscillators. Set them by choice of R and C to 100Hz, 7Hz or whatever you need.

Each will give you a 3v square wave at the inverter output, or a roughly 1v triangle wave at the capacitor. The inverter output voltage will faithfully follow the battery voltage, which you can measure with a DVM if you need to know the amplitude of the signal very accurately. The triangle wave will not follow the battery voltage, so is less useful for accuracy.

Now divide down these high voltages with resistive dividers, until you get the mV, uV or nV that you need for the EEG input. To get 3uV from 3V, a 10^6:1 divider is needed. You could use 10meg series with 10 ohm. Any noise generated by the 10meg resistor must be compared against the 3v signal that is dropped across it, a negligible effect. If you want to use smaller resistor values, then you could cascade several (say) 10k:10ohm pairs, each getting 1000:1 ratio. The tolerances of these resistors are largely irrelevant. If each was 1%, that gives you 2% on the ratio, so outputting 3uV +/- 2%.

If you use an EEG receiver to instrument a real human head, then there are a number of noise sources with which you must up put. EM fields from mains wiring and radio signals. DC contact potentials caused by differences in body chemistry and contact material. Johnson noise from resistors. All of these will be present as well when using a battery powered signal generator as I describe, no worse, in fact the body contact one should be better.

If your EEG recevier has a low enough noise level to pull a 100nV signal off a human head, it will do the same with a 100nV signal from a calibration signal generator. If it hasn't, it won't.

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  • \$\begingroup\$ What about the risk of noise at the point of contact between the EEG device and the device? WOnt the noise overwhelm this low voltage, low frequency signal? \$\endgroup\$ – Minaj Jun 25 '16 at 8:21
  • \$\begingroup\$ I'm a bit rusty with my noise analysis skills, but wouldn't such dividers risk to introduce too much noise? To reach a nV level voltage from that battery voltage level you'd need very high divider ratios (order of 1:10^8), so assuming practical resistor values (around ohms in the lower leg) you'll end up with tens of megs in the upper leg. Wouldn't those high resistance values introduce lots of thermal noise? Moreover it could be useful to point out to the OP he'll need very high precision components: Even an 0.1% tolerance on the upper leg would swamp the lower leg completely. \$\endgroup\$ – Lorenzo Donati Jun 25 '16 at 8:45
  • \$\begingroup\$ Rusty indeed you are. I've updated the answer to address those points. \$\endgroup\$ – Neil_UK Jun 25 '16 at 12:46
  • \$\begingroup\$ Source impedance may be a problem, depending on the amplifier being built \$\endgroup\$ – Scott Seidman Jun 25 '16 at 16:28

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