I have a circuit which has an input of voltage of the order of micro. I want to convert it into volt order. I am using an instrumentation amplifier AD620 with gain of 1000 to first convert it into milli order. And then I am using TL084cn as a comparator whose one input is the milli order voltage and the other is grounded. The +Vcc and -Vcc is 10 V. Also the frequency of operation is between 8-40 Hz. The question is that theoretically this is possible but before I implement it practically I would want to know whether I could face some issue for such sudden abrupt change in the order of the voltage. Also, keep in mind please that I don't actually care about the shape of the signal cause in my project the information is in frequency and not in amplitude.
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I'm not sure the AD650 is well suited to this task; its input offset is low, but still on the order of tens of microvolts, and if your input is on the order of microvolts the input offset will be a concern. – Hearth Feb 05 '19 at 15:15
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Please,Check the link in the below comment and let me know what you think about it – user183710 Feb 05 '19 at 17:06
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Below as in the comment below the other answer – user183710 Feb 05 '19 at 17:07
3 Answers
If you use comparator then you may not get one to one mapping of input voltage and output voltage. If your input is always +Ve voltage then your output of comparator will always be stuck at +Vcc as other end of your comparator is grounded. You may need low noise high gain amplifier to do so. Also note that high gain will amplify noise in the signal too.
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I'm trying to recreate this circuit and instead of 12 ohm resistance on the instrumentation amplifier I'm using 56ohm one so that I can amplify the signal by approx 1000 factor – user183710 Feb 05 '19 at 15:20
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My input is not always positive..it's a sinusoidal signal so during -ve half cycle it should reach -Vcc right? – user183710 Feb 06 '19 at 07:33
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@user183710 The opamps in circuit is not acting as comparator, rather it acts as filter. May be you can learn about these filter by searching internet about "active filters". Yes you can recreate the circuit,but i suggest first understand the working of active filters and so. And you are correct, for sinusoidal signal the output of comparator in -ve half cycle will always be -Vcc(assuming -ve terminal of comparator is grounded). – Creative Feb 06 '19 at 12:41
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I know about the filters because even I am using these kind of filters. I am just using these opamps to shoot the voltage to Vcc level because I am passing the signal though filters of 3 orders as a pair. The opamps are not the issue , the main issue I am facing is why won't the comparator after an instrumentation amplifier won't do one to one mapping? Why is it practically so difficult to amplify the signal from order of micro to milli and then shoot it to volts? – user183710 Feb 06 '19 at 12:45
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@user183710 How will you recreate the shape of input signal using comparator? Lets assume input to your instrumentation amplifier is a pure 10Hz sinusoidal signal. if output of instrumentation amplifier is fed to the comparator, then the output of comparator will be square wave not sinusoidal. So you lose the shape of the input signal. – Creative Feb 06 '19 at 12:59
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I don't mind losing the shape of the signal cause the information is in the frequency and not the amplitude. I only to measure the frequency. Probably this might clear your doubts – user183710 Feb 06 '19 at 13:06
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I'm sorry I didn't mention this earlier.. probably would have saved a lot conversion – user183710 Feb 06 '19 at 13:11
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@user183710 Its fine. Shapes and frequencies of signals are related. What if input to your instrumentation amplifier is 1*sin(10t)+0.1*sin(100t) signal? What will be the output of comparator? Will you be able to retrieve both the frequencies? – Creative Feb 06 '19 at 13:26
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In EEG the input will only have maximum frequency of 40hz and probably some little high frequency noise or 60hz noise which will be filtered out in the later part of the circuit. The first step would be to amplify all types of signals coming in and then filter it out. So to answer your question I don't see why both the 10 and 100 hz frequency signals won't be amplified. All the signals should be ..am I right in assuming this? – user183710 Feb 06 '19 at 13:29
Common opamps have Rnoise of 1kohm, perhaps 10kohm; these opamps will in 100 cycle-per-second (100 Hertz) bandwidth produce 4nV * sqrt(100) or 12nV * sqrt(100) equivalent input noise, or 40 nanovolt for the 1Kohm Rnoise opamp and 120 nanovolt for the 10kohm Rnoise opamp.
If you amplify these noise levels by 5,000,000X to produce logic levels, you will have 200 mVRMS or about (6.2X more, for 1ppm noise peaks) 1.2 volts riding on the 5 volt logic output, assuming no hysteresis in the comparator --- this for the 1kohm Rnoise opmap.
For the 10kohm Rnoise opamp, the output noise voltages will be 3.16 or sqrt(10) larger, and the 1ppm noise peaks will be 1.2 volts * 3.1 == 3.7 volts, for a 5 volt logic output.
Your goal is achievable, but spend the extra power (in that first amplifier) and use an amplifier with 1 or 2 nV/rtHz (60 to 240 ohms Rnoise) noise density.
I used 100Hz to ensure your 40Hz requirement can be achieved.
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In addition to the other excellent observations, input offset voltage of the AD620 is 50 microvolts max. This is the order of your signal, if not bigger. After you amplify it by 1000, you may have a 50 millivolt offset, in either direction. You will need to deal with this.
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I was thinking of adding a capacitor near the feedback/gain resistor. Although I don't know what value of capacitor should I use. Do you know how to calculate also Any other suggestions /techniques? – user183710 Feb 06 '19 at 14:28
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@user183710 -- this would be much easier if you posted a tentative design. A gain of 1,000 in one step is often NOT the right thing to do, for a variety of reasons -- for example, w/ the AD620, you're taking a 35 dB hit in CMR at a gain of 1000! The conversation should go along the lines of "why do you need to do this" – Scott Seidman Feb 06 '19 at 14:58