Microvolt signal generator - Page 1
Forum section: electronics / projects

author: planc
author_rank: Regular Contributor
timestamp: June 27, 2023, 12:39:04 am
content: Hi, I owned a collection of about 5 EEG units from one manufacturer with different versions (from their prototype unit to advanced unit). I want to see how well each amplifier behaves. I googled for microvolt signal generator and there is only one hardware available in the market that can do that. It costs $900 so I may just build one from scratch.If I'll build a circuit from scratch, how easy it is? To get clean signal, would it require millions of dollars of laboratory equipment like that used by Intel?The objective would be to create clean 10 microvolt, 30 microvolt sine or square wave. Can you do that with a 555 timer and voltage divider resistors without noise? I actually tried it but it is so noisy and my oscilloscope being only 1mV in the vertical can't distinguish the microvolts.

author: moffy
author_rank: Super Contributor
timestamp: June 27, 2023, 02:23:13 am
content: You could just use a resistive divider on the output of an ordinary signal generator to get whatever low voltages you need. e.g. 1V through a 1k/1R network is 1mV, through a 1Meg/1R network it is 1uV. It will have the noise of a 1R resistor RMS summed with the output noise of the signal generator divided by the division factor. For your scope you would need a low noise preamp.

author: Conrad Hoffman
author_rank: Super Contributor
timestamp: June 27, 2023, 02:51:24 am
content: Don't know anything about the waveforms or the impedance involved, which I'm sure matters. Maybe wind a simple ratio transformer. That would get you any tiny voltage you want, with almost zero noise.

author: daisizhou
author_rank: Super Contributor
timestamp: June 27, 2023, 02:57:13 am
content: This is an ECG signal simulator, which is the same as mine

author: planc
author_rank: Regular Contributor
timestamp: June 27, 2023, 02:58:39 am
content: Quote from: moffy on June 27, 2023, 02:23:13 amI can only find 10 Ohm resistor. Difficult to find 1 Ohm. How does 10 Meg/10R differ to 1 Meg/1R in terms of noises? What will produce more noise, 1R or 10R?I just can't find a used Netech Minisim EEG signal generator (NOT ECG simulator!) at Ebay or anywhere. Anyone can help me look for one for budget of around $200 top.

author: SiliconWizard
author_rank: Super Contributor
timestamp: June 27, 2023, 03:48:17 am
content: You can additionally buffer the output of the attenuator with an opamp. You'll get some offset, but EEG measurements are made differentially anyway and support some amount of common-mode voltage.You may find this reading interesting:https://iopscience.iop.org/article/10.1088/1741-2552/abc7f0

author: moffy
author_rank: Super Contributor
timestamp: June 27, 2023, 03:55:06 am
content: Quote from: planc on June 27, 2023, 02:58:39 amThe 10R resistor will be 10^0.5 or 3.16 times noisier than the 1R resistor, it is pretty low noise. Resistor noise is equal to: E = (4*k*T*R*(f1-f2)^0.5 which for 10R resistor is = 0.4nV/(Hz)^0.5.

author: S57UUU
author_rank: Regular Contributor
timestamp: June 27, 2023, 04:21:07 am
content: The datasheet mentions floating outputs. The simplest way to do that is a transformer.A transformer going down to 0.1Hz sounds scary, but then you only need microvolts. A 50Hz transformer with 6V output, for example, can give 1mV down to 50*0.001/6 = 0.0083Hz.Other ways of making floating outputs would require DC-DC converters, which are noisy.Edit: correct formula

author: S57UUU
author_rank: Regular Contributor
timestamp: June 27, 2023, 04:41:21 am
content: The output impedance of the transformer will be low, so to simulate human head impedance, you will need to add a series resistor to the secondary. How high is the human head impedance? No idea, probably a few kohm to 100kohm, will depend a lot on how moist is the skin. You can simply measure it, with a battery operated (floating) ohmmeter connected to your EEG electrodes.PS a normal broadband oscilloscope preamplifier will be quite useless to see these signals. I suggest you make one with a low 1/f noise chopper amp, and a 10Hz lowpass filter. Maybe just use the same amp in a sallen-key LP type circuit.

