Voltage to frequency converter

Discussion in 'General Electronics Chat' started by Godiskind, Feb 28, 2015.

  1. Godiskind

    Thread Starter New Member

    Nov 21, 2013
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    Hello, I am very new to such a good forum as this. I have a question to ask; I am using an AD620ANZ amplifier and I am getting the correct output voltage that I expected. However, there is an additional offset of about 7vdc and I do not know why and where it has come from. Does anyone know why I am getting this dc offset (7vdc)? I am using 9v power supply. Also, I am using the V-F converter (lm331) and by the time I connect the my desired output voltage+ that particular dc offset; the V-F converter converts the dc offset to frequency but seems to ignore my actual desired input voltage. I added a dc offset to that AC signal in order for it to work with the V/F converter but to no avail.
    Thank you very much in advance.
     
  2. Wendy

    Moderator

    Mar 24, 2008
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    Welcome to AAC!

    A thread belongs to the OP (original poster). Trying to take over someone elses thread is called hijacking, which is not allowed at All About Circuits. I have therefore given you a thread of your very own.

    This was split from Voltage to frequency converter
     
  3. Godiskind

    Thread Starter New Member

    Nov 21, 2013
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    Thank you very much @ Bill Marsden. I really look forward to another good reply from anyone. Thank you very much in advance.
     
  4. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    Can you post a schematic of the current set-up?
    That way we can see how to correct the offset.

    Bertus
     
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  5. Godiskind

    Thread Starter New Member

    Nov 21, 2013
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    @bertus. Thank you very much for your swift and reliable reply. I am very grateful. Please note that the purpose of the CA3140EZ amplifier is to create a differential input for the INA (AD620ANZ) and the purpose of the voltage divider circuit is to further divide the voltage (3.8v) from the signal generator in order to generate voltages in the milli volt region. This is for an ECG project. Thanks for your assistance in advance. Apologies, the circuit was actually powered by a 9v dual power supply

    upload_2015-2-28_5-54-52.png
     
  6. bertus

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    Apr 5, 2008
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  7. Godiskind

    Thread Starter New Member

    Nov 21, 2013
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    @bertus thank you very much for your helpful reply. Yes. I am doing so in order for me to generate a differential input voltages for the AD620ANZ INA. That way, the difference between the inputs of the AD620ANZ will be multiplied by 1052 and will give me approximately 2v at its output. I do not know why I am getting an offset voltage (approximately 7vdc) at the output of the AD620ANZ. If I assume ignoring the offset voltage; by the time I get to the V-F converter: the reciprocal of that particular dc offset is what I get in frequencies. Thank you very much for taking out time to critically look into my situation. I look forward to getting another useful response (regarding how to remove the dc offset and also why the dc offset is present).
     
  8. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    Did you see the ECG circuit on page 12 of the datasheet?

    [​IMG]

    Bertus
     
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  9. Godiskind

    Thread Starter New Member

    Nov 21, 2013
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    @ Bertus. Yes I did see it. I am not applying the circuit on my body at the moment. That is why I created differential voltage at the input of the AD620ANZ INA. I think the right leg driver circuit (AD705J) is added there in order to get a high CMR, to name a few. At the output of the AD620ANZ; I understand that there need to be filter there. I see. Just to confirm; does the high pass filter removes the DC component? (Yes, I think so). On this basis, I have created an 8th order Butterworth active bandpass filter [with cutoff frequencies of 10mHz and 250Hz]: [​IMG]. When I connected the bandpass filter circuit to the input of the AD620ANZ; the signal at the output was totally and almost eliminated (I think it was just noise that appear at the output) by the filter. The reason why I chose those frequencies, is because that is the frequency range for ECG signal in general. Where from 0.01Hz - 150Hz is for Adult and from 150Hz - 250Hz is for children. My assumption is that with these circuit, the dc offset will be eliminated but the other scenario I found myself in is that there seemed to be no signal at the output of the filter.

    I am very grateful for the good replies that you are giving me. Thank you very much in advance for more of these.
     
  10. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    In the data sheet the filtering is done AFTER the AD620.
    In your schematic the 741's are shown.
    I may hope you are using a better opamp for the circuit.