author: moffy
author_rank: Super Contributor
timestamp: June 27, 2023, 04:46:49 am
content: Quote from: S57UUU on June 27, 2023, 04:21:07 amBatteries make low noise floating supplies.

author: planc
author_rank: Regular Contributor
timestamp: June 27, 2023, 10:35:40 pm
content: Guys. What are the advantage or use of floating output? What happens if the outputs are not floating?Also why do you need to buffer the output? Is it not the output is continuous and not in bursts?Many thanks.

author: PCB.Wiz
author_rank: Super Contributor
timestamp: June 27, 2023, 11:14:51 pm
content: Quote from: planc on June 27, 2023, 10:35:40 pmFloating here means isolated from the mains, hence your original example uses a Battery.Quote from: planc on June 27, 2023, 10:35:40 pmI'm not sure a buffer is needed, just a low impedance divider - you might start with 2.5V full scale and divide by 1000Quote from: planc on June 27, 2023, 10:35:40 pmThat's why you measure on the upstream side of the divider and know it is 1000x the signal on the pinsA 555 can only do square waves.For 0.1 ~ 60Hz the simplest would be a small MCU with a DAC, doing waveform playback. Then you can cover Sine, Square and ABR outputs.The SiLabs EFM8 series have good DACs

author: planc
author_rank: Regular Contributor
timestamp: June 28, 2023, 02:05:48 am
content: Quote from: moffy on June 27, 2023, 02:23:13 amWhat do you mean by division factor? Did you mean the oscilloscope vertical voltage scale like 1mV, 10mV? Can you give example of the division factor?For noise free, what else can be used that doesn't have resistive divider? For example, if the NeTech Minisim EEG signal generator can create 10 microvolt without resistive divider, how can it do it that uses battery as power source?

author: rfclown
author_rank: Frequent Contributor
timestamp: June 28, 2023, 02:48:38 am
content: Not sure why you think they don't have a resistor divider. That's the easy way to do it.I breadboarded an EEG thing years ago. I used an 8 pin PIC running at 5V for the ADC and UART interface. For a calibration signal, I used a PIC output to generate a 5V square wave. Two 1Meg resistors in series to 100 ohms for a voltage divider (other end of 100 ohms was to a 2.5V virtual ground). That gave me a +/- 125 uV calibration signal.

author: planc
author_rank: Regular Contributor
timestamp: June 28, 2023, 03:11:45 am
content: 555 timer and resistive divider look cheap. Their unit costs $900 and no other companies manufacture it so it sounds state of the art. Is there no chip that can directly output 30 microvolt (or the range) without using any resistive divider? Why not?

author: rfclown
author_rank: Frequent Contributor
timestamp: June 28, 2023, 04:10:51 am
content: And EEG units cost how much? The parts in my schematic cost less than $20 for this one channel unit. (I sampled all the semiconductors, so it didn't cost me anything.) If you crack open their unit, the parts cost wouldn't be much. The people that use this stuff don't roll their own and the price is going to be what the market is willing to pay. If you want to make your own for hobby use, there is no great cost involved. The largest cost is the case if you want to make it look nice. Their thing has a lot of flexibility, not just a square wave, so there is more than a 555, but the electronics cost could still be in single or low double digits dollar wise. The mechanics are where the cost is.If there is no chip that directly outputs 30 microvolts, it's because no one wants to buy one. There is no need for one. Want a lower voltage? Use a divider. What's in a chip? Mostly transistors, resistors and very small capacitors, all of which are worse than discrete parts.On your previous question about the floating thing, as PCB.Wiz said, it is an issue of being isolated from the mains for safety reasons. Making it battery powered solves that problem.And don't fret about the resistor noise. Moffy gave the formula. For a 100 ohm resistor at 25C (room temperature), the RMS noise is 0.01 uV in a 60 Hz bandwidth (where I put the corner frequency in my EEG amp). As others pointed out, the reason you can't see this with the oscilloscope is due to the front end of the scope (scope probe plus input circuit) and the fact that it has a very wide bandwidth. That's why the input stage of an EEG machine has an instrumentation amplifier followed by lots of gain, and a low bandwidth.

author: Someone
author_rank: Super Contributor
timestamp: June 28, 2023, 10:34:11 am
content: Quote from: rfclown on June 28, 2023, 04:10:51 amThey do exist and there is a (small but high value) market for such devices, but its orders of magnitude in complexity and cost outside of what the OP is asking for.

author: David Hess
author_rank: Super Contributor
timestamp: June 28, 2023, 01:00:53 pm
content: Resistor dividers work for creating accurate microvolt level signals, however care must be taken to prevent common mode noise from corrupting the output. Connecting the lower resistor to the ground at the input will do this with the higher resistor operating as a current source so the microvolt level signal will follow the ground noise at the input.

author: mawyatt
author_rank: Super Contributor
timestamp: June 28, 2023, 02:26:24 pm
content: Recall awhile back that some folks took an inexpensive AWG (Feeltech, Uni-T, Juntek, or something) and added a LiON battery pack to make it portable. If this AWG has the usual available waveforms (our SDG2042X does), some of which are medical in nature, then with a simple but carefully assembled resistive divider one could create an effective floating microvolt level AWG capable of producing all sorts of waveforms, including medical types.Best,

author: planc
author_rank: Regular Contributor
timestamp: June 28, 2023, 11:05:09 pm
content: It may be easy to create microvolt output using resistive divider, but the challenge may be to produce noise free microvolt output. Because if it is easy to build noise free microvolt output, how come no other manufacturers have produced EEG simulator? The product Netech MiniSim EEG was made before the year 2000 and selling now for $900. And it is the only kind in the world. No other challenger. If it is cheap to build one. Doesn't any person or company want to earn 800 dollars for one unit? Has anyone of you actually build one with very clean microvolt output? Or something only Netech can do because clean output is really very tricky and require sophisticated circuit that not even Intel or Apple know how?

author: PCB.Wiz
author_rank: Super Contributor
timestamp: June 28, 2023, 11:32:06 pm
content: Quote from: planc on June 28, 2023, 11:05:09 pmYou are over thinking this.The market here is very small, and it is also a medical instrument, so the price here is no surprise at all.It is not because it is technically complex or difficult.

author: planc
author_rank: Regular Contributor
timestamp: June 29, 2023, 05:09:40 am
content: There are several dozens of ECG simulator manufacturers that produced 1 mV sine wave products. I wonder why, is it bec ECG for the heart is 1000x much more common?What would produce cleaner microvolt sine wave. An ECG simulator with 1mV output and 1K/IR resistive divider producing microvolt output or a circuit built from scratch that has 1Meg/1R resistive divider that can produce microvolt?How many of you have actually build circuit that can produce microvolt output? Pls share your experiences and parts you used. Thanks.

author: Vgkid
author_rank: Super Contributor
timestamp: June 29, 2023, 07:59:30 am
content: If you were going to use a sine wave, I would agree that as Conrad suggested, a transformer would be easier to use, but not as good for Non sinusoidal sources. A resistive divider would be easier , to implement. Remember attenuator's are used for audio frequencies as well . If output impedance doesn't have to be 50 ohms, you could use any ratio you desired.Looking into the Caddock 1776, even one of those may work, but checking the math, the output impedance would be an issue(unless accuracy at higher ranges is not an issue), though you can go with the 50ohm output impedance and divide on a ratio of that for ease.I have many DC microvolt/nanovolt sources though, though that has tradeoffs vs AC.