    Bertus
     
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  11. Godiskind

    Thread Starter New Member

    Nov 21, 2013
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    I intend sending the signal through, first, and then introduce the isolation circuit [from the patient]. These are the simulation results I got from the aforementioned 8th order active bandpass filter:
    Low pass corner frequency- [​IMG]

    High pass corner frequency - [​IMG]

    and finally, the 8th order active bandpass filter -
    [​IMG]
     
  12. Godiskind

    Thread Starter New Member

    Nov 21, 2013
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    I am using the CA3140 opamp in practice because of its ability to work under low power supply voltages. Apologies; this is the schematic that I am using at present- [​IMG]
    Thank you very much in advance for your good assistance concerning why I am getting very low voltage [or no voltage at all] at the output of the filter. I think the issue of the dc offset should be overcome because of the presence of the filter. I do not know why I am at the time, not getting any signal at the output of the filter.
     
  13. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    Can the offset be influenced with the ref connection (pin 5) of the AD620?
    (I am not that familiar with instrumentation opamps).

    Bertus
     
  14. Godiskind

    Thread Starter New Member

    Nov 21, 2013
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    @bertus. I do not know. I think it may influence it. However, the datasheet connects pin 5 to ground. I also think that the filter would remove the dc component present. But by the time I connected the filter; it seems as if I have lost my input signal. When I checked to see whether the components (V-F converter and F-V converter - LM331, and optocoupler -4N35) using DC signal; are working; they were doing what is expected of them. But by the time I changed my signal to AC; and connected it to the V-F converter, then to the optocoupler and finally to the F-V converter, I discovered that it is not working properly because of my observation (the dc offset) at the output of the AD620ANZ. I then moved on to connecting the filter to see what the outcome will be: but to my surprise it appears as if the output signal was just noise and that the input signal [to the filter] is lost.

    Thank you very much for your good assistance. I still look forward to seeing your invaluable suggestions and those of the other gentle individual(s) that can assist me. Thank you very much in advance.
     
  15. Godiskind

    Thread Starter New Member

    Nov 21, 2013
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    Hello, when looking for corner frequencies; do you find those of the second order filter at -6db or -3dB? I noticed that -3dB is for the first order and -6dB is for the second order (unity gain). Thank you very much in advance.
     
  16. bertus

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  17. Godiskind

    Thread Starter New Member

    Nov 21, 2013
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    @ bertus! Thank you very much for the swift and very helpful suggestions and directions that you have given me. I am very grateful.

    I have a 4N35 optocoupler and want to transmit dc voltage from a power supply and use the transmitted dc voltage to power an amplifier. I noticed that when I connect [from 15Vdc; 1.3Kohms to the input(pin 1) of the 4N35 and pin2(cathode) to ground] the ouput (at the collector which has a pull up resistor of 1k5 ohms) the ouput dc voltage using a multimeter is less than 1Vdc. However, when I either connect the pin 2(cathode of the 4n35) TO POWER SUPPLY or leave it floating; I get almost half the power supply voltage (6vdc when trying to transmit 15Vdc). I want to optically isolate the power supply from say an amplifier. That is why I am using the 4N35. I have a circuit (step up dc voltage circuit that uses 555 timer) to double the halved input voltage.

    I do not know why I am having half of the input dc voltage at the output of the 4N35 when I connect the cathode to either power supply or leave it floating. I do not know why I am having less than 1vdc at the output of the 4N35 when I connect the cathode(pin2 of the 4N35) to ground.
    My expectation was that since the CTR of the 4N35 for 100% is 10mA; Vs - Vf)/I = R. Therefore 15v-1.7)/10mA = 1k3 ohms. The output dcvoltage should be the same as that of the input voltage.
    Thank you very much sir in advance for your swift and invaluable response.

    Regards
     
  18. Godiskind

    Thread Starter New Member

    Nov 21, 2013
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    Hello, Thank you very much in advance for the response to my message above
     
  19. Godiskind

    Thread Starter New Member

    Nov 21, 2013
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    @bertus, thank you very much for your previous posts. Do you have another invaluable suggestions to give me again? Thank you very much in advance for your reply.
     
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