author: mawyatt
author_rank: Super Contributor
timestamp: June 29, 2023, 01:20:40 pm
content: 50 years ago we needed to create tiny test signals for a device called MSP (Miles Signal Processor) an advanced signal processor to replace the MAID signal processor which is used with a buried line Miles Magnetostrictive cable. The signal frequency range was 0.1~100Hz and needed to be in the sub-microvolt range.Our solution was an HP3300 Function Generator followed by a precision resistive voltage divider, recall we used a double divider to keep the impedance levels low and minimize capacitive coupling and external noise (mains) pickup. Not Rocket Science then nor now, we probably just used 0.1% metal film RN60C or E resistors back then.Edit: If you use quality metal film resistors the low frequency noise (without a DC current bias) referenced to the divider input for high division ranges (>100/1) will be:VNoise = Squareroot(ENBW) * Squareroot(4kTR),where ENBW is effective noise bandwidth (3dB BW * pi/2 for 1st order LPF), R is the shunt resistor in the divider as "seen" from the divider input, k is Boltzmann Constant 1.38E-23 J/K and T is temperature in Kelvin.So a divider with a 1 ohm shunt will produce 1.28nv rms with 100Hz ENBW @ 298K, 10 ohms shunt 4.06nv rms, 100 ohms 12.8nv rms and so on. Likely you'll have trouble measuring this low a level, but can be assured that this is what the effective noise contributed by the divider will be guaranteed by design. Note this is typical thermal induced noise, which will have a ~6 peak to peak times the rms value.Anyway, hope this helps.Best,

author: Vovk_Z
author_rank: Super Contributor
timestamp: June 29, 2023, 03:46:22 pm
content: Quote from: planc on June 29, 2023, 05:09:40 amI have kind-of-well-known Victor's ultra low-distortion 1kHz oscillator and quite good sound card (ASUS Xonar Essence STX). Have couple tests of microvolt dividers. I don't remember exact test setup (1M / 1 R divider ? and 10M / 1R divider ? possibly). Here are two mesurements in my attachments. 0 dB is 2 V RMS, -120 dB must be 2 uV RMS, and -140 dB must be 0.2 uV RMS.

author: David Hess
author_rank: Super Contributor
timestamp: June 29, 2023, 05:36:51 pm
content: Quote from: planc on June 29, 2023, 05:09:40 amI have done it a couple of times in different ways.The first is the way I described, with a resistive divider but with the bottom of the divider located *at* the input so common mode ground noise is removed. The top resistor then acts as a current source and the voltage across the lower resistor follows the ground noise, removing it.The other way I have done it is with coaxial attenuators between the signal source and input. This does not remove the common mode ground noise so it must be kept to low levels. At high levels of attenuation, double shielded cable is required to prevent leakage around the attenuators. I used this method for calibrating Tektronix AM502 and 7A22 differential amplifiers which have a sensitivity of 10 microvolts per division, oscilloscope inputs at the millivolt level, and RF receivers which have microvolt sensitivity.

author: planc
author_rank: Regular Contributor
timestamp: June 30, 2023, 10:46:14 pm
content: Note that in EEG the inputs need differential signal. My waveform generator has only 1 signal with respect to ground. I manage to find 1 Ohm and 1 MegaOhm resistor by buying a set with 130 values. How do you connect it to the EEG input?

author: planc
author_rank: Regular Contributor
timestamp: July 01, 2023, 01:08:33 am
content: This is my 3-in-1 oscilloscope/generator. It can allow 1 volt output.This is the EEG input circuitIt accepts 2 pairs of electrodes so one set is 3 wire. For channel 1, it's Blue, Yellow, Black.Black is the ground. Blue, Yellow are the 2 differential inputs of the LT1168.The generator has 2 wire output. So I'll just connect them to the 2 differential inputs of the EEG? How about the ground. Should I also connect the ground of the generator to the ground of the differential?My concern is let's say the LT1168 expects microvolts input and it is exposed to the full 3 volts of the generator (let's say by connection mistake). Would the circuit suffer a catastrophic meltdown? Or would it just clip? What part of the datasheet can you find maximum inputs allowed for the LT1168? It's one of my 5 EEG collection and I don't want to ruin it with no possibility of repair or shipping it to the manufacturer since it's a discontinued vintage unit.By the way. Why are there diodes in the input. What do you think are their functions? Thanks.

author: daisizhou
author_rank: Super Contributor
timestamp: July 01, 2023, 02:54:55 am
content: Would you be able to send the rest of the picture below? I want to see the full structure

author: Vovk_Z
author_rank: Super Contributor
timestamp: July 01, 2023, 07:32:30 am
content: Quote from: planc on May 08, 1975, 09:42:51 pmYou need to read some basic literature about hiw opamp work.It can be "The art of Electronics", this book has good covering of opamps too.

author: radiolistener
author_rank: Super Contributor
timestamp: July 01, 2023, 07:43:04 am
content: Quote from: planc on June 27, 2023, 12:39:04 amYou can use almost any AWG for that for example FY6800 or SDG2042X. In order to get microvolt level, just put proper attenuator on the output. For example if there is 1 Vpk on AWG output and you put two 40 dB attenuators in series (total 80 dB), the output will be 100 uVpk, in order to get 1 uV you're needs to setup 0.01 V output on your AWG.But since you're needs microvolt levels, this is hard to acheieve precise ampitudes, due to leakages. These leakages happens anywhere - through outer surface for cable shield, through space, through connectors, etc... Fighting with shielding and signal leakages requires to use expensive equipment for measurement and is not easy task. If you don't want to spend a lot of time to fight with complicated issue related to it, it's better to buy such device for 900 USD, because you will spend much more to develop something like that which can produce reliable results and can be used for measurement and calibration purposes...

author: planc
author_rank: Regular Contributor
timestamp: July 02, 2023, 12:18:29 am
content: I set up the generator and used 3 values of the resistive divider, 1 Ohm, 10 Ohm, 30 Ohm divided by 1 MegaOhm to produce 1 microvolt, 10 microvolt, 30 microvoltThis is 30 microvolt at 10 Hz.This is 10 microvolt at 10 Hz.This is 1 microvolt at 10 Hz.The following waveform is when the EEG doesn't have any wire or electrodes connected to the unit.My 2 questions.1. In the last image, is it not it is the amplifier noises? but the noises are supposed to be in the range of 1 microvolt. Why are the noises amplitude reaching 30 microvolts when the inputs were left hanging or no wire or electrodes connected?2. In the 1 microvolt image, you can't distinguish any signal. Well. I wanted to get a generator to study and the compare the CMRR (Common Mode Rejection Ratio) of the amplifiers used in the 5 EEG units (different versions). Microvolt signal generators don't seem to offer that possibility. So what should I get instead to study and compare amplifiers CMRR? The 5 EEG has different CMRR and I wanted to see the differences in that department.Thanks.

author: radiolistener
author_rank: Super Contributor
timestamp: July 02, 2023, 04:56:03 am
content: Quote from: planc on July 02, 2023, 12:18:29 amThe noise on amplifier input depends on a source impedance. Most of amplifiers has a high noise when there is high impedance on it's input. The same high noise can happens with short circuit on the input.Also note that using resistors and breadboard for attenuator is a bad idea, because all your wires receive RFI from environment and nearby equipment, such as FNIRSI, mains cables and other. Attenuator should be within shielded box and connected with a shielded cable with no holes and open wires. It's better to buy SMA attenuators, they are in a metal enclosure, so there is no need for additional shielding and small enough to minimize RFI reception.Also there is a good idea to use RF chokes on a shielded cable to eliminate common mode current noises.

author: planc
author_rank: Regular Contributor
timestamp: July 02, 2023, 08:40:55 am
content: If the inputs are short circuited. Impedance is low. Why is the noise simlar to high impedance inputs?I tried it. I shorted the inputs and the noises disappeared and there was only a straight line.Also I realised that using signal generators can't make one distinguish the CMRR of the units. From a CMRR of 90 dB and 120 dB. The waveform at 30 microvolt and 10 microvolt were identical at 2 Hz.What kind of signal generator that I need that can inject common mode noises to compare the CMRR of 2 different amplifiers?

author: radiolistener
author_rank: Super Contributor
timestamp: July 02, 2023, 05:31:22 pm
content: Quote from: planc on July 02, 2023, 08:40:55 amit depends on amplifier type, some of them have higher noise figure with low source impedance, some of them have higher noise figure with high source impedance.Also note that short circuit is a kind of magnetic loop antenna, which receive a lot of noise from nearby electronic equipment and mains line. For a clean experiment, you're needs to put your short circuit and input frontent inside a good shielded enclosure and put it away from any wires or electronic equipment for at least several meters. It should be powered from a battery to make sure that there is no common mode currents through wires or RFI received through near field coupling.If you want to minimize noise, it's better to use shielded terminator with impedance that is well matched with your device input impedance. If your device has high impedance input (such as 1 MΩ) you can leave it open. But for low impedance input (such as 50 Ω or 600 Ω) it's important to use terminator instead of open input.As I said before, measurement for so low signals (about 1 uVrms) is not so easy and requires special measures and measurement methods. Using proper AWG here is not enough, because every wire geometry, shielding and electromagnetic environment have significant effect on measurement results.Common issue with obtaining microvolt signal level is leakage from signal source (through ground, through signal wires, or with near field coupling) or even through attenuator. This is why it worth to buy proper shielded cables, good quality attenuators and good AWG with low unwanted emissions to avoid such issues.For example, when my PSG9080 AWG output is open and disabled (nothing is connected to it), I still can receive it on nearby VHF receiver which has MDS sensitivity about several microvolts...As you can see, when you're working with microvolt signal level, the signal path is not obvious and it can bypass your attenuator through near field coupling. Especially if you're don't use shielding at all, like in your resistive divider on breadboard. This is why you can see the same amplitude for expected 1 uV and for 10 uV. It just means that you receive not signal from your attenuator, but signal leakage which bypass your attenuator.Another point here is to use proper attenuator. Not a simple resistive divider which output impedance depends on divide ratio. Your attenuator needs to have proper input and output impedance. You're needs to use Pi or Tee attenuator, which can be calculated here for example:https://www.everythingrf.com/rf-calculators/pi-attenuator-calculatorFor example if you want voltage ratio 1:100 attenuator, this is 40 dB attenuation. Pi attenuator for 50 Ω impedance needs R1=51 Ω and R2=2500 Ω.But if your device input has different than 50 Ω impedance, you're needs to recalculate it. And take into account your cable characteristic impedance.To reduce possible leakage it's better to split 40 dB into two 20 dB attenuators connected in series.Also you're needs to take into account required bandwidth. By adding low pass filter you can cut off significant power of high frequency noise, so your measurement noise floor will be better.

author: planc
author_rank: Regular Contributor
timestamp: July 03, 2023, 12:42:42 am
content: Thanks for your tips. Last year I spent 3 weeks bewildered with certain waveforms. And later found out it was AC interference causing aliasing in the output from the wire at the back of the table. I have since avoided AC source by turning all sources off or capacitive coupling in my body (I used grounded ESD pads, etc.).When you were mentioning about high impedance input (such as 1 MΩ). Were you referring to the amplifier?But is it not most amplifier have 10 GigaOhm to 1000Giga Ohm input impedance? Are there really onin the MegaOhms only? In the case of my EEG units. There were 2 specs.1. Input Impedance 10 GigaOhms, CMRR 100 dB2. Input Impedance 1000 GigaOhms, CMRR 120 dBI initially wanted to get a microvolt signal generator because I thought the sine waves output can distinguish between the common mode noises. But it can't.I know this needs another thread. But can you at least give tips how to test for CMRR. Like can I connect the generator 30 microvolt, 10 Hz outputs to both Active and Reference electrode inputs (the differential inputs to either the LT1168 or AD620 chips) and maybe check the output for noises? What equipments must I need to distinguish and compare 10 Gigaohms, CMRR 100 dB vs 1000 Gigaohms, CMRR 120 dB? Thanks.

author: AYN
author_rank: Newbie
timestamp: July 05, 2023, 02:00:07 am
content: I am using the DAC+1M/1K series resistor voltage divider of ESP32 to achieve an mV signal generator. Generate a single ECG signal. After amplifying by 400 through AD8232, the relevant signal was measured. After calculation, it has been proven that the full range (3V) partial voltage generation of DAC can generate mV signals. The uV signal can be achieved by reducing the amplitude of the DAC signal. All power supplies use lithium batteries (signal generators, amplifiers, and oscilloscopes) to avoid interference. This circuit is very simple. I hope it will be helpful to you.